3.685 \(\int \frac{1}{(e \cos (c+d x))^{3/2} \sqrt{a+i a \tan (c+d x)}} \, dx\)

Optimal. Leaf size=495 \[ -\frac{i \sqrt{2} \sqrt{a} \sec (c+d x) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{a-i a \tan (c+d x)} \sqrt{e \cos (c+d x)}}{\sqrt{a} \sqrt{e}}\right )}{d e^{3/2} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{i \sqrt{2} \sqrt{a} \sec (c+d x) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{a-i a \tan (c+d x)} \sqrt{e \cos (c+d x)}}{\sqrt{a} \sqrt{e}}\right )}{d e^{3/2} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{i \sqrt{a} \sec (c+d x) \log \left (-\sqrt{2} \sqrt{a} \sqrt{a-i a \tan (c+d x)} \sqrt{e \cos (c+d x)}+\sqrt{e} \cos (c+d x) (a-i a \tan (c+d x))+a \sqrt{e}\right )}{\sqrt{2} d e^{3/2} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{i \sqrt{a} \sec (c+d x) \log \left (\sqrt{2} \sqrt{a} \sqrt{a-i a \tan (c+d x)} \sqrt{e \cos (c+d x)}+\sqrt{e} \cos (c+d x) (a-i a \tan (c+d x))+a \sqrt{e}\right )}{\sqrt{2} d e^{3/2} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}} \]

[Out]

((-I)*Sqrt[2]*Sqrt[a]*ArcTan[1 - (Sqrt[2]*Sqrt[e*Cos[c + d*x]]*Sqrt[a - I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e])]*
Sec[c + d*x])/(d*e^(3/2)*Sqrt[a - I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) + (I*Sqrt[2]*Sqrt[a]*ArcTan[1
+ (Sqrt[2]*Sqrt[e*Cos[c + d*x]]*Sqrt[a - I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e])]*Sec[c + d*x])/(d*e^(3/2)*Sqrt[a
- I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) + (I*Sqrt[a]*Log[a*Sqrt[e] - Sqrt[2]*Sqrt[a]*Sqrt[e*Cos[c + d
*x]]*Sqrt[a - I*a*Tan[c + d*x]] + Sqrt[e]*Cos[c + d*x]*(a - I*a*Tan[c + d*x])]*Sec[c + d*x])/(Sqrt[2]*d*e^(3/2
)*Sqrt[a - I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (I*Sqrt[a]*Log[a*Sqrt[e] + Sqrt[2]*Sqrt[a]*Sqrt[e*C
os[c + d*x]]*Sqrt[a - I*a*Tan[c + d*x]] + Sqrt[e]*Cos[c + d*x]*(a - I*a*Tan[c + d*x])]*Sec[c + d*x])/(Sqrt[2]*
d*e^(3/2)*Sqrt[a - I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])

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Rubi [A]  time = 0.332187, antiderivative size = 495, 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 = {3514, 3513, 297, 1162, 617, 204, 1165, 628} \[ -\frac{i \sqrt{2} \sqrt{a} \sec (c+d x) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{a-i a \tan (c+d x)} \sqrt{e \cos (c+d x)}}{\sqrt{a} \sqrt{e}}\right )}{d e^{3/2} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{i \sqrt{2} \sqrt{a} \sec (c+d x) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{a-i a \tan (c+d x)} \sqrt{e \cos (c+d x)}}{\sqrt{a} \sqrt{e}}\right )}{d e^{3/2} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{i \sqrt{a} \sec (c+d x) \log \left (-\sqrt{2} \sqrt{a} \sqrt{a-i a \tan (c+d x)} \sqrt{e \cos (c+d x)}+\sqrt{e} \cos (c+d x) (a-i a \tan (c+d x))+a \sqrt{e}\right )}{\sqrt{2} d e^{3/2} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{i \sqrt{a} \sec (c+d x) \log \left (\sqrt{2} \sqrt{a} \sqrt{a-i a \tan (c+d x)} \sqrt{e \cos (c+d x)}+\sqrt{e} \cos (c+d x) (a-i a \tan (c+d x))+a \sqrt{e}\right )}{\sqrt{2} d e^{3/2} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/((e*Cos[c + d*x])^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]),x]

