\(\int (a+i a \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{3/2} \, dx\) [1143]

Optimal result

Integrand size = 32, antiderivative size = 329 \[ \int (a+i a \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{3/2} \, dx=-\frac {\sqrt [4]{-1} a^{5/2} (c-3 i d) \left (c^2+18 i c d+15 d^2\right ) \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{8 d^{3/2} f}-\frac {4 i \sqrt {2} a^{5/2} (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f}+\frac {a^2 \left (c^2+14 i c d+19 d^2\right ) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{8 d f}+\frac {a^2 (c+13 i d) \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 d f}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f} \] Output:

-1/8*(-1)^(1/4)*a^(5/2)*(c-3*I*d)*(c^2+18*I*c*d+15*d^2)*arctanh((-1)^(3/4) 
*d^(1/2)*(a+I*a*tan(f*x+e))^(1/2)/a^(1/2)/(c+d*tan(f*x+e))^(1/2))/d^(3/2)/ 
f-4*I*2^(1/2)*a^(5/2)*(c-I*d)^(3/2)*arctanh(2^(1/2)*a^(1/2)*(c+d*tan(f*x+e 
))^(1/2)/(c-I*d)^(1/2)/(a+I*a*tan(f*x+e))^(1/2))/f+1/8*a^2*(c^2+14*I*c*d+1 
9*d^2)*(a+I*a*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(1/2)/d/f+1/12*a^2*(c+13* 
I*d)*(a+I*a*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(3/2)/d/f-1/3*a^2*(a+I*a*ta 
n(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(5/2)/d/f
 

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(1526\) vs. \(2(329)=658\).

Time = 6.85 (sec) , antiderivative size = 1526, normalized size of antiderivative = 4.64 \[ \int (a+i a \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{3/2} \, dx =\text {Too large to display} \] Input:

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

Output:

(d*(a + I*a*Tan[e + f*x])^(5/2)*Sqrt[c + d*Tan[e + f*x]])/(3*f) - ((((3*I) 
/16)*(7*c - (5*I)*d)*(I*a*c - a*d)^3*Sqrt[(I*a)/(-((a^2*c)/(I*a*c - a*d)) 
- (I*a^2*d)/(I*a*c - a*d))]*(-((a^2*c)/(I*a*c - a*d)) - (I*a^2*d)/(I*a*c - 
 a*d))^3*Sqrt[(I*a*(c + d*Tan[e + f*x]))/(I*a*c - a*d)]*Sqrt[1 + (I*a*d*(a 
 + I*a*Tan[e + f*x]))/((I*a*c - a*d)*(-((a^2*c)/(I*a*c - a*d)) - (I*a^2*d) 
/(I*a*c - a*d)))]*(((2*I)*a*d*(a + I*a*Tan[e + f*x]))/((I*a*c - a*d)*(-((a 
^2*c)/(I*a*c - a*d)) - (I*a^2*d)/(I*a*c - a*d))) + (4*a^2*d^2*(a + I*a*Tan 
[e + f*x])^2)/(3*(I*a*c - a*d)^2*(-((a^2*c)/(I*a*c - a*d)) - (I*a^2*d)/(I* 
a*c - a*d))^2) - (2*(-1)^(1/4)*Sqrt[a]*Sqrt[d]*ArcSinh[((-1)^(1/4)*Sqrt[a] 
*Sqrt[d]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[I*a*c - a*d]*Sqrt[-((a^2*c)/(I* 
a*c - a*d)) - (I*a^2*d)/(I*a*c - a*d)])]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt 
[I*a*c - a*d]*Sqrt[-((a^2*c)/(I*a*c - a*d)) - (I*a^2*d)/(I*a*c - a*d)]*Sqr 
t[1 + (I*a*d*(a + I*a*Tan[e + f*x]))/((I*a*c - a*d)*(-((a^2*c)/(I*a*c - a* 
d)) - (I*a^2*d)/(I*a*c - a*d)))])))/(a^2*d^2*f*Sqrt[a + I*a*Tan[e + f*x]]* 
Sqrt[c + d*Tan[e + f*x]]) + (a*(-1/2*(a^2*(7*c - (5*I)*d)*d) - (I/2)*a^2*( 
6*c^2 - (5*I)*c*d - d^2))*(2*a*((-2*Sqrt[2]*ArcTan[(Sqrt[-(a*c) + I*a*d]*S 
qrt[a + I*a*Tan[e + f*x]])/(Sqrt[2]*a*Sqrt[c + d*Tan[e + f*x]])])/Sqrt[-(a 
*c) + I*a*d] - (2*(-1)^(3/4)*Sqrt[c + I*d]*Sqrt[(c/(c + I*d) + (I*d)/(c + 
I*d))^(-1)]*Sqrt[c/(c + I*d) + (I*d)/(c + I*d)]*ArcSin[((-1)^(1/4)*Sqrt[d] 
*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c + I*d]*Sqrt[c/(c + I*d) + ...
 

