三角函数高次有理分式的不定积分

葡萄糖

\[ \color{black}{\int \dfrac{\cos^3x+\sin^3x}{1+\sin^4x}{\rm\,d}x} \]
\begin{align*}
&&\color{black}{=\,\,}&\color{black}{\frac{1}{4\sqrt{1+\sqrt{2}\,}}\arctan\left(\frac{\sqrt{2\left(1+\sqrt{2}\,\right)}+2\cos x}{\sqrt{2\left(\sqrt{2}-1\right)}}\right)-\frac{1}{4\sqrt{1+\sqrt{2}\,}}\arctan\left(\frac{\sqrt{2\left(1+\sqrt{2}\,\right)}-2\cos x}{\sqrt{2\left(\sqrt{2}-1\right)}}\right)}\\
&&&\color{black}{+\frac{\sqrt{1+\sqrt{2}\,}}{8}\ln\left(\frac{\cos^2x-\sqrt{1+\sqrt{2}\,}\cos x+\sqrt2}{\cos^2x+\sqrt{1+\sqrt{2}\,}\cos x+\sqrt2}\right)+\frac{1}{2\sqrt2\,}\ln\left(\frac{\sin^2x+\sqrt{2}\sin x+1}{\sin^2x-\sqrt{2}\sin x+1}\right)+C}
\end{align*}
还可以化为更简洁的形式吗？

mathematica

\[\left(-\frac{1}{4}+\frac{i}{4}\right) \sqrt[4]{-1} \left(-\tanh ^{-1}\left(\sqrt[4]{-1} \sin (x)\right)+\tanh ^{-1}\left((-1)^{3/4} \sin (x)\right)+\sqrt[8]{-1} \sqrt[4]{2} \left(-i \tan ^{-1}\left(\frac{(-1)^{7/8} \tan \left(\frac{x}{2}\right)+(-1)^{5/8}}{\sqrt[4]{2}}\right)+(-1)^{3/4} \tanh ^{-1}\left(\frac{(-1)^{5/8} \left(\tan \left(\frac{x}{2}\right)+\sqrt[4]{-1}\right)}{\sqrt[4]{2}}\right)+(-1)^{3/4} \tanh ^{-1}\left(\frac{(-1)^{7/8} \left((-1)^{3/4} \tan \left(\frac{x}{2}\right)+1\right)}{\sqrt[4]{2}}\right)+\tanh ^{-1}\left(\frac{\sqrt[8]{-1}-(-1)^{3/8} \tan \left(\frac{x}{2}\right)}{\sqrt[4]{2}}\right)\right)-\tanh ^{-1}\left(\sqrt[4]{-1} \csc (x)\right)+\tanh ^{-1}\left((-1)^{3/4} \csc (x)\right)\right)\]

mathematica

有这个表达式就很不错了，要求那么高干啥

zeroieme

反正切部分似乎可以合并成，我没有讨论区间
\(\frac{\tan ^{-1}\left(\frac{\sqrt{2 \left(\sqrt{2}-1\right)} \cos (x)}{\sqrt{2}-\cos ^2(x)}\right)}{4 \sqrt{1+\sqrt{2}}}\)

chyanog

1. Factor[Total/@GatherBy[List@@Int[(Cos[x]^3+Sin[x]^3)/(1+Sin[x]^4),x],Round[LeafCount@#,3]&]]/.{ArcTan[x_]-ArcTan[y_]:>ArcTan[(x-y)/(1+x y)//Simplify],Log[x_]-Log[y_]:>Log[x/y//Simplify]}//Total
   
2. D[%,x]==(Cos[x]^3+Sin[x]^3)/(1+Sin[x]^4)//FullSimplify

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将arctan和log合并一下：

$-\frac{\sqrt{-1+\sqrt{2}}}{4} \tan ^{-1}\left(\frac{\sqrt{2+2 \sqrt{2}} \cos (x)}{-2-\sqrt{2}+\left(1+\sqrt{2}\right) \cos ^2(x)}\right)+\frac{\sqrt{1+\sqrt{2}}}{8} \log \left(\frac{\sqrt{2}-\sqrt{2+2 \sqrt{2}} \cos (x)+\cos ^2(x)}{\sqrt{2}+\sqrt{2+2 \sqrt{2}} \cos (x)+\cos ^2(x)}\right)-\frac{\sqrt{2}}{4} \log \left(\frac{1-\sqrt{2} \sin (x)+\sin ^2(x)}{1+\sqrt{2} \sin (x)+\sin ^2(x)}\right)$

zeroieme

arctan内还可以再简化一点，但这个回复要点是楼主的结果是错的，害我验算到现在。幸好5#结果是对的。
\(\frac{\log \left(\frac{\sin ^2(x)+\sqrt{2} \sin (x)+1}{\sin ^2(x)-\sqrt{2} \sin (x)+1}\right)}{2 \sqrt{2}}+\frac{1}{8} \sqrt{1+\sqrt{2}} \log \left(\frac{\cos ^2(x)-\color{red}{\sqrt{2+2 \sqrt{2}}} \cos (x)+\sqrt{2}}{\cos ^2(x)+\color{red}{\sqrt{2+2 \sqrt{2}}} \cos (x)+\sqrt{2}}\right)+\frac{1}{4} \sqrt{\sqrt{2}-1} \tan ^{-1}\left(\frac{\sqrt{2 \left(\sqrt{2}-1\right)} \cos (x)}{\sqrt{2}-\cos ^2(x)}\right)\)

葡萄糖

zeroieme 发表于 2018-12-21 02:37
arctan内还可以再简化一点，但这个回复要点是楼主的结果是错的，害我验算到现在。幸好5#结果是对的。
...

