\(\int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{5/2} \, dx\) [408]

Optimal result

Integrand size = 30, antiderivative size = 612 \[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{5/2} \, dx=\frac {15 i a^3 (e \sec (c+d x))^{3/2}}{8 d \sqrt {a+i a \tan (c+d x)}}-\frac {15 i a^{7/2} e^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right ) \sec (c+d x)}{8 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {15 i a^{7/2} e^{3/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right ) \sec (c+d x)}{8 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {15 i a^{7/2} e^{3/2} \log \left (a-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))\right ) \sec (c+d x)}{16 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {15 i a^{7/2} e^{3/2} \log \left (a+\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))\right ) \sec (c+d x)}{16 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {3 i a^2 (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}}{3 d} \]

[Out]

15/8*I*a^3*(e*sec(d*x+c))^(3/2)/d/(a+I*a*tan(d*x+c))^(1/2)-15/16*I*a^(7/2)*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*2^(1/2)/(a-I*a*tan(d*x+c))^(1/2)/(a+I*a*tan(
d*x+c))^(1/2)+15/16*I*a^(7/2)*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*2^(1/2)/(a-I*a*tan(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2)+15/32*I*a^(7/2)*e^(3/2)*ln(a-2^
(1/2)*a^(1/2)*e^(1/2)*(a-I*a*tan(d*x+c))^(1/2)/(e*sec(d*x+c))^(1/2)+cos(d*x+c)*(a-I*a*tan(d*x+c)))*sec(d*x+c)/
d*2^(1/2)/(a-I*a*tan(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2)-15/32*I*a^(7/2)*e^(3/2)*ln(a+2^(1/2)*a^(1/2)*e^(1/
2)*(a-I*a*tan(d*x+c))^(1/2)/(e*sec(d*x+c))^(1/2)+cos(d*x+c)*(a-I*a*tan(d*x+c)))*sec(d*x+c)/d*2^(1/2)/(a-I*a*ta
n(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2)+3/4*I*a^2*(e*sec(d*x+c))^(3/2)*(a+I*a*tan(d*x+c))^(1/2)/d+1/3*I*a*(e*
sec(d*x+c))^(3/2)*(a+I*a*tan(d*x+c))^(3/2)/d

Rubi [A] (verified)

Time = 0.88 (sec) , antiderivative size = 612, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3579, 3580, 3576, 303, 1176, 631, 210, 1179, 642} \[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {15 i a^{7/2} e^{3/2} \sec (c+d x) \arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{8 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {15 i a^{7/2} e^{3/2} \sec (c+d x) \arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{8 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {15 i a^{7/2} e^{3/2} \sec (c+d x) \log \left (-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{16 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {15 i a^{7/2} e^{3/2} \sec (c+d x) \log \left (\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{16 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {15 i a^3 (e \sec (c+d x))^{3/2}}{8 d \sqrt {a+i a \tan (c+d x)}}+\frac {3 i a^2 \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}}{4 d}+\frac {i a (a+i a \tan (c+d x))^{3/2} (e \sec (c+d x))^{3/2}}{3 d} \]

[In]

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

[Out]