Rubi [A] (verified)

Time = 2.09 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.05, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.531, Rules used = {3042, 4039, 27, 3042, 4080, 27, 3042, 4080, 27, 3042, 4084, 3042, 4027, 221, 4082, 66, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4039

\(\displaystyle \frac {a \int \frac {1}{2} \sqrt {i \tan (e+f x) a+a} (a (i c+11 d)+a (c+13 i d) \tan (e+f x)) (c+d \tan (e+f x))^{3/2}dx}{3 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a \int \sqrt {i \tan (e+f x) a+a} (a (i c+11 d)+a (c+13 i d) \tan (e+f x)) (c+d \tan (e+f x))^{3/2}dx}{6 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a \int \sqrt {i \tan (e+f x) a+a} (a (i c+11 d)+a (c+13 i d) \tan (e+f x)) (c+d \tan (e+f x))^{3/2}dx}{6 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}\)

\(\Big \downarrow \) 4080

\(\displaystyle \frac {a \left (\frac {\int \frac {3}{2} \sqrt {i \tan (e+f x) a+a} \sqrt {c+d \tan (e+f x)} \left (\left (18 c d+i \left (c^2-13 d^2\right )\right ) a^2+\left (c^2+14 i d c+19 d^2\right ) \tan (e+f x) a^2\right )dx}{2 a}+\frac {a (c+13 i d) \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}\right )}{6 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a \left (\frac {3 \int \sqrt {i \tan (e+f x) a+a} \sqrt {c+d \tan (e+f x)} \left (\left (18 c d+i \left (c^2-13 d^2\right )\right ) a^2+\left (c^2+14 i d c+19 d^2\right ) \tan (e+f x) a^2\right )dx}{4 a}+\frac {a (c+13 i d) \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}\right )}{6 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a \left (\frac {3 \int \sqrt {i \tan (e+f x) a+a} \sqrt {c+d \tan (e+f x)} \left (\left (18 c d+i \left (c^2-13 d^2\right )\right ) a^2+\left (c^2+14 i d c+19 d^2\right ) \tan (e+f x) a^2\right )dx}{4 a}+\frac {a (c+13 i d) \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}\right )}{6 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}\)

\(\Big \downarrow \) 4080

\(\displaystyle \frac {a \left (\frac {3 \left (\frac {\int \frac {\sqrt {i \tan (e+f x) a+a} \left (\left (i c^3+49 d c^2-59 i d^2 c-19 d^3\right ) a^3+(c-3 i d) \left (c^2+18 i d c+15 d^2\right ) \tan (e+f x) a^3\right )}{2 \sqrt {c+d \tan (e+f x)}}dx}{a}+\frac {a^2 \left (c^2+14 i c d+19 d^2\right ) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}\right )}{4 a}+\frac {a (c+13 i d) \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}\right )}{6 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a \left (\frac {3 \left (\frac {\int \frac {\sqrt {i \tan (e+f x) a+a} \left (\left (i c^3+49 d c^2-59 i d^2 c-19 d^3\right ) a^3+(c-3 i d) \left (c^2+18 i d c+15 d^2\right ) \tan (e+f x) a^3\right )}{\sqrt {c+d \tan (e+f x)}}dx}{2 a}+\frac {a^2 \left (c^2+14 i c d+19 d^2\right ) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}\right )}{4 a}+\frac {a (c+13 i d) \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}\right )}{6 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a \left (\frac {3 \left (\frac {\int \frac {\sqrt {i \tan (e+f x) a+a} \left (\left (i c^3+49 d c^2-59 i d^2 c-19 d^3\right ) a^3+(c-3 i d) \left (c^2+18 i d c+15 d^2\right ) \tan (e+f x) a^3\right )}{\sqrt {c+d \tan (e+f x)}}dx}{2 a}+\frac {a^2 \left (c^2+14 i c d+19 d^2\right ) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}\right )}{4 a}+\frac {a (c+13 i d) \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}\right )}{6 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}\)

