论坛 › 难题征解 › 求 x^3/(( x^3-a)^2 + b^2) 的不定积分

葡萄糖 发表于 2019-2-4 18:26:39

求 x^3/(( x^3-a)^2 + b^2) 的不定积分

\[ \color{black}{{\large\int}\frac{x^3}{\left(x^3-a\right)^2+b^2}\mathrm{d}x} \]
推广自问题：
[提问] 求 x^2/(( x^2-a)^2 + b^2) 的不定积分
https://bbs.emath.ac.cn/thread-5715-1-2.html

mathe 发表于 2019-2-5 11:54:13

显然同a+bi的三次方根和它们的共轭相关，https://www.wolframalpha.com/input/?i=integrate+x%5E3%2F((+x%5E3-a)%5E2+%2B+b%5E2)

葡萄糖 发表于 2019-2-6 17:27:27

本帖最后由 葡萄糖 于 2019-2-6 18:58 编辑

mathe 发表于 2019-2-5 11:54
显然同a+bi的三次方根和它们的共轭相关， ...

新春愉快！

目前只找到这两个有初等的原函数：
\[\color{black}{{\large\int}\frac{x^2}{\left(x^3-1\right)^2+1}\mathrm{d}x}\]
\[\color{black}{{\large\int}\frac{x^5}{\left(x^3-1\right)^2+1}\mathrm{d}x}\]

链接好像打不开:Q:

\[{\large\int}\frac{x^3}{\left(x^3-1\right)^2+1}\mathrm{d}x\]

\begin{align*}
&&x_1&=\dfrac{\sqrt{3}+1}{2\sqrt{2}}+\dfrac{\sqrt{3}-1}{2\sqrt{2}}\mathit{i}\\
&&x_2&=-\dfrac{\sqrt{3}-1}{2\sqrt{2}}+\dfrac{\sqrt{3}+1}{2\sqrt{2}}\mathit{i}\\
&&x_3&=-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}\mathit{i}\\
&&x_4&=-\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}\mathit{i}\\
&&x_5&=-\dfrac{\sqrt{3}-1}{2\sqrt{2}}-\dfrac{\sqrt{3}+1}{2\sqrt{2}}\mathit{i}\\
&&x_6&=\dfrac{\sqrt{3}+1}{2\sqrt{2}}-\dfrac{\sqrt{3}-1}{2\sqrt{2}}\mathit{i}\\
\\
&&&x^6-2x^3+2=0
\end{align*}
\begin{align*}
&&{x_1}\!^3-1&=+\,\mathit{i}\\
&&{x_2}\!^3-1&=-\,\mathit{i}\\
&&{x_3}\!^3-1&=+\,\mathit{i}\\
&&{x_4}\!^3-1&=-\,\mathit{i}\\
&&{x_5}\!^3-1&=+\,\mathit{i}\\
&&{x_6}\!^3-1&=-\,\mathit{i}\\
\\
\end{align*}

wayne 发表于 2019-2-6 18:13:38

Mathematica直接返回的结果是：
1/6 RootSum #1)/(-a + #1^3) &]
$a^2 + b^2 - 2 a x^3 + x^6 = 0$的六个根是$t_k$。于是积分后，原函数表达就是： \[\frac{1}{6}\sum_{k=1}^6\frac{t_k \log (x-t_k)}{t_k^3-a}\]

mathe 发表于 2019-2-7 13:45:40

如果我们假设复数$(r\e^{i\theta_1})^3=a+bi$,并且$\theta_2=\theta_1+{2\pi}/3,\theta_3=\theta_1+{4\pi}/3$
于是$(x^3-a)^2+b^2=0$的六个根就是$r\e^{i\theta_1},r\e^{-i\theta_1},r\e^{i\theta_2},r\e^{-i\theta_2},r\e^{i\theta_3},r\e^{-i\theta_3}$
我们另外记$u_1=\cos(\theta_1),u_2=\cos(\theta_2),u_3=\cos(\theta_3)$,于是$u_1+u_2+u_3=0$
通过待定系数，假设
${x^3}/{(x^3-a)^2+b^2}={s_1x+t_1}/{x^2-2ru_1x+r^2}+{s_2x+t_2}/{x^2-2ru_2x+r^2}+{s_3x+t_3}/{x^2-2ru_3x+r^2}$
可以得出$t_1=t_2=t_3=0$，且
\(\begin{cases}s_1+s_2+s_3=0\\u_1s_1+u_2s_2+u_3s_3=0\\u_2u_3s_1+u_3u_1s_2+u_1u_2s_3=\frac1{4r^2}\end{cases}\)
可以求得
\(\begin{cases}s_1=\frac{1}{4r^2(2u_1^2+u_2u_3)}\\
s_2=\frac{1}{4r^2(2u_2^2+u_1u_3)}\\
s_3=\frac{1}{4r^2(2u_3^2+u_1u_2)}
\end{cases}\)
于是题目变成求积分
\[\sum_{h=1}^3 \int\frac{s_h(x-ru_h)+rs_hu_h}{(x-ru_h)^2+r^2(1-u_h^2)}dx=\sum_{h=1}^3 (\frac12 s_h \ln((x-ru_h^2)^2+r^2(1-u_h^2))+\frac{rs_hu_h}{r\sqrt{1-u_h^2}}\tan^{-1}(\frac{x-ru_h}{r\sqrt{1-u_h^2}}))\]

