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

Optimal. Leaf size=175 \[ -\frac {22 a^3 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {22 i a^3 (e \sec (c+d x))^{3/2}}{15 d}+\frac {22 a^3 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{5 d}+\frac {22 i \left (a^3+i a^3 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{35 d}+\frac {2 i a (a+i a \tan (c+d x))^2 (e \sec (c+d x))^{3/2}}{7 d} \]

[Out]

22/15*I*a^3*(e*sec(d*x+c))^(3/2)/d-22/5*a^3*e^2*(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/cos(d*x+c)^(1/2)/(e*sec(d*x+c))^(1/2)+22/5*a^3*e*sin(d*x+c)*(e*sec(d*x+c))^(1/2)/d+2
/7*I*a*(e*sec(d*x+c))^(3/2)*(a+I*a*tan(d*x+c))^2/d+22/35*I*(e*sec(d*x+c))^(3/2)*(a^3+I*a^3*tan(d*x+c))/d

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Rubi [A]  time = 0.19, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3498, 3486, 3768, 3771, 2639} \[ -\frac {22 a^3 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {22 i a^3 (e \sec (c+d x))^{3/2}}{15 d}+\frac {22 a^3 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{5 d}+\frac {22 i \left (a^3+i a^3 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{35 d}+\frac {2 i a (a+i a \tan (c+d x))^2 (e \sec (c+d x))^{3/2}}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-22*a^3*e^2*EllipticE[(c + d*x)/2, 2])/(5*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (((22*I)/15)*a^3*(e*Se
c[c + d*x])^(3/2))/d + (22*a^3*e*Sqrt[e*Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (((2*I)/7)*a*(e*Sec[c + d*x])^(3/2
)*(a + I*a*Tan[c + d*x])^2)/d + (((22*I)/35)*(e*Sec[c + d*x])^(3/2)*(a^3 + I*a^3*Tan[c + d*x]))/d

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 3498

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

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]

Rubi steps

\begin {align*} \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^3 \, dx &=\frac {2 i a (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2}{7 d}+\frac {1}{7} (11 a) \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2 \, dx\\ &=\frac {2 i a (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2}{7 d}+\frac {22 i (e \sec (c+d x))^{3/2} \left (a^3+i a^3 \tan (c+d x)\right )}{35 d}+\frac {1}{5} \left (11 a^2\right ) \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x)) \, dx\\ &=\frac {22 i a^3 (e \sec (c+d x))^{3/2}}{15 d}+\frac {2 i a (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2}{7 d}+\frac {22 i (e \sec (c+d x))^{3/2} \left (a^3+i a^3 \tan (c+d x)\right )}{35 d}+\frac {1}{5} \left (11 a^3\right ) \int (e \sec (c+d x))^{3/2} \, dx\\ &=\frac {22 i a^3 (e \sec (c+d x))^{3/2}}{15 d}+\frac {22 a^3 e \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 i a (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2}{7 d}+\frac {22 i (e \sec (c+d x))^{3/2} \left (a^3+i a^3 \tan (c+d x)\right )}{35 d}-\frac {1}{5} \left (11 a^3 e^2\right ) \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx\\ &=\frac {22 i a^3 (e \sec (c+d x))^{3/2}}{15 d}+\frac {22 a^3 e \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 i a (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2}{7 d}+\frac {22 i (e \sec (c+d x))^{3/2} \left (a^3+i a^3 \tan (c+d x)\right )}{35 d}-\frac {\left (11 a^3 e^2\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\\ &=-\frac {22 a^3 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {22 i a^3 (e \sec (c+d x))^{3/2}}{15 d}+\frac {22 a^3 e \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 i a (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2}{7 d}+\frac {22 i (e \sec (c+d x))^{3/2} \left (a^3+i a^3 \tan (c+d x)\right )}{35 d}\\ \end {align*}

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Mathematica [C]  time = 2.73, size = 129, normalized size = 0.74 \[ \frac {a^3 (1+i \tan (c+d x)) (e \sec (c+d x))^{3/2} \left (77 i e^{-2 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{5/2} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )-308 i \cos (2 (c+d x))+17 \tan (c+d x)+77 \sin (3 (c+d x)) \sec (c+d x)-116 i\right )}{210 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(e*Sec[c + d*x])^(3/2)*(1 + I*Tan[c + d*x])*(-116*I - (308*I)*Cos[2*(c + d*x)] + ((77*I)*(1 + E^((2*I)*(c
+ d*x)))^(5/2)*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/E^((2*I)*(c + d*x)) + 77*Sec[c + d*x]*
Sin[3*(c + d*x)] + 17*Tan[c + d*x]))/(210*d)

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ \frac {\sqrt {2} {\left (-462 i \, a^{3} e e^{\left (7 i \, d x + 7 i \, c\right )} - 574 i \, a^{3} e e^{\left (5 i \, d x + 5 i \, c\right )} - 506 i \, a^{3} e e^{\left (3 i \, d x + 3 i \, c\right )} - 154 i \, a^{3} e e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 105 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} {\rm integral}\left (\frac {11 i \, \sqrt {2} a^{3} e \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{5 \, d}, x\right )}{105 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(sqrt(2)*(-462*I*a^3*e*e^(7*I*d*x + 7*I*c) - 574*I*a^3*e*e^(5*I*d*x + 5*I*c) - 506*I*a^3*e*e^(3*I*d*x +
3*I*c) - 154*I*a^3*e*e^(I*d*x + I*c))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + 105*(d*e^(6*
I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)*integral(11/5*I*sqrt(2)*a^3*e*sqrt(e/(
e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c)/d, x))/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*
d*e^(2*I*d*x + 2*I*c) + d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.99, size = 392, normalized size = 2.24 \[ \frac {2 a^{3} \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (231 i \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-231 i \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+231 i \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-231 i \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+140 i \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-231 \left (\cos ^{4}\left (d x +c \right )\right )+294 \left (\cos ^{3}\left (d x +c \right )\right )-15 i \sin \left (d x +c \right )-63 \cos \left (d x +c \right )\right ) \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}}}{105 d \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*a^3/d*(1+cos(d*x+c))^2*(-1+cos(d*x+c))^2*(231*I*cos(d*x+c)^4*sin(d*x+c)*(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)-231*I*cos(d*x+c)^4*sin(d*x+c)*(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)+231*I*cos(d*x+c)^3*s
in(d*x+c)*(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)
-231*I*cos(d*x+c)^3*sin(d*x+c)*(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)+140*I*cos(d*x+c)^2*sin(d*x+c)-231*cos(d*x+c)^4+294*cos(d*x+c)^3-15*I*sin(d*x+c)-63*cos(d
*x+c))*(e/cos(d*x+c))^(3/2)/cos(d*x+c)^2/sin(d*x+c)^5

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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