\(\int \frac {\tan ^5(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx\) [546]

Optimal result

Integrand size = 23, antiderivative size = 291 \[ \int \frac {\tan ^5(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{(a-i b)^{5/2} d}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{(a+i b)^{5/2} d}-\frac {2 a^2 \tan ^3(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {4 a^2 \left (a^2+2 b^2\right ) \tan ^2(c+d x)}{b^2 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}-\frac {4 a \left (8 a^4+15 a^2 b^2+4 b^4\right ) \sqrt {a+b \tan (c+d x)}}{3 b^4 \left (a^2+b^2\right )^2 d}+\frac {2 \left (8 a^4+15 a^2 b^2+b^4\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b^3 \left (a^2+b^2\right )^2 d} \] Output:

-arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/(a-I*b)^(5/2)/d-arctanh((a+ 
b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/(a+I*b)^(5/2)/d-2/3*a^2*tan(d*x+c)^3/b/ 
(a^2+b^2)/d/(a+b*tan(d*x+c))^(3/2)-4*a^2*(a^2+2*b^2)*tan(d*x+c)^2/b^2/(a^2 
+b^2)^2/d/(a+b*tan(d*x+c))^(1/2)-4/3*a*(8*a^4+15*a^2*b^2+4*b^4)*(a+b*tan(d 
*x+c))^(1/2)/b^4/(a^2+b^2)^2/d+2/3*(8*a^4+15*a^2*b^2+b^4)*tan(d*x+c)*(a+b* 
tan(d*x+c))^(1/2)/b^3/(a^2+b^2)^2/d
 

Mathematica [A] (verified)

Time = 3.78 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.21 \[ \int \frac {\tan ^5(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\frac {2 \left (-\frac {3 b \left (-2 a b^2+a^2 \sqrt {-b^2}+\left (-b^2\right )^{3/2}\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-\sqrt {-b^2}}}\right )}{2 \sqrt {-b^2} \left (a^2+b^2\right )^2 \sqrt {a-\sqrt {-b^2}}}-\frac {3 b \left (2 a b^2+a^2 \sqrt {-b^2}+\left (-b^2\right )^{3/2}\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+\sqrt {-b^2}}}\right )}{2 \sqrt {-b^2} \left (a^2+b^2\right )^2 \sqrt {a+\sqrt {-b^2}}}+\frac {a^3 \left (8 a^2+7 b^2\right )}{b^3 \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {6 a \tan ^2(c+d x)}{b (a+b \tan (c+d x))^{3/2}}+\frac {\tan ^3(c+d x)}{(a+b \tan (c+d x))^{3/2}}-\frac {3 a^2 \left (8 a^4+15 a^2 b^2+5 b^4\right )}{b^3 \left (a^2+b^2\right )^2 \sqrt {a+b \tan (c+d x)}}\right )}{3 b d} \] Input:

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

Output:

(2*((-3*b*(-2*a*b^2 + a^2*Sqrt[-b^2] + (-b^2)^(3/2))*ArcTanh[Sqrt[a + b*Ta 
n[c + d*x]]/Sqrt[a - Sqrt[-b^2]]])/(2*Sqrt[-b^2]*(a^2 + b^2)^2*Sqrt[a - Sq 
rt[-b^2]]) - (3*b*(2*a*b^2 + a^2*Sqrt[-b^2] + (-b^2)^(3/2))*ArcTanh[Sqrt[a 
 + b*Tan[c + d*x]]/Sqrt[a + Sqrt[-b^2]]])/(2*Sqrt[-b^2]*(a^2 + b^2)^2*Sqrt 
[a + Sqrt[-b^2]]) + (a^3*(8*a^2 + 7*b^2))/(b^3*(a^2 + b^2)*(a + b*Tan[c + 
d*x])^(3/2)) - (6*a*Tan[c + d*x]^2)/(b*(a + b*Tan[c + d*x])^(3/2)) + Tan[c 
 + d*x]^3/(a + b*Tan[c + d*x])^(3/2) - (3*a^2*(8*a^4 + 15*a^2*b^2 + 5*b^4) 
)/(b^3*(a^2 + b^2)^2*Sqrt[a + b*Tan[c + d*x]])))/(3*b*d)
 

