求一不定积分
咱们论坛之前出现过各种大量的，定积分的case，其技巧性往往也特别的多。这次来一个不定积分\[\int\frac{\sqrt{x^2+1}}{x^3+1}\dif x\] 本帖最后由 .·.·. 于 2018-12-2 01:24 编辑
这是FullSimplify的结果
\(\frac{1}{6} \left(-2 \sqrt{2} \log \left(\sqrt{2} \sqrt{x^2+1}-x+1\right)+i \left(\log \left(((x-1) x+1) \left(2 \sqrt{x^2+1}-2 i \sqrt{3} \sqrt{x^2+1}+x \left(\sqrt{x^2+1}-i \sqrt{3} \sqrt{x^2+1}+\left(\sqrt{3}-3 i\right) x-2 i\right)-3 i+\sqrt{3}\right)\right)-\log \left(((x-1) x+1) \left(2 \sqrt{x^2+1}+2 i \sqrt{3} \sqrt{x^2+1}+x \left(\sqrt{x^2+1}+i \sqrt{3} \sqrt{x^2+1}+\left(\sqrt{3}+3 i\right) x+2 i\right)+3 i+\sqrt{3}\right)\right)\right)-2 i \tanh ^{-1}\left(\frac{x \left(4 \sqrt{3} \sqrt{x^2+1}+x \left(-4 \sqrt{3} \sqrt{x^2+1}+2 x \left(2 \sqrt{3} \sqrt{x^2+1}+3 x+i \sqrt{3}-3\right)+i \sqrt{3}+9\right)+2 i \sqrt{3}-6\right)}{8+x \left(x \left(2 x \left(x-2 i \sqrt{3}-2\right)+i \sqrt{3}+9\right)-4 i \sqrt{3}-4\right)}\right)+2 i \tanh ^{-1}\left(\frac{x \left(6 \sqrt{x^2+1}+2 i \sqrt{3} \sqrt{x^2+1}+x \left(-6 \sqrt{x^2+1}-2 i \sqrt{3} \sqrt{x^2+1}+x \left(6 \sqrt{x^2+1}+2 i \sqrt{3} \sqrt{x^2+1}+3 \left(\sqrt{3}+i\right) x-6 i-2 \sqrt{3}\right)+3 i+5 \sqrt{3}\right)-6 i-2 \sqrt{3}\right)}{x \left(x \left(x \left(\left(\sqrt{3}+i\right) x+4 i-4 \sqrt{3}\right)+3 i+5 \sqrt{3}\right)+4 i-4 \sqrt{3}\right)+4 \left(\sqrt{3}+i\right)}\right)+2 \sqrt{2} \log (x+1)\right)\)
看了一下用时In:= Timing/(x^3+1),x]]]
Out= {110.766,1/6 (2 I ArcTanh[(x (-6 I-2 Sqrt+6 Sqrt+2 I Sqrt Sqrt+x (3 I+5 Sqrt-6 Sqrt-2 I Sqrt Sqrt+x (-6 I-2 Sqrt+3 (I+Sqrt) x+6 Sqrt+2 I Sqrt Sqrt))))/(4 (I+Sqrt)+x (4 I-4 Sqrt+x (3 I+5 Sqrt+x (4 I-4 Sqrt+(I+Sqrt) x))))]-2 I ArcTanh[(x (-6+2 I Sqrt+4 Sqrt Sqrt+x (9+I Sqrt-4 Sqrt Sqrt+2 x (-3+I Sqrt+3 x+2 Sqrt Sqrt))))/(8+x (-4-4 I Sqrt+x (9+I Sqrt+2 x (-2-2 I Sqrt+x))))]+2 Sqrt Log-2 Sqrt Log Sqrt]+I (Log[(1+(-1+x) x) (-3 I+Sqrt+2 Sqrt-2 I Sqrt Sqrt+x (-2 I+(-3 I+Sqrt) x+Sqrt-I Sqrt Sqrt))]-Log[(1+(-1+x) x) (3 I+Sqrt+2 Sqrt+2 I Sqrt Sqrt+x (2 I+(3 I+Sqrt) x+Sqrt+I Sqrt Sqrt))]))}如果不化简In:= Timing/(x^3+1),x]]
