\(\int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2} \, dx\) [401]

Optimal result

Integrand size = 30, antiderivative size = 571 \[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2} \, dx=\frac {5 i a^2 (e \sec (c+d x))^{3/2}}{4 d \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a^{5/2} e^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right ) \sec (c+d x)}{4 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {5 i a^{5/2} e^{3/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right ) \sec (c+d x)}{4 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {5 i a^{5/2} e^{3/2} \log \left (a-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))\right ) \sec (c+d x)}{8 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a^{5/2} e^{3/2} \log \left (a+\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))\right ) \sec (c+d x)}{8 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {i a (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}{2 d} \]

[Out]

5/4*I*a^2*(e*sec(d*x+c))^(3/2)/d/(a+I*a*tan(d*x+c))^(1/2)-5/8*I*a^(5/2)*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)+5/8*I*a^(5/2)*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)+5/16*I*a^(5/2)*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)-5/16*I*a^(5/2)*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)+1/2*I*a*(e*sec(d*x+c))^(3/2)*(a+I*a*tan(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 571, normalized size of antiderivative = 1.00, number of steps used = 13, 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} \[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {5 i a^{5/2} e^{3/2} \sec (c+d x) \arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{4 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {5 i a^{5/2} e^{3/2} \sec (c+d x) \arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{4 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {5 i a^{5/2} e^{3/2} \sec (c+d x) \log \left (-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{8 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a^{5/2} e^{3/2} \sec (c+d x) \log \left (\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{8 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {5 i a^2 (e \sec (c+d x))^{3/2}}{4 d \sqrt {a+i a \tan (c+d x)}}+\frac {i a \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}}{2 d} \]

[In]

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

[Out]

(((5*I)/4)*a^2*(e*Sec[c + d*x])^(3/2))/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (((5*I)/4)*a^(5/2)*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]]) + (((5*I)/4)*a^(5/2)*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]]) + (((5*I)/8)*a^(5/2)*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]]) - (((5*I)/8)*a^(5/2)*e^(3/2)*Log[a + (Sqrt[2]*Sqrt[a]*Sqrt[e]*Sqrt[a - I*a*T
an[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]]) + ((I/2)*a*(e*Sec[c + d*x])^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/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} \sqrt {a+i a \tan (c+d x)}}{2 d}+\frac {1}{4} (5 a) \int (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)} \, dx \\ & = \frac {5 i a^2 (e \sec (c+d x))^{3/2}}{4 d \sqrt {a+i a \tan (c+d x)}}+\frac {i a (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}{2 d}+\frac {1}{8} \left (5 a^2\right ) \int \frac {(e \sec (c+d x))^{3/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx \\ & = \frac {5 i a^2 (e \sec (c+d x))^{3/2}}{4 d \sqrt {a+i a \tan (c+d x)}}+\frac {i a (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}{2 d}+\frac {\left (5 a^2 e \sec (c+d x)\right ) \int \sqrt {e \sec (c+d x)} \sqrt {a-i a \tan (c+d x)} \, dx}{8 \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \\ & = \frac {5 i a^2 (e \sec (c+d x))^{3/2}}{4 d \sqrt {a+i a \tan (c+d