葡萄糖 发表于 2019-2-8 11:56:10

本帖最后由 葡萄糖 于 2019-2-8 19:04 编辑

mathe 发表于 2019-2-7 13:45
如果我们假设复数$(r\e^{i\theta_1})^3=a+bi$,并且$\theta_2=\theta_1+{2\pi}/3,\theta_3=\theta_1+{4\pi}/3$ ...

找一个例子，结果挺漂亮的
\begin{gather*}
\Large{\left(x^3-1\right)^2+1}\\
\\
\bigg(x^2-2\sqrt{\,2\,}\cos\left(\frac{\pi}{12}\right)x+\sqrt{\,2\,}\bigg)
\bigg(x^2-2\sqrt{\,2\,}\cos\left(\frac{\pi}{12}+\frac{2\pi}{3}\right)x+\sqrt{\,2\,}\bigg)
\bigg(x^2-2\sqrt{\,2\,}\cos\left(\frac{\pi}{12}+\frac{4\pi}{3}\right)x+\sqrt{\,2\,}\bigg) \\
\bigg(x^2-2\sqrt{\,2\,}\cos\left(\frac{\pi}{12}\right)x+\sqrt{\,2\,}\bigg)
\bigg(x^2+2\sqrt{\,2\,}\cos\left(\,\frac{\pi}{4}\,\right)x+\sqrt{\,2\,}\bigg)
\bigg(x^2+2\sqrt{\,2\,}\cos\left(\frac{5\pi}{12}\right)x+\sqrt{\,2\,}\bigg)
\end{gather*}
\[ \Large{\color{black}{\frac{x^3}{\left(x^3-1\right)^2+1}}} \]

\begin{gather*}
\dfrac{\frac{\sqrt{\,2\,}}{3}\cos\left(\frac{5\pi}{12}\right)x}{x^2-2\sqrt{\,2\,}\cos\left(\frac{\pi}{12}\right)x+\sqrt{\,2\,}}
+\dfrac{\frac{\sqrt{\,2\,}}{3}\cos\left(\frac{\pi}{4}\right)x}{x^2+2\sqrt{\,2\,}\cos\left(\frac{\pi}{4}\right)x+\sqrt{\,2\,}}
-\dfrac{\frac{\sqrt{\,2\,}}{3}\cos\left(\frac{\pi}{12}\right)x}{x^2+2\sqrt{\,2\,}\cos\left(\frac{5\pi}{12}\right)x+\sqrt{\,2\,}}\\

\dfrac{\frac{\sqrt{\,2\,}}{3}\cos\left(\frac{5\pi}{12}\right)x}{\left(x-\sqrt{\,2\,}\cos\frac{\pi}{12}\right)^2+\left(\sqrt{\,2\,}\sin\frac{\pi}{12}\right)^2}
+\dfrac{\frac{\sqrt{\,2\,}}{3}\cos\left(\frac{\pi}{4}\right)x}{\left(x+\sqrt{\,2\,}\cos\frac{\pi}{4}\right)^2+\left(\sqrt{\,2\,}\sin\frac{\pi}{4}\right)^2}
-\dfrac{\frac{\sqrt{\,2\,}}{3}\cos\left(\frac{\pi}{12}\right)x}{\left(x+\sqrt{\,2\,}\cos\frac{5\pi}{12}\right)^2+\left(\sqrt{\,2\,}\sin\frac{5\pi}{12}\right)^2}\\