Out= {2.32813,(1/(18 Sqrt))((1/Sqrt])2 (3-I Sqrt) ArcTan[(x (6+2 I Sqrt+(-3-3 I Sqrt) x^3-2 I Sqrt] Sqrt+x^2 (6+2 I Sqrt-2 I Sqrt] Sqrt)+I x (3 I-5 Sqrt+2 Sqrt] Sqrt)))/(4 (-I+Sqrt)-4 (I+Sqrt) x+(-3 I+5 Sqrt) x^2-4 (I+Sqrt) x^3+(-I+Sqrt) x^4)]-(1/Sqrt])2 I (-3 I+Sqrt) ArcTan[(x (-6+2 I Sqrt+(3-3 I Sqrt) x^3-2 I Sqrt] Sqrt+x^2 (-6+2 I Sqrt-2 I Sqrt] Sqrt)+x (3-5 I Sqrt+2 I Sqrt] Sqrt)))/(4 (I+Sqrt)-4 (-I+Sqrt) x+(3 I+5 Sqrt) x^2-4 (-I+Sqrt) x^3+(I+Sqrt) x^4)]+12 Log+((-3 I+Sqrt) Log)/Sqrt]+((3 I+Sqrt) Log)/Sqrt]-12 Log Sqrt]-(1/Sqrt])(3 I+Sqrt) Log[(1-x+x^2) (3 I+Sqrt+(3 I+Sqrt) x^2+2 I Sqrt] Sqrt+I x (2+Sqrt] Sqrt))]-(1/Sqrt])(-3 I+Sqrt) Log[(1-x+x^2) (-3 I+Sqrt+(-3 I+Sqrt) x^2-2 I Sqrt] Sqrt-I x (2+Sqrt] Sqrt))])}普通化简In:= Timing/(x^3+1),x]]]
Out= {3.57813,(1/(18 Sqrt))((1/Sqrt])2 (3-I Sqrt) ArcTan[(x (6+2 I Sqrt+(-3-3 I Sqrt) x^3-2 I Sqrt] Sqrt+x^2 (6+2 I Sqrt-2 I Sqrt] Sqrt)+I x (3 I-5 Sqrt+2 Sqrt] Sqrt)))/(4 (-I+Sqrt)-4 (I+Sqrt) x+(-3 I+5 Sqrt) x^2-4 (I+Sqrt) x^3+(-I+Sqrt) x^4)]-(1/Sqrt])2 I (-3 I+Sqrt) ArcTan[(x (-6+2 I Sqrt+(3-3 I Sqrt) x^3-2 I Sqrt] Sqrt+x^2 (-6+2 I Sqrt-2 I Sqrt] Sqrt)+x (3-5 I Sqrt+2 I Sqrt] Sqrt)))/(4 (I+Sqrt)-4 (-I+Sqrt) x+(3 I+5 Sqrt) x^2-4 (-I+Sqrt) x^3+(I+Sqrt) x^4)]+12 Log+((-3 I+Sqrt) Log)/Sqrt]+((3 I+Sqrt) Log)/Sqrt]-12 Log Sqrt]-(1/Sqrt])(3 I+Sqrt) Log[(1-x+x^2) (3 I+Sqrt+(3 I+Sqrt) x^2+2 I Sqrt] Sqrt+I x (2+Sqrt] Sqrt))]-(1/Sqrt])(-3 I+Sqrt) Log[(1-x+x^2) (-3 I+Sqrt+(-3 I+Sqrt) x^2-2 I Sqrt] Sqrt-I x (2+Sqrt] Sqrt))])}Expand
