3.211 \(\int \frac{(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{13/2}} \, dx\)

Optimal. Leaf size=155 \[ \frac{14 a^3 \sin (c+d x)}{117 d e^5 (e \sec (c+d x))^{3/2}}-\frac{28 i \left (a^3+i a^3 \tan (c+d x)\right )}{117 d e^2 (e \sec (c+d x))^{9/2}}+\frac{14 a^3 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{39 d e^6 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}-\frac{2 i (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}} \]

[Out]

(14*a^3*EllipticE[(c + d*x)/2, 2])/(39*d*e^6*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (14*a^3*Sin[c + d*x])/
(117*d*e^5*(e*Sec[c + d*x])^(3/2)) - (((2*I)/13)*(a + I*a*Tan[c + d*x])^3)/(d*(e*Sec[c + d*x])^(13/2)) - (((28
*I)/117)*(a^3 + I*a^3*Tan[c + d*x]))/(d*e^2*(e*Sec[c + d*x])^(9/2))

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Rubi [A]  time = 0.147979, antiderivative size = 155, 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 = {3497, 3496, 3769, 3771, 2639} \[ \frac{14 a^3 \sin (c+d x)}{117 d e^5 (e \sec (c+d x))^{3/2}}-\frac{28 i \left (a^3+i a^3 \tan (c+d x)\right )}{117 d e^2 (e \sec (c+d x))^{9/2}}+\frac{14 a^3 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{39 d e^6 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}-\frac{2 i (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^3/(e*Sec[c + d*x])^(13/2),x]

[Out]

(14*a^3*EllipticE[(c + d*x)/2, 2])/(39*d*e^6*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (14*a^3*Sin[c + d*x])/
(117*d*e^5*(e*Sec[c + d*x])^(3/2)) - (((2*I)/13)*(a + I*a*Tan[c + d*x])^3)/(d*(e*Sec[c + d*x])^(13/2)) - (((28
*I)/117)*(a^3 + I*a^3*Tan[c + d*x]))/(d*e^2*(e*Sec[c + d*x])^(9/2))

Rule 3497

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)/(a*f*m), x] + Dist[(a*(m + n))/(m*d^2), Int[(d*Sec[e + f*x])^(m + 2)*(
a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n, 0] && LtQ[m, -
1] && IntegersQ[2*m, 2*n]

Rule 3496

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(2*b*(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1))/(f*m), x] - Dist[(b^2*(m + 2*n - 2))/(d^2*m), Int[(d*Sec[e + f
*x])^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n,
1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILt
Q[m, 0] && LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && IntegerQ[2*m]

Rule 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[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{(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{13/2}} \, dx &=-\frac{2 i (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}}+\frac{(7 a) \int \frac{(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{9/2}} \, dx}{13 e^2}\\ &=-\frac{2 i (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}}-\frac{28 i \left (a^3+i a^3 \tan (c+d x)\right )}{117 d e^2 (e \sec (c+d x))^{9/2}}+\frac{\left (35 a^3\right ) \int \frac{1}{(e \sec (c+d x))^{5/2}} \, dx}{117 e^4}\\ &=\frac{14 a^3 \sin (c+d x)}{117 d e^5 (e \sec (c+d x))^{3/2}}-\frac{2 i (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}}-\frac{28 i \left (a^3+i a^3 \tan (c+d x)\right )}{117 d e^2 (e \sec (c+d x))^{9/2}}+\frac{\left (7 a^3\right ) \int \frac{1}{\sqrt{e \sec (c+d x)}} \, dx}{39 e^6}\\ &=\frac{14 a^3 \sin (c+d x)}{117 d e^5 (e \sec (c+d x))^{3/2}}-\frac{2 i (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}}-\frac{28 i \left (a^3+i a^3 \tan (c+d x)\right )}{117 d e^2 (e \sec (c+d x))^{9/2}}+\frac{\left (7 a^3\right ) \int \sqrt{\cos (c+d x)} \, dx}{39 e^6 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}\\ &=\frac{14 a^3 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{39 d e^6 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{14 a^3 \sin (c+d x)}{117 d e^5 (e \sec (c+d x))^{3/2}}-\frac{2 i (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}}-\frac{28 i \left (a^3+i a^3 \tan (c+d x)\right )}{117 d e^2 (e \sec (c+d x))^{9/2}}\\ \end{align*}

Mathematica [C]  time = 6.36057, size = 155, normalized size = 1. \[ -\frac{a^3 e^{-4 i (c+d x)} (\tan (c+d x)-i)^3 \left (112 e^{2 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-34 e^{2 i (c+d x)}+124 e^{4 i (c+d x)}+50 e^{6 i (c+d x)}+9 e^{8 i (c+d x)}-117\right )}{936 d e^4 (e \sec (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^3/(e*Sec[c + d*x])^(13/2),x]

[Out]

-(a^3*(-117 - 34*E^((2*I)*(c + d*x)) + 124*E^((4*I)*(c + d*x)) + 50*E^((6*I)*(c + d*x)) + 9*E^((8*I)*(c + d*x)
) + 112*E^((2*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x)
)])*(-I + Tan[c + d*x])^3)/(936*d*e^4*E^((4*I)*(c + d*x))*(e*Sec[c + d*x])^(5/2))

