\(\int \frac {\tan ^{\frac {9}{2}}(c+d x)}{a+b \tan (c+d x)} \, dx\) [582]

Optimal result

Integrand size = 23, antiderivative size = 248 \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{a+b \tan (c+d x)} \, dx=-\frac {(a+b) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}+\frac {(a+b) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}-\frac {2 a^{9/2} \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{b^{7/2} \left (a^2+b^2\right ) d}-\frac {(a-b) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\tan (c+d x)}}{1+\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}+\frac {2 \left (a^2-b^2\right ) \sqrt {\tan (c+d x)}}{b^3 d}-\frac {2 a \tan ^{\frac {3}{2}}(c+d x)}{3 b^2 d}+\frac {2 \tan ^{\frac {5}{2}}(c+d x)}{5 b d} \] Output:

1/2*(a+b)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)/(a^2+b^2)/d+1/2*(a+b 
)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)/(a^2+b^2)/d-2*a^(9/2)*arctan( 
b^(1/2)*tan(d*x+c)^(1/2)/a^(1/2))/b^(7/2)/(a^2+b^2)/d-1/2*(a-b)*arctanh(2^ 
(1/2)*tan(d*x+c)^(1/2)/(1+tan(d*x+c)))*2^(1/2)/(a^2+b^2)/d+2*(a^2-b^2)*tan 
(d*x+c)^(1/2)/b^3/d-2/3*a*tan(d*x+c)^(3/2)/b^2/d+2/5*tan(d*x+c)^(5/2)/b/d
 

Mathematica [A] (verified)

Time = 1.44 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.98 \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {-\frac {15 \left (2 \sqrt {2} b^{7/2} (a+b) \left (\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-\arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )\right )+8 a^{9/2} \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )+\sqrt {2} b^{7/2} (-a+b) \left (\log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )-\log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )\right )+8 \sqrt {b} \left (-a^2+b^2\right ) \left (a^2+b^2\right ) \sqrt {\tan (c+d x)}\right )}{b^{3/2} \left (a^2+b^2\right )}-40 a \tan ^{\frac {3}{2}}(c+d x)+24 b \tan ^{\frac {5}{2}}(c+d x)}{60 b^2 d} \] Input:

Integrate[Tan[c + d*x]^(9/2)/(a + b*Tan[c + d*x]),x]
 

Output:

((-15*(2*Sqrt[2]*b^(7/2)*(a + b)*(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]] - 
 ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]) + 8*a^(9/2)*ArcTan[(Sqrt[b]*Sqrt[ 
Tan[c + d*x]])/Sqrt[a]] + Sqrt[2]*b^(7/2)*(-a + b)*(Log[1 - Sqrt[2]*Sqrt[T 
an[c + d*x]] + Tan[c + d*x]] - Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c 
+ d*x]]) + 8*Sqrt[b]*(-a^2 + b^2)*(a^2 + b^2)*Sqrt[Tan[c + d*x]]))/(b^(3/2 
)*(a^2 + b^2)) - 40*a*Tan[c + d*x]^(3/2) + 24*b*Tan[c + d*x]^(5/2))/(60*b^ 
2*d)
 

Rubi [A] (verified)

Time = 1.53 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.17, number of steps used = 27, number of rules used = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.130, Rules used = {3042, 4049, 27, 3042, 4130, 27, 3042, 4131, 27, 3042, 4136, 25, 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.

\(\displaystyle \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{a+b \tan (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x)^{9/2}}{a+b \tan (c+d x)}dx\)

\(\Big \downarrow \) 4049

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4130

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4131

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4136

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4017

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1482

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

\(\Big \downarrow \) 4117

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 218

\(\displaystyle \frac {2 \tan ^{\frac {5}{2}}(c+d x)}{5 b d}-\frac {\frac {2 a \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {2 \left (a^2-b^2\right ) \sqrt {\tan (c+d x)}}{b d}-\frac {\frac {2 a^{9/2} \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {b} d \left (a^2+b^2\right )}-\frac {2 b^3 \left (\frac {1}{2} (a+b) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\frac {1}{2} (a-b) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{b}}{b}}{b}\)

Input:

Int[Tan[c + d*x]^(9/2)/(a + b*Tan[c + d*x]),x]
 

