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]

6/5*I*a^3/d/e^2/(e*sec(d*x+c))^(1/2)-6/5*a^3*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2
*d*x+1/2*c),2^(1/2))/d/e^2/cos(d*x+c)^(1/2)/(e*sec(d*x+c))^(1/2)-4/5*I*a*(a+I*a*tan(d*x+c))^2/d/(e*sec(d*x+c))
^(5/2)

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Rubi [A]  time = 0.10, antiderivative size = 111, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [C]  time = 1.45, size = 108, normalized size = 0.97 \[ -\frac {4 i a^3 e^{2 i (c+d x)} \left (-\sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )+e^{2 i (c+d x)}+1\right )}{5 d e^2 \left (1+e^{2 i (c+d x)}\right ) \sqrt {e \sec (c+d x)}} \]

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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fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ \frac {\sqrt {2} {\left (-2 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 i \, a^{3} e^{\left (3 i \, d x + 3 i \, c\right )} + 4 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i \, a^{3} e^{\left (i \, d x + i \, c\right )} + 6 i \, a^{3}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 5 \, {\left (d e^{3} e^{\left (i \, d x + i \, c\right )} - d e^{3}\right )} {\rm integral}\left (\frac {\sqrt {2} {\left (3 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 6 i \, a^{3} e^{\left (i \, d x + i \, c\right )} + 3 i \, a^{3}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{5 \, {\left (d e^{3} e^{\left (3 i \, d x + 3 i \, c\right )} - 2 \, d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + d e^{3} e^{\left (i \, d x + i \, c\right )}\right )}}, x\right )}{5 \, {\left (d e^{3} e^{\left (i \, d x + i \, c\right )} - d e^{3}\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}}}\,{d x} \]

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)

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maple [B]  time = 0.96, size = 1086, normalized size = 9.78 \[ \text {result too large to display} \]

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*(1+cos(d*x+c))*(-1+cos(d*x+c))^2*(-24*I*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)
*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*cos(d*x+c)*sin(d*x+c)+12*I*(1/
(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*(-cos(d*x+c)
/(1+cos(d*x+c))^2)^(1/2)*sin(d*x+c)-12*I*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*
x+c)/(1+cos(d*x+c)))^(1/2)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)+24*I*(1/(1+cos(d*x+c)))^(1/2)*
(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(1/
2)*cos(d*x+c)*sin(d*x+c)+12*I*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticF(I*(-1+cos(d
*x+c))/sin(d*x+c),I)*cos(d*x+c)^2*sin(d*x+c)*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)+16*(-cos(d*x+c)/(1+cos(d*x+c
))^2)^(1/2)*cos(d*x+c)^5+16*I*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)^4*sin(d*x+c)-20*I*(-cos(d*x+c)/(
1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)*sin(d*x+c)+5*I*cos(d*x+c)*ln(-2*(2*cos(d*x+c)^2*(-cos(d*x+c)/(1+cos(d*x+c))^
2)^(1/2)-cos(d*x+c)^2+2*cos(d*x+c)-2*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)-1)/sin(d*x+c)^2)*sin(d*x+c)+16*(-cos
(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)^4-20*I*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)^2*sin(d*x+c)
-28*cos(d*x+c)^3*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)+16*I*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)^3*s
in(d*x+c)-12*I*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+
c),I)*cos(d*x+c)^2*sin(d*x+c)*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)-5*I*cos(d*x+c)*ln(-(2*cos(d*x+c)^2*(-cos(d*
x+c)/(1+cos(d*x+c))^2)^(1/2)-cos(d*x+c)^2+2*cos(d*x+c)-2*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)-1)/sin(d*x+c)^2)
*sin(d*x+c)-16*cos(d*x+c)^2*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)+12*cos(d*x+c)*(-cos(d*x+c)/(1+cos(d*x+c))^2)^
(1/2))/cos(d*x+c)^3/sin(d*x+c)^5/(e/cos(d*x+c))^(5/2)/(-cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)^3/(e/cos(c + d*x))^(5/2),x)

[Out]

int((a + a*tan(c + d*x)*1i)^3/(e/cos(c + d*x))^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - i a^{3} \left (\int \frac {i}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx + \int \left (- \frac {3 \tan {\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\right )\, dx + \int \frac {\tan ^{3}{\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx + \int \left (- \frac {3 i \tan ^{2}{\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\right )\, dx\right ) \]

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]

-I*a**3*(Integral(I/(e*sec(c + d*x))**(5/2), x) + Integral(-3*tan(c + d*x)/(e*sec(c + d*x))**(5/2), x) + Integ
ral(tan(c + d*x)**3/(e*sec(c + d*x))**(5/2), x) + Integral(-3*I*tan(c + d*x)**2/(e*sec(c + d*x))**(5/2), x))

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