\dfrac{\frac{\sqrt{\,2\,}}{3}\sin\left(\frac{\pi}{12}\right)x}{\left(x-\sqrt{\,2\,}\cos\frac{\pi}{12}\right)^2+\left(\sqrt{\,2\,}\sin\frac{\pi}{12}\right)^2}
+\dfrac{\frac{\sqrt{\,2\,}}{3}\sin\left(\frac{\pi}{4}\right)x}{\left(x+\sqrt{\,2\,}\cos\frac{\pi}{4}\right)^2+\left(\sqrt{\,2\,}\sin\frac{\pi}{4}\right)^2}
-\dfrac{\frac{\sqrt{\,2\,}}{3}\cos\left(\frac{\pi}{12}\right)x}{\left(x+\sqrt{\,2\,}\sin\frac{\pi}{12}\right)^2+\left(\sqrt{\,2\,}\cos\frac{\pi}{12}\right)^2}
\end{gather*}

\begin{align*}
&&&{\Large\int}\dfrac{\frac{\sqrt{\,2\,}}{3}\sin\left(\frac{\pi}{12}\right)x}{\left(x-\sqrt{\,2\,}\cos\frac{\pi}{12}\right)^2+\left(\sqrt{\,2\,}\sin\frac{\pi}{12}\right)^2}\large\mathrm{d}x\\
&&\large{=}\quad&\frac{\sqrt{\,2\,}}{3}\sin\left(\frac{\pi}{12}\right){\Large\int}\dfrac{x-\sqrt{\,2\,}\cos\frac{\pi}{12}+\sqrt{\,2\,}\cos\frac{\pi}{12}}{\left(x-\sqrt{\,2\,}\cos\frac{\pi}{12}\right)^2+\left(\sqrt{\,2\,}\sin\frac{\pi}{12}\right)^2}\large\mathrm{d}x\\
&&\large{=}\quad&\frac{\sqrt{\,2\,}}{6}\sin\left(\frac{\pi}{12}\right)\ln\left(x^2-2\sqrt{\,2\,}x\cos\frac{\pi}{12}+\sqrt{\,2\,}\right)\\
&&&\frac{\sqrt{\,2\,}}{3}\cos\left(\frac{\pi}{12}\right)\arctan\left(\frac{x-\sqrt{\,2\,}\cos\frac{\pi}{12}}{\sqrt{\,2\,}\sin\frac{\pi}{12}}\right)\\
\end{align*}

\begin{align*}   
&&&{\Large\int}\dfrac{\frac{\sqrt{\,2\,}}{3}\sin\left(\frac{\pi}{4}\right)x}{\left(x+\sqrt{\,2\,}\cos\frac{\pi}{4}\right)^2+\left(\sqrt{\,2\,}\sin\frac{\pi}{4}\right)^2}\large\mathrm{d}x\\   
&&\large{=}\quad&\frac{\sqrt{\,2\,}}{3}\sin\left(\frac{\pi}{4}\right){\Large\int}\dfrac{x+\sqrt{\,2\,}\cos\frac{\pi}{4}-\sqrt{\,2\,}\cos\frac{\pi}{4}}{\left(x+\sqrt{\,2\,}\cos\frac{\pi}{4}\right)^2+\left(\sqrt{\,2\,}\sin\frac{\pi}{4}\right)^2}\large\mathrm{d}x\\   
&&\large{=}\quad&+\frac{\sqrt{\,2\,}}{6}\sin\left(\frac{\pi}{4}\right)\ln\left(x^2+2\sqrt{\,2\,}x\cos\frac{\pi}{4}+\sqrt{\,2\,}\right)\\   
&&&-\frac{\sqrt{\,2\,}}{3}\cos\left(\frac{\pi}{4}\right)\arctan\left(\frac{x+\sqrt{\,2\,}\cos\frac{\pi}{4}}{\sqrt{\,2\,}\sin\frac{\pi}{4}}\right)\\   
\end{align*}

\begin{align*}
&&&-{\Large\int}\dfrac{\frac{\sqrt{\,2\,}}{3}\cos\left(\frac{\pi}{12}\right)x}{\left(x+\sqrt{\,2\,}\sin\frac{\pi}{12}\right)^2+\left(\sqrt{\,2\,}\cos\frac{\pi}{12}\right)^2}\large\mathrm{d}x\\
&&\large{=}\,\,&-\frac{\sqrt{\,2\,}}{3}\cos\left(\frac{\pi}{12}\right){\Large\int}\dfrac{x+\sqrt{\,2\,}\sin\frac{\pi}{12}-\sqrt{\,2\,}\sin\frac{\pi}{12}}{\left(x+\sqrt{\,2\,}\sin\frac{\pi}{12}\right)^2+\left(\sqrt{\,2\,}\cos\frac{\pi}{12}\right)^2}\large\mathrm{d}x\\
&&\large{=}\,\,&-\frac{\sqrt{\,2\,}}{6}\cos\left(\frac{\pi}{12}\right)\ln\left(x^2+2\sqrt{\,2\,}x\sin\frac{\pi}{12}+\sqrt{\,2\,}\right)\\
&&&+\frac{\sqrt{\,2\,}}{3}\sin\left(\frac{\pi}{12}\right)\arctan\left(\frac{x+\sqrt{\,2\,}\sin\frac{\pi}{12}}{\sqrt{\,2\,}\cos\frac{\pi}{12}}\right)\\
\end{align*}

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