(((15*I)/8)*a^3*(e*Sec[c + d*x])^(3/2))/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (((15*I)/8)*a^(7/2)*e^(3/2)*ArcTan[1
- (Sqrt[2]*Sqrt[e]*Sqrt[a - I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e*Sec[c + d*x]])]*Sec[c + d*x])/(Sqrt[2]*d*Sqrt[a
- I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) + (((15*I)/8)*a^(7/2)*e^(3/2)*ArcTan[1 + (Sqrt[2]*Sqrt[e]*Sqr
t[a - I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e*Sec[c + d*x]])]*Sec[c + d*x])/(Sqrt[2]*d*Sqrt[a - I*a*Tan[c + d*x]]*S
qrt[a + I*a*Tan[c + d*x]]) + (((15*I)/16)*a^(7/2)*e^(3/2)*Log[a - (Sqrt[2]*Sqrt[a]*Sqrt[e]*Sqrt[a - I*a*Tan[c
+ d*x]])/Sqrt[e*Sec[c + d*x]] + Cos[c + d*x]*(a - I*a*Tan[c + d*x])]*Sec[c + d*x])/(Sqrt[2]*d*Sqrt[a - I*a*Tan
[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (((15*I)/16)*a^(7/2)*e^(3/2)*Log[a + (Sqrt[2]*Sqrt[a]*Sqrt[e]*Sqrt[a
- I*a*Tan[c + d*x]])/Sqrt[e*Sec[c + d*x]] + Cos[c + d*x]*(a - I*a*Tan[c + d*x])]*Sec[c + d*x])/(Sqrt[2]*d*Sqrt
[a - I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) + (((3*I)/4)*a^2*(e*Sec[c + d*x])^(3/2)*Sqrt[a + I*a*Tan[c
+ d*x]])/d + ((I/3)*a*(e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x])^(3/2))/d

Rule 210

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 303

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

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[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])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[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 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[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 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 3576

Int[Sqrt[(d_.)*sec[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-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 3579

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] + Dist[a*((m + 2*n - 2)/(m + n - 1)), Int[(
d*Sec[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 3580

