3.204 \(\int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=139 \[ \frac {6 i a^3 \sqrt {e \sec (c+d x)}}{d}+\frac {6 i \left (a^3+i a^3 \tan (c+d x)\right ) \sqrt {e \sec (c+d x)}}{5 d}+\frac {6 a^3 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{d}+\frac {2 i a (a+i a \tan (c+d x))^2 \sqrt {e \sec (c+d x)}}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3498, 3486, 3771, 2641} \[ \frac {6 i a^3 \sqrt {e \sec (c+d x)}}{d}+\frac {6 i \left (a^3+i a^3 \tan (c+d x)\right ) \sqrt {e \sec (c+d x)}}{5 d}+\frac {6 a^3 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{d}+\frac {2 i a (a+i a \tan (c+d x))^2 \sqrt {e \sec (c+d x)}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((6*I)*a^3*Sqrt[e*Sec[c + d*x]])/d + (6*a^3*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])
/d + (((2*I)/5)*a*Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^2)/d + (((6*I)/5)*Sqrt[e*Sec[c + d*x]]*(a^3 + I*
a^3*Tan[c + d*x]))/d

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(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 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 \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^3 \, dx &=\frac {2 i a \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^2}{5 d}+\frac {1}{5} (9 a) \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^2 \, dx\\ &=\frac {2 i a \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^2}{5 d}+\frac {6 i \sqrt {e \sec (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}+\left (3 a^2\right ) \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x)) \, dx\\ &=\frac {6 i a^3 \sqrt {e \sec (c+d x)}}{d}+\frac {2 i a \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^2}{5 d}+\frac {6 i \sqrt {e \sec (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}+\left (3 a^3\right ) \int \sqrt {e \sec (c+d x)} \, dx\\ &=\frac {6 i a^3 \sqrt {e \sec (c+d x)}}{d}+\frac {2 i a \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^2}{5 d}+\frac {6 i \sqrt {e \sec (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}+\left (3 a^3 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {6 i a^3 \sqrt {e \sec (c+d x)}}{d}+\frac {6 a^3 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{d}+\frac {2 i a \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^2}{5 d}+\frac {6 i \sqrt {e \sec (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\\ \end {align*}

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Mathematica [A]  time = 1.45, size = 79, normalized size = 0.57 \[ \frac {a^3 \sec ^2(c+d x) \sqrt {e \sec (c+d x)} \left (-5 \sin (2 (c+d x))+20 i \cos (2 (c+d x))+30 \cos ^{\frac {5}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+18 i\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Sec[c + d*x]^2*Sqrt[e*Sec[c + d*x]]*(18*I + (20*I)*Cos[2*(c + d*x)] + 30*Cos[c + d*x]^(5/2)*EllipticF[(c
+ d*x)/2, 2] - 5*Sin[2*(c + d*x)]))/(5*d)

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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ \frac {\sqrt {2} {\left (50 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 72 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 30 i \, a^{3}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 5 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} {\rm integral}\left (-\frac {3 i \, \sqrt {2} a^{3} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )}}{d}, x\right )}{5 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/5*(sqrt(2)*(50*I*a^3*e^(4*I*d*x + 4*I*c) + 72*I*a^3*e^(2*I*d*x + 2*I*c) + 30*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^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*integral(-3*I*sqrt
(2)*a^3*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^(4*I*d*x + 4*I*c) + 2*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 \[ \int \sqrt {e \sec \left (d x + c\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.98, size = 213, normalized size = 1.53 \[ \frac {2 a^{3} \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (15 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \left (\cos ^{3}\left (d x +c \right )\right )+15 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+20 i \left (\cos ^{2}\left (d x +c \right )\right )-5 \cos \left (d x +c \right ) \sin \left (d x +c \right )-i\right ) \sqrt {\frac {e}{\cos \left (d x +c \right )}}}{5 d \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*a^3/d*(1+cos(d*x+c))^2*(-1+cos(d*x+c))^2*(15*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)^3+15*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+20*I*cos(d*x+c)^2-5*cos(d*x+c)*sin(d*x+c)-I)*(
e/cos(d*x+c))^(1/2)/cos(d*x+c)^2/sin(d*x+c)^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))**(1/2)*(a+I*a*tan(d*x+c))**3,x)

[Out]

-I*a**3*(Integral(I*sqrt(e*sec(c + d*x)), x) + Integral(-3*sqrt(e*sec(c + d*x))*tan(c + d*x), x) + Integral(sq
rt(e*sec(c + d*x))*tan(c + d*x)**3, x) + Integral(-3*I*sqrt(e*sec(c + d*x))*tan(c + d*x)**2, x))

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