Output:

(2*Tan[c + d*x]^(5/2))/(5*b*d) - (-((-(((2*a^(9/2)*ArcTan[(Sqrt[b]*Sqrt[Ta 
n[c + d*x]])/Sqrt[a]])/(Sqrt[b]*(a^2 + b^2)*d) - (2*b^3*(((a + b)*(-(ArcTa 
n[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[Tan[c 
 + d*x]]]/Sqrt[2]))/2 - ((a - b)*(-1/2*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] 
+ Tan[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))/b) + (2*(a^2 - b^2)*Sqrt[Tan[c + d*x 
]])/(b*d))/b) + (2*a*Tan[c + d*x]^(3/2))/(3*b*d))/b
 

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 4049

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c 
+ d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[1/(d*(m + n - 1)) 
 Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n 
- 1) - b^2*(b*c*(m - 2) + a*d*(1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[ 
e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2, x], x], 
x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2 
, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || I 
ntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])) 
)
 

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 4130

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) 
+ (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_. 
) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*(a + b*Tan[e + f*x])^m*((c + d*Tan[ 
e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1))   Int[(a 
+ b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C 
*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f*x] - (C* 
m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, 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[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[ 
c, 0] && NeQ[a, 0])))
 

rule 4131

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[C*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n + 
1))), x] + Simp[1/(d*(m + n + 1))   Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d 
*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b 
- b*C)*(m + n + 1)*Tan[e + f*x] - C*m*(b*c - a*d)*Tan[e + f*x]^2, x], x], x 
] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 
+ b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ 
[m] || (EqQ[c, 0] && NeQ[a, 0])))
 

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 = 0.15 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.15

method	result	size
derivativedivides	\(\frac {\frac {\frac {2 b^{2} \tan \left (d x +c \right )^{\frac {5}{2}}}{5}-\frac {2 a b \tan \left (d x +c \right )^{\frac {3}{2}}}{3}+2 a^{2} \sqrt {\tan \left (d x +c \right )}-2 b^{2} \sqrt {\tan \left (d x +c \right )}}{b^{3}}+\frac {\frac {b \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}+\frac {a \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}}{a^{2}+b^{2}}-\frac {2 a^{5} \arctan \left (\frac {b \sqrt {\tan \left (d x +c \right )}}{\sqrt {a b}}\right )}{b^{3} \left (a^{2}+b^{2}\right ) \sqrt {a b}}}{d}\)	\(284\)
default	\(\frac {\frac {\frac {2 b^{2} \tan \left (d x +c \right )^{\frac {5}{2}}}{5}-\frac {2 a b \tan \left (d x +c \right )^{\frac {3}{2}}}{3}+2 a^{2} \sqrt {\tan \left (d x +c \right )}-2 b^{2} \sqrt {\tan \left (d x +c \right )}}{b^{3}}+\frac {\frac {b \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}+\frac {a \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}}{a^{2}+b^{2}}-\frac {2 a^{5} \arctan \left (\frac {b \sqrt {\tan \left (d x +c \right )}}{\sqrt {a b}}\right )}{b^{3} \left (a^{2}+b^{2}\right ) \sqrt {a b}}}{d}\)	\(284\)

Input:

int(tan(d*x+c)^(9/2)/(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

1/d*(2/b^3*(1/5*b^2*tan(d*x+c)^(5/2)-1/3*a*b*tan(d*x+c)^(3/2)+a^2*tan(d*x+ 
c)^(1/2)-b^2*tan(d*x+c)^(1/2))+2/(a^2+b^2)*(1/8*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*arc 
tan(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^(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/b^3*a^5/(a^2+b^2)/(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. 804 vs. \(2 (210) = 420\).

