\(\int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\) [442]

Optimal result

Integrand size = 35, antiderivative size = 397 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=-\frac {(i a-b)^{5/2} (i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\left (40 a^3 A b-320 a A b^3-5 a^4 B-240 a^2 b^2 B+128 b^4 B\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{64 b^{3/2} d}-\frac {(i a+b)^{5/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\left (40 a^2 A b-64 A b^3-5 a^3 B-112 a b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{64 b d}+\frac {\left (40 a A b-5 a^2 B-48 b^2 B\right ) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{96 b d}+\frac {(8 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{24 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d} \] Output:

-(I*a-b)^(5/2)*(I*A-B)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+ 
c))^(1/2))/d+1/64*(40*A*a^3*b-320*A*a*b^3-5*B*a^4-240*B*a^2*b^2+128*B*b^4) 
*arctanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/b^(3/2)/d-(I*a+b 
)^(5/2)*(I*A+B)*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1 
/2))/d+1/64*(40*A*a^2*b-64*A*b^3-5*B*a^3-112*B*a*b^2)*tan(d*x+c)^(1/2)*(a+ 
b*tan(d*x+c))^(1/2)/b/d+1/96*(40*A*a*b-5*B*a^2-48*B*b^2)*tan(d*x+c)^(1/2)* 
(a+b*tan(d*x+c))^(3/2)/b/d+1/24*(8*A*b-B*a)*tan(d*x+c)^(1/2)*(a+b*tan(d*x+ 
c))^(5/2)/b/d+1/4*B*tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(7/2)/b/d
 

Mathematica [A] (verified)

Time = 3.08 (sec) , antiderivative size = 411, normalized size of antiderivative = 1.04 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {-192 \sqrt [4]{-1} (-a+i b)^{5/2} b (i A+B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+192 (-1)^{3/4} (a+i b)^{5/2} b (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-3 \left (-40 a^2 A b+64 A b^3+5 a^3 B+112 a b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}-2 \left (-40 a A b+5 a^2 B+48 b^2 B\right ) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}+8 (8 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}+48 B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{7/2}-\frac {3 \sqrt {a} \left (-40 a^3 A b+320 a A b^3+5 a^4 B+240 a^2 b^2 B-128 b^4 B\right ) \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {b} \sqrt {a+b \tan (c+d x)}}}{192 b d} \] Input:

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

Output:

(-192*(-1)^(1/4)*(-a + I*b)^(5/2)*b*(I*A + B)*ArcTan[((-1)^(1/4)*Sqrt[-a + 
 I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] + 192*(-1)^(3/4)*(a + 
I*b)^(5/2)*b*(A + I*B)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]] 
)/Sqrt[a + b*Tan[c + d*x]]] - 3*(-40*a^2*A*b + 64*A*b^3 + 5*a^3*B + 112*a* 
b^2*B)*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]] - 2*(-40*a*A*b + 5*a^2* 
B + 48*b^2*B)*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(3/2) + 8*(8*A*b - a 
*B)*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(5/2) + 48*B*Sqrt[Tan[c + d*x] 
]*(a + b*Tan[c + d*x])^(7/2) - (3*Sqrt[a]*(-40*a^3*A*b + 320*a*A*b^3 + 5*a 
^4*B + 240*a^2*b^2*B - 128*b^4*B)*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqr 
t[a]]*Sqrt[1 + (b*Tan[c + d*x])/a])/(Sqrt[b]*Sqrt[a + b*Tan[c + d*x]]))/(1 
92*b*d)
 

Rubi [A] (verified)

Time = 2.53 (sec) , antiderivative size = 400, normalized size of antiderivative = 1.01, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.486, Rules used = {3042, 4090, 27, 3042, 4130, 27, 3042, 4130, 27, 3042, 4130, 27, 3042, 4138, 2035, 2257, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4090

