3.207 \(\int \frac{(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=111 \[ \frac{6 i a^3}{5 d e^2 \sqrt{e \sec (c+d x)}}-\frac{6 a^3 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}-\frac{4 i a (a+i a \tan (c+d x))^2}{5 d (e \sec (c+d x))^{5/2}} \]

[Out]

(((6*I)/5)*a^3)/(d*e^2*Sqrt[e*Sec[c + d*x]]) - (6*a^3*EllipticE[(c + d*x)/2, 2])/(5*d*e^2*Sqrt[Cos[c + d*x]]*S
qrt[e*Sec[c + d*x]]) - (((4*I)/5)*a*(a + I*a*Tan[c + d*x])^2)/(d*(e*Sec[c + d*x])^(5/2))

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Rubi [A]  time = 0.101171, antiderivative size = 111, 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 = {3496, 3486, 3771, 2639} \[ \frac{6 i a^3}{5 d e^2 \sqrt{e \sec (c+d x)}}-\frac{6 a^3 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}-\frac{4 i a (a+i a \tan (c+d x))^2}{5 d (e \sec (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^3/(e*Sec[c + d*x])^(5/2),x]

[Out]

(((6*I)/5)*a^3)/(d*e^2*Sqrt[e*Sec[c + d*x]]) - (6*a^3*EllipticE[(c + d*x)/2, 2])/(5*d*e^2*Sqrt[Cos[c + d*x]]*S
qrt[e*Sec[c + d*x]]) - (((4*I)/5)*a*(a + I*a*Tan[c + d*x])^2)/(d*(e*Sec[c + d*x])^(5/2))

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 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

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

Mathematica [C]  time = 1.25711, size = 108, normalized size = 0.97 \[ -\frac{4 i a^3 e^{2 i (c+d x)} \left (-\sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )+e^{2 i (c+d x)}+1\right )}{5 d e^2 \left (1+e^{2 i (c+d x)}\right ) \sqrt{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^3/(e*Sec[c + d*x])^(5/2),x]

[Out]

(((-4*I)/5)*a^3*E^((2*I)*(c + d*x))*(1 + 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))]))/(d*e^2*(1 + E^((2*I)*(c + d*x)))*Sqrt[e*Sec[c + d*x]])

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Maple [B]  time = 0.245, size = 1086, normalized size = 9.8 \begin{align*} \text{result too large to display} \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))^(5/2),x)

[Out]

-1/10*a^3/d*(cos(d*x+c)+1)*(cos(d*x+c)-1)^2*(-12*I*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(-cos(d*x+c)/(cos(d*x+c)+
1)^2)^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)*cos(d*x+c)^2*sin(d*x+c)+12*I*(-c
os(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(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)-12*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)/(cos(d*x+c)+1)^2)^(1/2)*sin(d*x+c)+24*I*cos(d*x+c)*sin(d*x+c)*Ellip
ticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/
(cos(d*x+c)+1))^(1/2)+16*I*cos(d*x+c)^3*sin(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)-20*I*cos(d*x+c)*sin(d*
x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)-24*I*cos(d*x+c)*sin(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(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)+16*I*cos(d*x+c
)^4*sin(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+16*cos(d*x+c)^5*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)-20*I*
cos(d*x+c)^2*sin(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+16*cos(d*x+c)^4*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1
/2)+12*I*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*Ellip
ticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)*cos(d*x+c)^2*sin(d*x+c)+5*I*ln(-2*(2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*
cos(d*x+c)^2-cos(d*x+c)^2-2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*cos(d*x+c)-1)/sin(d*x+c)^2)*cos(d*x+c)*sin(
d*x+c)-5*I*cos(d*x+c)*ln(-(2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)^2-cos(d*x+c)^2-2*(-cos(d*x+c)/(co
s(d*x+c)+1)^2)^(1/2)+2*cos(d*x+c)-1)/sin(d*x+c)^2)*sin(d*x+c)-28*cos(d*x+c)^3*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(
1/2)-16*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)^2+12*cos(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2))/
cos(d*x+c)^3/sin(d*x+c)^5/(e/cos(d*x+c))^(5/2)/(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)

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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{5}{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))^(5/2),x, algorithm="maxima")

[Out]

integrate((I*a*tan(d*x + c) + a)^3/(e*sec(d*x + c))^(5/2), x)

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

[Out]

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

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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))**(5/2),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{5}{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))^(5/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^3/(e*sec(d*x + c))^(5/2), x)