Time = 0.21 (sec) , antiderivative size = 1634, normalized size of antiderivative = 6.59 \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{a+b \tan (c+d x)} \, dx=\text {Too large to display} \] Input:

integrate(tan(d*x+c)^(9/2)/(a+b*tan(d*x+c)),x, algorithm="fricas")
 

Output:

[1/30*(30*a^4*sqrt(-a/b)*log(-(2*b*sqrt(-a/b)*sqrt(tan(d*x + c)) - b*tan(d 
*x + c) + a)/(b*tan(d*x + c) + a)) + 30*sqrt(1/2)*(a^2*b^3 + b^5)*d*sqrt(( 
a^2 + 2*a*b + b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*arctan(((a^4 + 2*a^2*b^2 
 + b^4)*d^2*sqrt((a^2 + 2*a*b + b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*sqrt(( 
a^2 - 2*a*b + b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2)) + 2*sqrt(1/2)*(a^3 - a^2 
*b + a*b^2 - b^3)*d*sqrt((a^2 + 2*a*b + b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2) 
)*sqrt(tan(d*x + c)))/(a^2 - b^2)) + 30*sqrt(1/2)*(a^2*b^3 + b^5)*d*sqrt(( 
a^2 + 2*a*b + b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*arctan(-((a^4 + 2*a^2*b^ 
2 + b^4)*d^2*sqrt((a^2 + 2*a*b + b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*sqrt( 
(a^2 - 2*a*b + b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2)) - 2*sqrt(1/2)*(a^3 - a^ 
2*b + a*b^2 - b^3)*d*sqrt((a^2 + 2*a*b + b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2 
))*sqrt(tan(d*x + c)))/(a^2 - b^2)) + 15*sqrt(1/2)*(a^2*b^3 + b^5)*d*sqrt( 
(a^2 - 2*a*b + b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*log(2*sqrt(1/2)*(a^2 + 
b^2)*d*sqrt((a^2 - 2*a*b + b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*sqrt(tan(d* 
x + c)) - (a - b)*tan(d*x + c) - a + b) - 15*sqrt(1/2)*(a^2*b^3 + b^5)*d*s 
qrt((a^2 - 2*a*b + b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*log(-2*sqrt(1/2)*(a 
^2 + b^2)*d*sqrt((a^2 - 2*a*b + b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*sqrt(t 
an(d*x + c)) - (a - b)*tan(d*x + c) - a + b) + 4*(15*a^4 - 15*b^4 + 3*(a^2 
*b^2 + b^4)*tan(d*x + c)^2 - 5*(a^3*b + a*b^3)*tan(d*x + c))*sqrt(tan(d*x 
+ c)))/((a^2*b^3 + b^5)*d), -1/30*(60*a^4*sqrt(a/b)*arctan(b*sqrt(a/b)*...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{a+b \tan (c+d x)} \, dx=\text {Timed out} \] Input:

integrate(tan(d*x+c)**(9/2)/(a+b*tan(d*x+c)),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.91 \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{a+b \tan (c+d x)} \, dx=-\frac {\frac {120 \, a^{5} \arctan \left (\frac {b \sqrt {\tan \left (d x + c\right )}}{\sqrt {a b}}\right )}{{\left (a^{2} b^{3} + b^{5}\right )} \sqrt {a b}} - \frac {15 \, {\left (2 \, \sqrt {2} {\left (a + b\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (a + b\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} {\left (a - b\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left (a - b\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )}}{a^{2} + b^{2}} - \frac {8 \, {\left (3 \, b^{2} \tan \left (d x + c\right )^{\frac {5}{2}} - 5 \, a b \tan \left (d x + c\right )^{\frac {3}{2}} + 15 \, {\left (a^{2} - b^{2}\right )} \sqrt {\tan \left (d x + c\right )}\right )}}{b^{3}}}{60 \, d} \] Input:

integrate(tan(d*x+c)^(9/2)/(a+b*tan(d*x+c)),x, algorithm="maxima")
 

Output:

-1/60*(120*a^5*arctan(b*sqrt(tan(d*x + c))/sqrt(a*b))/((a^2*b^3 + b^5)*sqr 
t(a*b)) - 15*(2*sqrt(2)*(a + b)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d 
*x + c)))) + 2*sqrt(2)*(a + b)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d 
*x + c)))) - sqrt(2)*(a - b)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) 
 + 1) + sqrt(2)*(a - b)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1 
))/(a^2 + b^2) - 8*(3*b^2*tan(d*x + c)^(5/2) - 5*a*b*tan(d*x + c)^(3/2) + 
15*(a^2 - b^2)*sqrt(tan(d*x + c)))/b^3)/d
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{a+b \tan (c+d x)} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(tan(d*x+c)^(9/2)/(a+b*tan(d*x+c)),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [B] (verification not implemented)