\(\displaystyle \frac {\int -\frac {(a+b \tan (c+d x))^{5/2} \left (-\left ((8 A b-a B) \tan ^2(c+d x)\right )+8 b B \tan (c+d x)+a B\right )}{2 \sqrt {\tan (c+d x)}}dx}{4 b}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}-\frac {\int \frac {(a+b \tan (c+d x))^{5/2} \left (-\left ((8 A b-a B) \tan ^2(c+d x)\right )+8 b B \tan (c+d x)+a B\right )}{\sqrt {\tan (c+d x)}}dx}{8 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}-\frac {\int \frac {(a+b \tan (c+d x))^{5/2} \left (-\left ((8 A b-a B) \tan (c+d x)^2\right )+8 b B \tan (c+d x)+a B\right )}{\sqrt {\tan (c+d x)}}dx}{8 b}\)

\(\Big \downarrow \) 4130

\(\displaystyle \frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}-\frac {\frac {1}{3} \int \frac {(a+b \tan (c+d x))^{3/2} \left (-\left (\left (-5 B a^2+40 A b a-48 b^2 B\right ) \tan ^2(c+d x)\right )+48 b (A b+a B) \tan (c+d x)+a (8 A b+5 a B)\right )}{2 \sqrt {\tan (c+d x)}}dx-\frac {(8 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 d}}{8 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}-\frac {\frac {1}{6} \int \frac {(a+b \tan (c+d x))^{3/2} \left (-\left (\left (-5 B a^2+40 A b a-48 b^2 B\right ) \tan ^2(c+d x)\right )+48 b (A b+a B) \tan (c+d x)+a (8 A b+5 a B)\right )}{\sqrt {\tan (c+d x)}}dx-\frac {(8 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 d}}{8 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}-\frac {\frac {1}{6} \int \frac {(a+b \tan (c+d x))^{3/2} \left (-\left (\left (-5 B a^2+40 A b a-48 b^2 B\right ) \tan (c+d x)^2\right )+48 b (A b+a B) \tan (c+d x)+a (8 A b+5 a B)\right )}{\sqrt {\tan (c+d x)}}dx-\frac {(8 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 d}}{8 b}\)

\(\Big \downarrow \) 4130

\(\displaystyle \frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}-\frac {\frac {1}{6} \left (\frac {1}{2} \int \frac {3 \sqrt {a+b \tan (c+d x)} \left (-\left (\left (-5 B a^3+40 A b a^2-112 b^2 B a-64 A b^3\right ) \tan ^2(c+d x)\right )+64 b \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)+a \left (5 B a^2+24 A b a-16 b^2 B\right )\right )}{2 \sqrt {\tan (c+d x)}}dx-\frac {\left (-5 a^2 B+40 a A b-48 b^2 B\right ) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\right )-\frac {(8 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 d}}{8 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}-\frac {\frac {1}{6} \left (\frac {3}{4} \int \frac {\sqrt {a+b \tan (c+d x)} \left (-\left (\left (-5 B a^3+40 A b a^2-112 b^2 B a-64 A b^3\right ) \tan ^2(c+d x)\right )+64 b \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)+a \left (5 B a^2+24 A b a-16 b^2 B\right )\right )}{\sqrt {\tan (c+d x)}}dx-\frac {\left (-5 a^2 B+40 a A b-48 b^2 B\right ) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\right )-\frac {(8 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 d}}{8 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}-\frac {\frac {1}{6} \left (\frac {3}{4} \int \frac {\sqrt {a+b \tan (c+d x)} \left (-\left (\left (-5 B a^3+40 A b a^2-112 b^2 B a-64 A b^3\right ) \tan (c+d x)^2\right )+64 b \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)+a \left (5 B a^2+24 A b a-16 b^2 B\right )\right )}{\sqrt {\tan (c+d x)}}dx-\frac {\left (-5 a^2 B+40 a A b-48 b^2 B\right ) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\right )-\frac {(8 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 d}}{8 b}\)

