\(\int \frac {(a+i a \tan (c+d x))^{5/2}}{\sqrt {e \sec (c+d x)}} \, dx\) [410]

Optimal result

Integrand size = 30, antiderivative size = 437 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\sqrt {e \sec (c+d x)}} \, dx=\frac {5 i a^{7/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right ) \sec (c+d x)}{\sqrt {2} d \sqrt {e} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a^{7/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right ) \sec (c+d x)}{\sqrt {2} d \sqrt {e} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {5 i a^{7/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)} \left (\sqrt {a}+\cos (c+d x) \left (\sqrt {a}-i \sqrt {a} \tan (c+d x)\right )\right )}\right ) \sec (c+d x)}{\sqrt {2} d \sqrt {e} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {10 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {e \sec (c+d x)}}+\frac {i a (a+i a \tan (c+d x))^{3/2}}{d \sqrt {e \sec (c+d x)}} \] Output:

5/2*I*a^(7/2)*arctan(1-2^(1/2)*e^(1/2)*(a-I*a*tan(d*x+c))^(1/2)/a^(1/2)/(e 
*sec(d*x+c))^(1/2))*sec(d*x+c)*2^(1/2)/d/e^(1/2)/(a-I*a*tan(d*x+c))^(1/2)/ 
(a+I*a*tan(d*x+c))^(1/2)-5/2*I*a^(7/2)*arctan(1+2^(1/2)*e^(1/2)*(a-I*a*tan 
(d*x+c))^(1/2)/a^(1/2)/(e*sec(d*x+c))^(1/2))*sec(d*x+c)*2^(1/2)/d/e^(1/2)/ 
(a-I*a*tan(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2)+5/2*I*a^(7/2)*arctanh(2^ 
(1/2)*e^(1/2)*(a-I*a*tan(d*x+c))^(1/2)/(e*sec(d*x+c))^(1/2)/(a^(1/2)+cos(d 
*x+c)*(a^(1/2)-I*a^(1/2)*tan(d*x+c))))*sec(d*x+c)*2^(1/2)/d/e^(1/2)/(a-I*a 
*tan(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2)-10*I*a^2*(a+I*a*tan(d*x+c))^(1 
/2)/d/(e*sec(d*x+c))^(1/2)+I*a*(a+I*a*tan(d*x+c))^(3/2)/d/(e*sec(d*x+c))^( 
1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 3.12 (sec) , antiderivative size = 359, normalized size of antiderivative = 0.82 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\sqrt {e \sec (c+d x)}} \, dx=\frac {e^2 (a+i a \tan (c+d x))^{5/2} \left (\frac {5 \text {arctanh}\left (\frac {\sqrt {1-i \cos (c)+\sin (c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}}{\sqrt {-1-i \cos (c)-\sin (c)} \sqrt {i+\tan \left (\frac {d x}{2}\right )}}\right ) \sqrt {1-i \cos (c)+\sin (c)} (\cos (3 c)-i \sin (3 c)) \sqrt {i+\tan \left (\frac {d x}{2}\right )}}{\sqrt {-1-i \cos (c)-\sin (c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}}-\frac {5 \text {arctanh}\left (\frac {\sqrt {1+i \cos (c)-\sin (c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}}{\sqrt {-1+i \cos (c)+\sin (c)} \sqrt {i+\tan \left (\frac {d x}{2}\right )}}\right ) \sqrt {1+i \cos (c)-\sin (c)} (\cos (3 c)-i \sin (3 c)) \sqrt {i+\tan \left (\frac {d x}{2}\right )}}{\sqrt {-1+i \cos (c)+\sin (c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}}-(\cos (2 c)-i \sin (2 c)) (9 i+\tan (c+d x))\right )}{d (e \sec (c+d x))^{5/2} (\cos (d x)+i \sin (d x))^2} \] Input:

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

Output:

(e^2*(a + I*a*Tan[c + d*x])^(5/2)*((5*ArcTanh[(Sqrt[1 - I*Cos[c] + Sin[c]] 
*Sqrt[I - Tan[(d*x)/2]])/(Sqrt[-1 - I*Cos[c] - Sin[c]]*Sqrt[I + Tan[(d*x)/ 
2]])]*Sqrt[1 - I*Cos[c] + Sin[c]]*(Cos[3*c] - I*Sin[3*c])*Sqrt[I + Tan[(d* 
x)/2]])/(Sqrt[-1 - I*Cos[c] - Sin[c]]*Sqrt[I - Tan[(d*x)/2]]) - (5*ArcTanh 
[(Sqrt[1 + I*Cos[c] - Sin[c]]*Sqrt[I - Tan[(d*x)/2]])/(Sqrt[-1 + I*Cos[c] 
+ Sin[c]]*Sqrt[I + Tan[(d*x)/2]])]*Sqrt[1 + I*Cos[c] - Sin[c]]*(Cos[3*c] - 
 I*Sin[3*c])*Sqrt[I + Tan[(d*x)/2]])/(Sqrt[-1 + I*Cos[c] + Sin[c]]*Sqrt[I 
- Tan[(d*x)/2]]) - (Cos[2*c] - I*Sin[2*c])*(9*I + Tan[c + d*x])))/(d*(e*Se 
c[c + d*x])^(5/2)*(Cos[d*x] + I*Sin[d*x])^2)
 

Rubi [A] (verified)

Time = 1.03 (sec) , antiderivative size = 454, normalized size of antiderivative = 1.04, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {3042, 3979, 3042, 3977, 3042, 3980, 3042, 3976, 826, 1476, 1082, 217, 1479, 25, 27, 1103}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3979

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3977

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3980

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3976

\(\displaystyle \frac {5}{2} a \left (-\frac {4 i a^3 e \sec (c+d x) \int \frac {\cos (c+d x) (a-i a \tan (c+d x))}{e \left (a^2+\cos ^2(c+d x) (a-i a \tan (c+d x))^2\right )}d\frac {\sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}}{d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {4 i a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {e \sec (c+d x)}}\right )+\frac {i a (a+i a \tan (c+d x))^{3/2}}{d \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 826