Time = 8.19 (sec) , antiderivative size = 4082, normalized size of antiderivative = 16.46 \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{a+b \tan (c+d x)} \, dx=\text {Too large to display} \] Input:

int(tan(c + d*x)^(9/2)/(a + b*tan(c + d*x)),x)
 

Output:

(log(((((((((256*b^3*tan(c + d*x)^(1/2)*(a^2 - b^2)*(a^2 + b^2)^2*(1i/(d^2 
*(a*1i - b)^2))^(1/2) - (128*a*b*(4*a^2 - b^2)*(a^2 + b^2)^2)/d)*(1i/(d^2* 
(a*1i - b)^2))^(1/2))/2 + (64*a*tan(c + d*x)^(1/2)*(8*a^10 - 7*b^10 + 2*a^ 
2*b^8 + a^4*b^6))/(b^4*d^2))*(1i/(d^2*(a*1i - b)^2))^(1/2))/2 - (32*a^2*(1 
2*a^8 + b^8 + a^2*b^6 + 16*a^4*b^4 - 16*a^6*b^2))/(b^4*d^3))*(1i/(d^2*(a*1 
i - b)^2))^(1/2))/2 + (32*tan(c + d*x)^(1/2)*(2*a^10 - b^10))/(b^5*d^4))*( 
1i/(d^2*(a*1i - b)^2))^(1/2))/2 - (32*a^5*(a^4 + b^4 - a^2*b^2))/(b^5*d^5) 
)*(-1/(b^2*d^2*1i - a^2*d^2*1i + 2*a*b*d^2))^(1/2))/2 - tan(c + d*x)^(1/2) 
*(2/(b*d) - (2*a^2)/(b^3*d)) - log(- ((((((((256*b^3*tan(c + d*x)^(1/2)*(a 
^2 - b^2)*(a^2 + b^2)^2*(1i/(d^2*(a*1i - b)^2))^(1/2) + (128*a*b*(4*a^2 - 
b^2)*(a^2 + b^2)^2)/d)*(1i/(d^2*(a*1i - b)^2))^(1/2))/2 + (64*a*tan(c + d* 
x)^(1/2)*(8*a^10 - 7*b^10 + 2*a^2*b^8 + a^4*b^6))/(b^4*d^2))*(1i/(d^2*(a*1 
i - b)^2))^(1/2))/2 + (32*a^2*(12*a^8 + b^8 + a^2*b^6 + 16*a^4*b^4 - 16*a^ 
6*b^2))/(b^4*d^3))*(1i/(d^2*(a*1i - b)^2))^(1/2))/2 + (32*tan(c + d*x)^(1/ 
2)*(2*a^10 - b^10))/(b^5*d^4))*(1i/(d^2*(a*1i - b)^2))^(1/2))/2 - (32*a^5* 
(a^4 + b^4 - a^2*b^2))/(b^5*d^5))*(-1/(4*(b^2*d^2*1i - a^2*d^2*1i + 2*a*b* 
d^2)))^(1/2) + atan((((((-1i/(4*(b^2*d^2 - a^2*d^2 + a*b*d^2*2i)))^(1/2)*( 
(32*(8*a^3*b^10*d^4 - 4*a*b^12*d^4 + 28*a^5*b^8*d^4 + 16*a^7*b^6*d^4))/(b^ 
5*d^5) - (32*tan(c + d*x)^(1/2)*(-1i/(4*(b^2*d^2 - a^2*d^2 + a*b*d^2*2i))) 
^(1/2)*(16*b^14*d^4 + 16*a^2*b^12*d^4 - 16*a^4*b^10*d^4 - 16*a^6*b^8*d^...
 

Reduce [F]

\[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{a+b \tan (c+d x)} \, dx=\int \frac {\sqrt {\tan \left (d x +c \right )}\, \tan \left (d x +c \right )^{4}}{a +\tan \left (d x +c \right ) b}d x \] Input:

int(tan(d*x+c)^(9/2)/(a+b*tan(d*x+c)),x)
 

Output:

int((sqrt(tan(c + d*x))*tan(c + d*x)**4)/(tan(c + d*x)*b + a),x)