\(\int \frac {(e \sec (c+d x))^{9/2}}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [430]

Optimal result

Integrand size = 30, antiderivative size = 325 \[ \int \frac {(e \sec (c+d x))^{9/2}}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {5 i e^{9/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} a^{5/2} d}+\frac {5 i e^{9/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} a^{5/2} d}-\frac {5 i e^{9/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)} \left (\sqrt {a}+\cos (c+d x) \left (\sqrt {a}+i \sqrt {a} \tan (c+d x)\right )\right )}\right )}{\sqrt {2} a^{5/2} d}+\frac {4 i e^2 (e \sec (c+d x))^{5/2}}{a d (a+i a \tan (c+d x))^{3/2}}+\frac {5 i e^4 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}{a^3 d} \] Output:

-5/2*I*e^(9/2)*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))*2^(1/2)/a^(5/2)/d+5/2*I*e^(9/2)*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))*2^(1/2)/a^(5/2) 
/d-5/2*I*e^(9/2)*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))))*2^(1/2)/ 
a^(5/2)/d+4*I*e^2*(e*sec(d*x+c))^(5/2)/a/d/(a+I*a*tan(d*x+c))^(3/2)+5*I*e^ 
4*(e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/a^3/d
 

Mathematica [A] (warning: unable to verify)

Time = 5.05 (sec) , antiderivative size = 370, normalized size of antiderivative = 1.14 \[ \int \frac {(e \sec (c+d x))^{9/2}}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {e^2 (e \sec (c+d x))^{5/2} (\cos (d x)+i \sin (d x))^3 \left (\cos (d x) (8 i \cos (2 c)-8 \sin (2 c))+\sec (c+d x) (i \cos (3 c)-\sin (3 c))+8 (\cos (2 c)+i \sin (2 c)) \sin (d x)+\frac {5 \left (\text {arctanh}\left (\frac {\sqrt {1-i \cos (c)+\sin (c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}}{\sqrt {-1-i \cos (c)-\sin (c)} \sqrt {i+\tan \left (\frac {d x}{2}\right )}}\right ) \sqrt {-1-i \cos (c)-\sin (c)} \sqrt {1+i \cos (c)-\sin (c)}-\text {arctanh}\left (\frac {\sqrt {1+i \cos (c)-\sin (c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}}{\sqrt {-1+i \cos (c)+\sin (c)} \sqrt {i+\tan \left (\frac {d x}{2}\right )}}\right ) \sqrt {1-i \cos (c)+\sin (c)} \sqrt {-1+i \cos (c)+\sin (c)}\right ) (\cos (3 c)+i \sin (3 c)) \sqrt {i+\tan \left (\frac {d x}{2}\right )}}{\sqrt {1+\cos (2 c)+i \sin (2 c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}}\right )}{d (a+i a \tan (c+d x))^{5/2}} \] Input:

Integrate[(e*Sec[c + d*x])^(9/2)/(a + I*a*Tan[c + d*x])^(5/2),x]
 

Output:

(e^2*(e*Sec[c + d*x])^(5/2)*(Cos[d*x] + I*Sin[d*x])^3*(Cos[d*x]*((8*I)*Cos 
[2*c] - 8*Sin[2*c]) + Sec[c + d*x]*(I*Cos[3*c] - Sin[3*c]) + 8*(Cos[2*c] + 
 I*Sin[2*c])*Sin[d*x] + (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]]*Sqrt[1 + I*Cos[c] - Sin[c]] - ArcTanh[(Sqrt[1 + I 
*Cos[c] - Sin[c]]*Sqrt[I - Tan[(d*x)/2]])/(Sqrt[-1 + I*Cos[c] + Sin[c]]*Sq 
rt[I + Tan[(d*x)/2]])]*Sqrt[1 - I*Cos[c] + Sin[c]]*Sqrt[-1 + I*Cos[c] + Si 
n[c]])*(Cos[3*c] + I*Sin[3*c])*Sqrt[I + Tan[(d*x)/2]])/(Sqrt[1 + Cos[2*c] 
+ I*Sin[2*c]]*Sqrt[I - Tan[(d*x)/2]])))/(d*(a + I*a*Tan[c + d*x])^(5/2))
 

Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 426, normalized size of antiderivative = 1.31, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {3042, 3981, 3042, 3982, 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.

