\(\int \frac {1}{\tan ^{\frac {5}{3}}(c+d x) (a+b \tan (c+d x))} \, dx\) [674]

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 = 525 \[ \int \frac {1}{\tan ^{\frac {5}{3}}(c+d x) (a+b \tan (c+d x))} \, dx=\frac {b \arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right ) d}-\frac {b \arctan \left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right ) d}+\frac {\sqrt {3} b^{8/3} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{\tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{a^{5/3} \left (a^2+b^2\right ) d}+\frac {\sqrt {3} a \arctan \left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{2 \left (a^2+b^2\right ) d}-\frac {b \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac {3 b^{8/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{\tan (c+d x)}\right )}{2 a^{5/3} \left (a^2+b^2\right ) d}+\frac {a \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{2 \left (a^2+b^2\right ) d}+\frac {\sqrt {3} b \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{4 \left (a^2+b^2\right ) d}-\frac {\sqrt {3} b \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{4 \left (a^2+b^2\right ) d}+\frac {b^{8/3} \log (a+b \tan (c+d x))}{2 a^{5/3} \left (a^2+b^2\right ) d}-\frac {a \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{4 \left (a^2+b^2\right ) d}-\frac {3 a}{2 \left (a^2+b^2\right ) d \tan ^{\frac {2}{3}}(c+d x)}-\frac {3 b^2}{2 a \left (a^2+b^2\right ) d \tan ^{\frac {2}{3}}(c+d x)} \] 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)*b^(8/3)*arctan(1/3*(a^(1/3)-2*b^ 
(1/3)*tan(d*x+c)^(1/3))*3^(1/2)/a^(1/3))/a^(5/3)/(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*b^(8/3)*ln(a^(1/3)+b^(1/3)*tan(d*x+c)^(1/3))/a^(5 
/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*b^(8/3)*ln(a+ 
b*tan(d*x+c))/a^(5/3)/(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-3/2*a/(a^2+b^2)/d/tan(d*x+c)^(2/3)-3/2*b^2/a/(a^2+b^2)/d 
/tan(d*x+c)^(2/3)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

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

Time = 0.16 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.20 \[ \int \frac {1}{\tan ^{\frac {5}{3}}(c+d x) (a+b \tan (c+d x))} \, dx=-\frac {3 \left (b^2 \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},1,\frac {1}{3},-\frac {b \tan (c+d x)}{a}\right )+a \left (a \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},1,\frac {2}{3},-\tan ^2(c+d x)\right )+2 b \operatorname {Hypergeometric2F1}\left (\frac {1}{6},1,\frac {7}{6},-\tan ^2(c+d x)\right ) \tan (c+d x)\right )\right )}{2 a \left (a^2+b^2\right ) d \tan ^{\frac {2}{3}}(c+d x)} \] Input:

Integrate[1/(Tan[c + d*x]^(5/3)*(a + b*Tan[c + d*x])),x]
 

Output:

(-3*(b^2*Hypergeometric2F1[-2/3, 1, 1/3, -((b*Tan[c + d*x])/a)] + a*(a*Hyp 
ergeometric2F1[-1/3, 1, 2/3, -Tan[c + d*x]^2] + 2*b*Hypergeometric2F1[1/6, 
 1, 7/6, -Tan[c + d*x]^2]*Tan[c + d*x])))/(2*a*(a^2 + b^2)*d*Tan[c + d*x]^ 
(2/3))
 

Rubi [A] (warning: unable to verify)

Time = 1.39 (sec) , antiderivative size = 383, normalized size of antiderivative = 0.73, number of steps used = 28, number of rules used = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.174, Rules used = {3042, 4057, 3042, 4012, 25, 3042, 4021, 3042, 3957, 266, 753, 27, 216, 807, 821, 16, 1142, 25, 1083, 217, 1103, 4117, 61, 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.

\(\displaystyle \int \frac {1}{\tan ^{\frac {5}{3}}(c+d x) (a+b \tan (c+d x))} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\tan (c+d x)^{5/3} (a+b \tan (c+d x))}dx\)

\(\Big \downarrow \) 4057

\(\displaystyle \frac {\int \frac {a-b \tan (c+d x)}{\tan ^{\frac {5}{3}}(c+d x)}dx}{a^2+b^2}+\frac {b^2 \int \frac {\tan ^2(c+d x)+1}{\tan ^{\frac {5}{3}}(c+d x) (a+b \tan (c+d x))}dx}{a^2+b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 \int \frac {\tan (c+d x)^2+1}{\tan (c+d x)^{5/3} (a+b \tan (c+d x))}dx}{a^2+b^2}+\frac {\int \frac {a-b \tan (c+d x)}{\tan (c+d x)^{5/3}}dx}{a^2+b^2}\)

