3.13.41 \(\int (a+b \tan (e+f x))^3 (c+d \tan (e+f x))^{5/2} \, dx\) [1241]

3.13.41.1 Optimal result

Integrand size = 27, antiderivative size = 322 \[ \int (a+b \tan (e+f x))^3 (c+d \tan (e+f x))^{5/2} \, dx=\frac {(i a+b)^3 (c-i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}-\frac {(i a-b)^3 (c+i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {2 \left (2 a^3 c d-6 a b^2 c d+3 a^2 b \left (c^2-d^2\right )-b^3 \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 \left (3 a^2 b c-b^3 c+a^3 d-3 a b^2 d\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 b \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {4 b^2 (b c-10 a d) (c+d \tan (e+f x))^{7/2}}{63 d^2 f}+\frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{7/2}}{9 d f} \]

output

(I*a+b)^3*(c-I*d)^(5/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/f-(I 
*a-b)^3*(c+I*d)^(5/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/f+2*(2 
*a^3*c*d-6*a*b^2*c*d+3*a^2*b*(c^2-d^2)-b^3*(c^2-d^2))*(c+d*tan(f*x+e))^(1/ 
2)/f+2/3*(a^3*d+3*a^2*b*c-3*a*b^2*d-b^3*c)*(c+d*tan(f*x+e))^(3/2)/f+2/5*b* 
(3*a^2-b^2)*(c+d*tan(f*x+e))^(5/2)/f-4/63*b^2*(-10*a*d+b*c)*(c+d*tan(f*x+e 
))^(7/2)/d^2/f+2/9*b^2*(a+b*tan(f*x+e))*(c+d*tan(f*x+e))^(7/2)/d/f
 

3.13.41.2 Mathematica [A] (verified)

Time = 6.22 (sec) , antiderivative size = 413, normalized size of antiderivative = 1.28 \[ \int (a+b \tan (e+f x))^3 (c+d \tan (e+f x))^{5/2} \, dx=\frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{7/2}}{9 d f}+\frac {2 \left (-\frac {2 b^2 (b c-10 a d) (c+d \tan (e+f x))^{7/2}}{7 d f}+\frac {i \left (\frac {9}{2} a \left (a^2-3 b^2\right ) d-\frac {9}{2} i b \left (3 a^2-b^2\right ) d\right ) \left (\frac {2}{5} (c+d \tan (e+f x))^{5/2}+(c-i d) \left (\frac {2}{3} (c+d \tan (e+f x))^{3/2}+(c-i d) \left (\frac {2 (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{-c+i d}+2 \sqrt {c+d \tan (e+f x)}\right )\right )\right )}{2 f}-\frac {i \left (\frac {9}{2} a \left (a^2-3 b^2\right ) d+\frac {9}{2} i b \left (3 a^2-b^2\right ) d\right ) \left (\frac {2}{5} (c+d \tan (e+f x))^{5/2}+(c+i d) \left (\frac {2}{3} (c+d \tan (e+f x))^{3/2}+(c+i d) \left (\frac {2 (c+i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{-c-i d}+2 \sqrt {c+d \tan (e+f x)}\right )\right )\right )}{2 f}\right )}{9 d} \]

input

Integrate[(a + b*Tan[e + f*x])^3*(c + d*Tan[e + f*x])^(5/2),x]
 

output

(2*b^2*(a + b*Tan[e + f*x])*(c + d*Tan[e + f*x])^(7/2))/(9*d*f) + (2*((-2* 
b^2*(b*c - 10*a*d)*(c + d*Tan[e + f*x])^(7/2))/(7*d*f) + ((I/2)*((9*a*(a^2 
 - 3*b^2)*d)/2 - ((9*I)/2)*b*(3*a^2 - b^2)*d)*((2*(c + d*Tan[e + f*x])^(5/ 
2))/5 + (c - I*d)*((2*(c + d*Tan[e + f*x])^(3/2))/3 + (c - I*d)*((2*(c - I 
*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(-c + I*d) + 2* 
Sqrt[c + d*Tan[e + f*x]]))))/f - ((I/2)*((9*a*(a^2 - 3*b^2)*d)/2 + ((9*I)/ 
2)*b*(3*a^2 - b^2)*d)*((2*(c + d*Tan[e + f*x])^(5/2))/5 + (c + I*d)*((2*(c 
 + d*Tan[e + f*x])^(3/2))/3 + (c + I*d)*((2*(c + I*d)^(3/2)*ArcTanh[Sqrt[c 
 + d*Tan[e + f*x]]/Sqrt[c + I*d]])/(-c - I*d) + 2*Sqrt[c + d*Tan[e + f*x]] 
))))/f))/(9*d)
 

3.13.41.3 Rubi [A] (warning: unable to verify)

Time = 2.09 (sec) , antiderivative size = 316, normalized size of antiderivative = 0.98, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 4049, 27, 3042, 4113, 3042, 4011, 3042, 4011, 3042, 4011, 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.

