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

Optimal. Leaf size=524 \[ -\frac{i a^{3/2} e^{3/2} \sec (c+d x) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{\sqrt{2} d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{i a^{3/2} e^{3/2} \sec (c+d x) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{\sqrt{2} d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{i a^{3/2} e^{3/2} \sec (c+d x) \log \left (-\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{2 \sqrt{2} d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{i a^{3/2} e^{3/2} \sec (c+d x) \log \left (\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{2 \sqrt{2} d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{i a (e \sec (c+d x))^{3/2}}{d \sqrt{a+i a \tan (c+d x)}} \]

[Out]

(I*a*(e*Sec[c + d*x])^(3/2))/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (I*a^(3/2)*e^(3/2)*ArcTan[1 - (Sqrt[2]*Sqrt[e]*S
qrt[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]]) + (I*a^(3/2)*e^(3/2)*ArcTan[1 + (Sqrt[2]*Sqrt[e]*Sqrt[a - I*a*Tan[c + d*x]])/(Sqr
t[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]]) +
((I/2)*a^(3/2)*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
]]) - ((I/2)*a^(3/2)*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]])

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Rubi [A]  time = 0.445326, antiderivative size = 524, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {3498, 3499, 3495, 297, 1162, 617, 204, 1165, 628} \[ -\frac{i a^{3/2} e^{3/2} \sec (c+d x) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{\sqrt{2} d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{i a^{3/2} e^{3/2} \sec (c+d x) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{\sqrt{2} d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{i a^{3/2} e^{3/2} \sec (c+d x) \log \left (-\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{2 \sqrt{2} d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{i a^{3/2} e^{3/2} \sec (c+d x) \log \left (\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{2 \sqrt{2} d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{i a (e \sec (c+d x))^{3/2}}{d \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I*a*(e*Sec[c + d*x])^(3/2))/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (I*a^(3/2)*e^(3/2)*ArcTan[1 - (Sqrt[2]*Sqrt[e]*S
qrt[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]]) + (I*a^(3/2)*e^(3/2)*ArcTan[1 + (Sqrt[2]*Sqrt[e]*Sqrt[a - I*a*Tan[c + d*x]])/(Sqr
t[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]]) +
((I/2)*a^(3/2)*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
]]) - ((I/2)*a^(3/2)*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]])

Rule 3498

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 3499

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]

Rule 3495

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 297

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 1162

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 617

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 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 1165

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 628

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]

