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

Optimal. Leaf size=364 \[ \frac{3 i a^{3/2} \sqrt{e} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{\sqrt{2} d}-\frac{3 i a^{3/2} \sqrt{e} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{\sqrt{2} d}-\frac{3 i a^{3/2} \sqrt{e} \log \left (-\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{2 \sqrt{2} d}+\frac{3 i a^{3/2} \sqrt{e} \log \left (\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{2 \sqrt{2} d}+\frac{i a \sqrt{a+i a \tan (c+d x)} \sqrt{e \sec (c+d x)}}{d} \]

[Out]

((3*I)*a^(3/2)*Sqrt[e]*ArcTan[1 - (Sqrt[2]*Sqrt[e]*Sqrt[a + I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e*Sec[c + d*x]])]
)/(Sqrt[2]*d) - ((3*I)*a^(3/2)*Sqrt[e]*ArcTan[1 + (Sqrt[2]*Sqrt[e]*Sqrt[a + I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e
*Sec[c + d*x]])])/(Sqrt[2]*d) - (((3*I)/2)*a^(3/2)*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])])/(Sqrt[2]*d) + (((3*I)/2)*a^(3/2)*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*Ta
n[c + d*x])])/(Sqrt[2]*d) + (I*a*Sqrt[e*Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/d

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Rubi [A]  time = 0.315121, antiderivative size = 364, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {3498, 3495, 297, 1162, 617, 204, 1165, 628} \[ \frac{3 i a^{3/2} \sqrt{e} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{\sqrt{2} d}-\frac{3 i a^{3/2} \sqrt{e} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{\sqrt{2} d}-\frac{3 i a^{3/2} \sqrt{e} \log \left (-\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{2 \sqrt{2} d}+\frac{3 i a^{3/2} \sqrt{e} \log \left (\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{2 \sqrt{2} d}+\frac{i a \sqrt{a+i a \tan (c+d x)} \sqrt{e \sec (c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*I)*a^(3/2)*Sqrt[e]*ArcTan[1 - (Sqrt[2]*Sqrt[e]*Sqrt[a + I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e*Sec[c + d*x]])]
)/(Sqrt[2]*d) - ((3*I)*a^(3/2)*Sqrt[e]*ArcTan[1 + (Sqrt[2]*Sqrt[e]*Sqrt[a + I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e
*Sec[c + d*x]])])/(Sqrt[2]*d) - (((3*I)/2)*a^(3/2)*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])])/(Sqrt[2]*d) + (((3*I)/2)*a^(3/2)*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*Ta
n[c + d*x])])/(Sqrt[2]*d) + (I*a*Sqrt[e*Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/d

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 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 \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2} \, dx &=\frac{i a \sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}}{d}+\frac{1}{2} (3 a) \int \sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)} \, dx\\ &=\frac{i a \sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}}{d}-\frac{\left (6 i a^2 e^2\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}\\ &=\frac{i a \sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}}{d}+\frac{\left (3 i a^2 e\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}-\frac{\left (3 i a^2 e\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}\\ &=\frac{i a \sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}}{d}-\frac{\left (3 i a^2\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}-\frac{\left (3 i a^2\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}-\frac{\left (3 i a^{3/2} \sqrt{e}\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}-\frac{\left (3 i a^{3/2} \sqrt{e}\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}\\ &=-\frac{3 i a^{3/2} \sqrt{e} \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 )}{2 \sqrt{2} d}+\frac{3 i a^{3/2} \sqrt{e} \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 )}{2 \sqrt{2} d}+\frac{i a \sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}}{d}-\frac{\left (3 i a^{3/2} \sqrt{e}\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}+\frac{\left (3 i a^{3/2} \sqrt{e}\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}\\ &=\frac{3 i a^{3/2} \sqrt{e} \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}-\frac{3 i a^{3/2} \sqrt{e} \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}-\frac{3 i a^{3/2} \sqrt{e} \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 )}{2 \sqrt{2} d}+\frac{3 i a^{3/2} \sqrt{e} \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 )}{2 \sqrt{2} d}+\frac{i a \sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}}{d}\\ \end{align*}