\(-\frac{1}{3} \sqrt{2} \log \left(\sqrt{2} \sqrt{x^2+1}-x+1\right)-\frac{1}{6} i \log \left(((x-1) x+1) \left(2 \sqrt{x^2+1}+2 i \sqrt{3} \sqrt{x^2+1}+x \left(\sqrt{x^2+1}+i \sqrt{3} \sqrt{x^2+1}+\left(\sqrt{3}+3 i\right) x+2 i\right)+3 i+\sqrt{3}\right)\right)+\frac{1}{6} i \log \left(((x-1) x+1) \left(2 \sqrt{x^2+1}-2 i \sqrt{3} \sqrt{x^2+1}+x \left(\sqrt{x^2+1}-i \sqrt{3} \sqrt{x^2+1}+\left(\sqrt{3}-3 i\right) x-2 i\right)-3 i+\sqrt{3}\right)\right)-\frac{1}{3} i \tanh ^{-1}\left(\frac{x \left(4 \sqrt{3} \sqrt{x^2+1}+x \left(-4 \sqrt{3} \sqrt{x^2+1}+2 x \left(2 \sqrt{3} \sqrt{x^2+1}+3 x+i \sqrt{3}-3\right)+i \sqrt{3}+9\right)+2 i \sqrt{3}-6\right)}{8+x \left(x \left(2 x \left(x-2 i \sqrt{3}-2\right)+i \sqrt{3}+9\right)-4 i \sqrt{3}-4\right)}\right)+\frac{1}{3} i \tanh ^{-1}\left(\frac{x \left(6 \sqrt{x^2+1}+2 i \sqrt{3} \sqrt{x^2+1}+x \left(-6 \sqrt{x^2+1}-2 i \sqrt{3} \sqrt{x^2+1}+x \left(6 \sqrt{x^2+1}+2 i \sqrt{3} \sqrt{x^2+1}+3 \left(\sqrt{3}+i\right) x-6 i-2 \sqrt{3}\right)+3 i+5 \sqrt{3}\right)-6 i-2 \sqrt{3}\right)}{x \left(x \left(x \left(\left(\sqrt{3}+i\right) x+4 i-4 \sqrt{3}\right)+3 i+5 \sqrt{3}\right)+4 i-4 \sqrt{3}\right)+4 \left(\sqrt{3}+i\right)}\right)+\frac{1}{3} \sqrt{2} \log (x+1)\)
ExpandAll
\(\frac{1}{3} \tan ^{-1}\left(\frac{3 i \sqrt{3} x^4}{i x^4+\sqrt{3} x^4+4 i x^3-4 \sqrt{3} x^3+3 i x^2+5 \sqrt{3} x^2+4 i x-4 \sqrt{3} x+4 i+4 \sqrt{3}}-\frac{3 x^4}{i x^4+\sqrt{3} x^4+4 i x^3-4 \sqrt{3} x^3+3 i x^2+5 \sqrt{3} x^2+4 i x-4 \sqrt{3} x+4 i+4 \sqrt{3}}+\frac{6 i \sqrt{x^2+1} x^3}{i x^4+\sqrt{3} x^4+4 i x^3-4 \sqrt{3} x^3+3 i x^2+5 \sqrt{3} x^2+4 i x-4 \sqrt{3} x+4 i+4 \sqrt{3}}-\frac{2 i \sqrt{3} x^3}{i x^4+\sqrt{3} x^4+4 i x^3-4 \sqrt{3} x^3+3 i x^2+5 \sqrt{3} x^2+4 i x-4 \sqrt{3} x+4 i+4 \sqrt{3}}-\frac{2 \sqrt{3} \sqrt{x^2+1} x^3}{i x^4+\sqrt{3} x^4+4 i x^3-4 \sqrt{3} x^3+3 i x^2+5 \sqrt{3} x^2+4 i x-4 \sqrt{3} x+4 