\(\Big \downarrow \) 4012

\(\displaystyle \frac {\int -\frac {b+a \tan (c+d x)}{\tan ^{\frac {2}{3}}(c+d x)}dx-\frac {3 a}{2 d \tan ^{\frac {2}{3}}(c+d x)}}{a^2+b^2}+\frac {b^2 \int \frac {\tan (c+d x)^2+1}{\tan (c+d x)^{5/3} (a+b \tan (c+d x))}dx}{a^2+b^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {-\int \frac {b+a \tan (c+d x)}{\tan ^{\frac {2}{3}}(c+d x)}dx-\frac {3 a}{2 d \tan ^{\frac {2}{3}}(c+d x)}}{a^2+b^2}+\frac {b^2 \int \frac {\tan (c+d x)^2+1}{\tan (c+d x)^{5/3} (a+b \tan (c+d x))}dx}{a^2+b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\int \frac {b+a \tan (c+d x)}{\tan (c+d x)^{2/3}}dx-\frac {3 a}{2 d \tan ^{\frac {2}{3}}(c+d x)}}{a^2+b^2}+\frac {b^2 \int \frac {\tan (c+d x)^2+1}{\tan (c+d x)^{5/3} (a+b \tan (c+d x))}dx}{a^2+b^2}\)

\(\Big \downarrow \) 4021

\(\displaystyle \frac {a \left (-\int \sqrt [3]{\tan (c+d x)}dx\right )-b \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)}dx-\frac {3 a}{2 d \tan ^{\frac {2}{3}}(c+d x)}}{a^2+b^2}+\frac {b^2 \int \frac {\tan (c+d x)^2+1}{\tan (c+d x)^{5/3} (a+b \tan (c+d x))}dx}{a^2+b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a \left (-\int \sqrt [3]{\tan (c+d x)}dx\right )-b \int \frac {1}{\tan (c+d x)^{2/3}}dx-\frac {3 a}{2 d \tan ^{\frac {2}{3}}(c+d x)}}{a^2+b^2}+\frac {b^2 \int \frac {\tan (c+d x)^2+1}{\tan (c+d x)^{5/3} (a+b \tan (c+d x))}dx}{a^2+b^2}\)

\(\Big \downarrow \) 3957

\(\displaystyle \frac {-\frac {a \int \frac {\sqrt [3]{\tan (c+d x)}}{\tan ^2(c+d x)+1}d\tan (c+d x)}{d}-\frac {b \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x) \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{d}-\frac {3 a}{2 d \tan ^{\frac {2}{3}}(c+d x)}}{a^2+b^2}+\frac {b^2 \int \frac {\tan (c+d x)^2+1}{\tan (c+d x)^{5/3} (a+b \tan (c+d x))}dx}{a^2+b^2}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {-\frac {3 a \int \frac {\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}-\frac {3 b \int \frac {1}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}-\frac {3 a}{2 d \tan ^{\frac {2}{3}}(c+d x)}}{a^2+b^2}+\frac {b^2 \int \frac {\tan (c+d x)^2+1}{\tan (c+d x)^{5/3} (a+b \tan (c+d x))}dx}{a^2+b^2}\)

\(\Big \downarrow \) 753

\(\displaystyle \frac {-\frac {3 a \int \frac {\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}-\frac {3 b \left (\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \int \frac {2-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{2 \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+2}{2 \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}d\sqrt [3]{\tan (c+d x)}\right )}{d}-\frac {3 a}{2 d \tan ^{\frac {2}{3}}(c+d x)}}{a^2+b^2}+\frac {b^2 \int \frac {\tan (c+d x)^2+1}{\tan (c+d x)^{5/3} (a+b \tan (c+d x))}dx}{a^2+b^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {3 a \int \frac {\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}-\frac {3 b \left (\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{6} \int \frac {2-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+2}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )}{d}-\frac {3 a}{2 d \tan ^{\frac {2}{3}}(c+d x)}}{a^2+b^2}+\frac {b^2 \int \frac {\tan (c+d x)^2+1}{\tan (c+d x)^{5/3} (a+b \tan (c+d x))}dx}{a^2+b^2}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {-\frac {3 a \int \frac {\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}-\frac {3 b \left (\frac {1}{6} \int \frac {2-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+2}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {3 a}{2 d \tan ^{\frac {2}{3}}(c+d x)}}{a^2+b^2}+\frac {b^2 \int \frac {\tan (c+d x)^2+1}{\tan (c+d x)^{5/3} (a+b \tan (c+d x))}dx}{a^2+b^2}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {-\frac {3 a \int \frac {\tan ^{\frac {2}{3}}(c+d x)}{\tan (c+d x)+1}d\tan ^{\frac {2}{3}}(c+d x)}{2 d}-\frac {3 b \left (\frac {1}{6} \int \frac {2-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+2}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {3 a}{2 d \tan ^{\frac {2}{3}}(c+d x)}}{a^2+b^2}+\frac {b^2 \int \frac {\tan (c+d x)^2+1}{\tan (c+d x)^{5/3} (a+b \tan (c+d x))}dx}{a^2+b^2}\)

