\(\int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^2} \, dx\) [593]

Optimal result

Integrand size = 23, antiderivative size = 257 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {\left (a^2+2 a b-b^2\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}-\frac {\left (a^2+2 a b-b^2\right ) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}+\frac {a^{3/2} \left (a^2+5 b^2\right ) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{b^{3/2} \left (a^2+b^2\right )^2 d}+\frac {\left (a^2-2 a b-b^2\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\tan (c+d x)}}{1+\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}-\frac {a^2 \sqrt {\tan (c+d x)}}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))} \] Output:

-1/2*(a^2+2*a*b-b^2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)/(a^2+b^2) 
^2/d-1/2*(a^2+2*a*b-b^2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)/(a^2+b 
^2)^2/d+a^(3/2)*(a^2+5*b^2)*arctan(b^(1/2)*tan(d*x+c)^(1/2)/a^(1/2))/b^(3/ 
2)/(a^2+b^2)^2/d+1/2*(a^2-2*a*b-b^2)*arctanh(2^(1/2)*tan(d*x+c)^(1/2)/(1+t 
an(d*x+c)))*2^(1/2)/(a^2+b^2)^2/d-a^2*tan(d*x+c)^(1/2)/b/(a^2+b^2)/d/(a+b* 
tan(d*x+c))
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.29 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.61 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {(-1)^{3/4} (-i a+b)^2 \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+\frac {a^{3/2} \left (a^2+5 b^2\right ) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{b^{3/2}}+(-1)^{3/4} (a-i b)^2 \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )-\frac {a^2 \left (a^2+b^2\right ) \sqrt {\tan (c+d x)}}{b (a+b \tan (c+d x))}}{\left (a^2+b^2\right )^2 d} \] Input:

Integrate[Tan[c + d*x]^(5/2)/(a + b*Tan[c + d*x])^2,x]
 

Output:

((-1)^(3/4)*((-I)*a + b)^2*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] + (a^(3/2 
)*(a^2 + 5*b^2)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/b^(3/2) + (- 
1)^(3/4)*(a - I*b)^2*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] - (a^2*(a^2 + 
b^2)*Sqrt[Tan[c + d*x]])/(b*(a + b*Tan[c + d*x])))/((a^2 + b^2)^2*d)
 

Rubi [A] (verified)

Time = 1.14 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.12, number of steps used = 21, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.870, Rules used = {3042, 4048, 27, 3042, 4136, 27, 3042, 4017, 27, 1482, 1476, 1082, 217, 1479, 25, 27, 1103, 4117, 73, 218}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4048

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4136

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4017

\(\displaystyle \frac {\frac {a^2 \left (a^2+5 b^2\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {4 \int \frac {b \left (2 a b+\left (a^2-b^2\right ) \tan (c+d x)\right )}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {a^2 \left (a^2+5 b^2\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {4 b \int \frac {2 a b+\left (a^2-b^2\right ) \tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\)

\(\Big \downarrow \) 1482

\(\displaystyle \frac {\frac {a^2 \left (a^2+5 b^2\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {4 b \left (\frac {1}{2} \left (a^2+2 a b-b^2\right ) \int \frac {\tan (c+d x)+1}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} \left (a^2-2 a b-b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d \left (a^2+b^2\right )}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {\frac {a^2 \left (a^2+5 b^2\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {4 b \left (\frac {1}{2} \left (a^2+2 a b-b^2\right ) \left (\frac {1}{2} \int \frac {1}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \int \frac {1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )-\frac {1}{2} \left (a^2-2 a b-b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d \left (a^2+b^2\right )}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {\frac {a^2 \left (a^2+5 b^2\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {4 b \left (\frac {1}{2} \left (a^2+2 a b-b^2\right ) \left (\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}\right )-\frac {1}{2} \left (a^2-2 a b-b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d \left (a^2+b^2\right )}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {a^2 \left (a^2+5 b^2\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {4 b \left (\frac {1}{2} \left (a^2+2 a b-b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\frac {1}{2} \left (a^2-2 a b-b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d \left (a^2+b^2\right )}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {\frac {a^2 \left (a^2+5 b^2\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {4 b \left (\frac {1}{2} \left (a^2+2 a b-b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\frac {1}{2} \left (a^2-2 a b-b^2\right ) \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {a^2 \left (a^2+5 b^2\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {4 b \left (\frac {1}{2} \left (a^2+2 a b-b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\frac {1}{2} \left (a^2-2 a b-b^2\right ) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {a^2 \left (a^2+5 b^2\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {4 b \left (\frac {1}{2} \left (a^2+2 a b-b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\frac {1}{2} \left (a^2-2 a b-b^2\right ) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\tan (c+d x)}+1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )\right )}{d \left (a^2+b^2\right )}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {\frac {a^2 \left (a^2+5 b^2\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {4 b \left (\frac {1}{2} \left (a^2+2 a b-b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\frac {1}{2} \left (a^2-2 a b-b^2\right ) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\)

