\(\int \frac {\sqrt [3]{\tan (c+d x)}}{a+b \tan (c+d x)} \, dx\) [672]

Optimal result
Mathematica [C] (verified)
Rubi [A] (warning: unable to verify)
Maple [A] (verified)
Fricas [C] (verification not implemented)
Sympy [F]
Maxima [F(-1)]
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [F]

Optimal result

Integrand size = 23, antiderivative size = 465 \[ \int \frac {\sqrt [3]{\tan (c+d x)}}{a+b \tan (c+d x)} \, dx=-\frac {b \arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right ) d}+\frac {b \arctan \left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right ) d}+\frac {\sqrt {3} \sqrt [3]{a} b^{2/3} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{\tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{\left (a^2+b^2\right ) d}-\frac {\sqrt {3} a \arctan \left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{2 \left (a^2+b^2\right ) d}+\frac {b \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac {3 \sqrt [3]{a} b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right ) d}-\frac {a \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{2 \left (a^2+b^2\right ) d}-\frac {\sqrt {3} b \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{4 \left (a^2+b^2\right ) d}+\frac {\sqrt {3} b \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{4 \left (a^2+b^2\right ) d}+\frac {\sqrt [3]{a} b^{2/3} \log (a+b \tan (c+d x))}{2 \left (a^2+b^2\right ) d}+\frac {a \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{4 \left (a^2+b^2\right ) d} \] Output:

1/2*b*arctan(-3^(1/2)+2*tan(d*x+c)^(1/3))/(a^2+b^2)/d+1/2*b*arctan(3^(1/2) 
+2*tan(d*x+c)^(1/3))/(a^2+b^2)/d+3^(1/2)*a^(1/3)*b^(2/3)*arctan(1/3*(a^(1/ 
3)-2*b^(1/3)*tan(d*x+c)^(1/3))*3^(1/2)/a^(1/3))/(a^2+b^2)/d-1/2*3^(1/2)*a* 
arctan(1/3*(1-2*tan(d*x+c)^(2/3))*3^(1/2))/(a^2+b^2)/d+b*arctan(tan(d*x+c) 
^(1/3))/(a^2+b^2)/d-3/2*a^(1/3)*b^(2/3)*ln(a^(1/3)+b^(1/3)*tan(d*x+c)^(1/3 
))/(a^2+b^2)/d-1/2*a*ln(1+tan(d*x+c)^(2/3))/(a^2+b^2)/d-1/4*3^(1/2)*b*ln(1 
-3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))/(a^2+b^2)/d+1/4*3^(1/2)*b*ln(1 
+3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))/(a^2+b^2)/d+1/2*a^(1/3)*b^(2/3 
)*ln(a+b*tan(d*x+c))/(a^2+b^2)/d+1/4*a*ln(1-tan(d*x+c)^(2/3)+tan(d*x+c)^(4 
/3))/(a^2+b^2)/d
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.26 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.44 \[ \int \frac {\sqrt [3]{\tan (c+d x)}}{a+b \tan (c+d x)} \, dx=\frac {2 \sqrt [3]{a} b^{2/3} \left (2 \sqrt {3} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{\tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )-2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{\tan (c+d x)}\right )+\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sqrt [3]{\tan (c+d x)}+b^{2/3} \tan ^{\frac {2}{3}}(c+d x)\right )\right )+12 b \operatorname {Hypergeometric2F1}\left (\frac {1}{6},1,\frac {7}{6},-\tan ^2(c+d x)\right ) \sqrt [3]{\tan (c+d x)}+3 a \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},-\tan ^2(c+d x)\right ) \tan ^{\frac {4}{3}}(c+d x)}{4 \left (a^2+b^2\right ) d} \] Input:

Integrate[Tan[c + d*x]^(1/3)/(a + b*Tan[c + d*x]),x]
 

Output:

(2*a^(1/3)*b^(2/3)*(2*Sqrt[3]*ArcTan[(a^(1/3) - 2*b^(1/3)*Tan[c + d*x]^(1/ 
3))/(Sqrt[3]*a^(1/3))] - 2*Log[a^(1/3) + b^(1/3)*Tan[c + d*x]^(1/3)] + Log 
[a^(2/3) - a^(1/3)*b^(1/3)*Tan[c + d*x]^(1/3) + b^(2/3)*Tan[c + d*x]^(2/3) 
]) + 12*b*Hypergeometric2F1[1/6, 1, 7/6, -Tan[c + d*x]^2]*Tan[c + d*x]^(1/ 
3) + 3*a*Hypergeometric2F1[2/3, 1, 5/3, -Tan[c + d*x]^2]*Tan[c + d*x]^(4/3 
))/(4*(a^2 + b^2)*d)
 

Rubi [A] (warning: unable to verify)