Rubi [A] (warning: unable to verify)

Time = 1.85 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.10, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.783, Rules used = {3042, 4048, 27, 3042, 4128, 27, 3042, 4130, 27, 3042, 4113, 3042, 4022, 3042, 4020, 25, 73, 221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4048

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4128

\(\displaystyle \frac {\frac {2 \int \frac {\tan (c+d x) \left (-2 a \tan (c+d x) b^3+\left (8 a^4+15 b^2 a^2+b^4\right ) \tan ^2(c+d x)+8 a^2 \left (a^2+2 b^2\right )\right )}{2 \sqrt {a+b \tan (c+d x)}}dx}{b \left (a^2+b^2\right )}-\frac {4 a^2 \left (a^2+2 b^2\right ) \tan ^2(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^3(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {\tan (c+d x) \left (-2 a \tan (c+d x) b^3+\left (8 a^4+15 b^2 a^2+b^4\right ) \tan ^2(c+d x)+8 a^2 \left (a^2+2 b^2\right )\right )}{\sqrt {a+b \tan (c+d x)}}dx}{b \left (a^2+b^2\right )}-\frac {4 a^2 \left (a^2+2 b^2\right ) \tan ^2(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^3(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\tan (c+d x) \left (-2 a \tan (c+d x) b^3+\left (8 a^4+15 b^2 a^2+b^4\right ) \tan (c+d x)^2+8 a^2 \left (a^2+2 b^2\right )\right )}{\sqrt {a+b \tan (c+d x)}}dx}{b \left (a^2+b^2\right )}-\frac {4 a^2 \left (a^2+2 b^2\right ) \tan ^2(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^3(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4130

\(\displaystyle \frac {\frac {\frac {2 \int -\frac {-3 \left (a^2-b^2\right ) \tan (c+d x) b^3+2 a \left (8 a^4+15 b^2 a^2+4 b^4\right ) \tan ^2(c+d x)+2 a \left (8 a^4+15 b^2 a^2+b^4\right )}{2 \sqrt {a+b \tan (c+d x)}}dx}{3 b}+\frac {2 \left (8 a^4+15 a^2 b^2+b^4\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}}{b \left (a^2+b^2\right )}-\frac {4 a^2 \left (a^2+2 b^2\right ) \tan ^2(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^3(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {2 \left (8 a^4+15 a^2 b^2+b^4\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}-\frac {\int \frac {-3 \left (a^2-b^2\right ) \tan (c+d x) b^3+2 a \left (8 a^4+15 b^2 a^2+4 b^4\right ) \tan ^2(c+d x)+2 a \left (8 a^4+15 b^2 a^2+b^4\right )}{\sqrt {a+b \tan (c+d x)}}dx}{3 b}}{b \left (a^2+b^2\right )}-\frac {4 a^2 \left (a^2+2 b^2\right ) \tan ^2(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^3(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {2 \left (8 a^4+15 a^2 b^2+b^4\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}-\frac {\int \frac {-3 \left (a^2-b^2\right ) \tan (c+d x) b^3+2 a \left (8 a^4+15 b^2 a^2+4 b^4\right ) \tan (c+d x)^2+2 a \left (8 a^4+15 b^2 a^2+b^4\right )}{\sqrt {a+b \tan (c+d x)}}dx}{3 b}}{b \left (a^2+b^2\right )}-\frac {4 a^2 \left (a^2+2 b^2\right ) \tan ^2(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^3(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4113