Rubi steps

\begin{align*} \int (e \sec (c+d x))^{3/2} \sqrt{a+i a \tan (c+d x)} \, dx &=\frac{i a (e \sec (c+d x))^{3/2}}{d \sqrt{a+i a \tan (c+d x)}}+\frac{1}{2} a \int \frac{(e \sec (c+d x))^{3/2}}{\sqrt{a+i a \tan (c+d x)}} \, dx\\ &=\frac{i a (e \sec (c+d x))^{3/2}}{d \sqrt{a+i a \tan (c+d x)}}+\frac{(a e \sec (c+d x)) \int \sqrt{e \sec (c+d x)} \sqrt{a-i a \tan (c+d x)} \, dx}{2 \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}}{d \sqrt{a+i a \tan (c+d x)}}+\frac{\left (2 i a^2 e^3 \sec (c+d x)\right ) \operatorname{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 )}{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}}{d \sqrt{a+i a \tan (c+d x)}}-\frac{\left (i a^2 e^2 \sec (c+d x)\right ) \operatorname{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 )}{d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{\left (i a^2 e^2 \sec (c+d x)\right ) \operatorname{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 )}{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}}{d \sqrt{a+i a \tan (c+d x)}}+\frac{\left (i a^2 e \sec (c+d x)\right ) \operatorname{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 )}{2 d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{\left (i a^2 e \sec (c+d x)\right ) \operatorname{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 )}{2 d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{\left (i a^{3/2} e^{3/2} \sec (c+d x)\right ) \operatorname{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 )}{2 \sqrt{2} d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{\left (i a^{3/2} e^{3/2} \sec (c+d x)\right ) \operatorname{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 )}{2 \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}}{d \sqrt{a+i a \tan (c+d x)}}+\frac{i a^{3/2} 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)}{2 \sqrt{2} d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{i a^{3/2} 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)}{2 \sqrt{2} d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{\left (i a^{3/2} e^{3/2} \sec (c+d x)\right ) \operatorname{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 )}{\sqrt{2} d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{\left (i a^{3/2} e^{3/2} \sec (c+d x)\right ) \operatorname{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 )}{\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}}{d \sqrt{a+i a \tan (c+d x)}}-\frac{i a^{3/2} e^{3/2} \tan ^{-1}\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)}{\sqrt{2} d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{i a^{3/2} e^{3/2} \tan ^{-1}\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)}{\sqrt{2} d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{i a^{3/2} 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)}{2 \sqrt{2} d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{i a^{3/2} 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)}{2 \sqrt{2} d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.79836, size = 373, normalized size = 0.71 \[ \frac{e (\cos (c)-i \sin (c)) \sqrt{a+i a \tan (c+d x)} \sqrt{e \sec (c+d x)} \left (\sqrt{\sin (c)+i \cos (c)-1} \left (\sqrt{-\sin (c)+i \cos (c)-1} \sqrt{\tan \left (\frac{d x}{2}\right )+i} \cos (c+d x) \tan ^{-1}\left (\frac{\sqrt{-\sin (c)+i \cos (c)-1} \sqrt{-\tan \left (\frac{d x}{2}\right )+i}}{\sqrt{-\sin (c)-i \cos (c)-1} \sqrt{\tan \left (\frac{d x}{2}\right )+i}}\right )+\sqrt{-\sin (c)-i \cos (c)-1} \sqrt{-\tan \left (\frac{d x}{2}\right )+i} (\sin (d x)+i \cos (d x))\right )-\sqrt{-\sin (c)-i \cos (c)-1} \sqrt{\sin (c)-i \cos (c)-1} \sqrt{\tan \left (\frac{d x}{2}\right )+i} \cos (c+d x) \tan ^{-1}\left (\frac{\sqrt{\sin (c)-i \cos (c)-1} \sqrt{-\tan \left (\frac{d x}{2}\right )+i}}{\sqrt{\sin (c)+i \cos (c)-1} \sqrt{\tan \left (\frac{d x}{2}\right )+i}}\right )\right )}{d \sqrt{-\sin (c)-i \cos (c)-1} \sqrt{\sin (c)+i \cos (c)-1} \sqrt{-\tan \left (\frac{d x}{2}\right )+i}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*Sqrt[e*Sec[c + d*x]]*(Cos[c] - I*Sin[c])*(-(ArcTan[(Sqrt[-1 - I*Cos[c] + Sin[c]]*Sqrt[I - Tan[(d*x)/2]])/(S
qrt[-1 + I*Cos[c] + Sin[c]]*Sqrt[I + Tan[(d*x)/2]])]*Cos[c + d*x]*Sqrt[-1 - I*Cos[c] - Sin[c]]*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]]*(I*Cos[d*x]
+ Sin[d*x])*Sqrt[I - Tan[(d*x)/2]] + ArcTan[(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]])]*Cos[c + d*x]*Sqrt[-1 + I*Cos[c] - Sin[c]]*Sqrt[I + Tan[(d*x)/2]]))
*Sqrt[a + I*a*Tan[c + d*x]])/(d*Sqrt[-1 - I*Cos[c] - Sin[c]]*Sqrt[-1 + I*Cos[c] + Sin[c]]*Sqrt[I - Tan[(d*x)/2
]])