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Maple [B]  time = 0.37, size = 380, normalized size = 2.5 \begin{align*}{\frac{2\,{a}^{3}}{117\,d \left ( \cos \left ( dx+c \right ) \right ) ^{7}\sin \left ( dx+c \right ) } \left ( -36\,i \left ( \cos \left ( dx+c \right ) \right ) ^{7}\sin \left ( dx+c \right ) -36\, \left ( \cos \left ( dx+c \right ) \right ) ^{8}+13\,i \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) +21\,i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) \cos \left ( dx+c \right ) \sin \left ( dx+c \right ) -21\,i{\it EllipticE} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) \cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}+31\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}+21\,i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) \sin \left ( dx+c \right ) -21\,i{\it EllipticE} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) \sin \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}-2\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}-14\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}+21\,\cos \left ( dx+c \right ) \right ) \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{-{\frac{13}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(13/2),x)

[Out]

2/117*a^3/d*(-36*I*cos(d*x+c)^7*sin(d*x+c)-36*cos(d*x+c)^8+13*I*cos(d*x+c)^5*sin(d*x+c)+21*I*(1/(cos(d*x+c)+1)
)^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)*cos(d*x+c)*sin(d*x+c)-21*I*
EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)*cos(d*x+c)*sin(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c
)+1))^(1/2)+31*cos(d*x+c)^6+21*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(cos(d
*x+c)-1)/sin(d*x+c),I)*sin(d*x+c)-21*I*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)*sin(d*x+c)*(1/(cos(d*x+c)+1))^
(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-2*cos(d*x+c)^4-14*cos(d*x+c)^2+21*cos(d*x+c))/cos(d*x+c)^7/sin(d*x+c)/
(e/cos(d*x+c))^(13/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\left (e \sec \left (d x + c\right )\right )^{\frac{13}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(13/2),x, algorithm="maxima")

[Out]

integrate((I*a*tan(d*x + c) + a)^3/(e*sec(d*x + c))^(13/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\sqrt{2}{\left (-9 i \, a^{3} e^{\left (9 i \, d x + 9 i \, c\right )} + 9 i \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} - 50 i \, a^{3} e^{\left (7 i \, d x + 7 i \, c\right )} + 50 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 124 i \, a^{3} e^{\left (5 i \, d x + 5 i \, c\right )} + 124 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 302 i \, a^{3} e^{\left (3 i \, d x + 3 i \, c\right )} - 34 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 219 i \, a^{3} e^{\left (i \, d x + i \, c\right )} - 117 i \, a^{3}\right )} \sqrt{\frac{e}{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 )} + 936 \,{\left (d e^{7} e^{\left (2 i \, d x + 2 i \, c\right )} - d e^{7} e^{\left (i \, d x + i \, c\right )}\right )}{\rm integral}\left (\frac{\sqrt{2}{\left (-7 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 14 i \, a^{3} e^{\left (i \, d x + i \, c\right )} - 7 i \, a^{3}\right )} \sqrt{\frac{e}{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 )}}{39 \,{\left (d e^{7} e^{\left (3 i \, d x + 3 i \, c\right )} - 2 \, d e^{7} e^{\left (2 i \, d x + 2 i \, c\right )} + d e^{7} e^{\left (i \, d x + i \, c\right )}\right )}}, x\right )}{936 \,{\left (d e^{7} e^{\left (2 i \, d x + 2 i \, c\right )} - d e^{7} e^{\left (i \, d x + i \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(13/2),x, algorithm="fricas")

[Out]

1/936*(sqrt(2)*(-9*I*a^3*e^(9*I*d*x + 9*I*c) + 9*I*a^3*e^(8*I*d*x + 8*I*c) - 50*I*a^3*e^(7*I*d*x + 7*I*c) + 50
*I*a^3*e^(6*I*d*x + 6*I*c) - 124*I*a^3*e^(5*I*d*x + 5*I*c) + 124*I*a^3*e^(4*I*d*x + 4*I*c) - 302*I*a^3*e^(3*I*
d*x + 3*I*c) - 34*I*a^3*e^(2*I*d*x + 2*I*c) - 219*I*a^3*e^(I*d*x + I*c) - 117*I*a^3)*sqrt(e/(e^(2*I*d*x + 2*I*
c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + 936*(d*e^7*e^(2*I*d*x + 2*I*c) - d*e^7*e^(I*d*x + I*c))*integral(1/39*sqrt(
2)*(-7*I*a^3*e^(2*I*d*x + 2*I*c) - 14*I*a^3*e^(I*d*x + I*c) - 7*I*a^3)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/
2*I*d*x + 1/2*I*c)/(d*e^7*e^(3*I*d*x + 3*I*c) - 2*d*e^7*e^(2*I*d*x + 2*I*c) + d*e^7*e^(I*d*x + I*c)), x))/(d*e
^7*e^(2*I*d*x + 2*I*c) - d*e^7*e^(I*d*x + 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((a+I*a*tan(d*x+c))**3/(e*sec(d*x+c))**(13/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\left (e \sec \left (d x + c\right )\right )^{\frac{13}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(13/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^3/(e*sec(d*x + c))^(13/2), x)