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*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

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Rubi [A]  time = 0.139344, antiderivative size = 139, normalized size of antiderivative = 1., 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 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 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 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]

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*}

Mathematica [A]  time = 1.24938, 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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Maple [A]  time = 0.276, size = 213, normalized size = 1.5 \begin{align*}{\frac{2\,{a}^{3} \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2} \left ( \cos \left ( dx+c \right ) -1 \right ) ^{2}}{5\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{4}} \left ( 15\,i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \left ( \cos \left ( dx+c \right ) \right ) ^{3}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) +15\,i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \left ( \cos \left ( dx+c \right ) \right ) ^{2}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) +20\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}-5\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) -i \right ) \sqrt{{\frac{e}{\cos \left ( dx+c \right ) }}}} \end{align*}

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{e \sec \left (d x + c\right )}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 )}} \end{align*}

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

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]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{e \sec \left (d x + c\right )}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}\,{d x} \end{align*}

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)