[Out]

((-I)*Sqrt[2]*Sqrt[a]*ArcTan[1 - (Sqrt[2]*Sqrt[e*Cos[c + d*x]]*Sqrt[a - I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e])]*
Sec[c + d*x])/(d*e^(3/2)*Sqrt[a - I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) + (I*Sqrt[2]*Sqrt[a]*ArcTan[1
+ (Sqrt[2]*Sqrt[e*Cos[c + d*x]]*Sqrt[a - I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e])]*Sec[c + d*x])/(d*e^(3/2)*Sqrt[a
- I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) + (I*Sqrt[a]*Log[a*Sqrt[e] - Sqrt[2]*Sqrt[a]*Sqrt[e*Cos[c + d
*x]]*Sqrt[a - I*a*Tan[c + d*x]] + Sqrt[e]*Cos[c + d*x]*(a - I*a*Tan[c + d*x])]*Sec[c + d*x])/(Sqrt[2]*d*e^(3/2
)*Sqrt[a - I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (I*Sqrt[a]*Log[a*Sqrt[e] + Sqrt[2]*Sqrt[a]*Sqrt[e*C
os[c + d*x]]*Sqrt[a - I*a*Tan[c + d*x]] + Sqrt[e]*Cos[c + d*x]*(a - I*a*Tan[c + d*x])]*Sec[c + d*x])/(Sqrt[2]*
d*e^(3/2)*Sqrt[a - I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])

Rule 3514

Int[1/((cos[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)*Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[1/
(d*Cos[e + f*x]*Sqrt[a - b*Tan[e + f*x]]*Sqrt[a + b*Tan[e + f*x]]), Int[Sqrt[a - b*Tan[e + f*x]]/Sqrt[d*Cos[e
+ f*x]], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0]