\(\Big \downarrow \) 4117

\(\displaystyle \frac {\frac {a^2 \left (a^2+5 b^2\right ) \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}d\tan (c+d x)}{d \left (a^2+b^2\right )}-\frac {4 b \left (\frac {1}{2} \left (a^2+2 a b-b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\frac {1}{2} \left (a^2-2 a b-b^2\right ) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {2 a^2 \left (a^2+5 b^2\right ) \int \frac {1}{a+b \tan (c+d x)}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}-\frac {4 b \left (\frac {1}{2} \left (a^2+2 a b-b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\frac {1}{2} \left (a^2-2 a b-b^2\right ) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {2 a^{3/2} \left (a^2+5 b^2\right ) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {b} d \left (a^2+b^2\right )}-\frac {4 b \left (\frac {1}{2} \left (a^2+2 a b-b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\frac {1}{2} \left (a^2-2 a b-b^2\right ) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\)

Input:

Int[Tan[c + d*x]^(5/2)/(a + b*Tan[c + d*x])^2,x]
 

Output:

((2*a^(3/2)*(a^2 + 5*b^2)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(S 
qrt[b]*(a^2 + b^2)*d) - (4*b*(((a^2 + 2*a*b - b^2)*(-(ArcTan[1 - Sqrt[2]*S 
qrt[Tan[c + d*x]]]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[ 
2]))/2 - ((a^2 - 2*a*b - b^2)*(-1/2*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + T 
an[c + d*x]]/Sqrt[2] + Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/ 
(2*Sqrt[2])))/2))/((a^2 + b^2)*d))/(2*b*(a^2 + b^2)) - (a^2*Sqrt[Tan[c + d 
*x]])/(b*(a^2 + b^2)*d*(a + b*Tan[c + d*x]))
 

Defintions of rubi rules used

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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

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 218

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

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 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 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

rule 1482

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
a*c, 2]}, Simp[(d*q + a*e)/(2*a*c)   Int[(q + c*x^2)/(a + c*x^4), x], x] + 
Simp[(d*q - a*e)/(2*a*c)   Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ[{a 
, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(- 
a)*c]
 

rule 3042

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

rule 4017

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ 
)]], x_Symbol] :> Simp[2/f   Subst[Int[(b*c + d*x^2)/(b^2 + x^4), x], x, Sq 
rt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] & 
& NeQ[c^2 + d^2, 0]
 

rule 4048

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

rule 4136

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) 
+ (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) 
+ (f_.)*(x_)]), x_Symbol] :> Simp[1/(a^2 + b^2)   Int[(c + d*Tan[e + f*x])^ 
n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Simp[ 
(A*b^2 - a*b*B + a^2*C)/(a^2 + b^2)   Int[(c + d*Tan[e + f*x])^n*((1 + Tan[ 
e + f*x]^2)/(a + b*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, 
 C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] & 
&  !GtQ[n, 0] &&  !LeQ[n, -1]
 