i+4 \sqrt{3}}+\frac{6 x^3}{i x^4+\sqrt{3} x^4+4 i x^3-4 \sqrt{3} x^3+3 i x^2+5 \sqrt{3} x^2+4 i x-4 \sqrt{3} x+4 i+4 \sqrt{3}}+\frac{5 i \sqrt{3} x^2}{i x^4+\sqrt{3} x^4+4 i x^3-4 \sqrt{3} x^3+3 i x^2+5 \sqrt{3} x^2+4 i x-4 \sqrt{3} x+4 i+4 \sqrt{3}}+\frac{2 \sqrt{3} \sqrt{x^2+1} x^2}{i x^4+\sqrt{3} x^4+4 i x^3-4 \sqrt{3} x^3+3 i x^2+5 \sqrt{3} x^2+4 i x-4 \sqrt{3} x+4 i+4 \sqrt{3}}-\frac{6 i \sqrt{x^2+1} x^2}{i x^4+\sqrt{3} x^4+4 i x^3-4 \sqrt{3} x^3+3 i x^2+5 \sqrt{3} x^2+4 i x-4 \sqrt{3} x+4 i+4 \sqrt{3}}-\frac{3 x^2}{i x^4+\sqrt{3} x^4+4 i x^3-4 \sqrt{3} x^3+3 i x^2+5 \sqrt{3} x^2+4 i x-4 \sqrt{3} x+4 i+4 \sqrt{3}}+\frac{6 i \sqrt{x^2+1} x}{i x^4+\sqrt{3} x^4+4 i x^3-4 \sqrt{3} x^3+3 i x^2+5 \sqrt{3} x^2+4 i x-4 \sqrt{3} x+4 i+4 \sqrt{3}}-\frac{2 i \sqrt{3} x}{i x^4+\sqrt{3} x^4+4 i x^3-4 \sqrt{3} x^3+3 i x^2+5 \sqrt{3} x^2+4 i x-4 \sqrt{3} x+4 i+4 \sqrt{3}}-\frac{2 \sqrt{3} \sqrt{x^2+1} x}{i x^4+\sqrt{3} x^4+4 i x^3-4 \sqrt{3} x^3+3 i x^2+5 \sqrt{3} x^2+4 i x-4 \sqrt{3} x+4 i+4 \sqrt{3}}+\frac{6 x}{i x^4+\sqrt{3} x^4+4 i x^3-4 \sqrt{3} x^3+3 i x^2+5 \sqrt{3} x^2+4 i x-4 \sqrt{3} x+4 i+4 \sqrt{3}}\right)+\frac{1}{3} i \tanh ^{-1}\left(-\frac{6 x^4}{2 x^4-4 i \sqrt{3} x^3-4 x^3+i \sqrt{3} x^2+9 x^2-4 i \sqrt{3} x-4 x+8}-\frac{2 i \sqrt{3} x^3}{2 x^4-4 i \sqrt{3} x^3-4 x^3+i \sqrt{3} x^2+9 x^2-4 i \sqrt{3} x-4 x+8}-\frac{4 \sqrt{3} \sqrt{x^2+1} x^3}{2 x^4-4 i \sqrt{3} x^3-4 x^3+i \sqrt{3} x^2+9 x^2-4 i \sqrt{3} x-4 x+8}+\frac{6 x^3}{2 x^4-4 i \sqrt{3} x^3-4 x^3+i \sqrt{3} x^2+9 x^2-4 i \sqrt{3} x-4 x+8}+\frac{4 \sqrt{3} \sqrt{x^2+1} x^2}{2 x^4-4 i \sqrt{3} x^3-4 x^3+i \sqrt{3} x^2+9 x^2-4 i \sqrt{3} x-4 x+8}-\frac{i \sqrt{3} x^2}{2 x^4-4 i \sqrt{3} x^3-4 x^3+i \sqrt{3} x^2+9 x^2-4 i \sqrt{3} x-4 x+8}-\frac{9 x^2}{2 x^4-4 i \sqrt{3} x^3-4 x^3+i \sqrt{3} x^2+9 x^2-4 i \sqrt{3} x-4 x+8}-\frac{2 i \sqrt{3} x}{2 x^4-4 i \sqrt{3} x^3-4 x^3+i \sqrt{3} x^2+9 x^2-4 i \sqrt{3} x-4 x+8}-\frac{4 \sqrt{3} \sqrt{x^2+1} x}{2 x^4-4 i \sqrt{3} x^3-4 x^3+i \sqrt{3} x^2+9 x^2-4 i \sqrt{3} x-4 x+8}+\frac{6 x}{2 x^4-4 i \sqrt{3} x^3-4 x^3+i \sqrt{3} x^2+9 x^2-4 i \sqrt{3} x-4 x+8}\right)-\frac{1}{6} i \log \left(3 i x^4+\sqrt{3} x^4-i x^3-\sqrt{3} x^3+i \sqrt{3} \sqrt{x^2+1} x^3+\sqrt{x^2+1} x^3+4 i x^2+2 \sqrt{3} x^2+i \sqrt{3} \sqrt{x^2+1} x^2+\sqrt{x^2+1} x^2-i x-\sqrt{3} x-i \sqrt{3} \sqrt{x^2+1} x-\sqrt{x^2+1} x+3 i+\sqrt{3}+2 i \sqrt{3} \sqrt{x^2+1}+2 \sqrt{x^2+1}\right)-\frac{1}{3} \sqrt{2} \log \left(-x+\sqrt{2} \sqrt{x^2+1}+1\right)+\frac{1}{6} i \log \left(-3 i x^4+\sqrt{3} x^4+i x^3-\sqrt{3} x^3-i \sqrt{3} \sqrt{x^2+1} x^3+\sqrt{x^2+1} x^3-4 i x^2+2 \sqrt{3} x^2-i \sqrt{3} \sqrt{x^2+1} x^2+\sqrt{x^2+1} x^2+i x-\sqrt{3} x+i \sqrt{3} \sqrt{x^2+1} x-\sqrt{x^2+1} x-3 i+\sqrt{3}-2 i \sqrt{3} \sqrt{x^2+1}+2 \sqrt{x^2+1}\right)+\frac{1}{3} \log (x+1) \sqrt{2}\)
然后用了一款神奇的软件(3) -> integrate(sqrt(x^2+1)/(x^3+1),x)
(3)
+------+
+-+ | 2 +-+ 2
+-+ (\|2+ x + 1)\|x+ 1+ (- x - 1)\|2- x- x - 2
\|2 log(----------------------------------------------------)
+------+
| 2 2
(x + 1)\|x+ 1- x- x
+
+------+
| 2 2
(2 x - 1)\|x+ 1- 2 x+ x - 1
- 2 atan(---------------------------------)
+------+
| 2
\|x+ 1- x - 1
/
3
Type: Union(Expression(Integer),...)
某软件 秒杀的结果是：
\[\frac{\sqrt{2} \log (\frac{(\sqrt{2} + x + 1) \sqrt{x^2 + 1} + (- x - 1)
\sqrt{2} - x^2 - x - 2}{(x + 1) \sqrt{x^2 + 1} - x^2 - x}) - 2 \text{atan}
(\frac{(2 x - 1) \sqrt{x^2 + 1} - 2 x^2 + x - 1}{\sqrt{x^2 + 1} - x - 1})}{3}\]
本帖最后由 .·.·. 于 2018-12-2 12:33 编辑
wayne 发表于 2018-12-2 09:29
某软件 秒杀的结果是：
\[\frac{\sqrt{2} \log (\frac{(\sqrt{2} + x + 1) \sqrt{x^2 + 1} + (- x - 1)
...
