3.268 \(\int \sqrt [3]{d \sec (e+f x)} (a+i a \tan (e+f x))^2 \, dx\)

Optimal. Leaf size=69 \[ \frac{12 i \sqrt [6]{2} a^2 \sqrt [3]{d \sec (e+f x)} \text{Hypergeometric2F1}\left (-\frac{7}{6},\frac{1}{6},\frac{7}{6},\frac{1}{2} (1-i \tan (e+f x))\right )}{f \sqrt [6]{1+i \tan (e+f x)}} \]

[Out]

((12*I)*2^(1/6)*a^2*Hypergeometric2F1[-7/6, 1/6, 7/6, (1 - I*Tan[e + f*x])/2]*(d*Sec[e + f*x])^(1/3))/(f*(1 +
I*Tan[e + f*x])^(1/6))

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Rubi [A]  time = 0.157643, antiderivative size = 69, 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 = {3505, 3523, 70, 69} \[ \frac{12 i \sqrt [6]{2} a^2 \sqrt [3]{d \sec (e+f x)} \text{Hypergeometric2F1}\left (-\frac{7}{6},\frac{1}{6},\frac{7}{6},\frac{1}{2} (1-i \tan (e+f x))\right )}{f \sqrt [6]{1+i \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(d*Sec[e + f*x])^(1/3)*(a + I*a*Tan[e + f*x])^2,x]

[Out]

((12*I)*2^(1/6)*a^2*Hypergeometric2F1[-7/6, 1/6, 7/6, (1 - I*Tan[e + f*x])/2]*(d*Sec[e + f*x])^(1/3))/(f*(1 +
I*Tan[e + f*x])^(1/6))

Rule 3505

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(d*S
ec[e + f*x])^m/((a + b*Tan[e + f*x])^(m/2)*(a - b*Tan[e + f*x])^(m/2)), Int[(a + b*Tan[e + f*x])^(m/2 + n)*(a
- b*Tan[e + f*x])^(m/2), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0]

Rule 3523

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist
[(a*c)/f, Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f,
m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*Simp[(b*c)/(b*c - a*d) + (b*d*x)/(b*c -
a*d), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

\begin{align*} \int \sqrt [3]{d \sec (e+f x)} (a+i a \tan (e+f x))^2 \, dx &=\frac{\sqrt [3]{d \sec (e+f x)} \int \sqrt [6]{a-i a \tan (e+f x)} (a+i a \tan (e+f x))^{13/6} \, dx}{\sqrt [6]{a-i a \tan (e+f x)} \sqrt [6]{a+i a \tan (e+f x)}}\\ &=\frac{\left (a^2 \sqrt [3]{d \sec (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{(a+i a x)^{7/6}}{(a-i a x)^{5/6}} \, dx,x,\tan (e+f x)\right )}{f \sqrt [6]{a-i a \tan (e+f x)} \sqrt [6]{a+i a \tan (e+f x)}}\\ &=\frac{\left (2 \sqrt [6]{2} a^3 \sqrt [3]{d \sec (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{\left (\frac{1}{2}+\frac{i x}{2}\right )^{7/6}}{(a-i a x)^{5/6}} \, dx,x,\tan (e+f x)\right )}{f \sqrt [6]{a-i a \tan (e+f x)} \sqrt [6]{\frac{a+i a \tan (e+f x)}{a}}}\\ &=\frac{12 i \sqrt [6]{2} a^2 \, _2F_1\left (-\frac{7}{6},\frac{1}{6};\frac{7}{6};\frac{1}{2} (1-i \tan (e+f x))\right ) \sqrt [3]{d \sec (e+f x)}}{f \sqrt [6]{1+i \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.965874, size = 128, normalized size = 1.86 \[ -\frac{3 a^2 e^{-2 i e} \sqrt [3]{d \sec (e+f x)} (\cos (2 (e+f x))+i \sin (2 (e+f x))) \left (7 i \sqrt [3]{1+e^{2 i (e+f x)}} \text{Hypergeometric2F1}\left (\frac{1}{6},\frac{1}{3},\frac{7}{6},-e^{2 i (e+f x)}\right )+\sec (e) \sin (f x) \sec (e+f x)+\tan (e)-8 i\right )}{4 f (\cos (f x)+i \sin (f x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Sec[e + f*x])^(1/3)*(a + I*a*Tan[e + f*x])^2,x]

[Out]

(-3*a^2*(d*Sec[e + f*x])^(1/3)*(Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)])*(-8*I + (7*I)*(1 + E^((2*I)*(e + f*x)))
^(1/3)*Hypergeometric2F1[1/6, 1/3, 7/6, -E^((2*I)*(e + f*x))] + Sec[e]*Sec[e + f*x]*Sin[f*x] + Tan[e]))/(4*E^(
(2*I)*e)*f*(Cos[f*x] + I*Sin[f*x])^2)

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Maple [F]  time = 0.132, size = 0, normalized size = 0. \begin{align*} \int \sqrt [3]{d\sec \left ( fx+e \right ) } \left ( a+ia\tan \left ( fx+e \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \sec \left (f x + e\right )\right )^{\frac{1}{3}}{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{2^{\frac{1}{3}}{\left (27 i \, a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 21 i \, a^{2}\right )} \left (\frac{d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac{1}{3}} e^{\left (\frac{1}{3} i \, f x + \frac{1}{3} i \, e\right )} + 4 \,{\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}{\rm integral}\left (-\frac{7 i \cdot 2^{\frac{1}{3}} a^{2} \left (\frac{d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac{1}{3}} e^{\left (-\frac{2}{3} i \, f x - \frac{2}{3} i \, e\right )}}{4 \, f}, x\right )}{4 \,{\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2^(1/3)*(27*I*a^2*e^(2*I*f*x + 2*I*e) + 21*I*a^2)*(d/(e^(2*I*f*x + 2*I*e) + 1))^(1/3)*e^(1/3*I*f*x + 1/3*
I*e) + 4*(f*e^(2*I*f*x + 2*I*e) + f)*integral(-7/4*I*2^(1/3)*a^2*(d/(e^(2*I*f*x + 2*I*e) + 1))^(1/3)*e^(-2/3*I
*f*x - 2/3*I*e)/f, x))/(f*e^(2*I*f*x + 2*I*e) + f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral((d*sec(e + f*x))**(1/3), x) + Integral(-(d*sec(e + f*x))**(1/3)*tan(e + f*x)**2, x) + Integral(
2*I*(d*sec(e + f*x))**(1/3)*tan(e + f*x), x))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \sec \left (f x + e\right )\right )^{\frac{1}{3}}{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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