\(\int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^{3/2}} \, dx\) [424]

Optimal result

Integrand size = 30, antiderivative size = 283 \[ \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {i \sqrt {2} e^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{a^{3/2} d}+\frac {i \sqrt {2} e^{5/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{a^{3/2} d}-\frac {i \sqrt {2} e^{5/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)} \left (\sqrt {a}+\cos (c+d x) \left (\sqrt {a}+i \sqrt {a} \tan (c+d x)\right )\right )}\right )}{a^{3/2} d}+\frac {4 i e^2 \sqrt {e \sec (c+d x)}}{a d \sqrt {a+i a \tan (c+d x)}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 3.68 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.19 \[ \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {e (e \sec (c+d x))^{3/2} (\cos (d x)+i \sin (d x))^2 \left (\cos (d x) (4 i \cos (c)-4 \sin (c))+4 (\cos (c)+i \sin (c)) \sin (d x)+\frac {2 \left (\text {arctanh}\left (\frac {\sqrt {1-i \cos (c)+\sin (c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}}{\sqrt {-1-i \cos (c)-\sin (c)} \sqrt {i+\tan \left (\frac {d x}{2}\right )}}\right ) \sqrt {-1-i \cos (c)-\sin (c)} \sqrt {1+i \cos (c)-\sin (c)}-\text {arctanh}\left (\frac {\sqrt {1+i \cos (c)-\sin (c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}}{\sqrt {-1+i \cos (c)+\sin (c)} \sqrt {i+\tan \left (\frac {d x}{2}\right )}}\right ) \sqrt {1-i \cos (c)+\sin (c)} \sqrt {-1+i \cos (c)+\sin (c)}\right ) (\cos (2 c)+i \sin (2 c)) \sqrt {i+\tan \left (\frac {d x}{2}\right )}}{\sqrt {1+\cos (2 c)+i \sin (2 c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}}\right )}{d (a+i a \tan (c+d x))^{3/2}} \] Input:

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

Output:

(e*(e*Sec[c + d*x])^(3/2)*(Cos[d*x] + I*Sin[d*x])^2*(Cos[d*x]*((4*I)*Cos[c 
] - 4*Sin[c]) + 4*(Cos[c] + I*Sin[c])*Sin[d*x] + (2*(ArcTanh[(Sqrt[1 - I*C 
os[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]]*Sqrt[I + Tan[(d*x)/2]])]*Sqrt[1 - I*Cos[c] + Sin[c] 
]*Sqrt[-1 + I*Cos[c] + Sin[c]])*(Cos[2*c] + I*Sin[2*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])^(3/2))
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 378, normalized size of antiderivative = 1.34, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3981, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3981

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3976

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

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

((4*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)))/(a*d) + ((4*I)*e^2*Sqrt[ 
e*Sec[c + d*x]])/(a*d*Sqrt[a + I*a*Tan[c + d*x]])
 

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]
 

Maple [A] (verified)

Time = 10.57 (sec) , antiderivative size = 275, normalized size of antiderivative = 0.97

method	result	size
default	\(\frac {\sqrt {e \sec \left (d x +c \right )}\, e^{2} \left (i \left (\cos \left (d x +c \right )+\sin \left (d x +c \right )+1\right ) \operatorname {arctanh}\left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )+i \left (-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1\right ) \operatorname {arctanh}\left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )+8 i \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+\left (\cos \left (d x +c \right )-\sin \left (d x +c \right )+1\right ) \operatorname {arctanh}\left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )+\left (\cos \left (d x +c \right )+\sin \left (d x +c \right )+1\right ) \operatorname {arctanh}\left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )\right )}{2 d \left (\cos \left (d x +c \right )+1\right ) a \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\)	\(275\)

Input:

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

Output:

1/2/d*(e*sec(d*x+c))^(1/2)*e^2*(I*(cos(d*x+c)+sin(d*x+c)+1)*arctanh(1/2*(c 
ot(d*x+c)-csc(d*x+c)-1)/(1/(cos(d*x+c)+1))^(1/2))+I*(-cos(d*x+c)+sin(d*x+c 
)-1)*arctanh(1/2/(1/(cos(d*x+c)+1))^(1/2)*(cot(d*x+c)-csc(d*x+c)+1))+8*I*( 
cos(d*x+c)+1)*(1/(cos(d*x+c)+1))^(1/2)+(cos(d*x+c)-sin(d*x+c)+1)*arctanh(1 
/2*(cot(d*x+c)-csc(d*x+c)-1)/(1/(cos(d*x+c)+1))^(1/2))+(cos(d*x+c)+sin(d*x 
+c)+1)*arctanh(1/2/(1/(cos(d*x+c)+1))^(1/2)*(cot(d*x+c)-csc(d*x+c)+1)))/(c 
os(d*x+c)+1)/a/(a*(1+I*tan(d*x+c)))^(1/2)/(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. 539 vs. \(2 (209) = 418\).