Rule 3513

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[cos[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Dist[(-4*b)/f
, Subst[Int[x^2/(a^2*d^2 + x^4), x], x, Sqrt[d*Cos[e + f*x]]*Sqrt[a + b*Tan[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 \frac{1}{(e \cos (c+d x))^{3/2} \sqrt{a+i a \tan (c+d x)}} \, dx &=\frac{\sec (c+d x) \int \frac{\sqrt{a-i a \tan (c+d x)}}{\sqrt{e \cos (c+d x)}} \, dx}{e \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}\\ &=\frac{(4 i a \sec (c+d x)) \operatorname{Subst}\left (\int \frac{x^2}{a^2 e^2+x^4} \, dx,x,\sqrt{e \cos (c+d x)} \sqrt{a-i a \tan (c+d x)}\right )}{d e \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}\\ &=-\frac{(2 i a \sec (c+d x)) \operatorname{Subst}\left (\int \frac{a e-x^2}{a^2 e^2+x^4} \, dx,x,\sqrt{e \cos (c+d x)} \sqrt{a-i a \tan (c+d x)}\right )}{d e \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{(2 i a \sec (c+d x)) \operatorname{Subst}\left (\int \frac{a e+x^2}{a^2 e^2+x^4} \, dx,x,\sqrt{e \cos (c+d x)} \sqrt{a-i a \tan (c+d x)}\right )}{d e \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}\\ &=\frac{\left (i \sqrt{a} \sec (c+d x)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2} \sqrt{a} \sqrt{e}+2 x}{-a e-\sqrt{2} \sqrt{a} \sqrt{e} x-x^2} \, dx,x,\sqrt{e \cos (c+d x)} \sqrt{a-i a \tan (c+d x)}\right )}{\sqrt{2} d e^{3/2} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{\left (i \sqrt{a} \sec (c+d x)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2} \sqrt{a} \sqrt{e}-2 x}{-a e+\sqrt{2} \sqrt{a} \sqrt{e} x-x^2} \, dx,x,\sqrt{e \cos (c+d x)} \sqrt{a-i a \tan (c+d x)}\right )}{\sqrt{2} d e^{3/2} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{(i a \sec (c+d x)) \operatorname{Subst}\left (\int \frac{1}{a e-\sqrt{2} \sqrt{a} \sqrt{e} x+x^2} \, dx,x,\sqrt{e \cos (c+d x)} \sqrt{a-i a \tan (c+d x)}\right )}{d e \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{(i a \sec (c+d x)) \operatorname{Subst}\left (\int \frac{1}{a e+\sqrt{2} \sqrt{a} \sqrt{e} x+x^2} \, dx,x,\sqrt{e \cos (c+d x)} \sqrt{a-i a \tan (c+d x)}\right )}{d e \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}\\ &=\frac{i \sqrt{a} \log \left (a \sqrt{e}-\sqrt{2} \sqrt{a} \sqrt{e \cos (c+d x)} \sqrt{a-i a \tan (c+d x)}+\sqrt{e} \cos (c+d x) (a-i a \tan (c+d x))\right ) \sec (c+d x)}{\sqrt{2} d e^{3/2} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{i \sqrt{a} \log \left (a \sqrt{e}+\sqrt{2} \sqrt{a} \sqrt{e \cos (c+d x)} \sqrt{a-i a \tan (c+d x)}+\sqrt{e} \cos (c+d x) (a-i a \tan (c+d x))\right ) \sec (c+d x)}{\sqrt{2} d e^{3/2} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{\left (i \sqrt{2} \sqrt{a} \sec (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt{e \cos (c+d x)} \sqrt{a-i a \tan (c+d x)}}{\sqrt{a} \sqrt{e}}\right )}{d e^{3/2} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{\left (i \sqrt{2} \sqrt{a} \sec (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt{e \cos (c+d x)} \sqrt{a-i a \tan (c+d x)}}{\sqrt{a} \sqrt{e}}\right )}{d e^{3/2} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}\\ &=-\frac{i \sqrt{2} \sqrt{a} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \cos (c+d x)} \sqrt{a-i a \tan (c+d x)}}{\sqrt{a} \sqrt{e}}\right ) \sec (c+d x)}{d e^{3/2} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{i \sqrt{2} \sqrt{a} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e \cos (c+d x)} \sqrt{a-i a \tan (c+d x)}}{\sqrt{a} \sqrt{e}}\right ) \sec (c+d x)}{d e^{3/2} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{i \sqrt{a} \log \left (a \sqrt{e}-\sqrt{2} \sqrt{a} \sqrt{e \cos (c+d x)} \sqrt{a-i a \tan (c+d x)}+\sqrt{e} \cos (c+d x) (a-i a \tan (c+d x))\right ) \sec (c+d x)}{\sqrt{2} d e^{3/2} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{i \sqrt{a} \log \left (a \sqrt{e}+\sqrt{2} \sqrt{a} \sqrt{e \cos (c+d x)} \sqrt{a-i a \tan (c+d x)}+\sqrt{e} \cos (c+d x) (a-i a \tan (c+d x))\right ) \sec (c+d x)}{\sqrt{2} d e^{3/2} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 9.51568, size = 209, normalized size = 0.42 \[ \frac{i e^{\frac{1}{2} i (c+d x)} \left (\log \left (-\sqrt{2} e^{\frac{1}{2} i (c+d x)}+e^{i (c+d x)}+1\right )-\log \left (\sqrt{2} e^{\frac{1}{2} i (c+d x)}+e^{i (c+d x)}+1\right )+2 \tan ^{-1}\left (1-\sqrt{2} e^{\frac{1}{2} i (c+d x)}\right )-2 \tan ^{-1}\left (1+\sqrt{2} e^{\frac{1}{2} i (c+d x)}\right )\right )}{\sqrt{2} d e \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{e e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((e*Cos[c + d*x])^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]),x]

[Out]

(I*E^((I/2)*(c + d*x))*(2*ArcTan[1 - Sqrt[2]*E^((I/2)*(c + d*x))] - 2*ArcTan[1 + Sqrt[2]*E^((I/2)*(c + d*x))]
+ Log[1 - Sqrt[2]*E^((I/2)*(c + d*x)) + E^(I*(c + d*x))] - Log[1 + Sqrt[2]*E^((I/2)*(c + d*x)) + E^(I*(c + d*x
))]))/(Sqrt[2]*d*e*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[(e*(1 + E^((2*I)*(c + d*x))))/
E^(I*(c + d*x))])