\(\displaystyle \frac {\frac {\frac {2 \left (8 a^4+15 a^2 b^2+b^4\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}-\frac {\int \frac {-6 a b^4-3 \left (a^2-b^2\right ) \tan (c+d x) b^3}{\sqrt {a+b \tan (c+d x)}}dx+\frac {4 a \left (8 a^4+15 a^2 b^2+4 b^4\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{3 b}}{b \left (a^2+b^2\right )}-\frac {4 a^2 \left (a^2+2 b^2\right ) \tan ^2(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^3(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {2 \left (8 a^4+15 a^2 b^2+b^4\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}-\frac {\int \frac {-6 a b^4-3 \left (a^2-b^2\right ) \tan (c+d x) b^3}{\sqrt {a+b \tan (c+d x)}}dx+\frac {4 a \left (8 a^4+15 a^2 b^2+4 b^4\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{3 b}}{b \left (a^2+b^2\right )}-\frac {4 a^2 \left (a^2+2 b^2\right ) \tan ^2(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^3(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4022

\(\displaystyle -\frac {2 a^2 \tan ^3(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {-\frac {4 a^2 \left (a^2+2 b^2\right ) \tan ^2(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {2 \left (8 a^4+15 a^2 b^2+b^4\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}-\frac {-\frac {3}{2} i b^3 (a-i b)^2 \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {3}{2} i b^3 (a+i b)^2 \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {4 a \left (8 a^4+15 a^2 b^2+4 b^4\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{3 b}}{b \left (a^2+b^2\right )}}{b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {2 a^2 \tan ^3(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {-\frac {4 a^2 \left (a^2+2 b^2\right ) \tan ^2(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {2 \left (8 a^4+15 a^2 b^2+b^4\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}-\frac {-\frac {3}{2} i b^3 (a-i b)^2 \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {3}{2} i b^3 (a+i b)^2 \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {4 a \left (8 a^4+15 a^2 b^2+4 b^4\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{3 b}}{b \left (a^2+b^2\right )}}{b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 4020

\(\displaystyle -\frac {2 a^2 \tan ^3(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {-\frac {4 a^2 \left (a^2+2 b^2\right ) \tan ^2(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {2 \left (8 a^4+15 a^2 b^2+b^4\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}-\frac {-\frac {3 b^3 (a+i b)^2 \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {3 b^3 (a-i b)^2 \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {4 a \left (8 a^4+15 a^2 b^2+4 b^4\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{3 b}}{b \left (a^2+b^2\right )}}{b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {2 a^2 \tan ^3(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {-\frac {4 a^2 \left (a^2+2 b^2\right ) \tan ^2(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {2 \left (8 a^4+15 a^2 b^2+b^4\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}-\frac {\frac {3 b^3 (a+i b)^2 \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}+\frac {3 b^3 (a-i b)^2 \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {4 a \left (8 a^4+15 a^2 b^2+4 b^4\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{3 b}}{b \left (a^2+b^2\right )}}{b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {2 a^2 \tan ^3(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {-\frac {4 a^2 \left (a^2+2 b^2\right ) \tan ^2(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {2 \left (8 a^4+15 a^2 b^2+b^4\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}-\frac {-\frac {3 i b^2 (a-i b)^2 \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{d}+\frac {3 i b^2 (a+i b)^2 \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{d}+\frac {4 a \left (8 a^4+15 a^2 b^2+4 b^4\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{3 b}}{b \left (a^2+b^2\right )}}{b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {2 a^2 \tan ^3(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {-\frac {4 a^2 \left (a^2+2 b^2\right ) \tan ^2(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {2 \left (8 a^4+15 a^2 b^2+b^4\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}-\frac {\frac {4 a \left (8 a^4+15 a^2 b^2+4 b^4\right ) \sqrt {a+b \tan (c+d x)}}{b d}+\frac {3 i b^3 (a+i b)^2 \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}-\frac {3 i b^3 (a-i b)^2 \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}}{3 b}}{b \left (a^2+b^2\right )}}{b \left (a^2+b^2\right )}\)

Input:

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

Output:

(-2*a^2*Tan[c + d*x]^3)/(3*b*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^(3/2)) + ( 
(-4*a^2*(a^2 + 2*b^2)*Tan[c + d*x]^2)/(b*(a^2 + b^2)*d*Sqrt[a + b*Tan[c + 
d*x]]) + ((2*(8*a^4 + 15*a^2*b^2 + b^4)*Tan[c + d*x]*Sqrt[a + b*Tan[c + d* 
x]])/(3*b*d) - (((3*I)*(a + I*b)^2*b^3*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]]) 
/(Sqrt[a - I*b]*d) - ((3*I)*(a - I*b)^2*b^3*ArcTan[Tan[c + d*x]/Sqrt[a + I 
*b]])/(Sqrt[a + I*b]*d) + (4*a*(8*a^4 + 15*a^2*b^2 + 4*b^4)*Sqrt[a + b*Tan 
[c + d*x]])/(b*d))/(3*b))/(b*(a^2 + b^2)))/(b*(a^2 + b^2))
 

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 221

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

rule 3042

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

rule 4020

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

rule 4022

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2   Int[(a + b*Tan[e + f*x])^m*( 
1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2   Int[(a + b*Tan[e + f*x])^m 
*(1 + I*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 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 4113

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

rule 4128

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[(A*d^2 + c*(c*C - B*d))*(a + b*Tan[e + 
 f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Sim 
p[1/(d*(n + 1)*(c^2 + d^2))   Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e 
 + f*x])^(n + 1)*Simp[A*d*(b*d*m - a*c*(n + 1)) + (c*C - B*d)*(b*c*m + a*d* 
(n + 1)) - d*(n + 1)*((A - C)*(b*c - a*d) + B*(a*c + b*d))*Tan[e + f*x] - b 
*(d*(B*c - A*d)*(m + n + 1) - C*(c^2*m - d^2*(n + 1)))*Tan[e + f*x]^2, x], 
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ 
[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
 

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])))
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2201\) vs. \(2(261)=522\).

Time = 0.26 (sec) , antiderivative size = 2202, normalized size of antiderivative = 7.57

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(2202\)
default	\(\text {Expression too large to display}\)	\(2202\)

Input:

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

Output:

-6/d/b^4*a*(a+b*tan(d*x+c))^(1/2)-1/2/d*b^2/(a^2+b^2)^3*ln(b*tan(d*x+c)+a- 
(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*( 
a^2+b^2)^(1/2)+2*a)^(1/2)*a-1/2/d*b^2/(a^2+b^2)^(7/2)*ln(b*tan(d*x+c)+a+(a 
+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^ 
2+b^2)^(1/2)+2*a)^(1/2)*a^2+1/2/d*b^2/(a^2+b^2)^(7/2)*ln(b*tan(d*x+c)+a-(a 
+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^ 
2+b^2)^(1/2)+2*a)^(1/2)*a^2+6/d*b^2/(a^2+b^2)^(7/2)/(2*(a^2+b^2)^(1/2)-2*a 
)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)-(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2 
*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^3+5/d*b^4/(a^2+b^2)^(7/2)/(2*(a^2+b^2)^(1/2 
)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)-(2*(a^2+b^2)^(1/2)+2*a)^(1/2 
))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a-2/d*b^2/(a^2+b^2)^(5/2)/(2*(a^2+b^2)^( 
1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)-(2*(a^2+b^2)^(1/2)+2*a)^( 
1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a+6/d*b^2/(a^2+b^2)^(7/2)/(2*(a^2+b^2 
)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a 
)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^3+5/d*b^4/(a^2+b^2)^(7/2)/(2*(a^ 
2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2 
)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a-2/d*b^2/(a^2+b^2)^(5/2)/(2* 
(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^( 
1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a+1/2/d*b^2/(a^2+b^2)^3*ln 
(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3401 vs. \(2 (257) = 514\).

