3.672 \(\int \frac{1}{(e \cos (c+d x))^{11/2} (a+i a \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=164 \[ \frac{14 \sin (c+d x) \cos ^5(c+d x)}{5 a^2 d (e \cos (c+d x))^{11/2}}+\frac{14 \sin (c+d x) \cos ^3(c+d x)}{15 a^2 d (e \cos (c+d x))^{11/2}}-\frac{4 i \cos ^2(c+d x)}{3 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \cos (c+d x))^{11/2}}-\frac{14 \cos ^{\frac{11}{2}}(c+d x) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^2 d (e \cos (c+d x))^{11/2}} \]

[Out]

(-14*Cos[c + d*x]^(11/2)*EllipticE[(c + d*x)/2, 2])/(5*a^2*d*(e*Cos[c + d*x])^(11/2)) + (14*Cos[c + d*x]^3*Sin
[c + d*x])/(15*a^2*d*(e*Cos[c + d*x])^(11/2)) + (14*Cos[c + d*x]^5*Sin[c + d*x])/(5*a^2*d*(e*Cos[c + d*x])^(11
/2)) - (((4*I)/3)*Cos[c + d*x]^2)/(d*(e*Cos[c + d*x])^(11/2)*(a^2 + I*a^2*Tan[c + d*x]))

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Rubi [A]  time = 0.190808, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3515, 3500, 3768, 3771, 2639} \[ \frac{14 \sin (c+d x) \cos ^5(c+d x)}{5 a^2 d (e \cos (c+d x))^{11/2}}+\frac{14 \sin (c+d x) \cos ^3(c+d x)}{15 a^2 d (e \cos (c+d x))^{11/2}}-\frac{4 i \cos ^2(c+d x)}{3 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \cos (c+d x))^{11/2}}-\frac{14 \cos ^{\frac{11}{2}}(c+d x) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^2 d (e \cos (c+d x))^{11/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((e*Cos[c + d*x])^(11/2)*(a + I*a*Tan[c + d*x])^2),x]

[Out]

(-14*Cos[c + d*x]^(11/2)*EllipticE[(c + d*x)/2, 2])/(5*a^2*d*(e*Cos[c + d*x])^(11/2)) + (14*Cos[c + d*x]^3*Sin
[c + d*x])/(15*a^2*d*(e*Cos[c + d*x])^(11/2)) + (14*Cos[c + d*x]^5*Sin[c + d*x])/(5*a^2*d*(e*Cos[c + d*x])^(11
/2)) - (((4*I)/3)*Cos[c + d*x]^2)/(d*(e*Cos[c + d*x])^(11/2)*(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 3768

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

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))^{11/2} (a+i a \tan (c+d x))^2} \, dx &=\frac{\int \frac{(e \sec (c+d x))^{11/2}}{(a+i a \tan (c+d x))^2} \, dx}{(e \cos (c+d x))^{11/2} (e \sec (c+d x))^{11/2}}\\ &=-\frac{4 i \cos ^2(c+d x)}{3 d (e \cos (c+d x))^{11/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{\left (7 e^2\right ) \int (e \sec (c+d x))^{7/2} \, dx}{3 a^2 (e \cos (c+d x))^{11/2} (e \sec (c+d x))^{11/2}}\\ &=\frac{14 \cos ^3(c+d x) \sin (c+d x)}{15 a^2 d (e \cos (c+d x))^{11/2}}-\frac{4 i \cos ^2(c+d x)}{3 d (e \cos (c+d x))^{11/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{\left (7 e^4\right ) \int (e \sec (c+d x))^{3/2} \, dx}{5 a^2 (e \cos (c+d x))^{11/2} (e \sec (c+d x))^{11/2}}\\ &=\frac{14 \cos ^3(c+d x) \sin (c+d x)}{15 a^2 d (e \cos (c+d x))^{11/2}}+\frac{14 \cos ^5(c+d x) \sin (c+d x)}{5 a^2 d (e \cos (c+d x))^{11/2}}-\frac{4 i \cos ^2(c+d x)}{3 d (e \cos (c+d x))^{11/2} \left (a^2+i a^2 \tan (c+d x)\right )}-\frac{\left (7 e^6\right ) \int \frac{1}{\sqrt{e \sec (c+d x)}} \, dx}{5 a^2 (e \cos (c+d x))^{11/2} (e \sec (c+d x))^{11/2}}\\ &=\frac{14 \cos ^3(c+d x) \sin (c+d x)}{15 a^2 d (e \cos (c+d x))^{11/2}}+\frac{14 \cos ^5(c+d x) \sin (c+d x)}{5 a^2 d (e \cos (c+d x))^{11/2}}-\frac{4 i \cos ^2(c+d x)}{3 d (e \cos (c+d x))^{11/2} \left (a^2+i a^2 \tan (c+d x)\right )}-\frac{\left (7 \cos ^{\frac{11}{2}}(c+d x)\right ) \int \sqrt{\cos (c+d x)} \, dx}{5 a^2 (e \cos (c+d x))^{11/2}}\\ &=-\frac{14 \cos ^{\frac{11}{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))^{11/2}}+\frac{14 \cos ^3(c+d x) \sin (c+d x)}{15 a^2 d (e \cos (c+d x))^{11/2}}+\frac{14 \cos ^5(c+d x) \sin (c+d x)}{5 a^2 d (e \cos (c+d x))^{11/2}}-\frac{4 i \cos ^2(c+d x)}{3 d (e \cos (c+d x))^{11/2} \left (a^2+i a^2 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 5.81227, size = 406, normalized size = 2.48 \[ \frac{2 \sqrt{2} \csc (c) e^{3 i c+2 i d x} (\cos (2 c)+i \sin (2 c)) \cos ^{\frac{7}{2}}(c+d x) (\cos (d x)+i \sin (d x))^2 \left (-\frac{1}{2} e^{-2 i c} \sqrt{e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \left (7 \left (1+e^{2 i (c+d x)}\right )^{5/2} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-e^{2 i (c+d x)}\right )+\left (-1+e^{2 i c}\right ) \left (56 e^{2 i (c+d x)}+21 e^{4 i (c+d x)}+47\right )\right )+42 \sqrt{2-2 i e^{i (c+d x)}} \sqrt{e^{i (c+d x)} \left (e^{i (c+d x)}-i\right )} \cos ^{\frac{5}{2}}(c+d x) F\left (\left .\sin ^{-1}\left (\sqrt{\sin (c+d x)-i \cos (c+d x)}\right )\right |-1\right )-42 \sqrt{2-2 i e^{i (c+d x)}} \sqrt{e^{i (c+d x)} \left (e^{i (c+d x)}-i\right )} \cos ^{\frac{5}{2}}(c+d x) E\left (\left .\sin ^{-1}\left (\sqrt{\sin (c+d x)-i \cos (c+d x)}\right )\right |-1\right )\right )}{15 d \left (1+e^{2 i (c+d x)}\right )^3 (a+i a \tan (c+d x))^2 (e \cos (c+d x))^{11/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((e*Cos[c + d*x])^(11/2)*(a + I*a*Tan[c + d*x])^2),x]