为什么这款软件到了你这里就这么好看了
还有
试过Mathematica做FullSimplify
发现这两个不定积分的确只是差一个常数
然鹅Mathematica根本没有意识到这一点
我限制了$x\in\text{Integers}$进行化简（里面还有一句Refine）都没有用 事实上，经过测试。这些积分都是有较为简洁但不简单的初等原函数的
\[\int\frac{\sqrt{px+q}}{ax+b}\dif x\]
\[\int\frac{\sqrt{px+q}}{cx^2+bx+c}\dif x\]
\[\int\frac{\sqrt{px^2+qx+r}}{ax+b}\dif x\]
\[\int\frac{\sqrt{px^2+qx+r}}{ax^2+bx+c}\dif x\]
还是认认真真的计算一下吧，$\int\frac{\sqrt{x^2+1}}{x^3+1}dx$
变量代换$x=tant$,得到 $\int\frac{dt}{cos^3t+sin^3t}$, 继续变量代换$s=tan(t/2)$, 得到$\int-\frac{2 \left(s^2+1\right)^2}{\left(s^2-2 s-1\right) \left(s^4+2 s^3+2 s^2-2 s+1\right)}ds = \frac{1}{3} \left(2 \tan ^{-1}\left(\frac{s^2+s}{1-s}\right)-\sqrt{2} \left(\log \left(-s+\sqrt{2}+1\right)-\log \left(s+\sqrt{2}-1\right)\right)\right)$ 令 `x=\sinh t`，有 `\dif x=\cosh t \dif t`，那么
\begin{align*}\int \frac{\sqrt{x^2+1}}{x^3+1}\dif x&=\int\frac{\cosh^2 t}{\sinh^3t+1}\dif t\\
&=\int\frac{1+\sinh^2 t}{1+\sinh^3t}\dif t\\
&=\frac 23\int\frac{\dif t}{1+\sinh t}+\frac 13\int\frac{1+\sinh t}{1-\sinh t+\sinh^2t}\dif t\end{align*}剩下就好办了。 楼上的学哥都是高手啊，这个怎么求啊，思考了很长时间未果\[\int^{\oo}_\sqrt{33}\frac1{\sqrt{x^3-11x^2+11x+121}}\dif x=\frac1{6\sqrt2\pi^2}\Gamma\left(\frac1{11}\right)\Gamma\left(\frac3{11}\right)\Gamma\left(\frac4{11}\right)\Gamma\left(\frac5{11}\right)\Gamma\left(\frac9{11}\right)\] 本帖最后由 chyanog 于 2018-12-3 16:11 编辑
int=Integrate/(x^3+1),x];
Assuming[0<x<1,
(ComplexExpand,TargetFunctions->{Re, Im}]/.(f:Cos|Sin):>FullSimplify]]Sqrt^2]]//Simplify)//.{ ArcTan+ ArcTan:>ArcTan[(x+y)/(1-x y)],ArcTan-ArcTan:>ArcTan[(x-y)/(1+x y)]}//FullSimplify]
这个化简起来有点费劲
$\frac{1}{3} \left(\sqrt{2} \log \left(\frac{\sqrt{2} \sqrt{x^2+1}+x-1}{x+1}\right)+\cot ^{-1}\left(\frac{((x-3) x+1) \left(x^2+x+1\right)}{(x-1) x \left(2 \left(\sqrt{3}-2 \sqrt{x^2+1}\right)+\sqrt{3} (x-1) x\right)+\sqrt{3}}\right)\right)$
用Mathematica的一个package Rubi(Rule-based Integration)可以直接得到很简洁的结果，顺便算一下bbs.emath.ac.cn/thread-2613-1-1.html (integrate(x/sqrt(x^4+10*x^2-96*x-71),x))
本帖最后由 kastin 于 2018-12-3 15:44 编辑
续上7楼：
\begin{align*}\frac 23\int\frac{\dif t}{1+\sinh t}+\frac 13\int\frac{1+\sinh t}{1-\sinh t+\sinh^2t}&=\frac{2\sqrt{2}}{3}\mathrm{arctanh}\frac{\tanh(t/2)-1}{\sqrt{2}}-\frac 23\arctan(\mathrm{sech}t-\tanh t)+c\\
&=\frac{\sqrt{2}}{3}\ln\frac{\sqrt{x^2+1}+x-\sqrt{2}+1}{\sqrt{x^2+1}+x+\sqrt{2}+1}+\frac 23\arctan\frac{x-1}{\sqrt{x^2+1}}+c\end{align*}