Time = 1.33 (sec) , antiderivative size = 379, normalized size of antiderivative = 0.82, number of steps used = 27, number of rules used = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.130, Rules used = {3042, 4057, 3042, 4011, 3042, 4021, 3042, 3957, 266, 753, 27, 216, 807, 821, 16, 1142, 25, 1083, 217, 1103, 4117, 60, 70, 16, 1082, 217}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\sqrt [3]{\tan (c+d x)}}{a+b \tan (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt [3]{\tan (c+d x)}}{a+b \tan (c+d x)}dx\)

\(\Big \downarrow \) 4057

\(\displaystyle \frac {b^2 \int \frac {\sqrt [3]{\tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}{a+b \tan (c+d x)}dx}{a^2+b^2}+\frac {\int \sqrt [3]{\tan (c+d x)} (a-b \tan (c+d x))dx}{a^2+b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 \int \frac {\sqrt [3]{\tan (c+d x)} \left (\tan (c+d x)^2+1\right )}{a+b \tan (c+d x)}dx}{a^2+b^2}+\frac {\int \sqrt [3]{\tan (c+d x)} (a-b \tan (c+d x))dx}{a^2+b^2}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {\int \frac {b+a \tan (c+d x)}{\tan ^{\frac {2}{3}}(c+d x)}dx-\frac {3 b \sqrt [3]{\tan (c+d x)}}{d}}{a^2+b^2}+\frac {b^2 \int \frac {\sqrt [3]{\tan (c+d x)} \left (\tan (c+d x)^2+1\right )}{a+b \tan (c+d x)}dx}{a^2+b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 \int \frac {\sqrt [3]{\tan (c+d x)} \left (\tan (c+d x)^2+1\right )}{a+b \tan (c+d x)}dx}{a^2+b^2}+\frac {\int \frac {b+a \tan (c+d x)}{\tan (c+d x)^{2/3}}dx-\frac {3 b \sqrt [3]{\tan (c+d x)}}{d}}{a^2+b^2}\)

\(\Big \downarrow \) 4021

\(\displaystyle \frac {a \int \sqrt [3]{\tan (c+d x)}dx+b \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)}dx-\frac {3 b \sqrt [3]{\tan (c+d x)}}{d}}{a^2+b^2}+\frac {b^2 \int \frac {\sqrt [3]{\tan (c+d x)} \left (\tan (c+d x)^2+1\right )}{a+b \tan (c+d x)}dx}{a^2+b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 \int \frac {\sqrt [3]{\tan (c+d x)} \left (\tan (c+d x)^2+1\right )}{a+b \tan (c+d x)}dx}{a^2+b^2}+\frac {a \int \sqrt [3]{\tan (c+d x)}dx+b \int \frac {1}{\tan (c+d x)^{2/3}}dx-\frac {3 b \sqrt [3]{\tan (c+d x)}}{d}}{a^2+b^2}\)

\(\Big \downarrow \) 3957

\(\displaystyle \frac {\frac {a \int \frac {\sqrt [3]{\tan (c+d x)}}{\tan ^2(c+d x)+1}d\tan (c+d x)}{d}+\frac {b \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x) \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{d}-\frac {3 b \sqrt [3]{\tan (c+d x)}}{d}}{a^2+b^2}+\frac {b^2 \int \frac {\sqrt [3]{\tan (c+d x)} \left (\tan (c+d x)^2+1\right )}{a+b \tan (c+d x)}dx}{a^2+b^2}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {\frac {3 a \int \frac {\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}+\frac {3 b \int \frac {1}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}-\frac {3 b \sqrt [3]{\tan (c+d x)}}{d}}{a^2+b^2}+\frac {b^2 \int \frac {\sqrt [3]{\tan (c+d x)} \left (\tan (c+d x)^2+1\right )}{a+b \tan (c+d x)}dx}{a^2+b^2}\)

\(\Big \downarrow \) 753

\(\displaystyle \frac {\frac {3 a \int \frac {\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}+\frac {3 b \left (\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \int \frac {2-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{2 \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+2}{2 \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}d\sqrt [3]{\tan (c+d x)}\right )}{d}-\frac {3 b \sqrt [3]{\tan (c+d x)}}{d}}{a^2+b^2}+\frac {b^2 \int \frac {\sqrt [3]{\tan (c+d x)} \left (\tan (c+d x)^2+1\right )}{a+b \tan (c+d x)}dx}{a^2+b^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3 a \int \frac {\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}+\frac {3 b \left (\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{6} \int \frac {2-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+2}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )}{d}-\frac {3 b \sqrt [3]{\tan (c+d x)}}{d}}{a^2+b^2}+\frac {b^2 \int \frac {\sqrt [3]{\tan (c+d x)} \left (\tan (c+d x)^2+1\right )}{a+b \tan (c+d x)}dx}{a^2+b^2}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {3 a \int \frac {\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}+\frac {3 b \left (\frac {1}{6} \int \frac {2-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+2}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {3 b \sqrt [3]{\tan (c+d x)}}{d}}{a^2+b^2}+\frac {b^2 \int \frac {\sqrt [3]{\tan (c+d x)} \left (\tan (c+d x)^2+1\right )}{a+b \tan (c+d x)}dx}{a^2+b^2}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {\frac {3 a \int \frac {\tan ^{\frac {2}{3}}(c+d x)}{\tan (c+d x)+1}d\tan ^{\frac {2}{3}}(c+d x)}{2 d}+\frac {3 b \left (\frac {1}{6} \int \frac {2-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+2}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {3 b \sqrt [3]{\tan (c+d x)}}{d}}{a^2+b^2}+\frac {b^2 \int \frac {\sqrt [3]{\tan (c+d x)} \left (\tan (c+d x)^2+1\right )}{a+b \tan (c+d x)}dx}{a^2+b^2}\)