\(\Big \downarrow \) 4084

\(\displaystyle \frac {a \left (\frac {3 \left (\frac {64 a^3 d (c-i d)^2 \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx+a^2 (3 d+i c) \left (c^2+18 i c d+15 d^2\right ) \int \frac {(a-i a \tan (e+f x)) \sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx}{2 a}+\frac {a^2 \left (c^2+14 i c d+19 d^2\right ) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}\right )}{4 a}+\frac {a (c+13 i d) \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}\right )}{6 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a \left (\frac {3 \left (\frac {64 a^3 d (c-i d)^2 \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx+a^2 (3 d+i c) \left (c^2+18 i c d+15 d^2\right ) \int \frac {(a-i a \tan (e+f x)) \sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx}{2 a}+\frac {a^2 \left (c^2+14 i c d+19 d^2\right ) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}\right )}{4 a}+\frac {a (c+13 i d) \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}\right )}{6 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}\)

\(\Big \downarrow \) 4027

\(\displaystyle \frac {a \left (\frac {3 \left (\frac {a^2 (3 d+i c) \left (c^2+18 i c d+15 d^2\right ) \int \frac {(a-i a \tan (e+f x)) \sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx-\frac {128 i a^5 d (c-i d)^2 \int \frac {1}{a (c-i d)-\frac {2 a^2 (c+d \tan (e+f x))}{i \tan (e+f x) a+a}}d\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {i \tan (e+f x) a+a}}}{f}}{2 a}+\frac {a^2 \left (c^2+14 i c d+19 d^2\right ) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}\right )}{4 a}+\frac {a (c+13 i d) \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}\right )}{6 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {a \left (\frac {3 \left (\frac {a^2 (3 d+i c) \left (c^2+18 i c d+15 d^2\right ) \int \frac {(a-i a \tan (e+f x)) \sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx-\frac {64 i \sqrt {2} a^{7/2} d (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f}}{2 a}+\frac {a^2 \left (c^2+14 i c d+19 d^2\right ) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}\right )}{4 a}+\frac {a (c+13 i d) \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}\right )}{6 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}\)

\(\Big \downarrow \) 4082

\(\displaystyle \frac {a \left (\frac {3 \left (\frac {\frac {a^4 (3 d+i c) \left (c^2+18 i c d+15 d^2\right ) \int \frac {1}{\sqrt {i \tan (e+f x) a+a} \sqrt {c+d \tan (e+f x)}}d\tan (e+f x)}{f}-\frac {64 i \sqrt {2} a^{7/2} d (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f}}{2 a}+\frac {a^2 \left (c^2+14 i c d+19 d^2\right ) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}\right )}{4 a}+\frac {a (c+13 i d) \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}\right )}{6 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {a \left (\frac {3 \left (\frac {\frac {2 a^4 (3 d+i c) \left (c^2+18 i c d+15 d^2\right ) \int \frac {1}{i a-\frac {d (i \tan (e+f x) a+a)}{c+d \tan (e+f x)}}d\frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}}{f}-\frac {64 i \sqrt {2} a^{7/2} d (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f}}{2 a}+\frac {a^2 \left (c^2+14 i c d+19 d^2\right ) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}\right )}{4 a}+\frac {a (c+13 i d) \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}\right )}{6 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {a \left (\frac {3 \left (\frac {\frac {2 (-1)^{3/4} a^{7/2} (3 d+i c) \left (c^2+18 i c d+15 d^2\right ) \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {d} f}-\frac {64 i \sqrt {2} a^{7/2} d (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f}}{2 a}+\frac {a^2 \left (c^2+14 i c d+19 d^2\right ) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}\right )}{4 a}+\frac {a (c+13 i d) \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 f}\right )}{6 d}-\frac {a^2 \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}\)