\(\displaystyle \int \frac {(e \sec (c+d x))^{9/2}}{(a+i a \tan (c+d x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(e \sec (c+d x))^{9/2}}{(a+i a \tan (c+d x))^{5/2}}dx\)

\(\Big \downarrow \) 3981

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3982

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3976

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

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

Int[(e*Sec[c + d*x])^(9/2)/(a + I*a*Tan[c + d*x])^(5/2),x]
 

Output:

((4*I)*e^2*(e*Sec[c + d*x])^(5/2))/(a*d*(a + I*a*Tan[c + d*x])^(3/2)) - (5 
*e^2*(((-2*I)*e^4*((-(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]*Sqr 
t[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]*Sq 
rt[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)))/d - (I*e^2*Sqrt[e 
*Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(a*d)))/a^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 3981

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

rule 3982

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

Maple [A] (verified)

Time = 11.13 (sec) , antiderivative size = 314, normalized size of antiderivative = 0.97

method	result	size
default	\(-\frac {\sqrt {e \sec \left (d x +c \right )}\, e^{4} \left (4 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \left (\sin \left (d x +c \right )+\tan \left (d x +c \right )\right )-36 i \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+\left (-5 \sin \left (d x +c \right )+5 \cos \left (d x +c \right )+5\right ) \operatorname {arctanh}\left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )+\left (5 \sin \left (d x +c \right )+5 \cos \left (d x +c \right )+5\right ) \operatorname {arctanh}\left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )+i \left (5 \sin \left (d x +c \right )+5 \cos \left (d x +c \right )+5\right ) \operatorname {arctanh}\left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )+i \left (5 \sin \left (d x +c \right )-5 \cos \left (d x +c \right )-5\right ) \operatorname {arctanh}\left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )\right )}{4 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, a^{2}}\)	\(314\)

Input:

int((e*sec(d*x+c))^(9/2)/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/4/d*(e*sec(d*x+c))^(1/2)*e^4/(cos(d*x+c)+1)/(a*(1+I*tan(d*x+c)))^(1/2)/ 
(1/(cos(d*x+c)+1))^(1/2)/a^2*(4*(1/(cos(d*x+c)+1))^(1/2)*(sin(d*x+c)+tan(d 
*x+c))-36*I*(cos(d*x+c)+1)*(1/(cos(d*x+c)+1))^(1/2)+(-5*sin(d*x+c)+5*cos(d 
*x+c)+5)*arctanh(1/2/(1/(cos(d*x+c)+1))^(1/2)*(-cot(d*x+c)+csc(d*x+c)+1))+ 
(5*sin(d*x+c)+5*cos(d*x+c)+5)*arctanh(1/2*(-cot(d*x+c)+csc(d*x+c)-1)/(1/(c 
os(d*x+c)+1))^(1/2))+I*(5*sin(d*x+c)+5*cos(d*x+c)+5)*arctanh(1/2/(1/(cos(d 
*x+c)+1))^(1/2)*(-cot(d*x+c)+csc(d*x+c)+1))+I*(5*sin(d*x+c)-5*cos(d*x+c)-5 
)*arctanh(1/2*(-cot(d*x+c)+csc(d*x+c)-1)/(1/(cos(d*x+c)+1))^(1/2)))
 

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. 541 vs. \(2 (243) = 486\).

Time = 0.12 (sec) , antiderivative size = 541, normalized size of antiderivative = 1.66 \[ \int \frac {(e \sec (c+d x))^{9/2}}{(a+i a \tan (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate((e*sec(d*x+c))^(9/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fric 
as")
 

Output:

-1/2*(sqrt(25*I*e^9/(a^5*d^2))*a^3*d*e^(I*d*x + I*c)*log(2/5*(sqrt(25*I*e^ 
9/(a^5*d^2))*a^3*d + 5*(e^4*e^(2*I*d*x + 2*I*c) + e^4)*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))/ 
e^4) - sqrt(25*I*e^9/(a^5*d^2))*a^3*d*e^(I*d*x + I*c)*log(-2/5*(sqrt(25*I* 
e^9/(a^5*d^2))*a^3*d - 5*(e^4*e^(2*I*d*x + 2*I*c) + e^4)*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) 
)/e^4) - sqrt(-25*I*e^9/(a^5*d^2))*a^3*d*e^(I*d*x + I*c)*log(2/5*(sqrt(-25 
*I*e^9/(a^5*d^2))*a^3*d + 5*(e^4*e^(2*I*d*x + 2*I*c) + e^4)*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))/e^4) + sqrt(-25*I*e^9/(a^5*d^2))*a^3*d*e^(I*d*x + I*c)*log(-2/5*(sqrt 
(-25*I*e^9/(a^5*d^2))*a^3*d - 5*(e^4*e^(2*I*d*x + 2*I*c) + e^4)*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))/e^4) + 4*(-5*I*e^4*e^(2*I*d*x + 2*I*c) - 4*I*e^4)*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))*e^(-I*d*x - I*c)/(a^3*d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(e \sec (c+d x))^{9/2}}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((e*sec(d*x+c))**(9/2)/(a+I*a*tan(d*x+c))**(5/2),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. 2449 vs. \(2 (243) = 486\).