x)}}+\frac {i a (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}{2 d}+\frac {\left (5 i a^3 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 )}{2 d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \\ & = \frac {5 i a^2 (e \sec (c+d x))^{3/2}}{4 d \sqrt {a+i a \tan (c+d x)}}+\frac {i a (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}{2 d}-\frac {\left (5 i a^3 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 )}{4 d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (5 i a^3 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 )}{4 d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \\ & = \frac {5 i a^2 (e \sec (c+d x))^{3/2}}{4 d \sqrt {a+i a \tan (c+d x)}}+\frac {i a (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}{2 d}+\frac {\left (5 i a^3 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 )}{8 d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (5 i a^3 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 )}{8 d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (5 i a^{5/2} 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 )}{8 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (5 i a^{5/2} 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 )}{8 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \\ & = \frac {5 i a^2 (e \sec (c+d x))^{3/2}}{4 d \sqrt {a+i a \tan (c+d x)}}+\frac {5 i a^{5/2} 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)}{8 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a^{5/2} 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)}{8 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {i a (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}{2 d}+\frac {\left (5 i a^{5/2} 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 )}{4 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {\left (5 i a^{5/2} 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 )}{4 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \\ & = \frac {5 i a^2 (e \sec (c+d x))^{3/2}}{4 d \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a^{5/2} 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)}{4 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {5 i a^{5/2} 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)}{4 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {5 i a^{5/2} 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)}{8 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a^{5/2} 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)}{8 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {i a (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.66 (sec) , antiderivative size = 375, normalized size of antiderivative = 0.66 \[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2} \, dx=\frac {\cos ^3(c+d x) (e \sec (c+d x))^{3/2} (\cos (d x)-i \sin (d x)) \left (2 \sec ^2(c+d x) (i \cos (c)+\sin (c))+5 \sec (c+d x) (i \cos (2 c+d x)+\sin (2 c+d x))+\frac {5 \left (\text {arctanh}\left (\frac {\sqrt {1+i \cos (c)-\sin (c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}}{\sqrt {-1+i \cos (c)+\sin (c)} \sqrt {i+\tan \left (\frac {d x}{2}\right )}}\right ) \sqrt {-1-i \cos (c)-\sin (c)} \sqrt {1+i \cos (c)-\sin (c)}-\text {arctanh}\left (\frac {\sqrt {1-i \cos (c)+\sin (c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}}{\sqrt {-1-i \cos (c)-\sin (c)} \sqrt {i+\tan \left (\frac {d x}{2}\right )}}\right ) \sqrt {1-i \cos (c)+\sin (c)} \sqrt {-1+i \cos (c)+\sin (c)}\right ) (\cos (2 c)-i \sin (2 c)) \sqrt {i+\tan \left (\frac {d x}{2}\right )}}{\sqrt {-1-i \cos (c)-\sin (c)} \sqrt {-1+i \cos (c)+\sin (c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}}\right ) (a+i a \tan (c+d x))^{3/2}}{4 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 10.54 (sec) , antiderivative size = 422, normalized size of antiderivative = 0.74