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Maple [A]  time = 0.391, size = 309, normalized size = 0.6 \begin{align*}{\frac{\cos \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) -1 \right ) ^{2}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) -1 \right ) } \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{3}{2}}}\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}} \left ( i{\it Artanh} \left ({\frac{\cos \left ( dx+c \right ) +1+\sin \left ( dx+c \right ) }{2}\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}} \right ) \cos \left ( dx+c \right ) -i{\it Artanh} \left ({\frac{\cos \left ( dx+c \right ) +1-\sin \left ( dx+c \right ) }{2}\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}} \right ) \cos \left ( dx+c \right ) +2\,i\sin \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}+{\it Artanh} \left ({\frac{\cos \left ( dx+c \right ) +1+\sin \left ( dx+c \right ) }{2}\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}} \right ) \cos \left ( dx+c \right ) +{\it Artanh} \left ({\frac{\cos \left ( dx+c \right ) +1-\sin \left ( dx+c \right ) }{2}\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}} \right ) \cos \left ( dx+c \right ) -2\,\cos \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}-2\,\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) \left ( \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 2.30355, size = 2531, normalized size = 4.83 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((16*sqrt(2)*e*cos(2*d*x + 2*c) + 16*I*sqrt(2)*e*sin(2*d*x + 2*c) + 16*sqrt(2)*e)*arctan2(sqrt(2)*cos(1/4*arc
tan2(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) + (16*sqrt(2)*e*cos(2*d*x + 2*c) + 16*I*sqrt(2)*e*sin(2*d*x + 2*c) + 16*sqrt(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) + (16*sqrt(2)*e*cos(2*d*x + 2*c) + 16*I*sqrt(2)*e*sin(2*d*x + 2*c) + 16*sqrt(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) + (16*sqrt(2)*e*cos(2*d*x + 2*c) + 16*I*sqrt(2)*e*sin(2*d*x + 2*c) + 16*sqrt(2)*e)*arctan2(sqrt(2)*c
os(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) + (16*I*sqrt(2)*e*cos(2*d*x + 2*c) - 16*sqrt(2)*e*sin(2*d*x + 2*c) + 16*I*sqrt(2)*e)*arctan2(sqr
t(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) + (-16*I*sqrt(2)*e*cos(2*d*x + 2*c) + 16*sqrt(2)*e*sin(2*d*x + 2*c) - 16*I*sqrt(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) - 128*e*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + (8*sqrt(2)*e*cos(2*d*x + 2*c
) + 8*I*sqrt(2)*e*sin(2*d*x + 2*c) + 8*sqrt(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*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*arct
an2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) - (8*sqrt(2)*e*cos(2*d*x + 2*c) + 8*I*sqrt(2)*e*sin(2*d*x + 2*c)
+ 8*sqrt(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*arct
an2(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*si
n(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) + (8*I*sqrt(2)*e*cos(2*d*x + 2*c) - 8*sqrt(2)*e*sin(2*d*x + 2*c) + 8*I*sqrt(2)*e)*log(2*cos(1/4*a
rctan2(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*s
qrt(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) + (-8*I*sqrt(2)*e*cos(2*d*x + 2*c) + 8*sqrt(2)*e*sin(2*d*x + 2*c) - 8*I*sqrt(2)*e)*log(2*c
os(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) + (8*I*sqrt(2)*e*cos(2*d*x + 2*c) - 8*sqrt(2)*e*sin(2*d*x + 2*c) + 8*I*sqrt(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) + (-8*I*sqrt(2)*e*cos(2*d*x + 2*c) + 8*sqrt(2)*e*sin(2*d*x + 2*c) - 8*I*sqr
t(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) - 128*I*e*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*sqr
t(a)*sqrt(e)/(d*(-64*I*cos(2*d*x + 2*c) + 64*sin(2*d*x + 2*c) - 64*I))

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Fricas [A]  time = 2.21191, size = 1427, normalized size = 2.72 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e \sec \left (d x + c\right )\right )^{\frac{3}{2}} \sqrt{i \, a \tan \left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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