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(3/2)/Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {i a (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {1}{2} (3 a) \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2} \, dx \\ & = \frac {3 i a^2 (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {1}{8} \left (15 a^2\right ) \int (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)} \, dx \\ & = \frac {15 i a^3 (e \sec (c+d x))^{3/2}}{8 d \sqrt {a+i a \tan (c+d x)}}+\frac {3 i a^2 (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {1}{16} \left (15 a^3\right ) \int \frac {(e \sec (c+d x))^{3/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx \\ & = \frac {15 i a^3 (e \sec (c+d x))^{3/2}}{8 d \sqrt {a+i a \tan (c+d x)}}+\frac {3 i a^2 (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {\left (15 a^3 e \sec (c+d x)\right ) \int \sqrt {e \sec (c+d x)} \sqrt {a-i a \tan (c+d x)} \, dx}{16 \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \\ & = \frac {15 i a^3 (e \sec (c+d x))^{3/2}}{8 d \sqrt {a+i a \tan (c+d x)}}+\frac {3 i a^2 (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {\left (15 i a^4 e^3 \sec (c+d x)\right ) \text {Subst}\left (\int \frac {x^2}{a^2+e^2 x^4} \, dx,x,\frac {\sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{4 d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \\ & = \frac {15 i a^3 (e \sec (c+d x))^{3/2}}{8 d \sqrt {a+i a \tan (c+d x)}}+\frac {3 i a^2 (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac {\left (15 i a^4 e^2 \sec (c+d x)\right ) \text {Subst}\left (\int \frac {a-e x^2}{a^2+e^2 x^4} \, dx,x,\frac {\sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{8 d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (15 i a^4 e^2 \sec (c+d x)\right ) \text {Subst}\left (\int \frac {a+e x^2}{a^2+e^2 x^4} \, dx,x,\frac {\sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{8 d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \\ & = \frac {15 i a^3 (e \sec (c+d x))^{3/2}}{8 d \sqrt {a+i a \tan (c+d x)}}+\frac {3 i a^2 (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {\left (15 i a^4 e \sec (c+d x)\right ) \text {Subst}\left (\int \frac {1}{\frac {a}{e}-\frac {\sqrt {2} \sqrt {a} x}{\sqrt {e}}+x^2} \, dx,x,\frac {\sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{16 d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (15 i a^4 e \sec (c+d x)\right ) \text {Subst}\left (\int \frac {1}{\frac {a}{e}+\frac {\sqrt {2} \sqrt {a} x}{\sqrt {e}}+x^2} \, dx,x,\frac {\sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{16 d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (15 i a^{7/2} e^{3/2} \sec (c+d x)\right ) \text {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt {a}}{\sqrt {e}}+2 x}{-\frac {a}{e}-\frac {\sqrt {2} \sqrt {a} x}{\sqrt {e}}-x^2} \, dx,x,\frac {\sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{16 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (15 i a^{7/2} e^{3/2} \sec (c+d x)\right ) \text {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt {a}}{\sqrt {e}}-2 x}{-\frac {a}{e}+\frac {\sqrt {2} \sqrt {a} x}{\sqrt {e}}-x^2} \, dx,x,\frac {\sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{16 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \\ & = \frac {15 i a^3 (e \sec (c+d x))^{3/2}}{8 d \sqrt {a+i a \tan (c+d x)}}+\frac {15 i a^{7/2} e^{3/2} \log \left (a-\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))\right ) \sec (c+d x)}{16 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {15 i a^{7/2} e^{3/2} \log \left (a+\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))\right ) \sec (c+d x)}{16 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {3 i a^2 (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {\left (15 i a^{7/2} e^{3/2} \sec (c+d x)\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{8 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {\left (15 i a^{7/2} e^{3/2} \sec (c+d x)\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{8 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \\ & = \frac {15 i a^3 (e \sec (c+d x))^{3/2}}{8 d \sqrt {a+i a \tan (c+d x)}}-\frac {15 i a^{7/2} 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)}{8 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {15 i a^{7/2} 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)}{8 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {15 i a^{7/2} e^{3/2} \log \left (a-\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))\right ) \sec (c+d x)}{16 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {15 i a^{7/2} e^{3/2} \log \left (a+\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))\right ) \sec (c+d x)}{16 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {3 i a^2 (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.20 (sec) , antiderivative size = 386, normalized size of antiderivative = 0.63 \[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{5/2} \, dx=\frac {\cos ^4(c+d x) (e \sec (c+d x))^{3/2} \left (\frac {1}{6} \sec ^3(c+d x) (63+79 \cos (2 (c+d x))+34 i \sin (2 (c+d x))) (i \cos (3 c+d x)+\sin (3 c+d x))+\frac {15 \left (\text {arctanh}\left (\frac {\sqrt {1+i \cos (c)-\sin (c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}}{\sqrt {-1+i \cos (c)+\sin (c)} \sqrt {i+\tan \left (\frac {d x}{2}\right )}}\right ) \sqrt {-1-i \cos (c)-\sin (c)} \sqrt {1+i \cos (c)-\sin (c)}-\text {arctanh}\left (\frac {\sqrt {1-i \cos (c)+\sin (c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}}{\sqrt {-1-i \cos (c)-\sin (c)} \sqrt {i+\tan \left (\frac {d x}{2}\right )}}\right ) \sqrt {1-i \cos (c)+\sin (c)} \sqrt {-1+i \cos (c)+\sin (c)}\right ) (\cos (3 c)-i \sin (3 c)) \sqrt {i+\tan \left (\frac {d x}{2}\right )}}{\sqrt {-1-i \cos (c)-\sin (c)} \sqrt {-1+i \cos (c)+\sin (c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}}\right ) (a+i a \tan (c+d x))^{5/2}}{8 d (\cos (d x)+i \sin (d x))^2} \]

[In]

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

[Out]

(Cos[c + d*x]^4*(e*Sec[c + d*x])^(3/2)*((Sec[c + d*x]^3*(63 + 79*Cos[2*(c + d*x)] + (34*I)*Sin[2*(c + d*x)])*(
I*Cos[3*c + d*x] + Sin[3*c + d*x]))/6 + (15*(ArcTanh[(Sqrt[1 + I*Cos[c] - Sin[c]]*Sqrt[I - Tan[(d*x)/2]])/(Sqr
t[-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[3*c] - I*Sin[3*c])*Sqrt[I + Tan[(d*x)/2]]
)/(Sqrt[-1 - I*Cos[c] - Sin[c]]*Sqrt[-1 + I*Cos[c] + Sin[c]]*Sqrt[I - Tan[(d*x)/2]]))*(a + I*a*Tan[c + d*x])^(
5/2))/(8*d*(Cos[d*x] + I*Sin[d*x])^2)