\(\displaystyle \int (a+b \tan (e+f x))^3 (c+d \tan (e+f x))^{5/2} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4049

\(\displaystyle \frac {2 \int -\frac {1}{2} (c+d \tan (e+f x))^{5/2} \left (-9 d a^3+7 b^2 d a+2 b^2 (b c-10 a d) \tan ^2(e+f x)+2 b^3 c-9 b \left (3 a^2-b^2\right ) d \tan (e+f x)\right )dx}{9 d}+\frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{7/2}}{9 d f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {\int (c+d \tan (e+f x))^{5/2} \left (-9 d a^3+7 b^2 d a+2 b^2 (b c-10 a d) \tan ^2(e+f x)+2 b^3 c-9 b \left (3 a^2-b^2\right ) d \tan (e+f x)\right )dx}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {\int (c+d \tan (e+f x))^{5/2} \left (-9 d a^3+7 b^2 d a+2 b^2 (b c-10 a d) \tan (e+f x)^2+2 b^3 c-9 b \left (3 a^2-b^2\right ) d \tan (e+f x)\right )dx}{9 d}\)

\(\Big \downarrow \) 4113

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {\int (c+d \tan (e+f x))^{5/2} \left (-9 a \left (a^2-3 b^2\right ) d-9 b \left (3 a^2-b^2\right ) \tan (e+f x) d\right )dx+\frac {4 b^2 (b c-10 a d) (c+d \tan (e+f x))^{7/2}}{7 d f}}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {\int (c+d \tan (e+f x))^{5/2} \left (-9 a \left (a^2-3 b^2\right ) d-9 b \left (3 a^2-b^2\right ) \tan (e+f x) d\right )dx+\frac {4 b^2 (b c-10 a d) (c+d \tan (e+f x))^{7/2}}{7 d f}}{9 d}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {\int (c+d \tan (e+f x))^{3/2} \left (-9 d \left (c a^3-3 b d a^2-3 b^2 c a+b^3 d\right )-9 d \left (d a^3+3 b c a^2-3 b^2 d a-b^3 c\right ) \tan (e+f x)\right )dx-\frac {18 b d \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {4 b^2 (b c-10 a d) (c+d \tan (e+f x))^{7/2}}{7 d f}}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {\int (c+d \tan (e+f x))^{3/2} \left (-9 d \left (c a^3-3 b d a^2-3 b^2 c a+b^3 d\right )-9 d \left (d a^3+3 b c a^2-3 b^2 d a-b^3 c\right ) \tan (e+f x)\right )dx-\frac {18 b d \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {4 b^2 (b c-10 a d) (c+d \tan (e+f x))^{7/2}}{7 d f}}{9 d}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {\int \sqrt {c+d \tan (e+f x)} \left (9 d \left (-\left (\left (c^2-d^2\right ) a^3\right )+6 b c d a^2+3 b^2 \left (c^2-d^2\right ) a-2 b^3 c d\right )-9 d \left (2 c d a^3+3 b \left (c^2-d^2\right ) a^2-6 b^2 c d a-b^3 \left (c^2-d^2\right )\right ) \tan (e+f x)\right )dx-\frac {18 b d \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {6 d \left (a^3 d+3 a^2 b c-3 a b^2 d-b^3 c\right ) (c+d \tan (e+f x))^{3/2}}{f}+\frac {4 b^2 (b c-10 a d) (c+d \tan (e+f x))^{7/2}}{7 d f}}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {\int \sqrt {c+d \tan (e+f x)} \left (9 d \left (-\left (\left (c^2-d^2\right ) a^3\right )+6 b c d a^2+3 b^2 \left (c^2-d^2\right ) a-2 b^3 c d\right )-9 d \left (2 c d a^3+3 b \left (c^2-d^2\right ) a^2-6 b^2 c d a-b^3 \left (c^2-d^2\right )\right ) \tan (e+f x)\right )dx-\frac {18 b d \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {6 d \left (a^3 d+3 a^2 b c-3 a b^2 d-b^3 c\right ) (c+d \tan (e+f x))^{3/2}}{f}+\frac {4 b^2 (b c-10 a d) (c+d \tan (e+f x))^{7/2}}{7 d f}}{9 d}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {\int \frac {-9 d (a c-b d) \left (a^2 c^2-3 b^2 c^2-8 a b d c-3 a^2 d^2+b^2 d^2\right )-9 d (b c+a d) \left (3 a^2 c^2-b^2 c^2-8 a b d c-a^2 d^2+3 b^2 d^2\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {18 b d \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {18 d \left (2 a^3 c d+3 a^2 b \left (c^2-d^2\right )-6 a b^2 c d-b^3 \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}-\frac {6 d \left (a^3 d+3 a^2 b c-3 a b^2 d-b^3 c\right ) (c+d \tan (e+f x))^{3/2}}{f}+\frac {4 b^2 (b c-10 a d) (c+d \tan (e+f x))^{7/2}}{7 d f}}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {\int \frac {-9 d (a c-b d) \left (a^2 c^2-3 b^2 c^2-8 a b d c-3 a^2 d^2+b^2 d^2\right )-9 d (b c+a d) \left (3 a^2 c^2-b^2 c^2-8 a b d c-a^2 d^2+3 b^2 d^2\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {18 b d \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {18 d \left (2 a^3 c d+3 a^2 b \left (c^2-d^2\right )-6 a b^2 c d-b^3 \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}-\frac {6 d \left (a^3 d+3 a^2 b c-3 a b^2 d-b^3 c\right ) (c+d \tan (e+f x))^{3/2}}{f}+\frac {4 b^2 (b c-10 a d) (c+d \tan (e+f x))^{7/2}}{7 d f}}{9 d}\)

