\(\int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \cos (c+d x))^{5/2}} \, dx\) [688]

Optimal result

Integrand size = 30, antiderivative size = 404 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \cos (c+d x))^{5/2}} \, dx=\frac {3 i \sqrt {a} e^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{4 \sqrt {2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac {3 i \sqrt {a} e^{5/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{4 \sqrt {2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}+\frac {3 i \sqrt {a} e^{5/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)} \left (\sqrt {a}+\cos (c+d x) \left (\sqrt {a}+i \sqrt {a} \tan (c+d x)\right )\right )}\right )}{4 \sqrt {2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}+\frac {i a}{2 d (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}-\frac {3 i \cos ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d (e \cos (c+d x))^{5/2}} \] Output:

3/8*I*a^(1/2)*e^(5/2)*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)/d/(e*cos(d*x+c))^(5/2)/(e*sec(d*x+c))^ 
(5/2)-3/8*I*a^(1/2)*e^(5/2)*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)/d/(e*cos(d*x+c))^(5/2)/(e*sec(d* 
x+c))^(5/2)+3/8*I*a^(1/2)*e^(5/2)*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)/d/(e*cos(d*x+c))^(5/2)/(e*sec(d*x+c))^(5/2)+1/2*I*a/d/(e* 
cos(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^(1/2)-3/4*I*cos(d*x+c)^2*(a+I*a*tan(d 
*x+c))^(1/2)/d/(e*cos(d*x+c))^(5/2)
 

Mathematica [A] (verified)

Time = 2.33 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.56 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {\cos (c+d x)} \left (\frac {3 i e^{-\frac {1}{2} i (2 c+5 d x)} \left (-e^{-2 i c}\right )^{3/4} \left (1+e^{2 i (c+d x)}\right )^2 \sqrt {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \left (\arctan \left (\frac {e^{\frac {i d x}{2}}}{\sqrt [4]{-e^{-2 i c}}}\right )-\text {arctanh}\left (\frac {e^{\frac {i d x}{2}}}{\sqrt [4]{-e^{-2 i c}}}\right )\right )}{4 \sqrt {2}}-3 i \cos ^{\frac {3}{2}}(c+d x)+2 \sqrt {\cos (c+d x)} (i \cos (c+d x)+\sin (c+d x))\right ) \sqrt {a+i a \tan (c+d x)}}{4 d (e \cos (c+d x))^{5/2}} \] Input:

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

Output:

(Sqrt[Cos[c + d*x]]*((((3*I)/4)*(-E^((-2*I)*c))^(3/4)*(1 + E^((2*I)*(c + d 
*x)))^2*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))]*(ArcTan[E^((I/2)*d 
*x)/(-E^((-2*I)*c))^(1/4)] - ArcTanh[E^((I/2)*d*x)/(-E^((-2*I)*c))^(1/4)]) 
)/(Sqrt[2]*E^((I/2)*(2*c + 5*d*x))) - (3*I)*Cos[c + d*x]^(3/2) + 2*Sqrt[Co 
s[c + d*x]]*(I*Cos[c + d*x] + Sin[c + d*x]))*Sqrt[a + I*a*Tan[c + d*x]])/( 
4*d*(e*Cos[c + d*x])^(5/2))
 

Rubi [A] (verified)

Time = 1.03 (sec) , antiderivative size = 445, normalized size of antiderivative = 1.10, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {3042, 3998, 3042, 3979, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3998

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3979

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3982

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3976

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

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

(((I/2)*a*(e*Sec[c + d*x])^(5/2))/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (3*a*(( 
(-2*I)*e^4*((-(ArcTan[1 - (Sqrt[2]*Sqrt[e]*Sqrt[a + I*a*Tan[c + d*x]])/(Sq 
rt[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]])]/(S 
qrt[2]*Sqrt[a]*Sqrt[e]))/(2*e) - (-1/2*Log[a - (Sqrt[2]*Sqrt[a]*Sqrt[e]*Sq 
rt[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]*S 
qrt[a + I*a*Tan[c + d*x]])/Sqrt[e*Sec[c + d*x]] + Cos[c + d*x]*(a + I*a*Ta 
n[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)))/4)/((e*Cos[c + d*x])^(5/2)*(e* 
Sec[c + d*x])^(5/2))
 

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 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 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]
 

rule 3998

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x 
_)])^(n_.), x_Symbol] :> Simp[(d*Cos[e + f*x])^m*(d*Sec[e + f*x])^m   Int[( 
a + b*Tan[e + f*x])^n/(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m 
, n}, x] &&  !IntegerQ[m]
 

Maple [A] (verified)

Time = 11.60 (sec) , antiderivative size = 291, normalized size of antiderivative = 0.72

method	result	size
default	\(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (3 i \cos \left (d x +c \right ) \operatorname {arctanh}\left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )-3 \cos \left (d x +c \right ) \operatorname {arctanh}\left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )+3 i \cos \left (d x +c \right ) \operatorname {arctanh}\left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )+3 \cos \left (d x +c \right ) \operatorname {arctanh}\left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )+2 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \left (3 \sin \left (d x +c \right )+2 \tan \left (d x +c \right )\right )+2 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \left (1+3 \cos \left (d x +c \right )-2 \sec \left (d x +c \right )\right )\right )}{8 d \left (-\sin \left (d x +c \right )+i \cos \left (d x +c \right )+i\right ) \sqrt {e \cos \left (d x +c \right )}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, e^{2}}\)	\(291\)

Input:

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

Output:

1/8/d*(a*(1+I*tan(d*x+c)))^(1/2)/(-sin(d*x+c)+I*cos(d*x+c)+I)/(e*cos(d*x+c 
))^(1/2)/(1/(cos(d*x+c)+1))^(1/2)/e^2*(3*I*cos(d*x+c)*arctanh(1/2/(1/(cos( 
d*x+c)+1))^(1/2)*(-cot(d*x+c)+csc(d*x+c)+1))-3*cos(d*x+c)*arctanh(1/2/(1/( 
cos(d*x+c)+1))^(1/2)*(-cot(d*x+c)+csc(d*x+c)+1))+3*I*cos(d*x+c)*arctanh(1/ 
2*(-cot(d*x+c)+csc(d*x+c)-1)/(1/(cos(d*x+c)+1))^(1/2))+3*cos(d*x+c)*arctan 
h(1/2*(-cot(d*x+c)+csc(d*x+c)-1)/(1/(cos(d*x+c)+1))^(1/2))+2*I*(1/(cos(d*x 
+c)+1))^(1/2)*(3*sin(d*x+c)+2*tan(d*x+c))+2*(1/(cos(d*x+c)+1))^(1/2)*(1+3* 
cos(d*x+c)-2*sec(d*x+c)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 569, normalized size of antiderivative = 1.41 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \cos (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

1/2*(sqrt(2)*sqrt(1/2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*sqrt(a/(e^(2*I*d*x 
+ 2*I*c) + 1))*(-3*I*e^(4*I*d*x + 4*I*c) + I*e^(2*I*d*x + 2*I*c))*e^(-1/2* 
I*d*x - 1/2*I*c) - (d*e^3*e^(4*I*d*x + 4*I*c) + 2*d*e^3*e^(2*I*d*x + 2*I*c 
) + d*e^3)*sqrt(9/16*I*a/(d^2*e^5))*log(4/3*I*d*e^3*sqrt(9/16*I*a/(d^2*e^5 
)) + sqrt(2)*sqrt(1/2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*sqrt(a/(e^(2*I*d*x 
+ 2*I*c) + 1))*e^(-1/2*I*d*x - 1/2*I*c)) + (d*e^3*e^(4*I*d*x + 4*I*c) + 2* 
d*e^3*e^(2*I*d*x + 2*I*c) + d*e^3)*sqrt(9/16*I*a/(d^2*e^5))*log(-4/3*I*d*e 
^3*sqrt(9/16*I*a/(d^2*e^5)) + sqrt(2)*sqrt(1/2)*sqrt(e*e^(2*I*d*x + 2*I*c) 
 + e)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(-1/2*I*d*x - 1/2*I*c)) - (d*e^3 
*e^(4*I*d*x + 4*I*c) + 2*d*e^3*e^(2*I*d*x + 2*I*c) + d*e^3)*sqrt(-9/16*I*a 
/(d^2*e^5))*log(4/3*I*d*e^3*sqrt(-9/16*I*a/(d^2*e^5)) + sqrt(2)*sqrt(1/2)* 
sqrt(e*e^(2*I*d*x + 2*I*c) + e)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(-1/2* 
I*d*x - 1/2*I*c)) + (d*e^3*e^(4*I*d*x + 4*I*c) + 2*d*e^3*e^(2*I*d*x + 2*I* 
c) + d*e^3)*sqrt(-9/16*I*a/(d^2*e^5))*log(-4/3*I*d*e^3*sqrt(-9/16*I*a/(d^2 
*e^5)) + sqrt(2)*sqrt(1/2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*sqrt(a/(e^(2*I* 
d*x + 2*I*c) + 1))*e^(-1/2*I*d*x - 1/2*I*c)))/(d*e^3*e^(4*I*d*x + 4*I*c) + 
 2*d*e^3*e^(2*I*d*x + 2*I*c) + d*e^3)
 

Sympy [F(-1)]

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

integrate((a+I*a*tan(d*x+c))**(1/2)/(e*cos(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. 2249 vs. \(2 (300) = 600\).

Time = 0.47 (sec) , antiderivative size = 2249, normalized size of antiderivative = 5.57 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \cos (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

-32*(6*(sqrt(2)*cos(4*d*x + 4*c) + 2*sqrt(2)*cos(2*d*x + 2*c) + I*sqrt(2)* 
sin(4*d*x + 4*c) + 2*I*sqrt(2)*sin(2*d*x + 2*c) + sqrt(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) + 6*(sqrt(2)*cos(4*d*x 
+ 4*c) + 2*sqrt(2)*cos(2*d*x + 2*c) + I*sqrt(2)*sin(4*d*x + 4*c) + 2*I*sqr 
t(2)*sin(2*d*x + 2*c) + sqrt(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) + 6*(sqrt(2)*cos(4*d*x + 4*c) + 2*sqrt(2)*cos(2* 
d*x + 2*c) + I*sqrt(2)*sin(4*d*x + 4*c) + 2*I*sqrt(2)*sin(2*d*x + 2*c) + s 
qrt(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) 
+ 6*(sqrt(2)*cos(4*d*x + 4*c) + 2*sqrt(2)*cos(2*d*x + 2*c) + I*sqrt(2)*sin 
(4*d*x + 4*c) + 2*I*sqrt(2)*sin(2*d*x + 2*c) + sqrt(2))*arctan2(sqrt(2)*co 
s(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 1, -sqrt(2)*sin(1/4*a 
rctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + 6*(-I*sqrt(2)*cos(4*d*x 
 + 4*c) - 2*I*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2)*sin(4*d*x + 4*c) + 2*sqrt 
(2)*sin(2*d*x + 2*c) - I*sqrt(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) + 6*(I*sqrt(2...
 

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(e)*sqrt(a)*int((sqrt(tan(c + d*x)*i + 1)*sqrt(cos(c + d*x)))/cos(c + 
 d*x)**3,x))/e**3