Maple [A] (verified)

Time = 9.46 (sec) , antiderivative size = 556, normalized size of antiderivative = 0.91

method	result	size
default	\(\frac {\left (-\frac {1}{96}-\frac {i}{96}\right ) \left (-\tan \left (d x +c \right )+i\right )^{2} \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {e \sec \left (d x +c \right )}\, \left (-45 \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+34 \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+45 i \left (\cos ^{3}\left (d x +c \right )\right ) \operatorname {arctanh}\left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )+8 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-79 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \left (\cos ^{2}\left (d x +c \right )\right )+45 \left (\cos ^{3}\left (d x +c \right )\right ) \operatorname {arctanh}\left (\frac {\cos \left (d x +c \right )-\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )-45 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \left (\cos ^{3}\left (d x +c \right )\right )-79 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \left (\cos ^{2}\left (d x +c \right )\right )-26 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+8 \sin \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-8 i \sin \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-34 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )-45 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \left (\cos ^{3}\left (d x +c \right )\right )+45 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-26 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+8 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\right ) \left (4 i \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+2 i \cos \left (d x +c \right ) \sin \left (d x +c \right )-4 \left (\cos ^{3}\left (d x +c \right )\right )-i \sin \left (d x +c \right )-2 \left (\cos ^{2}\left (d x +c \right )\right )+3 \cos \left (d x +c \right )+1\right ) a^{2} e}{d \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\)	\(556\)

[In]

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

[Out]

(-1/96-1/96*I)/d*(-tan(d*x+c)+I)^2*(a*(1+I*tan(d*x+c)))^(1/2)*(e*sec(d*x+c))^(1/2)*(-45*sin(d*x+c)*cos(d*x+c)^
2*(1/(cos(d*x+c)+1))^(1/2)+34*sin(d*x+c)*cos(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)+45*I*cos(d*x+c)^3*arctanh(1/2*(co
s(d*x+c)+sin(d*x+c)+1)/(cos(d*x+c)+1)/(1/(cos(d*x+c)+1))^(1/2))+8*I*(1/(cos(d*x+c)+1))^(1/2)-79*I*(1/(cos(d*x+
c)+1))^(1/2)*cos(d*x+c)^2+45*cos(d*x+c)^3*arctanh(1/2*(cos(d*x+c)-sin(d*x+c)+1)/(cos(d*x+c)+1)/(1/(cos(d*x+c)+
1))^(1/2))-45*(1/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^3-79*(1/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^2-26*(1/(cos(d*x+c)
+1))^(1/2)*cos(d*x+c)+8*sin(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)-8*I*sin(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)-34*I*(1/(c
os(d*x+c)+1))^(1/2)*cos(d*x+c)*sin(d*x+c)-45*I*(1/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^3+45*I*(1/(cos(d*x+c)+1))^(
1/2)*cos(d*x+c)^2*sin(d*x+c)-26*I*(1/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)+8*(1/(cos(d*x+c)+1))^(1/2))*(4*I*cos(d*x
+c)^2*sin(d*x+c)+2*I*cos(d*x+c)*sin(d*x+c)-4*cos(d*x+c)^3-I*sin(d*x+c)-2*cos(d*x+c)^2+3*cos(d*x+c)+1)*a^2*e/(c
os(d*x+c)+1)/(1/(cos(d*x+c)+1))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 635, normalized size of antiderivative = 1.04 \[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{5/2} \, dx=\frac {{\left (113 i \, a^{2} e e^{\left (4 i \, d x + 4 i \, c\right )} + 126 i \, a^{2} e e^{\left (2 i \, d x + 2 i \, c\right )} + 45 i \, a^{2} 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 )} + 6 \, \sqrt {\frac {225 i \, a^{5} e^{3}}{64 \, d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {2 \, {\left (15 \, {\left (a^{2} e e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} 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 )} + 8 i \, \sqrt {\frac {225 i \, a^{5} e^{3}}{64 \, d^{2}}} d\right )}}{15 \, a^{2} e}\right ) - 6 \, \sqrt {\frac {225 i \, a^{5} e^{3}}{64 \, d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {2 \, {\left (15 \, {\left (a^{2} e e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} 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 )} - 8 i \, \sqrt {\frac {225 i \, a^{5} e^{3}}{64 \, d^{2}}} d\right )}}{15 \, a^{2} e}\right ) + 6 \, \sqrt {-\frac {225 i \, a^{5} e^{3}}{64 \, d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {2 \, {\left (15 \, {\left (a^{2} e e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} 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 )} + 8 i \, \sqrt {-\frac {225 i \, a^{5} e^{3}}{64 \, d^{2}}} d\right )}}{15 \, a^{2} e}\right ) - 6 \, \sqrt {-\frac {225 i \, a^{5} e^{3}}{64 \, d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {2 \, {\left (15 \, {\left (a^{2} e e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} 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 )} - 8 i \, \sqrt {-\frac {225 i \, a^{5} e^{3}}{64 \, d^{2}}} d\right )}}{15 \, a^{2} e}\right )}{12 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