\(\Big \downarrow \) 821

\(\displaystyle \frac {\frac {3 a \left (\frac {1}{3} \int \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )d\tan ^{\frac {2}{3}}(c+d x)-\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\tan ^{\frac {2}{3}}(c+d x)\right )}{2 d}+\frac {3 b \left (\frac {1}{6} \int \frac {2-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+2}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {3 b \sqrt [3]{\tan (c+d x)}}{d}}{a^2+b^2}+\frac {b^2 \int \frac {\sqrt [3]{\tan (c+d x)} \left (\tan (c+d x)^2+1\right )}{a+b \tan (c+d x)}dx}{a^2+b^2}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {\frac {3 a \left (\frac {1}{3} \int \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )d\tan ^{\frac {2}{3}}(c+d x)-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{2 d}+\frac {3 b \left (\frac {1}{6} \int \frac {2-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+2}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {3 b \sqrt [3]{\tan (c+d x)}}{d}}{a^2+b^2}+\frac {b^2 \int \frac {\sqrt [3]{\tan (c+d x)} \left (\tan (c+d x)^2+1\right )}{a+b \tan (c+d x)}dx}{a^2+b^2}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {\frac {3 a \left (\frac {1}{3} \left (\frac {3}{2} \int 1d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{2} \int \left (2 \tan ^{\frac {2}{3}}(c+d x)-1\right )d\tan ^{\frac {2}{3}}(c+d x)\right )-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{2 d}+\frac {3 b \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{2} \sqrt {3} \int -\frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {3 b \sqrt [3]{\tan (c+d x)}}{d}}{a^2+b^2}+\frac {b^2 \int \frac {\sqrt [3]{\tan (c+d x)} \left (\tan (c+d x)^2+1\right )}{a+b \tan (c+d x)}dx}{a^2+b^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {3 a \left (\frac {1}{3} \left (\frac {3}{2} \int 1d\tan ^{\frac {2}{3}}(c+d x)-\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)\right )-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{2 d}+\frac {3 b \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {3 b \sqrt [3]{\tan (c+d x)}}{d}}{a^2+b^2}+\frac {b^2 \int \frac {\sqrt [3]{\tan (c+d x)} \left (\tan (c+d x)^2+1\right )}{a+b \tan (c+d x)}dx}{a^2+b^2}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {\frac {3 a \left (\frac {1}{3} \left (-3 \int \frac {1}{-2 \tan ^{\frac {2}{3}}(c+d x)-2}d\left (2 \tan ^{\frac {2}{3}}(c+d x)-1\right )-\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)\right )-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{2 d}+\frac {3 b \left (\frac {1}{6} \left (\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\int \frac {1}{-\tan ^{\frac {2}{3}}(c+d x)-1}d\left (2 \sqrt [3]{\tan (c+d x)}-\sqrt {3}\right )\right )+\frac {1}{6} \left (\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\int \frac {1}{-\tan ^{\frac {2}{3}}(c+d x)-1}d\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {3 b \sqrt [3]{\tan (c+d x)}}{d}}{a^2+b^2}+\frac {b^2 \int \frac {\sqrt [3]{\tan (c+d x)} \left (\tan (c+d x)^2+1\right )}{a+b \tan (c+d x)}dx}{a^2+b^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {3 a \left (\frac {1}{3} \left (\sqrt {3} \arctan \left (\frac {2 \tan ^{\frac {2}{3}}(c+d x)-1}{\sqrt {3}}\right )-\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)\right )-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{2 d}+\frac {3 b \left (\frac {1}{6} \left (\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )\right )+\frac {1}{6} \left (\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {3 b \sqrt [3]{\tan (c+d x)}}{d}}{a^2+b^2}+\frac {b^2 \int \frac {\sqrt [3]{\tan (c+d x)} \left (\tan (c+d x)^2+1\right )}{a+b \tan (c+d x)}dx}{a^2+b^2}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {b^2 \int \frac {\sqrt [3]{\tan (c+d x)} \left (\tan (c+d x)^2+1\right )}{a+b \tan (c+d x)}dx}{a^2+b^2}+\frac {\frac {3 a \left (\frac {\arctan \left (\frac {2 \tan ^{\frac {2}{3}}(c+d x)-1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{2 d}+\frac {3 b \left (\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (-\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )-\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )\right )+\frac {1}{6} \left (\arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )+\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )\right )\right )}{d}-\frac {3 b \sqrt [3]{\tan (c+d x)}}{d}}{a^2+b^2}\)