\(\Big \downarrow \) 4022

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {-\frac {9}{2} d (a+i b)^3 (c+i d)^3 \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {9}{2} d (a-i b)^3 (c-i d)^3 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx-\frac {18 b d \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {18 d \left (2 a^3 c d+3 a^2 b \left (c^2-d^2\right )-6 a b^2 c d-b^3 \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}-\frac {6 d \left (a^3 d+3 a^2 b c-3 a b^2 d-b^3 c\right ) (c+d \tan (e+f x))^{3/2}}{f}+\frac {4 b^2 (b c-10 a d) (c+d \tan (e+f x))^{7/2}}{7 d f}}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {-\frac {9}{2} d (a+i b)^3 (c+i d)^3 \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {9}{2} d (a-i b)^3 (c-i d)^3 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx-\frac {18 b d \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {18 d \left (2 a^3 c d+3 a^2 b \left (c^2-d^2\right )-6 a b^2 c d-b^3 \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}-\frac {6 d \left (a^3 d+3 a^2 b c-3 a b^2 d-b^3 c\right ) (c+d \tan (e+f x))^{3/2}}{f}+\frac {4 b^2 (b c-10 a d) (c+d \tan (e+f x))^{7/2}}{7 d f}}{9 d}\)

\(\Big \downarrow \) 4020

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {-\frac {9 i d (a-i b)^3 (c-i d)^3 \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}+\frac {9 i d (a+i b)^3 (c+i d)^3 \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {18 b d \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {18 d \left (2 a^3 c d+3 a^2 b \left (c^2-d^2\right )-6 a b^2 c d-b^3 \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}-\frac {6 d \left (a^3 d+3 a^2 b c-3 a b^2 d-b^3 c\right ) (c+d \tan (e+f x))^{3/2}}{f}+\frac {4 b^2 (b c-10 a d) (c+d \tan (e+f x))^{7/2}}{7 d f}}{9 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {\frac {9 i d (a-i b)^3 (c-i d)^3 \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {9 i d (a+i b)^3 (c+i d)^3 \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {18 b d \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {18 d \left (2 a^3 c d+3 a^2 b \left (c^2-d^2\right )-6 a b^2 c d-b^3 \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}-\frac {6 d \left (a^3 d+3 a^2 b c-3 a b^2 d-b^3 c\right ) (c+d \tan (e+f x))^{3/2}}{f}+\frac {4 b^2 (b c-10 a d) (c+d \tan (e+f x))^{7/2}}{7 d f}}{9 d}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {-\frac {9 (a+i b)^3 (c+i d)^3 \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}-\frac {9 (a-i b)^3 (c-i d)^3 \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}-\frac {18 b d \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {18 d \left (2 a^3 c d+3 a^2 b \left (c^2-d^2\right )-6 a b^2 c d-b^3 \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}-\frac {6 d \left (a^3 d+3 a^2 b c-3 a b^2 d-b^3 c\right ) (c+d \tan (e+f x))^{3/2}}{f}+\frac {4 b^2 (b c-10 a d) (c+d \tan (e+f x))^{7/2}}{7 d f}}{9 d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {-\frac {18 b d \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {18 d \left (2 a^3 c d+3 a^2 b \left (c^2-d^2\right )-6 a b^2 c d-b^3 \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}-\frac {6 d \left (a^3 d+3 a^2 b c-3 a b^2 d-b^3 c\right ) (c+d \tan (e+f x))^{3/2}}{f}-\frac {9 d (a-i b)^3 (c-i d)^{5/2} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}-\frac {9 d (a+i b)^3 (c+i d)^{5/2} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}+\frac {4 b^2 (b c-10 a d) (c+d \tan (e+f x))^{7/2}}{7 d f}}{9 d}\)