Time = 0.16 (sec) , antiderivative size = 3401, normalized size of antiderivative = 11.69 \[ \int \frac {\tan ^5(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F(-1)]

Timed out. \[ \int \frac {\tan ^5(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Giac [F(-1)]

Timed out. \[ \int \frac {\tan ^5(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 12.37 (sec) , antiderivative size = 4681, normalized size of antiderivative = 16.09 \[ \int \frac {\tan ^5(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

atan(((((1/(a^5*d^2 - b^5*d^2*1i + 5*a*b^4*d^2 - a^4*b*d^2*5i + a^2*b^3*d^ 
2*10i - 10*a^3*b^2*d^2))^(1/2)*(48*a*b^20*d^4 - ((1/(a^5*d^2 - b^5*d^2*1i 
+ 5*a*b^4*d^2 - a^4*b*d^2*5i + a^2*b^3*d^2*10i - 10*a^3*b^2*d^2))^(1/2)*(a 
 + b*tan(c + d*x))^(1/2)*(64*a*b^22*d^5 + 640*a^3*b^20*d^5 + 2880*a^5*b^18 
*d^5 + 7680*a^7*b^16*d^5 + 13440*a^9*b^14*d^5 + 16128*a^11*b^12*d^5 + 1344 
0*a^13*b^10*d^5 + 7680*a^15*b^8*d^5 + 2880*a^17*b^6*d^5 + 640*a^19*b^4*d^5 
 + 64*a^21*b^2*d^5))/4 + 368*a^3*b^18*d^4 + 1216*a^5*b^16*d^4 + 2240*a^7*b 
^14*d^4 + 2464*a^9*b^12*d^4 + 1568*a^11*b^10*d^4 + 448*a^13*b^8*d^4 - 64*a 
^15*b^6*d^4 - 80*a^17*b^4*d^4 - 16*a^19*b^2*d^4))/2 - ((a + b*tan(c + d*x) 
)^(1/2)*(320*a^4*b^14*d^3 - 16*b^18*d^3 + 1024*a^6*b^12*d^3 + 1440*a^8*b^1 
0*d^3 + 1024*a^10*b^8*d^3 + 320*a^12*b^6*d^3 - 16*a^16*b^2*d^3))/2)*(1/(a^ 
5*d^2 - b^5*d^2*1i + 5*a*b^4*d^2 - a^4*b*d^2*5i + a^2*b^3*d^2*10i - 10*a^3 
*b^2*d^2))^(1/2)*1i - (((1/(a^5*d^2 - b^5*d^2*1i + 5*a*b^4*d^2 - a^4*b*d^2 
*5i + a^2*b^3*d^2*10i - 10*a^3*b^2*d^2))^(1/2)*(48*a*b^20*d^4 + ((1/(a^5*d 
^2 - b^5*d^2*1i + 5*a*b^4*d^2 - a^4*b*d^2*5i + a^2*b^3*d^2*10i - 10*a^3*b^ 
2*d^2))^(1/2)*(a + b*tan(c + d*x))^(1/2)*(64*a*b^22*d^5 + 640*a^3*b^20*d^5 
 + 2880*a^5*b^18*d^5 + 7680*a^7*b^16*d^5 + 13440*a^9*b^14*d^5 + 16128*a^11 
*b^12*d^5 + 13440*a^13*b^10*d^5 + 7680*a^15*b^8*d^5 + 2880*a^17*b^6*d^5 + 
640*a^19*b^4*d^5 + 64*a^21*b^2*d^5))/4 + 368*a^3*b^18*d^4 + 1216*a^5*b^16* 
d^4 + 2240*a^7*b^14*d^4 + 2464*a^9*b^12*d^4 + 1568*a^11*b^10*d^4 + 448*...
 

Reduce [F]

\[ \int \frac {\tan ^5(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\int \frac {\sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{5}}{\tan \left (d x +c \right )^{3} b^{3}+3 \tan \left (d x +c \right )^{2} a \,b^{2}+3 \tan \left (d x +c \right ) a^{2} b +a^{3}}d x \] Input:

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

Output:

int((sqrt(tan(c + d*x)*b + a)*tan(c + d*x)**5)/(tan(c + d*x)**3*b**3 + 3*t 
an(c + d*x)**2*a*b**2 + 3*tan(c + d*x)*a**2*b + a**3),x)