\(\Big \downarrow \) 4117

\(\displaystyle \frac {b^2 \int \frac {\sqrt [3]{\tan (c+d x)}}{a+b \tan (c+d x)}d\tan (c+d x)}{d \left (a^2+b^2\right )}+\frac {\frac {3 a \left (\frac {\arctan \left (\frac {2 \tan ^{\frac {2}{3}}(c+d x)-1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{2 d}+\frac {3 b \left (\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (-\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )-\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )\right )+\frac {1}{6} \left (\arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )+\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )\right )\right )}{d}-\frac {3 b \sqrt [3]{\tan (c+d x)}}{d}}{a^2+b^2}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {b^2 \left (\frac {3 \sqrt [3]{\tan (c+d x)}}{b}-\frac {a \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x) (a+b \tan (c+d x))}d\tan (c+d x)}{b}\right )}{d \left (a^2+b^2\right )}+\frac {\frac {3 a \left (\frac {\arctan \left (\frac {2 \tan ^{\frac {2}{3}}(c+d x)-1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{2 d}+\frac {3 b \left (\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (-\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )-\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )\right )+\frac {1}{6} \left (\arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )+\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )\right )\right )}{d}-\frac {3 b \sqrt [3]{\tan (c+d x)}}{d}}{a^2+b^2}\)

\(\Big \downarrow \) 70

\(\displaystyle \frac {b^2 \left (\frac {3 \sqrt [3]{\tan (c+d x)}}{b}-\frac {a \left (\frac {3 \int \frac {1}{\frac {a^{2/3}}{b^{2/3}}-\frac {\sqrt [3]{\tan (c+d x)} \sqrt [3]{a}}{\sqrt [3]{b}}+\tan ^{\frac {2}{3}}(c+d x)}d\sqrt [3]{\tan (c+d x)}}{2 \sqrt [3]{a} b^{2/3}}+\frac {3 \int \frac {1}{\frac {\sqrt [3]{a}}{\sqrt [3]{b}}+\sqrt [3]{\tan (c+d x)}}d\sqrt [3]{\tan (c+d x)}}{2 a^{2/3} \sqrt [3]{b}}-\frac {\log (a+b \tan (c+d x))}{2 a^{2/3} \sqrt [3]{b}}\right )}{b}\right )}{d \left (a^2+b^2\right )}+\frac {\frac {3 a \left (\frac {\arctan \left (\frac {2 \tan ^{\frac {2}{3}}(c+d x)-1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{2 d}+\frac {3 b \left (\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (-\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )-\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )\right )+\frac {1}{6} \left (\arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )+\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )\right )\right )}{d}-\frac {3 b \sqrt [3]{\tan (c+d x)}}{d}}{a^2+b^2}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {b^2 \left (\frac {3 \sqrt [3]{\tan (c+d x)}}{b}-\frac {a \left (\frac {3 \int \frac {1}{\frac {a^{2/3}}{b^{2/3}}-\frac {\sqrt [3]{\tan (c+d x)} \sqrt [3]{a}}{\sqrt [3]{b}}+\tan ^{\frac {2}{3}}(c+d x)}d\sqrt [3]{\tan (c+d x)}}{2 \sqrt [3]{a} b^{2/3}}+\frac {3 \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{\tan (c+d x)}\right )}{2 a^{2/3} \sqrt [3]{b}}-\frac {\log (a+b \tan (c+d x))}{2 a^{2/3} \sqrt [3]{b}}\right )}{b}\right )}{d \left (a^2+b^2\right )}+\frac {\frac {3 a \left (\frac {\arctan \left (\frac {2 \tan ^{\frac {2}{3}}(c+d x)-1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{2 d}+\frac {3 b \left (\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (-\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )-\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )\right )+\frac {1}{6} \left (\arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )+\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )\right )\right )}{d}-\frac {3 b \sqrt [3]{\tan (c+d x)}}{d}}{a^2+b^2}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {b^2 \left (\frac {3 \sqrt [3]{\tan (c+d x)}}{b}-\frac {a \left (\frac {3 \int \frac {1}{-\tan ^{\frac {2}{3}}(c+d x)-3}d\left (1-\frac {2 \sqrt [3]{b} \sqrt [3]{\tan (c+d x)}}{\sqrt [3]{a}}\right )}{a^{2/3} \sqrt [3]{b}}+\frac {3 \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{\tan (c+d x)}\right )}{2 a^{2/3} \sqrt [3]{b}}-\frac {\log (a+b \tan (c+d x))}{2 a^{2/3} \sqrt [3]{b}}\right )}{b}\right )}{d \left (a^2+b^2\right )}+\frac {\frac {3 a \left (\frac {\arctan \left (\frac {2 \tan ^{\frac {2}{3}}(c+d x)-1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{2 d}+\frac {3 b \left (\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (-\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )-\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )\right )+\frac {1}{6} \left (\arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )+\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )\right )\right )}{d}-\frac {3 b \sqrt [3]{\tan (c+d x)}}{d}}{a^2+b^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {3 a \left (\frac {\arctan \left (\frac {2 \tan ^{\frac {2}{3}}(c+d x)-1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{2 d}+\frac {3 b \left (\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (-\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )-\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )\right )+\frac {1}{6} \left (\arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )+\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )\right )\right )}{d}-\frac {3 b \sqrt [3]{\tan (c+d x)}}{d}}{a^2+b^2}+\frac {b^2 \left (\frac {3 \sqrt [3]{\tan (c+d x)}}{b}-\frac {a \left (-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} \sqrt [3]{\tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{a^{2/3} \sqrt [3]{b}}+\frac {3 \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{\tan (c+d x)}\right )}{2 a^{2/3} \sqrt [3]{b}}-\frac {\log (a+b \tan (c+d x))}{2 a^{2/3} \sqrt [3]{b}}\right )}{b}\right )}{d \left (a^2+b^2\right )}\)