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.09

method	result	size
derivativedivides	\(\frac {\frac {-\frac {a b \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{2}+\frac {\left (-a^{2}+b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}}{\left (a^{2}+b^{2}\right )^{2}}+\frac {2 a^{2} \left (-\frac {\left (a^{2}+b^{2}\right ) \sqrt {\tan \left (d x +c \right )}}{2 b \left (a +b \tan \left (d x +c \right )\right )}+\frac {\left (a^{2}+5 b^{2}\right ) \arctan \left (\frac {b \sqrt {\tan \left (d x +c \right )}}{\sqrt {a b}}\right )}{2 b \sqrt {a b}}\right )}{\left (a^{2}+b^{2}\right )^{2}}}{d}\)	\(281\)
default	\(\frac {\frac {-\frac {a b \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{2}+\frac {\left (-a^{2}+b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}}{\left (a^{2}+b^{2}\right )^{2}}+\frac {2 a^{2} \left (-\frac {\left (a^{2}+b^{2}\right ) \sqrt {\tan \left (d x +c \right )}}{2 b \left (a +b \tan \left (d x +c \right )\right )}+\frac {\left (a^{2}+5 b^{2}\right ) \arctan \left (\frac {b \sqrt {\tan \left (d x +c \right )}}{\sqrt {a b}}\right )}{2 b \sqrt {a b}}\right )}{\left (a^{2}+b^{2}\right )^{2}}}{d}\)	\(281\)

Input:

int(tan(d*x+c)^(5/2)/(a+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

1/d*(2/(a^2+b^2)^2*(-1/4*a*b*2^(1/2)*(ln((tan(d*x+c)+2^(1/2)*tan(d*x+c)^(1 
/2)+1)/(tan(d*x+c)-2^(1/2)*tan(d*x+c)^(1/2)+1))+2*arctan(1+2^(1/2)*tan(d*x 
+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)))+1/8*(-a^2+b^2)*2^(1/2)*( 
ln((tan(d*x+c)-2^(1/2)*tan(d*x+c)^(1/2)+1)/(tan(d*x+c)+2^(1/2)*tan(d*x+c)^ 
(1/2)+1))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x 
+c)^(1/2))))+2*a^2/(a^2+b^2)^2*(-1/2*(a^2+b^2)/b*tan(d*x+c)^(1/2)/(a+b*tan 
(d*x+c))+1/2*(a^2+5*b^2)/b/(a*b)^(1/2)*arctan(b*tan(d*x+c)^(1/2)/(a*b)^(1/ 
2))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1498 vs. \(2 (228) = 456\).

Time = 0.30 (sec) , antiderivative size = 3021, normalized size of antiderivative = 11.75 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\text {Too large to display} \] Input:

integrate(tan(d*x+c)^(5/2)/(a+b*tan(d*x+c))^2,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\int \frac {\tan ^{\frac {5}{2}}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{2}}\, dx \] Input:

integrate(tan(d*x+c)**(5/2)/(a+b*tan(d*x+c))**2,x)
 

Output:

Integral(tan(c + d*x)**(5/2)/(a + b*tan(c + d*x))**2, x)
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.08 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {\frac {4 \, a^{2} \sqrt {\tan \left (d x + c\right )}}{a^{3} b + a b^{3} + {\left (a^{2} b^{2} + b^{4}\right )} \tan \left (d x + c\right )} - \frac {4 \, {\left (a^{4} + 5 \, a^{2} b^{2}\right )} \arctan \left (\frac {b \sqrt {\tan \left (d x + c\right )}}{\sqrt {a b}}\right )}{{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \sqrt {a b}} + \frac {2 \, \sqrt {2} {\left (a^{2} + 2 \, a b - b^{2}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (a^{2} + 2 \, a b - b^{2}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} {\left (a^{2} - 2 \, a b - b^{2}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left (a^{2} - 2 \, a b - b^{2}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}}}{4 \, d} \] Input:

integrate(tan(d*x+c)^(5/2)/(a+b*tan(d*x+c))^2,x, algorithm="maxima")
 

Output:

-1/4*(4*a^2*sqrt(tan(d*x + c))/(a^3*b + a*b^3 + (a^2*b^2 + b^4)*tan(d*x + 
c)) - 4*(a^4 + 5*a^2*b^2)*arctan(b*sqrt(tan(d*x + c))/sqrt(a*b))/((a^4*b + 
 2*a^2*b^3 + b^5)*sqrt(a*b)) + (2*sqrt(2)*(a^2 + 2*a*b - b^2)*arctan(1/2*s 
qrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 2*sqrt(2)*(a^2 + 2*a*b - b^2)*a 
rctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) - sqrt(2)*(a^2 - 2*a* 
b - b^2)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + sqrt(2)*(a^2 
 - 2*a*b - b^2)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1))/(a^4 
+ 2*a^2*b^2 + b^4))/d
 

Giac [F]

\[ \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\int { \frac {\tan \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:

integrate(tan(d*x+c)^(5/2)/(a+b*tan(d*x+c))^2,x, algorithm="giac")
 

Output:

undef
 

Mupad [B] (verification not implemented)

Time = 3.94 (sec) , antiderivative size = 11104, normalized size of antiderivative = 43.21 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\text {Too large to display} \] Input:

int(tan(c + d*x)^(5/2)/(a + b*tan(c + d*x))^2,x)
 

Output:

- atan(((1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b^ 
2*d^2)))^(1/2)*((1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 
6*a^2*b^2*d^2)))^(1/2)*((1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d 
^2*4i - 6*a^2*b^2*d^2)))^(1/2)*(((8*(96*a^2*b^14*d^4 + 480*a^4*b^12*d^4 + 
960*a^6*b^10*d^4 + 960*a^8*b^8*d^4 + 480*a^10*b^6*d^4 + 96*a^12*b^4*d^4))/ 
(b^9*d^5 + a^8*b*d^5 + 4*a^2*b^7*d^5 + 6*a^4*b^5*d^5 + 4*a^6*b^3*d^5) - (1 
6*tan(c + d*x)^(1/2)*(1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2* 
4i - 6*a^2*b^2*d^2)))^(1/2)*(32*b^18*d^4 + 160*a^2*b^16*d^4 + 288*a^4*b^14 
*d^4 + 160*a^6*b^12*d^4 - 160*a^8*b^10*d^4 - 288*a^10*b^8*d^4 - 160*a^12*b 
^6*d^4 - 32*a^14*b^4*d^4))/(b^9*d^4 + a^8*b*d^4 + 4*a^2*b^7*d^4 + 6*a^4*b^ 
5*d^4 + 4*a^6*b^3*d^4))*(1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d 
^2*4i - 6*a^2*b^2*d^2)))^(1/2) + (16*tan(c + d*x)^(1/2)*(60*a*b^13*d^2 + 8 
*a^13*b*d^2 + 52*a^3*b^11*d^2 + 128*a^5*b^9*d^2 + 424*a^7*b^7*d^2 + 380*a^ 
9*b^5*d^2 + 100*a^11*b^3*d^2))/(b^9*d^4 + a^8*b*d^4 + 4*a^2*b^7*d^4 + 6*a^ 
4*b^5*d^4 + 4*a^6*b^3*d^4)) + (8*(4*a*b^11*d^2 + 16*a^11*b*d^2 - 304*a^3*b 
^9*d^2 - 120*a^5*b^7*d^2 + 320*a^7*b^5*d^2 + 148*a^9*b^3*d^2))/(b^9*d^5 + 
a^8*b*d^5 + 4*a^2*b^7*d^5 + 6*a^4*b^5*d^5 + 4*a^6*b^3*d^5)) + (16*tan(c + 
d*x)^(1/2)*(a^10 - 2*b^10 - 4*a^2*b^8 - 27*a^4*b^6 + 15*a^6*b^4 + 9*a^8*b^ 
2))/(b^9*d^4 + a^8*b*d^4 + 4*a^2*b^7*d^4 + 6*a^4*b^5*d^4 + 4*a^6*b^3*d^4)) 
*1i - (1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b...
 

Reduce [F]

\[ \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\int \frac {\sqrt {\tan \left (d x +c \right )}\, \tan \left (d x +c \right )^{2}}{\tan \left (d x +c \right )^{2} b^{2}+2 \tan \left (d x +c \right ) a b +a^{2}}d x \] Input:

int(tan(d*x+c)^(5/2)/(a+b*tan(d*x+c))^2,x)
 

Output:

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