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

[Out]

1/12*((113*I*a^2*e*e^(4*I*d*x + 4*I*c) + 126*I*a^2*e*e^(2*I*d*x + 2*I*c) + 45*I*a^2*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) + 6*sqrt(225/64*I*a^5*e^3/d^2)*(d*e^(4*I*
d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log(2/15*(15*(a^2*e*e^(2*I*d*x + 2*I*c) + a^2*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) + 8*I*sqrt(225/64*I*a^5*e^3/d^2)*d)
/(a^2*e)) - 6*sqrt(225/64*I*a^5*e^3/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log(2/15*(15*(a
^2*e*e^(2*I*d*x + 2*I*c) + a^2*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) - 8*I*sqrt(225/64*I*a^5*e^3/d^2)*d)/(a^2*e)) + 6*sqrt(-225/64*I*a^5*e^3/d^2)*(d*e^(4*I*d*x + 4
*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log(2/15*(15*(a^2*e*e^(2*I*d*x + 2*I*c) + a^2*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) + 8*I*sqrt(-225/64*I*a^5*e^3/d^2)*d)/(a^2*
e)) - 6*sqrt(-225/64*I*a^5*e^3/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log(2/15*(15*(a^2*e*
e^(2*I*d*x + 2*I*c) + a^2*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) - 8*I*sqrt(-225/64*I*a^5*e^3/d^2)*d)/(a^2*e)))/(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d
)

Sympy [F(-1)]

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

[In]

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

[Out]

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. 3005 vs. \(2 (458) = 916\).

Time = 0.65 (sec) , antiderivative size = 3005, normalized size of antiderivative = 4.91 \[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{5/2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