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Maple [A]  time = 0.362, size = 232, normalized size = 0.5 \begin{align*}{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) -1 \right ) ^{2}}{ad \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) -1 \right ) }\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 ) +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 ) -{\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 ) +{\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 ) \right ) \left ( e\cos \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}} \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(1/(e*cos(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(1/2),x)

[Out]

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

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Maxima [A]  time = 3.32039, size = 964, normalized size = 1.95 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*I*sqrt(2)*arctan2(sqrt(2)*cos(1/2*d*x + 1/2*c) + 1, sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) + 2*I*sqrt(2)*ar
ctan2(sqrt(2)*cos(1/2*d*x + 1/2*c) + 1, -sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) + 2*I*sqrt(2)*arctan2(sqrt(2)*cos(1
/2*d*x + 1/2*c) - 1, sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) + 2*I*sqrt(2)*arctan2(sqrt(2)*cos(1/2*d*x + 1/2*c) - 1,
-sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) - 2*sqrt(2)*arctan2(sqrt(2)*sin(1/2*d*x + 1/2*c) + sin(d*x + c), sqrt(2)*c
os(1/2*d*x + 1/2*c) + cos(d*x + c) + 1) + 2*sqrt(2)*arctan2(-sqrt(2)*sin(1/2*d*x + 1/2*c) + sin(d*x + c), -sqr
t(2)*cos(1/2*d*x + 1/2*c) + cos(d*x + c) + 1) + I*sqrt(2)*log(2*sqrt(2)*sin(d*x + c)*sin(1/2*d*x + 1/2*c) + 2*
(sqrt(2)*cos(1/2*d*x + 1/2*c) + 1)*cos(d*x + c) + cos(d*x + c)^2 + 2*cos(1/2*d*x + 1/2*c)^2 + sin(d*x + c)^2 +
2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 1) - I*sqrt(2)*log(-2*sqrt(2)*sin(d*x + c)*sin(1/
2*d*x + 1/2*c) - 2*(sqrt(2)*cos(1/2*d*x + 1/2*c) - 1)*cos(d*x + c) + cos(d*x + c)^2 + 2*cos(1/2*d*x + 1/2*c)^2
+ sin(d*x + c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 1) - sqrt(2)*log(2*cos(1/2*d*x
+ 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2)
+ sqrt(2)*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)
*sin(1/2*d*x + 1/2*c) + 2) - sqrt(2)*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1
/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + sqrt(2)*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x +
1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2))/(sqrt(a)*d*e^(3/2))

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Fricas [A]  time = 2.29791, size = 973, normalized size = 1.97 \begin{align*} \frac{1}{2} \, \sqrt{\frac{4 i}{a d^{2} e^{3}}} \log \left (\frac{1}{2} i \, a d e^{2} \sqrt{\frac{4 i}{a d^{2} e^{3}}} + \sqrt{2} \sqrt{\frac{1}{2}} \sqrt{e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt{\frac{a}{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 )}\right ) - \frac{1}{2} \, \sqrt{\frac{4 i}{a d^{2} e^{3}}} \log \left (-\frac{1}{2} i \, a d e^{2} \sqrt{\frac{4 i}{a d^{2} e^{3}}} + \sqrt{2} \sqrt{\frac{1}{2}} \sqrt{e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt{\frac{a}{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 )}\right ) + \frac{1}{2} \, \sqrt{-\frac{4 i}{a d^{2} e^{3}}} \log \left (\frac{1}{2} i \, a d e^{2} \sqrt{-\frac{4 i}{a d^{2} e^{3}}} + \sqrt{2} \sqrt{\frac{1}{2}} \sqrt{e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt{\frac{a}{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 )}\right ) - \frac{1}{2} \, \sqrt{-\frac{4 i}{a d^{2} e^{3}}} \log \left (-\frac{1}{2} i \, a d e^{2} \sqrt{-\frac{4 i}{a d^{2} e^{3}}} + \sqrt{2} \sqrt{\frac{1}{2}} \sqrt{e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt{\frac{a}{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 )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(1/(e*cos(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 \frac{1}{\left (e \cos \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(1/(e*cos(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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