\(\Big \downarrow \) 4130

\(\displaystyle \frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}-\frac {\frac {1}{6} \left (\frac {3}{4} \left (\int \frac {-\left (\left (-5 B a^4+40 A b a^3-240 b^2 B a^2-320 A b^3 a+128 b^4 B\right ) \tan ^2(c+d x)\right )+128 b \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (5 B a^3+88 A b a^2-144 b^2 B a-64 A b^3\right )}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {\left (-5 a^3 B+40 a^2 A b-112 a b^2 B-64 A b^3\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )-\frac {\left (-5 a^2 B+40 a A b-48 b^2 B\right ) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\right )-\frac {(8 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 d}}{8 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}-\frac {\frac {1}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {-\left (\left (-5 B a^4+40 A b a^3-240 b^2 B a^2-320 A b^3 a+128 b^4 B\right ) \tan ^2(c+d x)\right )+128 b \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (5 B a^3+88 A b a^2-144 b^2 B a-64 A b^3\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {\left (-5 a^3 B+40 a^2 A b-112 a b^2 B-64 A b^3\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )-\frac {\left (-5 a^2 B+40 a A b-48 b^2 B\right ) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\right )-\frac {(8 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 d}}{8 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}-\frac {\frac {1}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {-\left (\left (-5 B a^4+40 A b a^3-240 b^2 B a^2-320 A b^3 a+128 b^4 B\right ) \tan (c+d x)^2\right )+128 b \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (5 B a^3+88 A b a^2-144 b^2 B a-64 A b^3\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {\left (-5 a^3 B+40 a^2 A b-112 a b^2 B-64 A b^3\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )-\frac {\left (-5 a^2 B+40 a A b-48 b^2 B\right ) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\right )-\frac {(8 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 d}}{8 b}\)

\(\Big \downarrow \) 4138

\(\displaystyle \frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}-\frac {\frac {1}{6} \left (\frac {3}{4} \left (\frac {\int \frac {-\left (\left (-5 B a^4+40 A b a^3-240 b^2 B a^2-320 A b^3 a+128 b^4 B\right ) \tan ^2(c+d x)\right )+128 b \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (5 B a^3+88 A b a^2-144 b^2 B a-64 A b^3\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{2 d}-\frac {\left (-5 a^3 B+40 a^2 A b-112 a b^2 B-64 A b^3\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )-\frac {\left (-5 a^2 B+40 a A b-48 b^2 B\right ) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\right )-\frac {(8 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 d}}{8 b}\)

\(\Big \downarrow \) 2035

\(\displaystyle \frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}-\frac {\frac {1}{6} \left (\frac {3}{4} \left (\frac {\int \frac {-\left (\left (-5 B a^4+40 A b a^3-240 b^2 B a^2-320 A b^3 a+128 b^4 B\right ) \tan ^2(c+d x)\right )+128 b \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (5 B a^3+88 A b a^2-144 b^2 B a-64 A b^3\right )}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\sqrt {\tan (c+d x)}}{d}-\frac {\left (-5 a^3 B+40 a^2 A b-112 a b^2 B-64 A b^3\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )-\frac {\left (-5 a^2 B+40 a A b-48 b^2 B\right ) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\right )-\frac {(8 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 d}}{8 b}\)

\(\Big \downarrow \) 2257

\(\displaystyle \frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}-\frac {\frac {1}{6} \left (\frac {3}{4} \left (\frac {\int \left (\frac {5 B a^4-40 A b a^3+240 b^2 B a^2+320 A b^3 a-128 b^4 B}{\sqrt {a+b \tan (c+d x)}}+\frac {128 \left (b \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right )+b \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)\right )}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}\right )d\sqrt {\tan (c+d x)}}{d}-\frac {\left (-5 a^3 B+40 a^2 A b-112 a b^2 B-64 A b^3\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )-\frac {\left (-5 a^2 B+40 a A b-48 b^2 B\right ) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\right )-\frac {(8 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 d}}{8 b}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}-\frac {-\frac {(8 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 d}+\frac {1}{6} \left (-\frac {\left (-5 a^2 B+40 a A b-48 b^2 B\right ) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}+\frac {3}{4} \left (-\frac {\left (-5 a^3 B+40 a^2 A b-112 a b^2 B-64 A b^3\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}+\frac {-\frac {\left (-5 a^4 B+40 a^3 A b-240 a^2 b^2 B-320 a A b^3+128 b^4 B\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b}}+64 b (-b+i a)^{5/2} (-B+i A) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+64 b (b+i a)^{5/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}\right )\right )}{8 b}\)