input

Int[(a + b*Tan[e + f*x])^3*(c + d*Tan[e + f*x])^(5/2),x]
 

output

(2*b^2*(a + b*Tan[e + f*x])*(c + d*Tan[e + f*x])^(7/2))/(9*d*f) - ((-9*(a 
- I*b)^3*(c - I*d)^(5/2)*d*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/f - (9*(a + 
 I*b)^3*(c + I*d)^(5/2)*d*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/f - (18*d*(2 
*a^3*c*d - 6*a*b^2*c*d + 3*a^2*b*(c^2 - d^2) - b^3*(c^2 - d^2))*Sqrt[c + d 
*Tan[e + f*x]])/f - (6*d*(3*a^2*b*c - b^3*c + a^3*d - 3*a*b^2*d)*(c + d*Ta 
n[e + f*x])^(3/2))/f - (18*b*(3*a^2 - b^2)*d*(c + d*Tan[e + f*x])^(5/2))/( 
5*f) + (4*b^2*(b*c - 10*a*d)*(c + d*Tan[e + f*x])^(7/2))/(7*d*f))/(9*d)
 

3.13.41.3.1 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 4011

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

rule 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 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 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]
 

3.13.41.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(4892\) vs. \(2(286)=572\).

Time = 1.15 (sec) , antiderivative size = 4893, normalized size of antiderivative = 15.20

method	result	size
parts	\(\text {Expression too large to display}\)	\(4893\)
derivativedivides	\(\text {Expression too large to display}\)	\(5053\)
default	\(\text {Expression too large to display}\)	\(5053\)

input

int((a+b*tan(f*x+e))^3*(c+d*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)
 

output

a^3*(2/3/f*d*(c+d*tan(f*x+e))^(3/2)+4/f*d*(c+d*tan(f*x+e))^(1/2)*c-1/4/f/d 
*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c 
^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*c^2+1/4/f*d*l 
n(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2 
+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)+1/4/f/d*ln(d*ta 
n(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^ 
(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^3-3/4/f*d*ln(d*tan(f*x+e)+c+(c+d*ta 
n(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2 
)^(1/2)+2*c)^(1/2)*c+3/f*d/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*ta 
n(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/ 
2))*c^2-1/f*d^3/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^( 
1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))-2/f*d/( 
2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2) 
^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*(c^2+d^2)^(1/2)*c+1/4/f/ 
d*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-( 
c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*c^2-1/4/f*d* 
ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^ 
2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)-1/4/f/d*ln((c+ 
d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2) 
^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^3+3/4/f*d*ln((c+d*tan(f*x+e))^(...
 

3.13.41.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 12155 vs. \(2 (280) = 560\).

Time = 7.01 (sec) , antiderivative size = 12155, normalized size of antiderivative = 37.75 \[ \int (a+b \tan (e+f x))^3 (c+d \tan (e+f x))^{5/2} \, dx=\text {Too large to display} \]

input

integrate((a+b*tan(f*x+e))^3*(c+d*tan(f*x+e))^(5/2),x, algorithm="fricas")
 

output

Too large to include
 

3.13.41.6 Sympy [F]

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

input

integrate((a+b*tan(f*x+e))**3*(c+d*tan(f*x+e))**(5/2),x)
 

output

Integral((a + b*tan(e + f*x))**3*(c + d*tan(e + f*x))**(5/2), x)
 

3.13.41.7 Maxima [F(-1)]

Timed out. \[ \int (a+b \tan (e+f x))^3 (c+d \tan (e+f x))^{5/2} \, dx=\text {Timed out} \]

input

integrate((a+b*tan(f*x+e))^3*(c+d*tan(f*x+e))^(5/2),x, algorithm="maxima")
 

output

Timed out
 

3.13.41.8 Giac [F(-1)]

Timed out. \[ \int (a+b \tan (e+f x))^3 (c+d \tan (e+f x))^{5/2} \, dx=\text {Timed out} \]

input

integrate((a+b*tan(f*x+e))^3*(c+d*tan(f*x+e))^(5/2),x, algorithm="giac")
 

output

Timed out
 

3.13.41.9 Mupad [F(-1)]

Timed out. \[ \int (a+b \tan (e+f x))^3 (c+d \tan (e+f x))^{5/2} \, dx=\text {Hanged} \]

input

int((a + b*tan(e + f*x))^3*(c + d*tan(e + f*x))^(5/2),x)
 

output

\text{Hanged}