[Out]

(2*Sqrt[2]*E^((3*I)*c + (2*I)*d*x)*Cos[c + d*x]^(7/2)*Csc[c]*(-42*Sqrt[2 - (2*I)*E^(I*(c + d*x))]*Sqrt[E^(I*(c
+ d*x))*(-I + E^(I*(c + d*x)))]*Cos[c + d*x]^(5/2)*EllipticE[ArcSin[Sqrt[(-I)*Cos[c + d*x] + Sin[c + d*x]]],
-1] + 42*Sqrt[2 - (2*I)*E^(I*(c + d*x))]*Sqrt[E^(I*(c + d*x))*(-I + E^(I*(c + d*x)))]*Cos[c + d*x]^(5/2)*Ellip
ticF[ArcSin[Sqrt[(-I)*Cos[c + d*x] + Sin[c + d*x]]], -1] - (Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))]*((
-1 + E^((2*I)*c))*(47 + 56*E^((2*I)*(c + d*x)) + 21*E^((4*I)*(c + d*x))) + 7*(1 + E^((2*I)*(c + d*x)))^(5/2)*H
ypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/(2*E^((2*I)*c)))*(Cos[2*c] + I*Sin[2*c])*(Cos[d*x] + I
*Sin[d*x])^2)/(15*d*(1 + E^((2*I)*(c + d*x)))^3*(e*Cos[c + d*x])^(11/2)*(a + I*a*Tan[c + d*x])^2)

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Maple [A]  time = 5.024, size = 321, normalized size = 2. \begin{align*}{\frac{2}{15\,{e}^{5}{a}^{2}d} \left ( -84\,\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+168\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}\cos \left ( 1/2\,dx+c/2 \right ) +84\,\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-168\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}\cos \left ( 1/2\,dx+c/2 \right ) +20\,i \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}-21\,{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}+36\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) -10\,i\sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) \left ( 4\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-4\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{-1} \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))^(11/2)/(a+I*a*tan(d*x+c))^2,x)

[Out]

2/15/(4*sin(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+1)/a^2/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1
/2)/e^5*(-84*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2
))*sin(1/2*d*x+1/2*c)^4+168*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+84*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1
/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2-168*sin(1/2*d*x+1/2*c)^4*cos
(1/2*d*x+1/2*c)+20*I*sin(1/2*d*x+1/2*c)^3-21*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)+36*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-10*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))^(11/2)/(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{\sqrt{\frac{1}{2}} \sqrt{e e^{\left (2 i \, d x + 2 i \, c\right )} + e}{\left (-84 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 224 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 188 i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-\frac{1}{2} i \, d x - \frac{1}{2} i \, c\right )} + 15 \,{\left (a^{2} d e^{6} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a^{2} d e^{6} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, a^{2} d e^{6} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d e^{6}\right )}{\rm integral}\left (\frac{14 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^{6} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d e^{6}\right )}}, x\right )}{15 \,{\left (a^{2} d e^{6} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a^{2} d e^{6} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, a^{2} d e^{6} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d e^{6}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cos(d*x+c))^(11/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/15*(sqrt(1/2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*(-84*I*e^(6*I*d*x + 6*I*c) - 224*I*e^(4*I*d*x + 4*I*c) - 188*I
*e^(2*I*d*x + 2*I*c))*e^(-1/2*I*d*x - 1/2*I*c) + 15*(a^2*d*e^6*e^(6*I*d*x + 6*I*c) + 3*a^2*d*e^6*e^(4*I*d*x +
4*I*c) + 3*a^2*d*e^6*e^(2*I*d*x + 2*I*c) + a^2*d*e^6)*integral(14/5*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^6*e^(2*I*d*x + 2*I*c) + a^2*d*e^6), x))/(a^2*d*e^6*e^(6*I*d*x + 6*I*c) + 3*
a^2*d*e^6*e^(4*I*d*x + 4*I*c) + 3*a^2*d*e^6*e^(2*I*d*x + 2*I*c) + a^2*d*e^6)

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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))**(11/2)/(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{11}{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))^(11/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

integrate(1/((e*cos(d*x + c))^(11/2)*(I*a*tan(d*x + c) + a)^2), x)