\(\Big \downarrow \) 821

\(\displaystyle \frac {-\frac {3 a \left (\frac {1}{3} \int \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )d\tan ^{\frac {2}{3}}(c+d x)-\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\tan ^{\frac {2}{3}}(c+d x)\right )}{2 d}-\frac {3 b \left (\frac {1}{6} \int \frac {2-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+2}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {3 a}{2 d \tan ^{\frac {2}{3}}(c+d x)}}{a^2+b^2}+\frac {b^2 \int \frac {\tan (c+d x)^2+1}{\tan (c+d x)^{5/3} (a+b \tan (c+d x))}dx}{a^2+b^2}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {-\frac {3 a \left (\frac {1}{3} \int \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )d\tan ^{\frac {2}{3}}(c+d x)-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{2 d}-\frac {3 b \left (\frac {1}{6} \int \frac {2-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+2}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {3 a}{2 d \tan ^{\frac {2}{3}}(c+d x)}}{a^2+b^2}+\frac {b^2 \int \frac {\tan (c+d x)^2+1}{\tan (c+d x)^{5/3} (a+b \tan (c+d x))}dx}{a^2+b^2}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {-\frac {3 a \left (\frac {1}{3} \left (\frac {3}{2} \int 1d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{2} \int \left (2 \tan ^{\frac {2}{3}}(c+d x)-1\right )d\tan ^{\frac {2}{3}}(c+d x)\right )-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{2 d}-\frac {3 b \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{2} \sqrt {3} \int -\frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {3 a}{2 d \tan ^{\frac {2}{3}}(c+d x)}}{a^2+b^2}+\frac {b^2 \int \frac {\tan (c+d x)^2+1}{\tan (c+d x)^{5/3} (a+b \tan (c+d x))}dx}{a^2+b^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {-\frac {3 a \left (\frac {1}{3} \left (\frac {3}{2} \int 1d\tan ^{\frac {2}{3}}(c+d x)-\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)\right )-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{2 d}-\frac {3 b \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {3 a}{2 d \tan ^{\frac {2}{3}}(c+d x)}}{a^2+b^2}+\frac {b^2 \int \frac {\tan (c+d x)^2+1}{\tan (c+d x)^{5/3} (a+b \tan (c+d x))}dx}{a^2+b^2}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {-\frac {3 a \left (\frac {1}{3} \left (-3 \int \frac {1}{-2 \tan ^{\frac {2}{3}}(c+d x)-2}d\left (2 \tan ^{\frac {2}{3}}(c+d x)-1\right )-\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)\right )-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{2 d}-\frac {3 b \left (\frac {1}{6} \left (\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\int \frac {1}{-\tan ^{\frac {2}{3}}(c+d x)-1}d\left (2 \sqrt [3]{\tan (c+d x)}-\sqrt {3}\right )\right )+\frac {1}{6} \left (\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\int \frac {1}{-\tan ^{\frac {2}{3}}(c+d x)-1}d\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {3 a}{2 d \tan ^{\frac {2}{3}}(c+d x)}}{a^2+b^2}+\frac {b^2 \int \frac {\tan (c+d x)^2+1}{\tan (c+d x)^{5/3} (a+b \tan (c+d x))}dx}{a^2+b^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {-\frac {3 a \left (\frac {1}{3} \left (\sqrt {3} \arctan \left (\frac {2 \tan ^{\frac {2}{3}}(c+d x)-1}{\sqrt {3}}\right )-\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)\right )-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{2 d}-\frac {3 b \left (\frac {1}{6} \left (\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )\right )+\frac {1}{6} \left (\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {3 a}{2 d \tan ^{\frac {2}{3}}(c+d x)}}{a^2+b^2}+\frac {b^2 \int \frac {\tan (c+d x)^2+1}{\tan (c+d x)^{5/3} (a+b \tan (c+d x))}dx}{a^2+b^2}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {b^2 \int \frac {\tan (c+d x)^2+1}{\tan (c+d x)^{5/3} (a+b \tan (c+d x))}dx}{a^2+b^2}+\frac {-\frac {3 a \left (\frac {\arctan \left (\frac {2 \tan ^{\frac {2}{3}}(c+d x)-1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{2 d}-\frac {3 a}{2 d \tan ^{\frac {2}{3}}(c+d x)}-\frac {3 b \left (\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (-\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )-\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )\right )+\frac {1}{6} \left (\arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )+\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )\right )\right )}{d}}{a^2+b^2}\)