Mathematica [A]  time = 1.74517, size = 348, normalized size = 0.96 \[ \frac{a e \sqrt{a+i a \tan (c+d x)} \left (\sqrt{-\sin (c)+i \cos (c)-1} \left (3 \sqrt{\sin (c)+i \cos (c)-1} \sqrt{\tan \left (\frac{d x}{2}\right )+i} \tan ^{-1}\left (\frac{\sqrt{\sin (c)-i \cos (c)-1} \sqrt{-\tan \left (\frac{d x}{2}\right )+i}}{\sqrt{\sin (c)+i \cos (c)-1} \sqrt{\tan \left (\frac{d x}{2}\right )+i}}\right )+i \sqrt{\sin (c)-i \cos (c)-1} \sqrt{-\tan \left (\frac{d x}{2}\right )+i} \sec (c+d x)\right )-3 \sqrt{-\sin (c)-i \cos (c)-1} \sqrt{\sin (c)-i \cos (c)-1} \sqrt{\tan \left (\frac{d x}{2}\right )+i} \tan ^{-1}\left (\frac{\sqrt{-\sin (c)+i \cos (c)-1} \sqrt{-\tan \left (\frac{d x}{2}\right )+i}}{\sqrt{-\sin (c)-i \cos (c)-1} \sqrt{\tan \left (\frac{d x}{2}\right )+i}}\right )\right )}{d \sqrt{-\sin (c)+i \cos (c)-1} \sqrt{\sin (c)-i \cos (c)-1} \sqrt{-\tan \left (\frac{d x}{2}\right )+i} \sqrt{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*e*(-3*ArcTan[(Sqrt[-1 + I*Cos[c] - Sin[c]]*Sqrt[I - Tan[(d*x)/2]])/(Sqrt[-1 - I*Cos[c] - Sin[c]]*Sqrt[I + T
an[(d*x)/2]])]*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]]*(I*Sec[c + d*x]*Sqrt[-1 - I*Cos[c] + Sin[c]]*Sqrt[I - Tan[(d*x)/2]] + 3*ArcTan[(Sqrt[-1 - I*C
os[c] + Sin[c]]*Sqrt[I - Tan[(d*x)/2]])/(Sqrt[-1 + I*Cos[c] + Sin[c]]*Sqrt[I + Tan[(d*x)/2]])]*Sqrt[-1 + I*Cos
[c] + Sin[c]]*Sqrt[I + Tan[(d*x)/2]]))*Sqrt[a + I*a*Tan[c + d*x]])/(d*Sqrt[e*Sec[c + d*x]]*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 = 0.336, size = 304, normalized size = 0.8 \begin{align*} -{\frac{a \left ( \cos \left ( dx+c \right ) -1 \right ) }{2\,d\sin \left ( dx+c \right ) \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) -1 \right ) }\sqrt{{\frac{e}{\cos \left ( dx+c \right ) }}}\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}} \left ( 3\,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 ) -3\,i\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 ) -2\,i\sin \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}-3\,{\it Artanh} \left ( 1/2\,\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \left ( \cos \left ( dx+c \right ) +1+\sin \left ( dx+c \right ) \right ) \right ) \cos \left ( dx+c \right ) -2\,\cos \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}-3\,{\it Artanh} \left ( 1/2\,\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \left ( \cos \left ( dx+c \right ) +1-\sin \left ( dx+c \right ) \right ) \right ) \cos \left ( dx+c \right ) -2\,\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ){\frac{1}{\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 2.34972, size = 2539, normalized size = 6.98 \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))^(1/2)*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

-((48*sqrt(2)*a*cos(2*d*x + 2*c) + 48*I*sqrt(2)*a*sin(2*d*x + 2*c) + 48*sqrt(2)*a)*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) + (48*sqrt(2)*a*cos(2*d*x + 2*c) + 48*I*sqrt(2)*a*sin(2*d*x + 2*c) + 48*sqrt(2)*a)*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) + (48*sqrt(2)*a*cos(2*d*x + 2*c) + 48*I*sqrt(2)*a*sin(2*d*x + 2*c) + 48*sqrt(2)*a)*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) + (48*sqrt(2)*a*cos(2*d*x + 2*c) + 48*I*sqrt(2)*a*sin(2*d*x + 2*c) + 48*sqrt(2)*a)*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) - (48*I*sqrt(2)*a*cos(2*d*x + 2*c) - 48*sqrt(2)*a*sin(2*d*x + 2*c) + 48*I*sqrt(2)*a)*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) - (-48*I*sqrt(2)*a*cos(2*d*x + 2*c) + 48*sqrt(2)*a*sin(2*d*x + 2*c) - 48*I*sqrt(2)*a)*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*a*cos(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - (24*sqrt(2)*a*cos(2*d*x + 2*
c) + 24*I*sqrt(2)*a*sin(2*d*x + 2*c) + 24*sqrt(2)*a)*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*a
rctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + (24*sqrt(2)*a*cos(2*d*x + 2*c) + 24*I*sqrt(2)*a*sin(2*d*x +
2*c) + 24*sqrt(2)*a)*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*co
s(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) - (-24*I*sqrt(2)*a*cos(2*d*x + 2*c) + 24*sqrt(2)*a*sin(2*d*x + 2*c) - 24*I*sqrt(2)*a)*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) - (24*I*sqrt(2)*a*cos(2*d*x + 2*c) - 24*sqrt(2)*a*sin(2*d*x + 2*c) + 24*I*sqrt(2
)*a)*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(si
n(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2) - (-24*I*sqrt(2)*a*cos(2*d*x + 2*c) + 24*sqrt(2)*a*sin(2*d*x + 2*c) -
24*I*sqrt(2)*a)*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) - (24*I*sqrt(2)*a*cos(2*d*x + 2*c) - 24*sqrt(2)*a*sin(2*d*x
+ 2*c) + 24*I*sqrt(2)*a)*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*a*sin(3/4*arctan2(sin(2*d*x + 2*c), cos(2
*d*x + 2*c))))*sqrt(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 = 2.13648, size = 1376, normalized size = 3.78 \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))^(1/2)*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/2*(4*I*a*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(
9*I*a^3*e/d^2)*d*log(1/3*(6*(a*e^(2*I*d*x + 2*I*c) + a)*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) + 2*I*sqrt(9*I*a^3*e/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/a)
+ sqrt(9*I*a^3*e/d^2)*d*log(1/3*(6*(a*e^(2*I*d*x + 2*I*c) + a)*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) - 2*I*sqrt(9*I*a^3*e/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2
*I*c)/a) - sqrt(-9*I*a^3*e/d^2)*d*log(2/3*(3*(a*e^(2*I*d*x + 2*I*c) + a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqr
t(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(3/2*I*d*x + 3/2*I*c) + I*sqrt(-9*I*a^3*e/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I
*d*x - 2*I*c)/a) + sqrt(-9*I*a^3*e/d^2)*d*log(2/3*(3*(a*e^(2*I*d*x + 2*I*c) + a)*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) - I*sqrt(-9*I*a^3*e/d^2)*d*e^(2*I*d*x + 2*I*c))
*e^(-2*I*d*x - 2*I*c)/a))/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))**(1/2)*(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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