\(\displaystyle \frac {5}{2} a \left (-\frac {4 i a^3 e \sec (c+d x) \left (\frac {\int \frac {a+\cos (c+d x) (a-i a \tan (c+d x))}{a^2+\cos ^2(c+d x) (a-i a \tan (c+d x))^2}d\frac {\sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}}{2 e}-\frac {\int \frac {a-\cos (c+d x) (a-i a \tan (c+d x))}{a^2+\cos ^2(c+d x) (a-i a \tan (c+d x))^2}d\frac {\sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}}{2 e}\right )}{d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {4 i a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {e \sec (c+d x)}}\right )+\frac {i a (a+i a \tan (c+d x))^{3/2}}{d \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {5}{2} a \left (-\frac {4 i a^3 e \sec (c+d x) \left (\frac {\frac {\int \frac {1}{\frac {a}{e}-\frac {\sqrt {2} \sqrt {a-i a \tan (c+d x)} \sqrt {a}}{\sqrt {e} \sqrt {e \sec (c+d x)}}+\frac {\cos (c+d x) (a-i a \tan (c+d x))}{e}}d\frac {\sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}}{2 e}+\frac {\int \frac {1}{\frac {a}{e}+\frac {\sqrt {2} \sqrt {a-i a \tan (c+d x)} \sqrt {a}}{\sqrt {e} \sqrt {e \sec (c+d x)}}+\frac {\cos (c+d x) (a-i a \tan (c+d x))}{e}}d\frac {\sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}}{2 e}}{2 e}-\frac {\int \frac {a-\cos (c+d x) (a-i a \tan (c+d x))}{a^2+\cos ^2(c+d x) (a-i a \tan (c+d x))^2}d\frac {\sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}}{2 e}\right )}{d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {4 i a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {e \sec (c+d x)}}\right )+\frac {i a (a+i a \tan (c+d x))^{3/2}}{d \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {5}{2} a \left (-\frac {4 i a^3 e \sec (c+d x) \left (\frac {\frac {\int \frac {1}{-\frac {\cos (c+d x) (a-i a \tan (c+d x))}{e}-1}d\left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}-\frac {\int \frac {1}{-\frac {\cos (c+d x) (a-i a \tan (c+d x))}{e}-1}d\left (\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}+1\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}}{2 e}-\frac {\int \frac {a-\cos (c+d x) (a-i a \tan (c+d x))}{a^2+\cos ^2(c+d x) (a-i a \tan (c+d x))^2}d\frac {\sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}}{2 e}\right )}{d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {4 i a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {e \sec (c+d x)}}\right )+\frac {i a (a+i a \tan (c+d x))^{3/2}}{d \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {5}{2} a \left (-\frac {4 i a^3 e \sec (c+d x) \left (\frac {\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}}{2 e}-\frac {\int \frac {a-\cos (c+d x) (a-i a \tan (c+d x))}{a^2+\cos ^2(c+d x) (a-i a \tan (c+d x))^2}d\frac {\sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}}{2 e}\right )}{d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {4 i a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {e \sec (c+d x)}}\right )+\frac {i a (a+i a \tan (c+d x))^{3/2}}{d \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {5}{2} a \left (-\frac {4 i a^3 e \sec (c+d x) \left (\frac {\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}}{2 e}-\frac {-\frac {\int -\frac {\sqrt {2} \sqrt {a}-\frac {2 \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}}{\sqrt {e} \left (\frac {a}{e}-\frac {\sqrt {2} \sqrt {a-i a \tan (c+d x)} \sqrt {a}}{\sqrt {e} \sqrt {e \sec (c+d x)}}+\frac {\cos (c+d x) (a-i a \tan (c+d x))}{e}\right )}d\frac {\sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}}{2 \sqrt {2} \sqrt {a} \sqrt {e}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {a}+\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{\sqrt {e} \left (\frac {a}{e}+\frac {\sqrt {2} \sqrt {a-i a \tan (c+d x)} \sqrt {a}}{\sqrt {e} \sqrt {e \sec (c+d x)}}+\frac {\cos (c+d x) (a-i a \tan (c+d x))}{e}\right )}d\frac {\sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}}{2 \sqrt {2} \sqrt {a} \sqrt {e}}}{2 e}\right )}{d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {4 i a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {e \sec (c+d x)}}\right )+\frac {i a (a+i a \tan (c+d x))^{3/2}}{d \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {5}{2} a \left (-\frac {4 i a^3 e \sec (c+d x) \left (\frac {\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}}{2 e}-\frac {\frac {\int \frac {\sqrt {2} \sqrt {a}-\frac {2 \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}}{\sqrt {e} \left (\frac {a}{e}-\frac {\sqrt {2} \sqrt {a-i a \tan (c+d x)} \sqrt {a}}{\sqrt {e} \sqrt {e \sec (c+d x)}}+\frac {\cos (c+d x) (a-i a \tan (c+d x))}{e}\right )}d\frac {\sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}}{2 \sqrt {2} \sqrt {a} \sqrt {e}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {a}+\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{\sqrt {e} \left (\frac {a}{e}+\frac {\sqrt {2} \sqrt {a-i a \tan (c+d x)} \sqrt {a}}{\sqrt {e} \sqrt {e \sec (c+d x)}}+\frac {\cos (c+d x) (a-i a \tan (c+d x))}{e}\right )}d\frac {\sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}}{2 \sqrt {2} \sqrt {a} \sqrt {e}}}{2 e}\right )}{d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {4 i a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {e \sec (c+d x)}}\right )+\frac {i a (a+i a \tan (c+d x))^{3/2}}{d \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {5}{2} a \left (-\frac {4 i a^3 e \sec (c+d x) \left (\frac {\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}}{2 e}-\frac {\frac {\int \frac {\sqrt {2} \sqrt {a}-\frac {2 \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}}{\frac {a}{e}-\frac {\sqrt {2} \sqrt {a-i a \tan (c+d x)} \sqrt {a}}{\sqrt {e} \sqrt {e \sec (c+d x)}}+\frac {\cos (c+d x) (a-i a \tan (c+d x))}{e}}d\frac {\sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}}{2 \sqrt {2} \sqrt {a} e}+\frac {\int \frac {\sqrt {a}+\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}}{\frac {a}{e}+\frac {\sqrt {2} \sqrt {a-i a \tan (c+d x)} \sqrt {a}}{\sqrt {e} \sqrt {e \sec (c+d x)}}+\frac {\cos (c+d x) (a-i a \tan (c+d x))}{e}}d\frac {\sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}}{2 \sqrt {a} e}}{2 e}\right )}{d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {4 i a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {e \sec (c+d x)}}\right )+\frac {i a (a+i a \tan (c+d x))^{3/2}}{d \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {5}{2} a \left (-\frac {4 i a^3 e \sec (c+d x) \left (\frac {\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}}{2 e}-\frac {\frac {\log \left (\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{2 \sqrt {2} \sqrt {a} \sqrt {e}}-\frac {\log \left (-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{2 \sqrt {2} \sqrt {a} \sqrt {e}}}{2 e}\right )}{d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {4 i a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {e \sec (c+d x)}}\right )+\frac {i a (a+i a \tan (c+d x))^{3/2}}{d \sqrt {e \sec (c+d x)}}\)