谢谢了！
整理一下，漂亮多了
\[\color{black}{\int \dfrac{\cos^3x+\sin^3x}{1+\sin^4x}{\rm\,d}x}\]
\begin{align*}

&&\color{black}{=\,\,}&\color{black}{+\frac{\sqrt{\sqrt{2}-1}}{4}\arctan\left(\frac{\sqrt{2\left(\sqrt{2}-1\right)}\cos (x)}{\sqrt{2}-\cos^2(x)}\right)
+\frac{1}{2\sqrt{2}}\ln\left(\frac{\sin^2(x)+\sqrt{2}\sin (x)+1}{\sin ^2(x)-\sqrt{2}\sin (x)+1}\right)}\\
&&&\color{black}{+\frac{\sqrt{1+\sqrt{2}\,}}{8}\ln\left(\frac
{\cos ^2(x)-\color{red}{\sqrt{2\left(1+\sqrt{2}\,\right)\,}}\cos(x)+\sqrt{2}}
{\cos ^2(x)+\color{red}{\sqrt{2\left(1+\sqrt{2}\,\right)\,}} \cos (x)+\sqrt{2}}\right)+C
}\\

\end{align*}

zeroieme

我就是觉得对数内可以分解，终于得到了
\(\frac{1}{4} \sqrt{\sqrt{2}-1} \tan ^{-1}\left(\frac{\sqrt{2 \left(\sqrt{2}-1\right)} \cos (x)}{\sqrt{2}-\cos ^2(x)}\right)+\frac{1}{8} \left(\sqrt{1+\sqrt{2}}+2 \sqrt{2}\right) \log \left(\frac{\sin (x)-\sqrt{1+\sqrt{2}} \cos (x)+\sqrt{2}+1}{-\sin (x)+\sqrt{1+\sqrt{2}} \cos (x)+\sqrt{2}+1}\right)+\frac{1}{8} \left(2 \sqrt{2}-\sqrt{1+\sqrt{2}}\right) \log \left(\frac{\sin (x)+\sqrt{1+\sqrt{2}} \cos (x)+\sqrt{2}+1}{-\sin (x)-\sqrt{1+\sqrt{2}} \cos (x)+\sqrt{2}+1}\right)\)

接着引入辅助角

\(\frac{1}{8} \left(2 \sqrt{2}-\sqrt{1+\sqrt{2}}\right) \log \left(\frac{1+\sqrt{2-\sqrt{2}} \cos \left(x-\cos ^{-1}\left(\frac{1}{\sqrt[4]{2}}\right)\right)}{1-\sqrt{2-\sqrt{2}} \cos \left(x-\cos ^{-1}\left(\frac{1}{\sqrt[4]{2}}\right)\right)}\right)+\frac{1}{8} \left(2 \sqrt{2}+\sqrt{1+\sqrt{2}}\right) \log \left(\frac{1-\sqrt{2-\sqrt{2}} \cos \left(x+\cos ^{-1}\left(\frac{1}{\sqrt[4]{2}}\right)\right)}{1+\sqrt{2-\sqrt{2}} \cos \left(x+\cos ^{-1}\left(\frac{1}{\sqrt[4]{2}}\right)\right)}\right)+\frac{1}{4} \sqrt{\sqrt{2}-1} \tan ^{-1}\left(\frac{\sqrt{2 \left(\sqrt{2}-1\right)} \cos (x)}{\sqrt{2}-\cos ^2(x)}\right)\)

zeroieme

在换元复数域完全分解时最对称，恢复实数化以及逆向换元时破坏了部分对称。

1. \[Integral](Cos[x]^3+Sin[x]^3)/(1+Sin[x]^4) \[DifferentialD]x//TrigToExp//#/.{x->2ArcTan[u]}/.{Log[X_]:>(X//ExpToTrig//TrigExpand//Factor//{Numerator[#],Denominator[#]}&//(If[MemberQ[Variables[#],u],{Coefficient[#,u,Exponent[#,u]],u/.Solve[#==0,u]//u-#&},{#}]//Flatten//Log//Total)&/@#&//#[[1]]-#[[2]]&//RootReduce/@#&)}&//Collect[#,Log[_],RootReduce]&//Select[#,Variables[#]!={}&]&

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1. Log[u+Root[1+4 #1^2+22 #1^4+4 #1^6+#1^8&,2]] Root[25-1088 #1^2+30720 #1^4-163840 #1^6+262144 #1^8&,1]+Log[u+Root[1+4 #1^2+22 #1^4+4 #1^6+#1^8&,1]] Root[25-1088 #1^2+30720 #1^4-163840 #1^6+262144 #1^8&,2]+Log[u+Root[1+4 #1^2+22 #1^4+4 #1^6+#1^8&,4]] Root[25-1088 #1^2+30720 #1^4-163840 #1^6+262144 #1^8&,3]+Log[u+Root[1+4 #1^2+22 #1^4+4 #1^6+#1^8&,3]] Root[25-1088 #1^2+30720 #1^4-163840 #1^6+262144 #1^8&,4]+Log[u+Root[1+4 #1^2+22 #1^4+4 #1^6+#1^8&,7]] Root[25-1088 #1^2+30720 #1^4-163840 #1^6+262144 #1^8&,5]+Log[u+Root[1+4 #1^2+22 #1^4+4 #1^6+#1^8&,8]] Root[25-1088 #1^2+30720 #1^4-163840 #1^6+262144 #1^8&,6]+Log[u+Root[1+4 #1^2+22 #1^4+4 #1^6+#1^8&,5]] Root[25-1088 #1^2+30720 #1^4-163840 #1^6+262144 #1^8&,7]+Log[u+Root[1+4 #1^2+22 #1^4+4 #1^6+#1^8&,6]] Root[25-1088 #1^2+30720 #1^4-163840 #1^6+262144 #1^8&,8]

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