192*(1808*a^2*e*cos(9/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2016*a^2*e*cos(5/4*arctan2(sin(2*d*x +
2*c), cos(2*d*x + 2*c))) + 720*a^2*e*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1808*I*a^2*e*sin(9
/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2016*I*a^2*e*sin(5/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2
*c))) + 720*I*a^2*e*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 90*(sqrt(2)*a^2*e*cos(6*d*x + 6*c)
+ 3*sqrt(2)*a^2*e*cos(4*d*x + 4*c) + 3*sqrt(2)*a^2*e*cos(2*d*x + 2*c) + I*sqrt(2)*a^2*e*sin(6*d*x + 6*c) + 3*I
*sqrt(2)*a^2*e*sin(4*d*x + 4*c) + 3*I*sqrt(2)*a^2*e*sin(2*d*x + 2*c) + sqrt(2)*a^2*e)*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) - 90*(sqrt(2)*a^2*e*cos(6*d*x + 6*c) + 3*sqrt(2)*a^2*e*cos(4*d*x + 4*c) + 3*sqrt(2)*a^2*e*cos(2*d*x + 2*
c) + I*sqrt(2)*a^2*e*sin(6*d*x + 6*c) + 3*I*sqrt(2)*a^2*e*sin(4*d*x + 4*c) + 3*I*sqrt(2)*a^2*e*sin(2*d*x + 2*c
) + sqrt(2)*a^2*e)*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) - 90*(sqrt(2)*a^2*e*cos(6*d*x + 6*c) + 3*sqrt(2)*a^2*e*cos(4
*d*x + 4*c) + 3*sqrt(2)*a^2*e*cos(2*d*x + 2*c) + I*sqrt(2)*a^2*e*sin(6*d*x + 6*c) + 3*I*sqrt(2)*a^2*e*sin(4*d*
x + 4*c) + 3*I*sqrt(2)*a^2*e*sin(2*d*x + 2*c) + sqrt(2)*a^2*e)*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) - 90*(sqrt(2)*a^2
*e*cos(6*d*x + 6*c) + 3*sqrt(2)*a^2*e*cos(4*d*x + 4*c) + 3*sqrt(2)*a^2*e*cos(2*d*x + 2*c) + I*sqrt(2)*a^2*e*si
n(6*d*x + 6*c) + 3*I*sqrt(2)*a^2*e*sin(4*d*x + 4*c) + 3*I*sqrt(2)*a^2*e*sin(2*d*x + 2*c) + sqrt(2)*a^2*e)*arct
an2(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) + 90*(-I*sqrt(2)*a^2*e*cos(6*d*x + 6*c) - 3*I*sqrt(2)*a^2*e*cos(4*d*x + 4*c) - 3*I*
sqrt(2)*a^2*e*cos(2*d*x + 2*c) + sqrt(2)*a^2*e*sin(6*d*x + 6*c) + 3*sqrt(2)*a^2*e*sin(4*d*x + 4*c) + 3*sqrt(2)
*a^2*e*sin(2*d*x + 2*c) - I*sqrt(2)*a^2*e)*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) + 90*(I*sqrt(2)*a^2*e*cos(6*d*x + 6*c) +
3*I*sqrt(2)*a^2*e*cos(4*d*x + 4*c) + 3*I*sqrt(2)*a^2*e*cos(2*d*x + 2*c) - sqrt(2)*a^2*e*sin(6*d*x + 6*c) - 3*s
qrt(2)*a^2*e*sin(4*d*x + 4*c) - 3*sqrt(2)*a^2*e*sin(2*d*x + 2*c) + I*sqrt(2)*a^2*e)*arctan2(-sqrt(2)*sin(1/4*a
rctan2(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)*c
os(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
) - 45*(sqrt(2)*a^2*e*cos(6*d*x + 6*c) + 3*sqrt(2)*a^2*e*cos(4*d*x + 4*c) + 3*sqrt(2)*a^2*e*cos(2*d*x + 2*c) +
I*sqrt(2)*a^2*e*sin(6*d*x + 6*c) + 3*I*sqrt(2)*a^2*e*sin(4*d*x + 4*c) + 3*I*sqrt(2)*a^2*e*sin(2*d*x + 2*c) +
sqrt(2)*a^2*e)*log(2*sqrt(2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(1/4*arctan2(sin(2*d*x +
2*c), cos(2*d*x + 2*c))) + 2*(sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1)*cos(1/2*arctan
2(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)))^2 + 2*cos(1/4*ar
ctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(