Input:

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

Output:

(I*a*(a + I*a*Tan[c + d*x])^(3/2))/(d*Sqrt[e*Sec[c + d*x]]) + (5*a*(((-4*I 
)*a^3*e*((-(ArcTan[1 - (Sqrt[2]*Sqrt[e]*Sqrt[a - I*a*Tan[c + d*x]])/(Sqrt[ 
a]*Sqrt[e*Sec[c + d*x]])]/(Sqrt[2]*Sqrt[a]*Sqrt[e])) + ArcTan[1 + (Sqrt[2] 
*Sqrt[e]*Sqrt[a - I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e*Sec[c + d*x]])]/(Sqrt 
[2]*Sqrt[a]*Sqrt[e]))/(2*e) - (-1/2*Log[a - (Sqrt[2]*Sqrt[a]*Sqrt[e]*Sqrt[ 
a - I*a*Tan[c + d*x]])/Sqrt[e*Sec[c + d*x]] + Cos[c + d*x]*(a - I*a*Tan[c 
+ d*x])]/(Sqrt[2]*Sqrt[a]*Sqrt[e]) + Log[a + (Sqrt[2]*Sqrt[a]*Sqrt[e]*Sqrt 
[a - I*a*Tan[c + d*x]])/Sqrt[e*Sec[c + d*x]] + Cos[c + d*x]*(a - I*a*Tan[c 
 + d*x])]/(2*Sqrt[2]*Sqrt[a]*Sqrt[e]))/(2*e))*Sec[c + d*x])/(d*Sqrt[a - I* 
a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - ((4*I)*a*Sqrt[a + I*a*Tan[c 
+ d*x]])/(d*Sqrt[e*Sec[c + d*x]])))/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 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 826

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 
2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s)   Int[(r + s*x^2)/(a + b*x^ 
4), x], x] - Simp[1/(2*s)   Int[(r - s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{ 
a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] 
 && AtomQ[SplitProduct[SumBaseQ, 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 3042

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

rule 3976

Int[Sqrt[(d_.)*sec[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.) 
*(x_)]], x_Symbol] :> Simp[-4*b*(d^2/f)   Subst[Int[x^2/(a^2 + d^2*x^4), x] 
, x, Sqrt[a + b*Tan[e + f*x]]/Sqrt[d*Sec[e + f*x]]], x] /; FreeQ[{a, b, d, 
e, f}, x] && EqQ[a^2 + b^2, 0]
 

rule 3977

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x 
_)])^(n_), x_Symbol] :> Simp[2*b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^( 
n - 1)/(f*m)), x] - Simp[b^2*((m + 2*n - 2)/(d^2*m))   Int[(d*Sec[e + f*x]) 
^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] 
&& EqQ[a^2 + b^2, 0] && GtQ[n, 1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || 
 (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILtQ[m, 0] & 
& LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) 
&& IntegerQ[2*m]
 

rule 3979

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

rule 3980

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(3/2)/Sqrt[(a_) + (b_.)*tan[(e_.) + (f_ 
.)*(x_)]], x_Symbol] :> Simp[d*(Sec[e + f*x]/(Sqrt[a - b*Tan[e + f*x]]*Sqrt 
[a + b*Tan[e + f*x]]))   Int[Sqrt[d*Sec[e + f*x]]*Sqrt[a - b*Tan[e + f*x]], 
 x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0]
 