method	result	size
default	\(\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) \left (-\tan \left (d x +c \right )+i\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {e \sec \left (d x +c \right )}\, \left (5 i \left (\cos ^{2}\left (d x +c \right )\right ) \operatorname {arctanh}\left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )-5 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \left (\cos ^{2}\left (d x +c \right )\right )+5 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )-7 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )-2 i \sin \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-5 \left (\cos ^{2}\left (d x +c \right )\right ) \operatorname {arctanh}\left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )-5 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \left (\cos ^{2}\left (d x +c \right )\right )-5 \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-2 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-7 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+2 \sin \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-2 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\right ) \left (2 i \left (\cos ^{2}\left (d x +c \right )\right )+2 \sin \left (d x +c \right ) \cos \left (d x +c \right )+i \cos \left (d x +c \right )+\sin \left (d x +c \right )-i\right ) a e}{d \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\)	\(422\)

[In]

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

[Out]

(1/16+1/16*I)/d*(-tan(d*x+c)+I)*(a*(1+I*tan(d*x+c)))^(1/2)*(e*sec(d*x+c))^(1/2)*(5*I*cos(d*x+c)^2*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))-5*I*(1/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^2+5
*I*(1/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)*sin(d*x+c)-7*I*(1/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)-2*I*sin(d*x+c)*(1/(c
os(d*x+c)+1))^(1/2)-5*cos(d*x+c)^2*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))-5*(1/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^2-5*sin(d*x+c)*cos(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)-2*I*(1/(cos(d*x+
c)+1))^(1/2)-7*(1/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)+2*sin(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)-2*(1/(cos(d*x+c)+1))^
(1/2))*(2*I*cos(d*x+c)^2+2*sin(d*x+c)*cos(d*x+c)+I*cos(d*x+c)+sin(d*x+c)-I)*a*e/(cos(d*x+c)+1)/(1/(cos(d*x+c)+
1))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 538, normalized size of antiderivative = 0.94 \[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2} \, dx=\frac {{\left (9 i \, a e e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i \, a 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 )} + \sqrt {\frac {25 i \, a^{3} e^{3}}{16 \, d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {2 \, {\left (5 \, {\left (a e e^{\left (2 i \, d x + 2 i \, c\right )} + a 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 )} + 4 i \, \sqrt {\frac {25 i \, a^{3} e^{3}}{16 \, d^{2}}} d\right )}}{5 \, a e}\right ) - \sqrt {\frac {25 i \, a^{3} e^{3}}{16 \, d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {2 \, {\left (5 \, {\left (a e e^{\left (2 i \, d x + 2 i \, c\right )} + a 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 )} - 4 i \, \sqrt {\frac {25 i \, a^{3} e^{3}}{16 \, d^{2}}} d\right )}}{5 \, a e}\right ) + \sqrt {-\frac {25 i \, a^{3} e^{3}}{16 \, d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {2 \, {\left (5 \, {\left (a e e^{\left (2 i \, d x + 2 i \, c\right )} + a 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 )} + 4 i \, \sqrt {-\frac {25 i \, a^{3} e^{3}}{16 \, d^{2}}} d\right )}}{5 \, a e}\right ) - \sqrt {-\frac {25 i \, a^{3} e^{3}}{16 \, d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {2 \, {\left (5 \, {\left (a e e^{\left (2 i \, d x + 2 i \, c\right )} + a 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 )} - 4 i \, \sqrt {-\frac {25 i \, a^{3} e^{3}}{16 \, d^{2}}} d\right )}}{5 \, a e}\right )}{2 \, {\left (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))^(3/2),x, algorithm="fricas")

[Out]

1/2*((9*I*a*e*e^(2*I*d*x + 2*I*c) + 5*I*a*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) + sqrt(25/16*I*a^3*e^3/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log(2/5*(5*(a*e*e^(2*I*d*x
+ 2*I*c) + a*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) +
4*I*sqrt(25/16*I*a^3*e^3/d^2)*d)/(a*e)) - sqrt(25/16*I*a^3*e^3/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log(2/5*(5*(a*
e*e^(2*I*d*x + 2*I*c) + a*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) - 4*I*sqrt(25/16*I*a^3*e^3/d^2)*d)/(a*e)) + sqrt(-25/16*I*a^3*e^3/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*
log(2/5*(5*(a*e*e^(2*I*d*x + 2*I*c) + a*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) + 4*I*sqrt(-25/16*I*a^3*e^3/d^2)*d)/(a*e)) - sqrt(-25/16*I*a^3*e^3/d^2)*(d*e^(2*I*d*x
+ 2*I*c) + d)*log(2/5*(5*(a*e*e^(2*I*d*x + 2*I*c) + a*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) - 4*I*sqrt(-25/16*I*a^3*e^3/d^2)*d)/(a*e)))/(d*e^(2*I*d*x + 2*I*c) + d
)

Sympy [F(-1)]

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

[In]

integrate((e*sec(d*x+c))**(3/2)*(a+I*a*tan(d*x+c))**(3/2),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. 2367 vs. \(2 (427) = 854\).

Time = 0.53 (sec) , antiderivative size = 2367, normalized size of antiderivative = 4.15 \[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

