三角函数高次有理分式的不定积分
\[ \color{black}{\int \dfrac{\cos^3x+\sin^3x}{1+\sin^4x}{\rm\,d}x} \]\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？ \[\left(-\frac{1}{4}+\frac{i}{4}\right) \sqrt{-1} \left(-\tanh ^{-1}\left(\sqrt{-1} \sin (x)\right)+\tanh ^{-1}\left((-1)^{3/4} \sin (x)\right)+\sqrt{-1} \sqrt{2} \left(-i \tan ^{-1}\left(\frac{(-1)^{7/8} \tan \left(\frac{x}{2}\right)+(-1)^{5/8}}{\sqrt{2}}\right)+(-1)^{3/4} \tanh ^{-1}\left(\frac{(-1)^{5/8} \left(\tan \left(\frac{x}{2}\right)+\sqrt{-1}\right)}{\sqrt{2}}\right)+(-1)^{3/4} \tanh ^{-1}\left(\frac{(-1)^{7/8} \left((-1)^{3/4} \tan \left(\frac{x}{2}\right)+1\right)}{\sqrt{2}}\right)+\tanh ^{-1}\left(\frac{\sqrt{-1}-(-1)^{3/8} \tan \left(\frac{x}{2}\right)}{\sqrt{2}}\right)\right)-\tanh ^{-1}\left(\sqrt{-1} \csc (x)\right)+\tanh ^{-1}\left((-1)^{3/4} \csc (x)\right)\right)\] 有这个表达式就很不错了，要求那么高干啥 反正切部分似乎可以合并成，我没有讨论区间
\(\frac{\tan ^{-1}\left(\frac{\sqrt{2 \left(\sqrt{2}-1\right)} \cos (x)}{\sqrt{2}-\cos ^2(x)}\right)}{4 \sqrt{1+\sqrt{2}}}\) Factor^3+Sin^3)/(1+Sin^4),x],Round&]]/.{ArcTan-ArcTan:>ArcTan[(x-y)/(1+x y)//Simplify],Log-Log:>Log}//Total
D[%,x]==(Cos^3+Sin^3)/(1+Sin^4)//FullSimplify
将arctan和log合并一下：
$-\frac{\sqrt{-1+\sqrt{2}}}{4} \tan ^{-1}\left(\frac{\sqrt{2+2 \sqrt{2}} \cos (x)}{-2-\sqrt{2}+\left(1+\sqrt{2}\right) \cos ^2(x)}\right)+\frac{\sqrt{1+\sqrt{2}}}{8} \log \left(\frac{\sqrt{2}-\sqrt{2+2 \sqrt{2}} \cos (x)+\cos ^2(x)}{\sqrt{2}+\sqrt{2+2 \sqrt{2}} \cos (x)+\cos ^2(x)}\right)-\frac{\sqrt{2}}{4} \log \left(\frac{1-\sqrt{2} \sin (x)+\sin ^2(x)}{1+\sqrt{2} \sin (x)+\sin ^2(x)}\right)$ 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。
\(\frac{\log \left(\frac{\sin ^2(x)+\sqrt{2} \sin (x)+1}{\sin ^2(x)-\sqrt{2} \sin (x)+1}\right)}{2 \sqrt{2}}+\frac{1}{8} \sqrt{1+\sqrt{2}} \log \left(\frac{\cos ^2(x)-\color{red}{\sqrt{2+2 \sqrt{2}}} \cos (x)+\sqrt{2}}{\cos ^2(x)+\color{red}{\sqrt{2+2 \sqrt{2}}} \cos (x)+\sqrt{2}}\right)+\frac{1}{4} \sqrt{\sqrt{2}-1} \tan ^{-1}\left(\frac{\sqrt{2 \left(\sqrt{2}-1\right)} \cos (x)}{\sqrt{2}-\cos ^2(x)}\right)\) 本帖最后由 葡萄糖 于 2018-12-21 12:45 编辑
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。
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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*} 本帖最后由 zeroieme 于 2018-12-21 19:52 编辑
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
\(\frac{1}{8} \left(2 \sqrt{2}-\sqrt{1+\sqrt{2}}\right) \log \left(\frac{1+\sqrt{2-\sqrt{2}} \cos \left(x-\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)\right)}{1-\sqrt{2-\sqrt{2}} \cos \left(x-\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)\right)}\right)+\frac{1}{8} \left(2 \sqrt{2}+\sqrt{1+\sqrt{2}}\right) \log \left(\frac{1-\sqrt{2-\sqrt{2}} \cos \left(x+\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)\right)}{1+\sqrt{2-\sqrt{2}} \cos \left(x+\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)\right)}\right)+\frac{1}{4} \sqrt{\sqrt{2}-1} \tan ^{-1}\left(\frac{\sqrt{2 \left(\sqrt{2}-1\right)} \cos (x)}{\sqrt{2}-\cos ^2(x)}\right)\) 本帖最后由 zeroieme 于 2018-12-22 09:54 编辑
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。
\(Cos^3+Sin^3)/(1+Sin^4) \x//TrigToExp//#/.{x->2ArcTan}/.{Log:>(X//ExpToTrig//TrigExpand//Factor//{Numerator[#],Denominator[#]}&//(If,u],{Coefficient[#,u,Exponent[#,u]],u/.Solve[#==0,u]//u-#&},{#}]//Flatten//Log//Total)&/@#&//#[]-#[]&//RootReduce/@#&)}&//Collect[#,Log,RootReduce]&//Select[#,Variables[#]!={}&]&
Log] Root+Log] Root+Log] Root+Log] Root+Log] Root+Log] Root+Log] Root+Log] Root
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