Input:

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

Output:

(B*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(7/2))/(4*b*d) - (-1/3*((8*A*b 
- a*B)*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(5/2))/d + (-1/2*((40*a*A*b 
 - 5*a^2*B - 48*b^2*B)*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(3/2))/d + 
(3*((64*(I*a - b)^(5/2)*b*(I*A - B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x 
]])/Sqrt[a + b*Tan[c + d*x]]] - ((40*a^3*A*b - 320*a*A*b^3 - 5*a^4*B - 240 
*a^2*b^2*B + 128*b^4*B)*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Ta 
n[c + d*x]]])/Sqrt[b] + 64*b*(I*a + b)^(5/2)*(I*A + B)*ArcTanh[(Sqrt[I*a + 
 b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d - ((40*a^2*A*b - 64*A 
*b^3 - 5*a^3*B - 112*a*b^2*B)*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]]) 
/d))/4)/6)/(8*b)
 

Defintions of rubi rules used

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 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2035

Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k   Subst 
[Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti 
onQ[m] && AlgebraicFunctionQ[Fx, x]
 

rule 2257

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol 
] :> Int[ExpandIntegrand[Px*(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a 
, c, d, e, q}, x] && PolyQ[Px, x] && IntegerQ[p]
 

rule 3042

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

rule 4090

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

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 4138

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] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, S 
imp[ff/f   Subst[Int[(a + b*ff*x)^m*(c + d*ff*x)^n*((A + B*ff*x + C*ff^2*x^ 
2)/(1 + ff^2*x^2)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f 
, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + 
 d^2, 0]
 

Maple [B] (warning: unable to verify)

result has leaf size over 500,000. Avoiding possible recursion issues.

Time = 5.22 (sec) , antiderivative size = 2656933, normalized size of antiderivative = 6692.53

\[\text {output too large to display}\]

Input:

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

Output:

result too large to display
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 17728 vs. \(2 (334) = 668\).

Time = 8.94 (sec) , antiderivative size = 35458, normalized size of antiderivative = 89.31 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \] Input:

integrate(tan(d*x+c)^(3/2)*(a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algo 
rithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

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

integrate(tan(d*x+c)^(3/2)*(a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algo 
rithm="maxima")
 

Output:

integrate((B*tan(d*x + c) + A)*(b*tan(d*x + c) + a)^(5/2)*tan(d*x + c)^(3/ 
2), x)
 

Giac [F]

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

integrate(tan(d*x+c)^(3/2)*(a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algo 
rithm="giac")
 

Output:

integrate((B*tan(d*x + c) + A)*(b*tan(d*x + c) + a)^(5/2)*tan(d*x + c)^(3/ 
2), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\left (\int \sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{4}d x \right ) b^{3}+3 \left (\int \sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{3}d x \right ) a \,b^{2}+3 \left (\int \sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{2}d x \right ) a^{2} b +\left (\int \sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )d x \right ) a^{3} \] Input:

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

Output:

int(sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*b + a)*tan(c + d*x)**4,x)*b**3 + 
3*int(sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*b + a)*tan(c + d*x)**3,x)*a*b** 
2 + 3*int(sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*b + a)*tan(c + d*x)**2,x)*a 
**2*b + int(sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*b + a)*tan(c + d*x),x)*a* 
*3