32*(144*a*e*cos(5/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 80*a*e*cos(1/4*arctan2(sin(2*d*x + 2*c), co
s(2*d*x + 2*c))) + 144*I*a*e*sin(5/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 80*I*a*e*sin(1/4*arctan2(s
in(2*d*x + 2*c), cos(2*d*x + 2*c))) - 10*(sqrt(2)*a*e*cos(4*d*x + 4*c) + 2*sqrt(2)*a*e*cos(2*d*x + 2*c) + I*sq
rt(2)*a*e*sin(4*d*x + 4*c) + 2*I*sqrt(2)*a*e*sin(2*d*x + 2*c) + sqrt(2)*a*e)*arctan2(sqrt(2)*cos(1/4*arctan2(s
in(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) - 1
0*(sqrt(2)*a*e*cos(4*d*x + 4*c) + 2*sqrt(2)*a*e*cos(2*d*x + 2*c) + I*sqrt(2)*a*e*sin(4*d*x + 4*c) + 2*I*sqrt(2
)*a*e*sin(2*d*x + 2*c) + sqrt(2)*a*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) - 10*(sqrt(2)*a*e*cos(4*d*x + 4*c) + 2*sq
rt(2)*a*e*cos(2*d*x + 2*c) + I*sqrt(2)*a*e*sin(4*d*x + 4*c) + 2*I*sqrt(2)*a*e*sin(2*d*x + 2*c) + sqrt(2)*a*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) - 10*(sqrt(2)*a*e*cos(4*d*x + 4*c) + 2*sqrt(2)*a*e*cos(2*d*x + 2*c) + I*sqrt(2)*
a*e*sin(4*d*x + 4*c) + 2*I*sqrt(2)*a*e*sin(2*d*x + 2*c) + sqrt(2)*a*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) + 10*(-I
*sqrt(2)*a*e*cos(4*d*x + 4*c) - 2*I*sqrt(2)*a*e*cos(2*d*x + 2*c) + sqrt(2)*a*e*sin(4*d*x + 4*c) + 2*sqrt(2)*a*
e*sin(2*d*x + 2*c) - I*sqrt(2)*a*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) + 10*(I*sqrt(2)*a*e*cos(4*d*x + 4*c) + 2*I*sqrt(
2)*a*e*cos(2*d*x + 2*c) - sqrt(2)*a*e*sin(4*d*x + 4*c) - 2*sqrt(2)*a*e*sin(2*d*x + 2*c) + I*sqrt(2)*a*e)*arcta
n2(-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), c
os(2*d*x + 2*c))) + 1) - 5*(sqrt(2)*a*e*cos(4*d*x + 4*c) + 2*sqrt(2)*a*e*cos(2*d*x + 2*c) + I*sqrt(2)*a*e*sin(
4*d*x + 4*c) + 2*I*sqrt(2)*a*e*sin(2*d*x + 2*c) + sqrt(2)*a*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*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) + 5*(sqrt(2)*a*e*cos(4*d*x + 4*c) + 2*sqrt(2)*a*e*c
os(2*d*x + 2*c) + I*sqrt(2)*a*e*sin(4*d*x + 4*c) + 2*I*sqrt(2)*a*e*sin(2*d*x + 2*c) + sqrt(2)*a*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*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), c
os(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) + 5*(-I*sqrt(2
)*a*e*cos(4*d*x + 4*c) - 2*I*sqrt(2)*a*e*cos(2*d*x + 2*c) + sqrt(2)*a*e*sin(4*d*x + 4*c) + 2*sqrt(2)*a*e*sin(2
*d*x + 2*c) - I*sqrt(2)*a*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*sq
rt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2) + 5*(I*sqrt(2)*a*e*cos(4*d*x + 4*c) + 2*I*sqrt
(2)*a*e*cos(2*d*x + 2*c) - sqrt(2)*a*e*sin(4*d*x + 4*c) - 2*sqrt(2)*a*e*sin(2*d*x + 2*c) + I*sqrt(2)*a*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) + 5*(-I*sqrt(2)*a*e*cos(4*d*x + 4*c) - 2*I*sqrt(2)*a*e*cos(2*d*x + 2*c) + sqrt(
2)*a*e*sin(4*d*x + 4*c) + 2*sqrt(2)*a*e*sin(2*d*x + 2*c) - I*sqrt(2)*a*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*arctan
2(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) +
5*(I*sqrt(2)*a*e*cos(4*d*x + 4*c) + 2*I*sqrt(2)*a*e*cos(2*d*x + 2*c) - sqrt(2)*a*e*sin(4*d*x + 4*c) - 2*sqrt(
2)*a*e*sin(2*d*x + 2*c) + I*sqrt(2)*a*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))*sqrt(a)*sqrt(e)/(d*(-1024*I*cos(4
*d*x + 4*c) - 2048*I*cos(2*d*x + 2*c) + 1024*sin(4*d*x + 4*c) + 2048*sin(2*d*x + 2*c) - 1024*I))

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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