\(\Big \downarrow \) 4117

\(\displaystyle \frac {b^2 \int \frac {1}{\tan ^{\frac {5}{3}}(c+d x) (a+b \tan (c+d x))}d\tan (c+d x)}{d \left (a^2+b^2\right )}+\frac {-\frac {3 a \left (\frac {\arctan \left (\frac {2 \tan ^{\frac {2}{3}}(c+d x)-1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{2 d}-\frac {3 a}{2 d \tan ^{\frac {2}{3}}(c+d x)}-\frac {3 b \left (\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (-\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )-\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )\right )+\frac {1}{6} \left (\arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )+\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )\right )\right )}{d}}{a^2+b^2}\)

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 70

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

Input:

Int[1/(Tan[c + d*x]^(5/3)*(a + b*Tan[c + d*x])),x]
 

Output:

(b^2*(-((b*(-((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))))/a) 
 - 3/(2*a*Tan[c + d*x]^(2/3))))/((a^2 + b^2)*d) + ((-3*a*(ArcTan[(-1 + 2*T 
an[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*a)/(2*d*Tan[c + 
 d*x]^(2/3)))/(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 61
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[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 4012
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/ 
(f*(m + 1)*(a^2 + b^2))), x] + Simp[1/(a^2 + b^2)   Int[(a + b*Tan[e + f*x] 
)^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a 
, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1 
]
 

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 = 350, normalized size of antiderivative = 0.67

method result size
derivativedivides \(\frac {\frac {-\frac {3 \left (-\sqrt {3}\, b +a \right ) \ln \left (1-\sqrt {3}\, \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}\right )}{4}-3 \left (2 b +\frac {\left (-\sqrt {3}\, b +a \right ) \sqrt {3}}{2}\right ) \arctan \left (-\sqrt {3}+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right )+\frac {3 \left (-\sqrt {3}\, b -a \right ) \ln \left (1+\sqrt {3}\, \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}\right )}{4}+3 \left (-2 b -\frac {\left (-\sqrt {3}\, b -a \right ) \sqrt {3}}{2}\right ) \arctan \left (\sqrt {3}+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right )}{3 a^{2}+3 b^{2}}-\frac {3 \left (\frac {\ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (\tan \left (d x +c \right )^{\frac {2}{3}}-\left (\frac {a}{b}\right )^{\frac {1}{3}} \tan \left (d x +c \right )^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \tan \left (d x +c \right )^{\frac {1}{3}}}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right ) b^{3}}{\left (a^{2}+b^{2}\right ) a}-\frac {3}{2 a \tan \left (d x +c \right )^{\frac {2}{3}}}+\frac {\frac {3 a \ln \left (1+\tan \left (d x +c \right )^{\frac {2}{3}}\right )}{2}-3 b \arctan \left (\tan \left (d x +c \right )^{\frac {1}{3}}\right )}{3 a^{2}+3 b^{2}}}{d}\) \(350\)
default \(\frac {\frac {-\frac {3 \left (-\sqrt {3}\, b +a \right ) \ln \left (1-\sqrt {3}\, \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}\right )}{4}-3 \left (2 b +\frac {\left (-\sqrt {3}\, b +a \right ) \sqrt {3}}{2}\right ) \arctan \left (-\sqrt {3}+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right )+\frac {3 \left (-\sqrt {3}\, b -a \right ) \ln \left (1+\sqrt {3}\, \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}\right )}{4}+3 \left (-2 b -\frac {\left (-\sqrt {3}\, b -a \right ) \sqrt {3}}{2}\right ) \arctan \left (\sqrt {3}+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right )}{3 a^{2}+3 b^{2}}-\frac {3 \left (\frac {\ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (\tan \left (d x +c \right )^{\frac {2}{3}}-\left (\frac {a}{b}\right )^{\frac {1}{3}} \tan \left (d x +c \right )^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \tan \left (d x +c \right )^{\frac {1}{3}}}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right ) b^{3}}{\left (a^{2}+b^{2}\right ) a}-\frac {3}{2 a \tan \left (d x +c \right )^{\frac {2}{3}}}+\frac {\frac {3 a \ln \left (1+\tan \left (d x +c \right )^{\frac {2}{3}}\right )}{2}-3 b \arctan \left (\tan \left (d x +c \right )^{\frac {1}{3}}\right )}{3 a^{2}+3 b^{2}}}{d}\) \(350\)