1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2
*c))) + 1) + 45*(sqrt(2)*a^2*e*cos(6*d*x + 6*c) + 3*sqrt(2)*a^2*e*cos(4*d*x + 4*c) + 3*sqrt(2)*a^2*e*cos(2*d*x
+ 2*c) + I*sqrt(2)*a^2*e*sin(6*d*x + 6*c) + 3*I*sqrt(2)*a^2*e*sin(4*d*x + 4*c) + 3*I*sqrt(2)*a^2*e*sin(2*d*x
+ 2*c) + sqrt(2)*a^2*e)*log(-2*sqrt(2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(1/4*arctan2(si
n(2*d*x + 2*c), cos(2*d*x + 2*c))) - 2*(sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 1)*cos(
1/2*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)))^2 + 2*
cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^
2 + 2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 - 2*sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos
(2*d*x + 2*c))) + 1) + 45*(-I*sqrt(2)*a^2*e*cos(6*d*x + 6*c) - 3*I*sqrt(2)*a^2*e*cos(4*d*x + 4*c) - 3*I*sqrt(2
)*a^2*e*cos(2*d*x + 2*c) + sqrt(2)*a^2*e*sin(6*d*x + 6*c) + 3*sqrt(2)*a^2*e*sin(4*d*x + 4*c) + 3*sqrt(2)*a^2*e
*sin(2*d*x + 2*c) - I*sqrt(2)*a^2*e)*log(2*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/4*
arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))
) + 2*sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2) + 45*(I*sqrt(2)*a^2*e*cos(6*d*x + 6*c)
+ 3*I*sqrt(2)*a^2*e*cos(4*d*x + 4*c) + 3*I*sqrt(2)*a^2*e*cos(2*d*x + 2*c) - sqrt(2)*a^2*e*sin(6*d*x + 6*c) -
3*sqrt(2)*a^2*e*sin(4*d*x + 4*c) - 3*sqrt(2)*a^2*e*sin(2*d*x + 2*c) + I*sqrt(2)*a^2*e)*log(2*cos(1/4*arctan2(s
in(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sqrt(2)*c
os(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 2*sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x +
2*c))) + 2) + 45*(-I*sqrt(2)*a^2*e*cos(6*d*x + 6*c) - 3*I*sqrt(2)*a^2*e*cos(4*d*x + 4*c) - 3*I*sqrt(2)*a^2*e*c
os(2*d*x + 2*c) + sqrt(2)*a^2*e*sin(6*d*x + 6*c) + 3*sqrt(2)*a^2*e*sin(4*d*x + 4*c) + 3*sqrt(2)*a^2*e*sin(2*d*
x + 2*c) - I*sqrt(2)*a^2*e)*log(2*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/4*arctan2(s
in(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 - 2*sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2*sqr
t(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2) + 45*(I*sqrt(2)*a^2*e*cos(6*d*x + 6*c) + 3*I*sq
rt(2)*a^2*e*cos(4*d*x + 4*c) + 3*I*sqrt(2)*a^2*e*cos(2*d*x + 2*c) - sqrt(2)*a^2*e*sin(6*d*x + 6*c) - 3*sqrt(2)
*a^2*e*sin(4*d*x + 4*c) - 3*sqrt(2)*a^2*e*sin(2*d*x + 2*c) + I*sqrt(2)*a^2*e)*log(2*cos(1/4*arctan2(sin(2*d*x
+ 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 - 2*sqrt(2)*cos(1/4*ar
ctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 2*sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) +
2))*sqrt(a)*sqrt(e)/(d*(-36864*I*cos(6*d*x + 6*c) - 110592*I*cos(4*d*x + 4*c) - 110592*I*cos(2*d*x + 2*c) + 36
864*sin(6*d*x + 6*c) + 110592*sin(4*d*x + 4*c) + 110592*sin(2*d*x + 2*c) - 36864*I))

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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