3.209 \(\int \frac{(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{9/2}} \, dx\)

Optimal. Leaf size=124 \[ -\frac{4 i \left (a^3+i a^3 \tan (c+d x)\right )}{15 d e^2 (e \sec (c+d x))^{5/2}}+\frac{2 a^3 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d e^4 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}-\frac{2 i (a+i a \tan (c+d x))^3}{9 d (e \sec (c+d x))^{9/2}} \]

[Out]

(2*a^3*EllipticE[(c + d*x)/2, 2])/(15*d*e^4*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) - (((2*I)/9)*(a + I*a*Tan
[c + d*x])^3)/(d*(e*Sec[c + d*x])^(9/2)) - (((4*I)/15)*(a^3 + I*a^3*Tan[c + d*x]))/(d*e^2*(e*Sec[c + d*x])^(5/
2))

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Rubi [A]  time = 0.126562, antiderivative size = 124, 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 = {3497, 3496, 3771, 2639} \[ -\frac{4 i \left (a^3+i a^3 \tan (c+d x)\right )}{15 d e^2 (e \sec (c+d x))^{5/2}}+\frac{2 a^3 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d e^4 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}-\frac{2 i (a+i a \tan (c+d x))^3}{9 d (e \sec (c+d x))^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^3/(e*Sec[c + d*x])^(9/2),x]

[Out]

(2*a^3*EllipticE[(c + d*x)/2, 2])/(15*d*e^4*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) - (((2*I)/9)*(a + I*a*Tan
[c + d*x])^3)/(d*(e*Sec[c + d*x])^(9/2)) - (((4*I)/15)*(a^3 + I*a^3*Tan[c + d*x]))/(d*e^2*(e*Sec[c + d*x])^(5/
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 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))^{9/2}} \, dx &=-\frac{2 i (a+i a \tan (c+d x))^3}{9 d (e \sec (c+d x))^{9/2}}+\frac{a \int \frac{(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{5/2}} \, dx}{3 e^2}\\ &=-\frac{2 i (a+i a \tan (c+d x))^3}{9 d (e \sec (c+d x))^{9/2}}-\frac{4 i \left (a^3+i a^3 \tan (c+d x)\right )}{15 d e^2 (e \sec (c+d x))^{5/2}}+\frac{a^3 \int \frac{1}{\sqrt{e \sec (c+d x)}} \, dx}{15 e^4}\\ &=-\frac{2 i (a+i a \tan (c+d x))^3}{9 d (e \sec (c+d x))^{9/2}}-\frac{4 i \left (a^3+i a^3 \tan (c+d x)\right )}{15 d e^2 (e \sec (c+d x))^{5/2}}+\frac{a^3 \int \sqrt{\cos (c+d x)} \, dx}{15 e^4 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}\\ &=\frac{2 a^3 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d e^4 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}-\frac{2 i (a+i a \tan (c+d x))^3}{9 d (e \sec (c+d x))^{9/2}}-\frac{4 i \left (a^3+i a^3 \tan (c+d x)\right )}{15 d e^2 (e \sec (c+d x))^{5/2}}\\ \end{align*}

Mathematica [C]  time = 2.40854, size = 118, normalized size = 0.95 \[ -\frac{a^3 e^{-2 i (c+d x)} (\tan (c+d x)-i)^3 \left (4 \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )+16 e^{2 i (c+d x)}+5 e^{4 i (c+d x)}+11\right )}{90 d e^2 (e \sec (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^3/(e*Sec[c + d*x])^(9/2),x]

[Out]

-(a^3*(11 + 16*E^((2*I)*(c + d*x)) + 5*E^((4*I)*(c + d*x)) + 4*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)/(90*d*e^2*E^((2*I)*(c + d*x))*(e*Sec[c + d*x])^(
5/2))

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Maple [B]  time = 0.283, size = 370, normalized size = 3. \begin{align*} -{\frac{2\,{a}^{3}}{45\,d \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) } \left ( 20\,i \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) -3\,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 ) +3\,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 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 ) +20\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}-9\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) -3\,i\sin \left ( dx+c \right ){\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}+3\,i\sin \left ( dx+c \right ){\it EllipticE} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}-19\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}+2\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-3\,\cos \left ( dx+c \right ) \right ) \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{-{\frac{9}{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))^(9/2),x)

[Out]

-2/45*a^3/d*(20*I*cos(d*x+c)^5*sin(d*x+c)-3*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*Ellip
ticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)*cos(d*x+c)*sin(d*x+c)+3*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)
+1))^(1/2)*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)*cos(d*x+c)*sin(d*x+c)+20*cos(d*x+c)^6-9*I*cos(d*x+c)^3*sin
(d*x+c)-3*I*sin(d*x+c)*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+
c)+1))^(1/2)+3*I*sin(d*x+c)*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos
(d*x+c)+1))^(1/2)-19*cos(d*x+c)^4+2*cos(d*x+c)^2-3*cos(d*x+c))/cos(d*x+c)^5/sin(d*x+c)/(e/cos(d*x+c))^(9/2)

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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{9}{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))^(9/2),x, algorithm="maxima")

[Out]

integrate((I*a*tan(d*x + c) + a)^3/(e*sec(d*x + c))^(9/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\sqrt{2}{\left (-5 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 5 i \, a^{3} e^{\left (5 i \, d x + 5 i \, c\right )} - 16 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 16 i \, a^{3} e^{\left (3 i \, d x + 3 i \, c\right )} - 23 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 11 i \, a^{3} e^{\left (i \, d x + i \, c\right )} - 12 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 )} + 90 \,{\left (d e^{5} e^{\left (i \, d x + i \, c\right )} - d e^{5}\right )}{\rm integral}\left (\frac{\sqrt{2}{\left (-i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i \, a^{3} e^{\left (i \, d x + i \, c\right )} - 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 )}}{15 \,{\left (d e^{5} e^{\left (3 i \, d x + 3 i \, c\right )} - 2 \, d e^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + d e^{5} e^{\left (i \, d x + i \, c\right )}\right )}}, x\right )}{90 \,{\left (d e^{5} e^{\left (i \, d x + i \, c\right )} - d e^{5}\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))^(9/2),x, algorithm="fricas")

[Out]

1/90*(sqrt(2)*(-5*I*a^3*e^(6*I*d*x + 6*I*c) + 5*I*a^3*e^(5*I*d*x + 5*I*c) - 16*I*a^3*e^(4*I*d*x + 4*I*c) + 16*
I*a^3*e^(3*I*d*x + 3*I*c) - 23*I*a^3*e^(2*I*d*x + 2*I*c) + 11*I*a^3*e^(I*d*x + I*c) - 12*I*a^3)*sqrt(e/(e^(2*I
*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + 90*(d*e^5*e^(I*d*x + I*c) - d*e^5)*integral(1/15*sqrt(2)*(-I*a^3
*e^(2*I*d*x + 2*I*c) - 2*I*a^3*e^(I*d*x + I*c) - 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^5*e^(3*I*d*x + 3*I*c) - 2*d*e^5*e^(2*I*d*x + 2*I*c) + d*e^5*e^(I*d*x + I*c)), x))/(d*e^5*e^(I*d*x + I
*c) - d*e^5)

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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))**(9/2),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{9}{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))^(9/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^3/(e*sec(d*x + c))^(9/2), x)