Time = 0.11 (sec) , antiderivative size = 539, normalized size of antiderivative = 1.90 \[ \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:

integrate((e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fric 
as")
 

Output:

-1/2*(a^2*d*sqrt(4*I*e^5/(a^3*d^2))*e^(I*d*x + I*c)*log((a^2*d*sqrt(4*I*e^ 
5/(a^3*d^2)) + 2*(e^2*e^(2*I*d*x + 2*I*c) + e^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))/e^2) - 
 a^2*d*sqrt(4*I*e^5/(a^3*d^2))*e^(I*d*x + I*c)*log(-(a^2*d*sqrt(4*I*e^5/(a 
^3*d^2)) - 2*(e^2*e^(2*I*d*x + 2*I*c) + e^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))/e^2) - a^2 
*d*sqrt(-4*I*e^5/(a^3*d^2))*e^(I*d*x + I*c)*log((a^2*d*sqrt(-4*I*e^5/(a^3* 
d^2)) + 2*(e^2*e^(2*I*d*x + 2*I*c) + e^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))/e^2) + a^2*d* 
sqrt(-4*I*e^5/(a^3*d^2))*e^(I*d*x + I*c)*log(-(a^2*d*sqrt(-4*I*e^5/(a^3*d^ 
2)) - 2*(e^2*e^(2*I*d*x + 2*I*c) + e^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))/e^2) + 8*(-I*e^ 
2*e^(2*I*d*x + 2*I*c) - I*e^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))*e^(-I*d*x - I*c)/(a^2*d)
 

Sympy [F(-1)]

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

integrate((e*sec(d*x+c))**(5/2)/(a+I*a*tan(d*x+c))**(3/2),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. 778 vs. \(2 (209) = 418\).

Time = 0.36 (sec) , antiderivative size = 778, normalized size of antiderivative = 2.75 \[ \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/4*(2*I*sqrt(2)*e^2*arctan2(sqrt(2)*cos(1/2*d*x + 1/2*c) + 1, sqrt(2)*si 
n(1/2*d*x + 1/2*c) + 1) + 2*I*sqrt(2)*e^2*arctan2(sqrt(2)*cos(1/2*d*x + 1/ 
2*c) + 1, -sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) + 2*I*sqrt(2)*e^2*arctan2(sqr 
t(2)*cos(1/2*d*x + 1/2*c) - 1, sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) + 2*I*sqr 
t(2)*e^2*arctan2(sqrt(2)*cos(1/2*d*x + 1/2*c) - 1, -sqrt(2)*sin(1/2*d*x + 
1/2*c) + 1) + 2*sqrt(2)*e^2*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) - 2*sqrt(2)*e^2*ar 
ctan2(-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) + I*sqrt(2)*e^2*log(2*sqrt(2)*sin(d*x + c)*sin 
(1/2*d*x + 1/2*c) + 2*(sqrt(2)*cos(1/2*d*x + 1/2*c) + 1)*cos(d*x + c) + co 
s(d*x + c)^2 + 2*cos(1/2*d*x + 1/2*c)^2 + sin(d*x + c)^2 + 2*sin(1/2*d*x + 
 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 1) - I*sqrt(2)*e^2*log(-2*sqr 
t(2)*sin(d*x + c)*sin(1/2*d*x + 1/2*c) - 2*(sqrt(2)*cos(1/2*d*x + 1/2*c) - 
 1)*cos(d*x + c) + cos(d*x + c)^2 + 2*cos(1/2*d*x + 1/2*c)^2 + sin(d*x + c 
)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 1) + sqr 
t(2)*e^2*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) - sqrt(2)*e^ 
2*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) + sqrt(2)*e^2*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*...
 

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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