Input:

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

Output:

-1/3*(a^2*Sqrt[a + I*a*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(5/2))/(d*f) + ( 
a*((a*(c + (13*I)*d)*Sqrt[a + I*a*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(3/2) 
)/(2*f) + (3*(((2*(-1)^(3/4)*a^(7/2)*(I*c + 3*d)*(c^2 + (18*I)*c*d + 15*d^ 
2)*ArcTanh[((-1)^(3/4)*Sqrt[d]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c 
 + d*Tan[e + f*x]])])/(Sqrt[d]*f) - ((64*I)*Sqrt[2]*a^(7/2)*(c - I*d)^(3/2 
)*d*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])/(Sqrt[c - I*d]*Sqrt 
[a + I*a*Tan[e + f*x]])])/f)/(2*a) + (a^2*(c^2 + (14*I)*c*d + 19*d^2)*Sqrt 
[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])/f))/(4*a)))/(6*d)
 

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 66

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 
2   Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre 
eQ[{a, b, c, d}, x] &&  !GtQ[c - a*(d/b), 0]
 

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 4027

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

rule 4039

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[a/(d*(m + n - 1)) 
Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + 
a*d*(m + 2*n) + (a*c*(m - 2) + b*d*(3*m + 2*n - 4))*Tan[e + f*x], x], x], x 
] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 
 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] 
 && (IntegerQ[m] || IntegersQ[2*m, 2*n])
 

rule 4080

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

rule 4082

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[b*(B/f)   Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x]], 
x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[ 
a^2 + b^2, 0] && EqQ[A*b + a*B, 0]
 

rule 4084

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1517 vs. \(2 (265 ) = 530\).

Time = 0.41 (sec) , antiderivative size = 1518, normalized size of antiderivative = 4.61

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1518\)
default	\(\text {Expression too large to display}\)	\(1518\)

Input:

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

Output:

1/96/f*(a*(1+I*tan(f*x+e)))^(1/2)*(c+d*tan(f*x+e))^(1/2)*a^2*(-16*2^(1/2)* 
d^2*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d) 
^(1/2)*tan(f*x+e)^2-28*2^(1/2)*c*d*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))* 
(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*tan(f*x+e)+3*I*2^(1/2)*(-a*(I*d-c))^ 
(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+ 
e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c^3+96*I*(I*a*d)^(1/2)*ln(( 
3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2) 
*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(I+tan(f*x+e)))*a*d^2-96*I*2 
^(1/2)*(-a*(I*d-c))^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f 
*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c*d+207 
*I*2^(1/2)*(-a*(I*d-c))^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*t 
an(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c*d 
^2-96*I*2^(1/2)*(-a*(I*d-c))^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*( 
c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))* 
a*d^2+136*I*(I*a*d)^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*( 
1+I*tan(f*x+e)))^(1/2)*c*d-45*2^(1/2)*(-a*(I*d-c))^(1/2)*ln(1/2*(2*I*a*d*t 
an(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2 
)+a*d)/(I*a*d)^(1/2))*a*c^2*d+135*2^(1/2)*(-a*(I*d-c))^(1/2)*ln(1/2*(2*I*a 
*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^ 
(1/2)+a*d)/(I*a*d)^(1/2))*a*d^3+96*I*(I*a*d)^(1/2)*ln((3*a*c+I*a*tan(f*...
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1472 vs. \(2 (253) = 506\).

