3.670 \(\int \frac{1}{(e \cos (c+d x))^{7/2} (a+i a \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=122 \[ -\frac{6 \sin (c+d x) \cos ^3(c+d x)}{a^2 d (e \cos (c+d x))^{7/2}}+\frac{4 i \cos ^2(c+d x)}{d \left (a^2+i a^2 \tan (c+d x)\right ) (e \cos (c+d x))^{7/2}}+\frac{6 \cos ^{\frac{7}{2}}(c+d x) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d (e \cos (c+d x))^{7/2}} \]

[Out]

(6*Cos[c + d*x]^(7/2)*EllipticE[(c + d*x)/2, 2])/(a^2*d*(e*Cos[c + d*x])^(7/2)) - (6*Cos[c + d*x]^3*Sin[c + d*
x])/(a^2*d*(e*Cos[c + d*x])^(7/2)) + ((4*I)*Cos[c + d*x]^2)/(d*(e*Cos[c + d*x])^(7/2)*(a^2 + I*a^2*Tan[c + d*x
]))

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Rubi [A]  time = 0.171009, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 5, 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} \[ -\frac{6 \sin (c+d x) \cos ^3(c+d x)}{a^2 d (e \cos (c+d x))^{7/2}}+\frac{4 i \cos ^2(c+d x)}{d \left (a^2+i a^2 \tan (c+d x)\right ) (e \cos (c+d x))^{7/2}}+\frac{6 \cos ^{\frac{7}{2}}(c+d x) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d (e \cos (c+d x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((e*Cos[c + d*x])^(7/2)*(a + I*a*Tan[c + d*x])^2),x]

[Out]

(6*Cos[c + d*x]^(7/2)*EllipticE[(c + d*x)/2, 2])/(a^2*d*(e*Cos[c + d*x])^(7/2)) - (6*Cos[c + d*x]^3*Sin[c + d*
x])/(a^2*d*(e*Cos[c + d*x])^(7/2)) + ((4*I)*Cos[c + d*x]^2)/(d*(e*Cos[c + d*x])^(7/2)*(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))^{7/2} (a+i a \tan (c+d x))^2} \, dx &=\frac{\int \frac{(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^2} \, dx}{(e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2}}\\ &=\frac{4 i \cos ^2(c+d x)}{d (e \cos (c+d x))^{7/2} \left (a^2+i a^2 \tan (c+d x)\right )}-\frac{\left (3 e^2\right ) \int (e \sec (c+d x))^{3/2} \, dx}{a^2 (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2}}\\ &=-\frac{6 \cos ^3(c+d x) \sin (c+d x)}{a^2 d (e \cos (c+d x))^{7/2}}+\frac{4 i \cos ^2(c+d x)}{d (e \cos (c+d x))^{7/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{\left (3 e^4\right ) \int \frac{1}{\sqrt{e \sec (c+d x)}} \, dx}{a^2 (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2}}\\ &=-\frac{6 \cos ^3(c+d x) \sin (c+d x)}{a^2 d (e \cos (c+d x))^{7/2}}+\frac{4 i \cos ^2(c+d x)}{d (e \cos (c+d x))^{7/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{\left (3 \cos ^{\frac{7}{2}}(c+d x)\right ) \int \sqrt{\cos (c+d x)} \, dx}{a^2 (e \cos (c+d x))^{7/2}}\\ &=\frac{6 \cos ^{\frac{7}{2}}(c+d x) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d (e \cos (c+d x))^{7/2}}-\frac{6 \cos ^3(c+d x) \sin (c+d x)}{a^2 d (e \cos (c+d x))^{7/2}}+\frac{4 i \cos ^2(c+d x)}{d (e \cos (c+d x))^{7/2} \left (a^2+i a^2 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 0.984883, size = 255, normalized size = 2.09 \[ \frac{-2 \sin (c+d x)+10 i \cos (c+d x)-6 i (\cos (c+d x)-i \sin (c+d x)) \sqrt{\sin (c+d x)-i \cos (c+d x)+1} \sqrt{\sin (c+d x)+i \sin (2 (c+d x))-i \cos (c+d x)+\cos (2 (c+d x))} F\left (\left .\sin ^{-1}\left (\sqrt{\sin (c+d x)-i \cos (c+d x)}\right )\right |-1\right )+6 \sqrt{\sin (c+d x)-i \cos (c+d x)+1} (\sin (c+d x)+i \cos (c+d x)) \sqrt{\sin (c+d x)+i \sin (2 (c+d x))-i \cos (c+d x)+\cos (2 (c+d x))} E\left (\left .\sin ^{-1}\left (\sqrt{\sin (c+d x)-i \cos (c+d x)}\right )\right |-1\right )}{a^2 d e^3 \sqrt{e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((e*Cos[c + d*x])^(7/2)*(a + I*a*Tan[c + d*x])^2),x]

[Out]

((10*I)*Cos[c + d*x] - 2*Sin[c + d*x] - (6*I)*EllipticF[ArcSin[Sqrt[(-I)*Cos[c + d*x] + Sin[c + d*x]]], -1]*(C
os[c + d*x] - I*Sin[c + d*x])*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)]] + 6*EllipticE[ArcSin[Sqrt[(-I)*Cos[c + d*x] + Sin[c + d*x]]], -1]*Sqrt[
1 - I*Cos[c + d*x] + Sin[c + d*x]]*(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)]])/(a^2*d*e^3*Sqrt[e*Cos[c + d*x]])

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Maple [A]  time = 1.97, size = 135, normalized size = 1.1 \begin{align*} -2\,{\frac{4\,i \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}-3\,{\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}}+2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) -2\,i\sin \left ( 1/2\,dx+c/2 \right ) }{{e}^{3}{a}^{2}\sin \left ( 1/2\,dx+c/2 \right ) \sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}e+e}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*cos(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^2,x)

[Out]

-2/e^3/a^2/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*(4*I*sin(1/2*d*x+1/2*c)^3-3*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)+2*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))^(7/2)/(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 (12 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 8 i\right )} e^{\left (-\frac{1}{2} i \, d x - \frac{1}{2} i \, c\right )} +{\left (a^{2} d e^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d e^{4}\right )}{\rm integral}\left (-\frac{6 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 )}}{a^{2} d e^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d e^{4}}, x\right )}{a^{2} d e^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cos(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

(sqrt(1/2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*(12*I*e^(2*I*d*x + 2*I*c) + 8*I)*e^(-1/2*I*d*x - 1/2*I*c) + (a^2*d*
e^4*e^(2*I*d*x + 2*I*c) + a^2*d*e^4)*integral(-6*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^4*e^(2*I*d*x + 2*I*c) + a^2*d*e^4), x))/(a^2*d*e^4*e^(2*I*d*x + 2*I*c) + a^2*d*e^4)

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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))**(7/2)/(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{7}{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))^(7/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

integrate(1/((e*cos(d*x + c))^(7/2)*(I*a*tan(d*x + c) + a)^2), x)