Maple [A] (warning: unable to verify)

Time = 7.94 (sec) , antiderivative size = 305, normalized size of antiderivative = 0.70

method	result	size
default	\(\frac {a^{2} \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (16 \sin \left (d x +c \right )-2 \tan \left (d x +c \right )+i \left (-16 \cos \left (d x +c \right )-18-2 \sec \left (d x +c \right )\right )-5 \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )-5 \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )+5 i \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )-5 i \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )\right )}{2 d \left (1+\cos \left (d x +c \right )+i \sin \left (d x +c \right )\right ) \sqrt {e \sec \left (d x +c \right )}}\)	\(305\)

Input:

int((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/2/d*a^2*(a*(1+I*tan(d*x+c)))^(1/2)/(1+cos(d*x+c)+I*sin(d*x+c))/(e*sec(d* 
x+c))^(1/2)*(16*sin(d*x+c)-2*tan(d*x+c)+I*(-16*cos(d*x+c)-18-2*sec(d*x+c)) 
-5*(cos(d*x+c)+1)*(1/(cos(d*x+c)+1))^(1/2)*arctanh(1/2*(-cot(d*x+c)+csc(d* 
x+c)-1)/(1/(cos(d*x+c)+1))^(1/2))-5*(cos(d*x+c)+1)*(1/(cos(d*x+c)+1))^(1/2 
)*arctanh(1/2/(1/(cos(d*x+c)+1))^(1/2)*(-cot(d*x+c)+csc(d*x+c)+1))+5*I*(co 
s(d*x+c)+1)*(1/(cos(d*x+c)+1))^(1/2)*arctanh(1/2/(1/(cos(d*x+c)+1))^(1/2)* 
(-cot(d*x+c)+csc(d*x+c)+1))-5*I*(cos(d*x+c)+1)*(1/(cos(d*x+c)+1))^(1/2)*ar 
ctanh(1/2*(-cot(d*x+c)+csc(d*x+c)-1)/(1/(cos(d*x+c)+1))^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 484, normalized size of antiderivative = 1.11 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\sqrt {e \sec (c+d x)}} \, dx=-\frac {\sqrt {\frac {25 i \, a^{5}}{d^{2} e}} d e \log \left (\frac {2 \, {\left (5 \, {\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + i \, \sqrt {\frac {25 i \, a^{5}}{d^{2} e}} d e\right )}}{5 \, a^{2}}\right ) - \sqrt {\frac {25 i \, a^{5}}{d^{2} e}} d e \log \left (\frac {2 \, {\left (5 \, {\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} - i \, \sqrt {\frac {25 i \, a^{5}}{d^{2} e}} d e\right )}}{5 \, a^{2}}\right ) + \sqrt {-\frac {25 i \, a^{5}}{d^{2} e}} d e \log \left (\frac {2 \, {\left (5 \, {\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + i \, \sqrt {-\frac {25 i \, a^{5}}{d^{2} e}} d e\right )}}{5 \, a^{2}}\right ) - \sqrt {-\frac {25 i \, a^{5}}{d^{2} e}} d e \log \left (\frac {2 \, {\left (5 \, {\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} - i \, \sqrt {-\frac {25 i \, a^{5}}{d^{2} e}} d e\right )}}{5 \, a^{2}}\right ) + 4 \, {\left (4 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i \, a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{2 \, d e} \] Input:

integrate((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(1/2),x, algorithm="fric 
as")
 

Output:

-1/2*(sqrt(25*I*a^5/(d^2*e))*d*e*log(2/5*(5*(a^2*e^(2*I*d*x + 2*I*c) + a^2 
)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1 
/2*I*d*x + 1/2*I*c) + I*sqrt(25*I*a^5/(d^2*e))*d*e)/a^2) - sqrt(25*I*a^5/( 
d^2*e))*d*e*log(2/5*(5*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*sqrt(a/(e^(2*I*d*x 
+ 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) - 
 I*sqrt(25*I*a^5/(d^2*e))*d*e)/a^2) + sqrt(-25*I*a^5/(d^2*e))*d*e*log(2/5* 
(5*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt( 
e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + I*sqrt(-25*I*a^5/(d 
^2*e))*d*e)/a^2) - sqrt(-25*I*a^5/(d^2*e))*d*e*log(2/5*(5*(a^2*e^(2*I*d*x 
+ 2*I*c) + a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I 
*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) - I*sqrt(-25*I*a^5/(d^2*e))*d*e)/a^2) + 
4*(4*I*a^2*e^(2*I*d*x + 2*I*c) + 5*I*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1) 
)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c))/(d*e)
 

Sympy [F(-1)]

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

integrate((a+I*a*tan(d*x+c))**(5/2)/(e*sec(d*x+c))**(1/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2013 vs. \(2 (331) = 662\).

Time = 0.33 (sec) , antiderivative size = 2013, normalized size of antiderivative = 4.61 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\sqrt {e \sec (c+d x)}} \, dx=\text {Too large to display} \] Input:

integrate((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(1/2),x, algorithm="maxi 
ma")
 

Output:

8*(10*(sqrt(2)*a^2*cos(2*d*x + 2*c) + I*sqrt(2)*a^2*sin(2*d*x + 2*c) + sqr 
t(2)*a^2)*arctan2(sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2* 
c))) + 1, sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1 
) + 10*(sqrt(2)*a^2*cos(2*d*x + 2*c) + I*sqrt(2)*a^2*sin(2*d*x + 2*c) + sq 
rt(2)*a^2)*arctan2(sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2 
*c))) + 1, -sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 
 1) + 10*(sqrt(2)*a^2*cos(2*d*x + 2*c) + I*sqrt(2)*a^2*sin(2*d*x + 2*c) + 
sqrt(2)*a^2)*arctan2(sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 
 2*c))) - 1, sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) 
+ 1) + 10*(sqrt(2)*a^2*cos(2*d*x + 2*c) + I*sqrt(2)*a^2*sin(2*d*x + 2*c) + 
 sqrt(2)*a^2)*arctan2(sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x 
+ 2*c))) - 1, -sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)) 
) + 1) - 10*(-I*sqrt(2)*a^2*cos(2*d*x + 2*c) + sqrt(2)*a^2*sin(2*d*x + 2*c 
) - I*sqrt(2)*a^2)*arctan2(sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2 
*d*x + 2*c))) + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))), sqrt 
(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + cos(1/2*arctan2 
(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) - 10*(I*sqrt(2)*a^2*cos(2*d*x + 
 2*c) - sqrt(2)*a^2*sin(2*d*x + 2*c) + I*sqrt(2)*a^2)*arctan2(-sqrt(2)*sin 
(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(1/2*arctan2(sin(2* 
d*x + 2*c), cos(2*d*x + 2*c))), -sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*...
 

Giac [F(-2)]

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

integrate((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(1/2),x, algorithm="giac 
")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument TypeDone
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(1/2),x)
 

Output:

(sqrt(e)*sqrt(a)*a**2*( - int((sqrt(sec(c + d*x))*sqrt(tan(c + d*x)*i + 1) 
*tan(c + d*x)**2)/sec(c + d*x),x) + 2*int((sqrt(sec(c + d*x))*sqrt(tan(c + 
 d*x)*i + 1)*tan(c + d*x))/sec(c + d*x),x)*i + int((sqrt(sec(c + d*x))*sqr 
t(tan(c + d*x)*i + 1))/sec(c + d*x),x)))/e