Input:

int(1/tan(d*x+c)^(5/3)/(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)+ta 
n(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^3/(a 
^2+b^2)/a-3/2/a/tan(d*x+c)^(2/3)+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 = 2.19 (sec) , antiderivative size = 74494, normalized size of antiderivative = 141.89 \[ \int \frac {1}{\tan ^{\frac {5}{3}}(c+d x) (a+b \tan (c+d x))} \, dx=\text {Too large to display} \] Input:

integrate(1/tan(d*x+c)^(5/3)/(a+b*tan(d*x+c)),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {1}{\tan ^{\frac {5}{3}}(c+d x) (a+b \tan (c+d x))} \, dx=\int \frac {1}{\left (a + b \tan {\left (c + d x \right )}\right ) \tan ^{\frac {5}{3}}{\left (c + d x \right )}}\, dx \] Input:

integrate(1/tan(d*x+c)**(5/3)/(a+b*tan(d*x+c)),x)
 

Output:

Integral(1/((a + b*tan(c + d*x))*tan(c + d*x)**(5/3)), x)
 

Maxima [F(-1)]

Timed out. \[ \int \frac {1}{\tan ^{\frac {5}{3}}(c+d x) (a+b \tan (c+d x))} \, dx=\text {Timed out} \] Input:

integrate(1/tan(d*x+c)^(5/3)/(a+b*tan(d*x+c)),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 457, normalized size of antiderivative = 0.87 \[ \int \frac {1}{\tan ^{\frac {5}{3}}(c+d x) (a+b \tan (c+d x))} \, dx=\frac {b^{3} \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^{4} d + a^{2} b^{2} d} - \frac {3 \, \left (-a b^{2}\right )^{\frac {1}{3}} b^{2} \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^{4} + \sqrt {3} a^{2} b^{2}\right )} d} - \frac {\left (-a b^{2}\right )^{\frac {1}{3}} b^{2} \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^{4} + a^{2} b^{2}\right )} 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}{2 \, a d \tan \left (d x + c\right )^{\frac {2}{3}}} \] Input:

integrate(1/tan(d*x+c)^(5/3)/(a+b*tan(d*x+c)),x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 2.91 (sec) , antiderivative size = 3159, normalized size of antiderivative = 6.02 \[ \int \frac {1}{\tan ^{\frac {5}{3}}(c+d x) (a+b \tan (c+d x))} \, dx=\text {Too large to display} \] Input:

int(1/(tan(c + d*x)^(5/3)*(a + b*tan(c + d*x))),x)
 

Output:

(log(tan(c + d*x)^(1/3) + 1i)*1i)/(2*(a*d*1i - b*d)) + log(tan(c + d*x)^(1 
/3)*1i + 1)/(2*(a*d - b*d*1i)) + 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 + 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 + 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*(tan(c + d*x)^(1/3)*(31492 
8*a^13*b^13*d^11 - 419904*a^15*b^11*d^11 + 118098*a^17*b^9*d^11 + 39366*a^ 
19*b^7*d^11 + 39366*a^21*b^5*d^11 + 13122*a^23*b^3*d^11) - 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 + 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*(tan(c + d*x 
)^(1/3)*(419904*a^14*b^15*d^14 + 1259712*a^16*b^13*d^14 + 944784*a^18*b^11 
*d^14 - 419904*a^20*b^9*d^14 - 629856*a^22*b^7*d^14 + 104976*a^26*b^3*d^14 
) - 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)*(419904*a^16*b^14*d^15 + 1259712*a^18*b^12*d^15 + 839808*a^20*b^10*d^1 
5 - 839808*a^22*b^8*d^15 - 1259712*a^24*b^6*d^15 - 419904*a^26*b^4*d^15...
 

Reduce [F]

\[ \int \frac {1}{\tan ^{\frac {5}{3}}(c+d x) (a+b \tan (c+d x))} \, dx=\int \frac {1}{\tan \left (d x +c \right )^{\frac {8}{3}} b +\tan \left (d x +c \right )^{\frac {5}{3}} a}d x \] Input:

int(1/tan(d*x+c)^(5/3)/(a+b*tan(d*x+c)),x)
 

Output:

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