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

Optimal. Leaf size=92 \[ \frac{2 \cos ^{\frac{3}{2}}(c+d x) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^2 d (e \cos (c+d x))^{3/2}}+\frac{4 i \cos ^2(c+d x)}{5 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \cos (c+d x))^{3/2}} \]

[Out]

(2*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2])/(5*a^2*d*(e*Cos[c + d*x])^(3/2)) + (((4*I)/5)*Cos[c + d*x]^2)
/(d*(e*Cos[c + d*x])^(3/2)*(a^2 + I*a^2*Tan[c + d*x]))

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Rubi [A]  time = 0.15422, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3515, 3500, 3771, 2639} \[ \frac{2 \cos ^{\frac{3}{2}}(c+d x) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^2 d (e \cos (c+d x))^{3/2}}+\frac{4 i \cos ^2(c+d x)}{5 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2])/(5*a^2*d*(e*Cos[c + d*x])^(3/2)) + (((4*I)/5)*Cos[c + d*x]^2)
/(d*(e*Cos[c + d*x])^(3/2)*(a^2 + I*a^2*Tan[c + d*x]))

Rule 3515

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(d*Co
s[e + f*x])^m*(d*Sec[e + f*x])^m, Int[(a + b*Tan[e + f*x])^n/(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e,
f, m, n}, x] &&  !IntegerQ[m]

Rule 3500

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(2*d^2
*(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 1))/(b*f*(m + 2*n)), x] - Dist[(d^2*(m - 2))/(b^2*(m + 2*n
)), Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a
^2 + b^2, 0] && LtQ[n, -1] && ((ILtQ[n/2, 0] && IGtQ[m - 1/2, 0]) || EqQ[n, -2] || IGtQ[m + n, 0] || (Integers
Q[n, m + 1/2] && GtQ[2*m + n + 1, 0])) && IntegerQ[2*m]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rubi steps

\begin{align*} \int \frac{1}{(e \cos (c+d x))^{3/2} (a+i a \tan (c+d x))^2} \, dx &=\frac{\int \frac{(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^2} \, dx}{(e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}}\\ &=\frac{4 i \cos ^2(c+d x)}{5 d (e \cos (c+d x))^{3/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{e^2 \int \frac{1}{\sqrt{e \sec (c+d x)}} \, dx}{5 a^2 (e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}}\\ &=\frac{4 i \cos ^2(c+d x)}{5 d (e \cos (c+d x))^{3/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{\cos ^{\frac{3}{2}}(c+d x) \int \sqrt{\cos (c+d x)} \, dx}{5 a^2 (e \cos (c+d x))^{3/2}}\\ &=\frac{2 \cos ^{\frac{3}{2}}(c+d x) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^2 d (e \cos (c+d x))^{3/2}}+\frac{4 i \cos ^2(c+d x)}{5 d (e \cos (c+d x))^{3/2} \left (a^2+i a^2 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 0.675349, size = 244, normalized size = 2.65 \[ \frac{(\sin (c+d x)+i \cos (c+d x)) \left (i \sin (2 (c+d x))+3 \cos (2 (c+d x))-2 \sqrt{\sin (c+d x)-i \cos (c+d x)+1} \sqrt{\sin (c+d x)+i \sin (2 (c+d x))-i \cos (c+d x)+\cos (2 (c+d x))} F\left (\left .\sin ^{-1}\left (\sqrt{\sin (c+d x)-i \cos (c+d x)}\right )\right |-1\right )+2 \sqrt{\sin (c+d x)-i \cos (c+d x)+1} \sqrt{\sin (c+d x)+i \sin (2 (c+d x))-i \cos (c+d x)+\cos (2 (c+d x))} E\left (\left .\sin ^{-1}\left (\sqrt{\sin (c+d x)-i \cos (c+d x)}\right )\right |-1\right )+3\right )}{5 a^2 d e \sqrt{e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I*Cos[c + d*x] + Sin[c + d*x])*(3 + 3*Cos[2*(c + d*x)] + 2*EllipticE[ArcSin[Sqrt[(-I)*Cos[c + d*x] + Sin[c +
d*x]]], -1]*Sqrt[1 - I*Cos[c + d*x] + Sin[c + d*x]]*Sqrt[(-I)*Cos[c + d*x] + Cos[2*(c + d*x)] + Sin[c + d*x]
+ I*Sin[2*(c + d*x)]] - 2*EllipticF[ArcSin[Sqrt[(-I)*Cos[c + d*x] + Sin[c + d*x]]], -1]*Sqrt[1 - I*Cos[c + d*x
] + Sin[c + d*x]]*Sqrt[(-I)*Cos[c + d*x] + Cos[2*(c + d*x)] + Sin[c + d*x] + I*Sin[2*(c + d*x)]] + I*Sin[2*(c
+ d*x)]))/(5*a^2*d*e*Sqrt[e*Cos[c + d*x]])

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Maple [A]  time = 2.392, size = 207, normalized size = 2.3 \begin{align*} -{\frac{2}{5\,{a}^{2}ed} \left ( 16\,i \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}-16\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}\cos \left ( 1/2\,dx+c/2 \right ) -24\,i \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}+16\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}\cos \left ( 1/2\,dx+c/2 \right ) +12\,i \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}-{\it EllipticE} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) ,\sqrt{2} \right ) \sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}-4\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) -2\,i\sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}e+e}}}} \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))^2,x)

[Out]

-2/5/e/a^2/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*(16*I*sin(1/2*d*x+1/2*c)^7-16*sin(1/2*d*x+1/
2*c)^6*cos(1/2*d*x+1/2*c)-24*I*sin(1/2*d*x+1/2*c)^5+16*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+12*I*sin(1/2*d*
x+1/2*c)^3-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)
-4*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-2*I*sin(1/2*d*x+1/2*c))/d

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \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))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (5 \, a^{2} d e^{2} e^{\left (2 i \, d x + 2 i \, c\right )}{\rm integral}\left (-\frac{2 i \, \sqrt{\frac{1}{2}} \sqrt{e e^{\left (2 i \, d x + 2 i \, c\right )} + e} e^{\left (\frac{1}{2} i \, d x + \frac{1}{2} i \, c\right )}}{5 \,{\left (a^{2} d e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d e^{2}\right )}}, x\right ) + \sqrt{\frac{1}{2}} \sqrt{e e^{\left (2 i \, d x + 2 i \, c\right )} + e}{\left (4 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i\right )} e^{\left (-\frac{1}{2} i \, d x - \frac{1}{2} i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{5 \, a^{2} d e^{2}} \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))^2,x, algorithm="fricas")

[Out]

1/5*(5*a^2*d*e^2*e^(2*I*d*x + 2*I*c)*integral(-2/5*I*sqrt(1/2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*e^(1/2*I*d*x +
1/2*I*c)/(a^2*d*e^2*e^(2*I*d*x + 2*I*c) + a^2*d*e^2), x) + sqrt(1/2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*(4*I*e^(2
*I*d*x + 2*I*c) + 2*I)*e^(-1/2*I*d*x - 1/2*I*c))*e^(-2*I*d*x - 2*I*c)/(a^2*d*e^2)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \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))**2,x)

[Out]

Exception raised: AttributeError

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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}}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}}\,{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))^2,x, algorithm="giac")

[Out]

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