\(\int \frac {(e \sec (c+d x))^{3/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [417]

Optimal result

Integrand size = 30, antiderivative size = 361 \[ \int \frac {(e \sec (c+d x))^{3/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {i \sqrt {2} \sqrt {a} e^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right ) \sec (c+d x)}{d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {i \sqrt {2} \sqrt {a} e^{3/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right ) \sec (c+d x)}{d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {i \sqrt {2} \sqrt {a} e^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)} \left (\sqrt {a}+\cos (c+d x) \left (\sqrt {a}-i \sqrt {a} \tan (c+d x)\right )\right )}\right ) \sec (c+d x)}{d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 1.64 (sec) , antiderivative size = 302, normalized size of antiderivative = 0.84 \[ \int \frac {(e \sec (c+d x))^{3/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {2 e \sqrt {e \sec (c+d x)} \left (\text {arctanh}\left (\frac {\sqrt {1+i \cos (c)-\sin (c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}}{\sqrt {-1+i \cos (c)+\sin (c)} \sqrt {i+\tan \left (\frac {d x}{2}\right )}}\right ) \sqrt {-1-i \cos (c)-\sin (c)} \sqrt {1+i \cos (c)-\sin (c)}-\text {arctanh}\left (\frac {\sqrt {1-i \cos (c)+\sin (c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}}{\sqrt {-1-i \cos (c)-\sin (c)} \sqrt {i+\tan \left (\frac {d x}{2}\right )}}\right ) \sqrt {1-i \cos (c)+\sin (c)} \sqrt {-1+i \cos (c)+\sin (c)}\right ) (\cos (d x)+i \sin (d x)) \sqrt {i+\tan \left (\frac {d x}{2}\right )}}{d \sqrt {-1-i \cos (c)-\sin (c)} \sqrt {-1+i \cos (c)+\sin (c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )} \sqrt {a+i a \tan (c+d x)}} \] Input:

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

Output:

(2*e*Sqrt[e*Sec[c + d*x]]*(ArcTanh[(Sqrt[1 + I*Cos[c] - Sin[c]]*Sqrt[I - T 
an[(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]]*Sqr 
t[I + Tan[(d*x)/2]])]*Sqrt[1 - I*Cos[c] + Sin[c]]*Sqrt[-1 + I*Cos[c] + Sin 
[c]])*(Cos[d*x] + I*Sin[d*x])*Sqrt[I + Tan[(d*x)/2]])/(d*Sqrt[-1 - I*Cos[c 
] - Sin[c]]*Sqrt[-1 + I*Cos[c] + Sin[c]]*Sqrt[I - Tan[(d*x)/2]]*Sqrt[a + I 
*a*Tan[c + d*x]])
 

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 373, normalized size of antiderivative = 1.03, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3980

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3976

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

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

((4*I)*a*e^3*((-(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 + (Sq 
rt[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*T 
an[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]])
 

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 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 = 10.14 (sec) , antiderivative size = 232, normalized size of antiderivative = 0.64

method	result	size
default	\(\frac {\left (-\csc \left (d x +c \right )^{2} \left (1-\cos \left (d x +c \right )\right )^{2}+1\right ) \sec \left (d x +c \right ) e \sqrt {e \sec \left (d x +c \right )}\, \left (\cos \left (d x +c \right )+1\right ) \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )+i\right ) \left (i \operatorname {arctanh}\left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )+i \operatorname {arctanh}\left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )+\operatorname {arctanh}\left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )-\operatorname {arctanh}\left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )\right )}{4 d \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}\)	\(232\)

Input:

int((e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/4/d*(-csc(d*x+c)^2*(1-cos(d*x+c))^2+1)*sec(d*x+c)*e*(e*sec(d*x+c))^(1/2) 
*(cos(d*x+c)+1)/(1/(cos(d*x+c)+1))^(1/2)/(a*(1+I*tan(d*x+c)))^(1/2)*(cot(d 
*x+c)-csc(d*x+c)+I)*(I*arctanh(1/2*(cot(d*x+c)-csc(d*x+c)-1)/(1/(cos(d*x+c 
)+1))^(1/2))+I*arctanh(1/2/(1/(cos(d*x+c)+1))^(1/2)*(cot(d*x+c)-csc(d*x+c) 
+1))+arctanh(1/2*(cot(d*x+c)-csc(d*x+c)-1)/(1/(cos(d*x+c)+1))^(1/2))-arcta 
nh(1/2/(1/(cos(d*x+c)+1))^(1/2)*(cot(d*x+c)-csc(d*x+c)+1)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 385, normalized size of antiderivative = 1.07 \[ \int \frac {(e \sec (c+d x))^{3/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {1}{2} \, \sqrt {\frac {4 i \, e^{3}}{a d^{2}}} \log \left (\frac {2 \, {\left (e e^{\left (2 i \, d x + 2 i \, c\right )} + e\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + i \, a d \sqrt {\frac {4 i \, e^{3}}{a d^{2}}}}{e}\right ) - \frac {1}{2} \, \sqrt {\frac {4 i \, e^{3}}{a d^{2}}} \log \left (\frac {2 \, {\left (e e^{\left (2 i \, d x + 2 i \, c\right )} + e\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} - i \, a d \sqrt {\frac {4 i \, e^{3}}{a d^{2}}}}{e}\right ) + \frac {1}{2} \, \sqrt {-\frac {4 i \, e^{3}}{a d^{2}}} \log \left (\frac {2 \, {\left (e e^{\left (2 i \, d x + 2 i \, c\right )} + e\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + i \, a d \sqrt {-\frac {4 i \, e^{3}}{a d^{2}}}}{e}\right ) - \frac {1}{2} \, \sqrt {-\frac {4 i \, e^{3}}{a d^{2}}} \log \left (\frac {2 \, {\left (e e^{\left (2 i \, d x + 2 i \, c\right )} + e\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} - i \, a d \sqrt {-\frac {4 i \, e^{3}}{a d^{2}}}}{e}\right ) \] Input:

integrate((e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fric 
as")
 

Output:

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

Sympy [F]

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

integrate((e*sec(d*x+c))**(3/2)/(a+I*a*tan(d*x+c))**(1/2),x)
 

Output:

Integral((e*sec(c + d*x))**(3/2)/sqrt(I*a*(tan(c + d*x) - I)), x)
 

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. 726 vs. \(2 (271) = 542\).

Time = 0.31 (sec) , antiderivative size = 726, normalized size of antiderivative = 2.01 \[ \int \frac {(e \sec (c+d x))^{3/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\text {Too large to display} \] Input:

integrate((e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxi 
ma")
 

Output:

-1/4*(2*I*sqrt(2)*e*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*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*arctan2(sqrt(2)*c 
os(1/2*d*x + 1/2*c) - 1, sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) + 2*I*sqrt(2)*e 
*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*arctan2(sqrt(2)*sin(1/2*d*x + 1/2*c) + sin(d*x + c), sqr 
t(2)*cos(1/2*d*x + 1/2*c) + cos(d*x + c) + 1) + 2*sqrt(2)*e*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) + co 
s(d*x + c) + 1) + I*sqrt(2)*e*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) + 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) - I*sqrt(2)*e*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) 
 + 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) - sqrt(2)*e*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*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*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(...
 

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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