Time = 0.37 (sec) , antiderivative size = 2449, normalized size of antiderivative = 7.54 \[ \int \frac {(e \sec (c+d x))^{9/2}}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate((e*sec(d*x+c))^(9/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxi 
ma")
 

Output:

-(64*e^4*cos(1/2*d*x + 1/2*c)^2 + 64*e^4*sin(1/2*d*x + 1/2*c)^2 + 16*e^4 - 
 10*(-I*sqrt(2)*e^4*cos(3/2*d*x + 3/2*c) - I*sqrt(2)*e^4*cos(1/2*d*x + 1/2 
*c) - sqrt(2)*e^4*sin(3/2*d*x + 3/2*c) + sqrt(2)*e^4*sin(1/2*d*x + 1/2*c)) 
*arctan2(sqrt(2)*sin(1/2*d*x + 1/2*c) + sin(d*x + c), sqrt(2)*cos(1/2*d*x 
+ 1/2*c) + cos(d*x + c) + 1) - 10*(I*sqrt(2)*e^4*cos(3/2*d*x + 3/2*c) + I* 
sqrt(2)*e^4*cos(1/2*d*x + 1/2*c) + sqrt(2)*e^4*sin(3/2*d*x + 3/2*c) - sqrt 
(2)*e^4*sin(1/2*d*x + 1/2*c))*arctan2(-sqrt(2)*sin(1/2*d*x + 1/2*c) + sin( 
d*x + c), -sqrt(2)*cos(1/2*d*x + 1/2*c) + cos(d*x + c) + 1) - (10*sqrt(2)* 
e^4*arctan2(sqrt(2)*cos(1/2*d*x + 1/2*c) + 1, sqrt(2)*sin(1/2*d*x + 1/2*c) 
 + 1) + 10*sqrt(2)*e^4*arctan2(sqrt(2)*cos(1/2*d*x + 1/2*c) + 1, -sqrt(2)* 
sin(1/2*d*x + 1/2*c) + 1) + 10*sqrt(2)*e^4*arctan2(sqrt(2)*cos(1/2*d*x + 1 
/2*c) - 1, sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) + 10*sqrt(2)*e^4*arctan2(sqrt 
(2)*cos(1/2*d*x + 1/2*c) - 1, -sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) - 5*I*sqr 
t(2)*e^4*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt( 
2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 5*I*sqrt(2 
)*e^4*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)* 
cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 5*I*sqrt(2)*e 
^4*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos 
(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 5*I*sqrt(2)*e^4* 
log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos...
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(e \sec (c+d x))^{9/2}}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((e*sec(d*x+c))^(9/2)/(a+I*a*tan(d*x+c))^(5/2),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. \[ \int \frac {(e \sec (c+d x))^{9/2}}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{9/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \] Input:

int((e/cos(c + d*x))^(9/2)/(a + a*tan(c + d*x)*1i)^(5/2),x)
 

Output:

int((e/cos(c + d*x))^(9/2)/(a + a*tan(c + d*x)*1i)^(5/2), x)
 

Reduce [F]

\[ \int \frac {(e \sec (c+d x))^{9/2}}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {\sqrt {e}\, \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{4}}{\sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}-2 \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) i -\sqrt {\tan \left (d x +c \right ) i +1}}d x \right ) e^{4}}{\sqrt {a}\, a^{2}} \] Input:

int((e*sec(d*x+c))^(9/2)/(a+I*a*tan(d*x+c))^(5/2),x)
 

Output:

( - sqrt(e)*int((sqrt(sec(c + d*x))*sec(c + d*x)**4)/(sqrt(tan(c + d*x)*i 
+ 1)*tan(c + d*x)**2 - 2*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)*i - sqrt(ta 
n(c + d*x)*i + 1)),x)*e**4)/(sqrt(a)*a**2)