Time = 0.17 (sec) , antiderivative size = 1472, normalized size of antiderivative = 4.47 \[ \int (a+i a \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{3/2} \, dx=\text {Too large to display} \] Input:

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

Output:

1/48*(96*sqrt(2)*(d*f*e^(4*I*f*x + 4*I*e) + 2*d*f*e^(2*I*f*x + 2*I*e) + d* 
f)*sqrt(-(a^5*c^3 - 3*I*a^5*c^2*d - 3*a^5*c*d^2 + I*a^5*d^3)/f^2)*log((sqr 
t(2)*f*sqrt(-(a^5*c^3 - 3*I*a^5*c^2*d - 3*a^5*c*d^2 + I*a^5*d^3)/f^2)*e^(I 
*f*x + I*e) + sqrt(2)*(-I*a^2*c - a^2*d + (-I*a^2*c - a^2*d)*e^(2*I*f*x + 
2*I*e))*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e 
) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/(-I*a^2*c - a^ 
2*d)) - 96*sqrt(2)*(d*f*e^(4*I*f*x + 4*I*e) + 2*d*f*e^(2*I*f*x + 2*I*e) + 
d*f)*sqrt(-(a^5*c^3 - 3*I*a^5*c^2*d - 3*a^5*c*d^2 + I*a^5*d^3)/f^2)*log(-( 
sqrt(2)*f*sqrt(-(a^5*c^3 - 3*I*a^5*c^2*d - 3*a^5*c*d^2 + I*a^5*d^3)/f^2)*e 
^(I*f*x + I*e) - sqrt(2)*(-I*a^2*c - a^2*d + (-I*a^2*c - a^2*d)*e^(2*I*f*x 
 + 2*I*e))*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2* 
I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/(-I*a^2*c - 
 a^2*d)) - 2*sqrt(2)*((3*a^2*c^2 - 82*I*a^2*c*d - 91*a^2*d^2)*e^(5*I*f*x + 
 5*I*e) + 2*(3*a^2*c^2 - 68*I*a^2*c*d - 49*a^2*d^2)*e^(3*I*f*x + 3*I*e) + 
3*(a^2*c^2 - 18*I*a^2*c*d - 13*a^2*d^2)*e^(I*f*x + I*e))*sqrt(((c - I*d)*e 
^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*f* 
x + 2*I*e) + 1)) + 3*(d*f*e^(4*I*f*x + 4*I*e) + 2*d*f*e^(2*I*f*x + 2*I*e) 
+ d*f)*sqrt((I*a^5*c^6 - 30*a^5*c^5*d - 87*I*a^5*c^4*d^2 - 1980*a^5*c^3*d^ 
3 + 6111*I*a^5*c^2*d^4 + 6210*a^5*c*d^5 - 2025*I*a^5*d^6)/(d^3*f^2))*log(( 
2*d^2*f*sqrt((I*a^5*c^6 - 30*a^5*c^5*d - 87*I*a^5*c^4*d^2 - 1980*a^5*c^...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int (a+i a \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{3/2} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(3*d-c>0)', see `assume?` for mor 
e details)
 

Giac [F(-2)]

Exception generated. \[ \int (a+i a \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{3/2} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument Typesym2poly/r2sym(const gen & e,const index_m & i,const vecteur 
& l) Erro
 

Mupad [F(-1)]

Timed out. \[ \int (a+i a \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{3/2} \, dx=\int {\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2} \,d x \] Input:

int((a + a*tan(e + f*x)*1i)^(5/2)*(c + d*tan(e + f*x))^(3/2),x)
 

Output:

int((a + a*tan(e + f*x)*1i)^(5/2)*(c + d*tan(e + f*x))^(3/2), x)
 

Reduce [F]

\[ \int (a+i a \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{3/2} \, dx=\sqrt {a}\, a^{2} \left (-\left (\int \sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )^{3}d x \right ) d -\left (\int \sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )^{2}d x \right ) c +2 \left (\int \sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )^{2}d x \right ) d i +2 \left (\int \sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )d x \right ) c i +\left (\int \sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )d x \right ) d +\left (\int \sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {d \tan \left (f x +e \right )+c}d x \right ) c \right ) \] Input:

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

Output:

sqrt(a)*a**2*( - int(sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*tan 
(e + f*x)**3,x)*d - int(sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)* 
tan(e + f*x)**2,x)*c + 2*int(sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d 
+ c)*tan(e + f*x)**2,x)*d*i + 2*int(sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + 
f*x)*d + c)*tan(e + f*x),x)*c*i + int(sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e 
+ f*x)*d + c)*tan(e + f*x),x)*d + int(sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e 
+ f*x)*d + c),x)*c)