Input:

Int[Tan[c + d*x]^(1/3)/(a + b*Tan[c + d*x]),x]
 

Output:

(b^2*(-((a*(-((Sqrt[3]*ArcTan[(1 - (2*b^(1/3)*Tan[c + d*x]^(1/3))/a^(1/3)) 
/Sqrt[3]])/(a^(2/3)*b^(1/3))) + (3*Log[a^(1/3) + b^(1/3)*Tan[c + d*x]^(1/3 
)])/(2*a^(2/3)*b^(1/3)) - Log[a + b*Tan[c + d*x]]/(2*a^(2/3)*b^(1/3))))/b) 
 + (3*Tan[c + d*x]^(1/3))/b))/((a^2 + b^2)*d) + ((3*a*(ArcTan[(-1 + 2*Tan[ 
c + d*x]^(2/3))/Sqrt[3]]/Sqrt[3] - Log[1 + Tan[c + d*x]^(2/3)]/3))/(2*d) + 
 (3*b*(ArcTan[Tan[c + d*x]^(1/3)]/3 + (-ArcTan[Sqrt[3] - 2*Tan[c + d*x]^(1 
/3)] - (Sqrt[3]*Log[1 - Sqrt[3]*Tan[c + d*x]^(1/3) + Tan[c + d*x]^(2/3)])/ 
2)/6 + (ArcTan[Sqrt[3] + 2*Tan[c + d*x]^(1/3)] + (Sqrt[3]*Log[1 + Sqrt[3]* 
Tan[c + d*x]^(1/3) + Tan[c + d*x]^(2/3)])/2)/6))/d - (3*b*Tan[c + d*x]^(1/ 
3))/d)/(a^2 + b^2)
 

Defintions of rubi rules used

rule 16
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 60
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 70
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[ 
{q = Rt[-(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q^2) 
, x] + (Simp[3/(2*b*q)   Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d*x)^(1 
/3)], x] + Simp[3/(2*b*q^2)   Subst[Int[1/(q + x), x], x, (c + d*x)^(1/3)], 
 x])] /; FreeQ[{a, b, c, d}, x] && NegQ[(b*c - a*d)/b]
 

rule 216
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 217
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 266
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 753
Int[((a_) + (b_.)*(x_)^(n_))^(-1), x_Symbol] :> Module[{r = Numerator[Rt[a/ 
b, n]], s = Denominator[Rt[a/b, n]], k, u, v}, Simp[u = Int[(r - s*Cos[(2*k 
 - 1)*(Pi/n)]*x)/(r^2 - 2*r*s*Cos[(2*k - 1)*(Pi/n)]*x + s^2*x^2), x] + Int[ 
(r + s*Cos[(2*k - 1)*(Pi/n)]*x)/(r^2 + 2*r*s*Cos[(2*k - 1)*(Pi/n)]*x + s^2* 
x^2), x]; 2*(r^2/(a*n))   Int[1/(r^2 + s^2*x^2), x] + 2*(r/(a*n))   Sum[u, 
{k, 1, (n - 2)/4}], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] && PosQ[a 
/b]
 

rule 807
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 821
Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Simp[-(3*Rt[a, 3]*Rt[b, 3])^(- 
1)   Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]*Rt[b, 3]) 
 Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2 
*x^2), x], x] /; FreeQ[{a, b}, x]
 

rule 1082
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1083
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 1103
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1142
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3957
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d   Subst[Int 
[x^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{b, c, d, n}, x] && 
!IntegerQ[n]
 

rule 4011
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Int 
[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x], x] 
, x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 
 0] && GtQ[m, 0]
 

rule 4021
Int[((b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Tan[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Tan[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x] && NeQ[c^ 
2 + d^2, 0] &&  !IntegerQ[2*m]
 

rule 4057
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)/((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[1/(c^2 + d^2)   Int[(a + b*Tan[e + f*x])^m 
*(c - d*Tan[e + f*x]), x], x] + Simp[d^2/(c^2 + d^2)   Int[(a + b*Tan[e + f 
*x])^m*((1 + Tan[e + f*x]^2)/(c + d*Tan[e + f*x])), x], x] /; FreeQ[{a, b, 
c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d 
^2, 0] &&  !IntegerQ[m]
 

rule 4117
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) 
+ (f_.)*(x_)])^(n_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
 Simp[A/f   Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x] /; 
FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]
 
Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 328, normalized size of antiderivative = 0.71

method result size
derivativedivides \(\frac {\frac {-\frac {3 \left (\sqrt {3}\, b -a \right ) \ln \left (1-\sqrt {3}\, \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}\right )}{4}-3 \left (-2 b +\frac {\left (\sqrt {3}\, b -a \right ) \sqrt {3}}{2}\right ) \arctan \left (-\sqrt {3}+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right )+\frac {3 \left (\sqrt {3}\, b +a \right ) \ln \left (1+\sqrt {3}\, \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}\right )}{4}+3 \left (2 b -\frac {\left (\sqrt {3}\, b +a \right ) \sqrt {3}}{2}\right ) \arctan \left (\sqrt {3}+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right )}{3 a^{2}+3 b^{2}}-\frac {3 \left (\frac {\ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (\tan \left (d x +c \right )^{\frac {2}{3}}-\left (\frac {a}{b}\right )^{\frac {1}{3}} \tan \left (d x +c \right )^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \tan \left (d x +c \right )^{\frac {1}{3}}}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right ) b a}{a^{2}+b^{2}}+\frac {-\frac {3 a \ln \left (1+\tan \left (d x +c \right )^{\frac {2}{3}}\right )}{2}+3 b \arctan \left (\tan \left (d x +c \right )^{\frac {1}{3}}\right )}{3 a^{2}+3 b^{2}}}{d}\) \(328\)
default \(\frac {\frac {-\frac {3 \left (\sqrt {3}\, b -a \right ) \ln \left (1-\sqrt {3}\, \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}\right )}{4}-3 \left (-2 b +\frac {\left (\sqrt {3}\, b -a \right ) \sqrt {3}}{2}\right ) \arctan \left (-\sqrt {3}+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right )+\frac {3 \left (\sqrt {3}\, b +a \right ) \ln \left (1+\sqrt {3}\, \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}\right )}{4}+3 \left (2 b -\frac {\left (\sqrt {3}\, b +a \right ) \sqrt {3}}{2}\right ) \arctan \left (\sqrt {3}+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right )}{3 a^{2}+3 b^{2}}-\frac {3 \left (\frac {\ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (\tan \left (d x +c \right )^{\frac {2}{3}}-\left (\frac {a}{b}\right )^{\frac {1}{3}} \tan \left (d x +c \right )^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \tan \left (d x +c \right )^{\frac {1}{3}}}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right ) b a}{a^{2}+b^{2}}+\frac {-\frac {3 a \ln \left (1+\tan \left (d x +c \right )^{\frac {2}{3}}\right )}{2}+3 b \arctan \left (\tan \left (d x +c \right )^{\frac {1}{3}}\right )}{3 a^{2}+3 b^{2}}}{d}\) \(328\)

Input:

int(tan(d*x+c)^(1/3)/(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

1/d*(3/(3*a^2+3*b^2)*(-1/4*(3^(1/2)*b-a)*ln(1-3^(1/2)*tan(d*x+c)^(1/3)+tan 
(d*x+c)^(2/3))-(-2*b+1/2*(3^(1/2)*b-a)*3^(1/2))*arctan(-3^(1/2)+2*tan(d*x+ 
c)^(1/3))+1/4*(3^(1/2)*b+a)*ln(1+3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3) 
)+(2*b-1/2*(3^(1/2)*b+a)*3^(1/2))*arctan(3^(1/2)+2*tan(d*x+c)^(1/3)))-3*(1 
/3/b/(a/b)^(2/3)*ln(tan(d*x+c)^(1/3)+(a/b)^(1/3))-1/6/b/(a/b)^(2/3)*ln(tan 
(d*x+c)^(2/3)-(a/b)^(1/3)*tan(d*x+c)^(1/3)+(a/b)^(2/3))+1/3/b/(a/b)^(2/3)* 
3^(1/2)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*tan(d*x+c)^(1/3)-1)))*b/(a^2+b^2 
)*a+3/(3*a^2+3*b^2)*(-1/2*a*ln(1+tan(d*x+c)^(2/3))+b*arctan(tan(d*x+c)^(1/ 
3))))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.84 (sec) , antiderivative size = 74435, normalized size of antiderivative = 160.08 \[ \int \frac {\sqrt [3]{\tan (c+d x)}}{a+b \tan (c+d x)} \, dx=\text {Too large to display} \] Input:

integrate(tan(d*x+c)^(1/3)/(a+b*tan(d*x+c)),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {\sqrt [3]{\tan (c+d x)}}{a+b \tan (c+d x)} \, dx=\int \frac {\sqrt [3]{\tan {\left (c + d x \right )}}}{a + b \tan {\left (c + d x \right )}}\, dx \] Input:

integrate(tan(d*x+c)**(1/3)/(a+b*tan(d*x+c)),x)
 

Output:

Integral(tan(c + d*x)**(1/3)/(a + b*tan(c + d*x)), x)
 

Maxima [F(-1)]

Timed out. \[ \int \frac {\sqrt [3]{\tan (c+d x)}}{a+b \tan (c+d x)} \, dx=\text {Timed out} \] Input:

integrate(tan(d*x+c)^(1/3)/(a+b*tan(d*x+c)),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 424, normalized size of antiderivative = 0.91 \[ \int \frac {\sqrt [3]{\tan (c+d x)}}{a+b \tan (c+d x)} \, dx=\frac {a b \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | -\left (-\frac {a}{b}\right )^{\frac {1}{3}} + \tan \left (d x + c\right )^{\frac {1}{3}} \right |}\right )}{a^{3} d + a b^{2} d} - \frac {{\left (\sqrt {3} a - b\right )} \arctan \left (\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}}\right )}{2 \, {\left (a^{2} d + b^{2} d\right )}} + \frac {{\left (\sqrt {3} a + b\right )} \arctan \left (-\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}}\right )}{2 \, {\left (a^{2} d + b^{2} d\right )}} + \frac {b \arctan \left (\tan \left (d x + c\right )^{\frac {1}{3}}\right )}{a^{2} d + b^{2} d} + \frac {a \log \left (\tan \left (d x + c\right )^{\frac {4}{3}} - \tan \left (d x + c\right )^{\frac {2}{3}} + 1\right )}{4 \, {\left (a^{2} d + b^{2} d\right )}} + \frac {3 \, b \log \left (\sqrt {3} \tan \left (d x + c\right )^{\frac {1}{3}} + \tan \left (d x + c\right )^{\frac {2}{3}} + 1\right )}{4 \, {\left (\sqrt {3} a^{2} d + \sqrt {3} b^{2} d\right )}} - \frac {3 \, b \log \left (-\sqrt {3} \tan \left (d x + c\right )^{\frac {1}{3}} + \tan \left (d x + c\right )^{\frac {2}{3}} + 1\right )}{4 \, {\left (\sqrt {3} a^{2} d + \sqrt {3} b^{2} d\right )}} - \frac {a \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} + 1\right )}{2 \, {\left (a^{2} d + b^{2} d\right )}} - \frac {3 \, \left (-a b^{2}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (\left (-\frac {a}{b}\right )^{\frac {1}{3}} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{{\left (\sqrt {3} a^{2} + \sqrt {3} b^{2}\right )} d} - \frac {\left (-a b^{2}\right )^{\frac {1}{3}} \log \left (\left (-\frac {a}{b}\right )^{\frac {2}{3}} + \left (-\frac {a}{b}\right )^{\frac {1}{3}} \tan \left (d x + c\right )^{\frac {1}{3}} + \tan \left (d x + c\right )^{\frac {2}{3}}\right )}{2 \, {\left (a^{2} + b^{2}\right )} d} \] Input:

integrate(tan(d*x+c)^(1/3)/(a+b*tan(d*x+c)),x, algorithm="giac")
 

Output:

a*b*(-a/b)^(1/3)*log(abs(-(-a/b)^(1/3) + tan(d*x + c)^(1/3)))/(a^3*d + a*b 
^2*d) - 1/2*(sqrt(3)*a - b)*arctan(sqrt(3) + 2*tan(d*x + c)^(1/3))/(a^2*d 
+ b^2*d) + 1/2*(sqrt(3)*a + b)*arctan(-sqrt(3) + 2*tan(d*x + c)^(1/3))/(a^ 
2*d + b^2*d) + b*arctan(tan(d*x + c)^(1/3))/(a^2*d + b^2*d) + 1/4*a*log(ta 
n(d*x + c)^(4/3) - tan(d*x + c)^(2/3) + 1)/(a^2*d + b^2*d) + 3/4*b*log(sqr 
t(3)*tan(d*x + c)^(1/3) + tan(d*x + c)^(2/3) + 1)/(sqrt(3)*a^2*d + sqrt(3) 
*b^2*d) - 3/4*b*log(-sqrt(3)*tan(d*x + c)^(1/3) + tan(d*x + c)^(2/3) + 1)/ 
(sqrt(3)*a^2*d + sqrt(3)*b^2*d) - 1/2*a*log(tan(d*x + c)^(2/3) + 1)/(a^2*d 
 + b^2*d) - 3*(-a*b^2)^(1/3)*arctan(1/3*sqrt(3)*((-a/b)^(1/3) + 2*tan(d*x 
+ c)^(1/3))/(-a/b)^(1/3))/((sqrt(3)*a^2 + sqrt(3)*b^2)*d) - 1/2*(-a*b^2)^( 
1/3)*log((-a/b)^(2/3) + (-a/b)^(1/3)*tan(d*x + c)^(1/3) + tan(d*x + c)^(2/ 
3))/((a^2 + b^2)*d)
 

Mupad [B] (verification not implemented)

Time = 8.92 (sec) , antiderivative size = 2111, normalized size of antiderivative = 4.54 \[ \int \frac {\sqrt [3]{\tan (c+d x)}}{a+b \tan (c+d x)} \, dx=\text {Too large to display} \] Input:

int(tan(c + d*x)^(1/3)/(a + b*tan(c + d*x)),x)
 

Output:

symsum(log(-root(32*a^2*b^2*d^4*z^4 + 16*b^4*d^4*z^4 + 16*a^4*d^4*z^4 - 16 
*a*b^2*d^3*z^3 - 16*a^3*d^3*z^3 - 4*b^2*d^2*z^2 + 12*a^2*d^2*z^2 - 4*a*d*z 
 + 1, z, k)*(root(32*a^2*b^2*d^4*z^4 + 16*b^4*d^4*z^4 + 16*a^4*d^4*z^4 - 1 
6*a*b^2*d^3*z^3 - 16*a^3*d^3*z^3 - 4*b^2*d^2*z^2 + 12*a^2*d^2*z^2 - 4*a*d* 
z + 1, z, k)*(root(32*a^2*b^2*d^4*z^4 + 16*b^4*d^4*z^4 + 16*a^4*d^4*z^4 - 
16*a*b^2*d^3*z^3 - 16*a^3*d^3*z^3 - 4*b^2*d^2*z^2 + 12*a^2*d^2*z^2 - 4*a*d 
*z + 1, z, k)^2*(root(32*a^2*b^2*d^4*z^4 + 16*b^4*d^4*z^4 + 16*a^4*d^4*z^4 
 - 16*a*b^2*d^3*z^3 - 16*a^3*d^3*z^3 - 4*b^2*d^2*z^2 + 12*a^2*d^2*z^2 - 4* 
a*d*z + 1, z, k)*((6561*(44*a^2*b^10*d^3 + 84*a^4*b^8*d^3 + 36*a^6*b^6*d^3 
 - 4*a^8*b^4*d^3))/d^6 + root(32*a^2*b^2*d^4*z^4 + 16*b^4*d^4*z^4 + 16*a^4 
*d^4*z^4 - 16*a*b^2*d^3*z^3 - 16*a^3*d^3*z^3 - 4*b^2*d^2*z^2 + 12*a^2*d^2* 
z^2 - 4*a*d*z + 1, z, k)^2*((6561*tan(c + d*x)^(1/3)*(64*a*b^13*d^6 + 240* 
a^3*b^11*d^6 + 320*a^5*b^9*d^6 + 160*a^7*b^7*d^6 - 16*a^11*b^3*d^6))/d^7 - 
 (6561*root(32*a^2*b^2*d^4*z^4 + 16*b^4*d^4*z^4 + 16*a^4*d^4*z^4 - 16*a*b^ 
2*d^3*z^3 - 16*a^3*d^3*z^3 - 4*b^2*d^2*z^2 + 12*a^2*d^2*z^2 - 4*a*d*z + 1, 
 z, k)*(64*a*b^14*d^6 + 192*a^3*b^12*d^6 + 128*a^5*b^10*d^6 - 128*a^7*b^8* 
d^6 - 192*a^9*b^6*d^6 - 64*a^11*b^4*d^6))/d^6)) - (6561*tan(c + d*x)^(1/3) 
*(50*a^2*b^9*d^3 - 58*a^4*b^7*d^3 + 22*a^6*b^5*d^3 + 2*a^8*b^3*d^3))/d^7) 
- (6561*(a*b^8 + 5*a^3*b^6))/d^6) + (6561*tan(c + d*x)^(1/3)*(a*b^7 - 2*a^ 
3*b^5))/d^7))*root(32*a^2*b^2*d^4*z^4 + 16*b^4*d^4*z^4 + 16*a^4*d^4*z^4...
 

Reduce [F]

\[ \int \frac {\sqrt [3]{\tan (c+d x)}}{a+b \tan (c+d x)} \, dx=\frac {3 \tan \left (d x +c \right )^{\frac {1}{3}}-\left (\int \frac {\tan \left (d x +c \right )^{\frac {1}{3}}}{\tan \left (d x +c \right )^{2} b +\tan \left (d x +c \right ) a}d x \right ) a d -\left (\int \frac {\tan \left (d x +c \right )^{\frac {7}{3}}}{a +\tan \left (d x +c \right ) b}d x \right ) b d -\left (\int \frac {\tan \left (d x +c \right )^{\frac {4}{3}}}{a +\tan \left (d x +c \right ) b}d x \right ) a d}{b d} \] Input:

int(tan(d*x+c)^(1/3)/(a+b*tan(d*x+c)),x)
 

Output:

(3*tan(c + d*x)**(1/3) - int(tan(c + d*x)**(1/3)/(tan(c + d*x)**2*b + tan( 
c + d*x)*a),x)*a*d - int((tan(c + d*x)**(1/3)*tan(c + d*x)**2)/(tan(c + d* 
x)*b + a),x)*b*d - int((tan(c + d*x)**(1/3)*tan(c + d*x))/(tan(c + d*x)*b 
+ a),x)*a*d)/(b*d)