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

Optimal. Leaf size=83 \[ -\frac{6 i 2^{5/6} \left (a^2+i a^2 \tan (e+f x)\right ) \text{Hypergeometric2F1}\left (-\frac{5}{6},-\frac{1}{6},\frac{5}{6},\frac{1}{2} (1-i \tan (e+f x))\right )}{f (1+i \tan (e+f x))^{5/6} \sqrt [3]{d \sec (e+f x)}} \]

[Out]

((-6*I)*2^(5/6)*Hypergeometric2F1[-5/6, -1/6, 5/6, (1 - I*Tan[e + f*x])/2]*(a^2 + I*a^2*Tan[e + f*x]))/(f*(d*S
ec[e + f*x])^(1/3)*(1 + I*Tan[e + f*x])^(5/6))

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Rubi [A]  time = 0.165748, antiderivative size = 83, 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} \[ -\frac{6 i 2^{5/6} \left (a^2+i a^2 \tan (e+f x)\right ) \text{Hypergeometric2F1}\left (-\frac{5}{6},-\frac{1}{6},\frac{5}{6},\frac{1}{2} (1-i \tan (e+f x))\right )}{f (1+i \tan (e+f x))^{5/6} \sqrt [3]{d \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-6*I)*2^(5/6)*Hypergeometric2F1[-5/6, -1/6, 5/6, (1 - I*Tan[e + f*x])/2]*(a^2 + I*a^2*Tan[e + f*x]))/(f*(d*S
ec[e + f*x])^(1/3)*(1 + I*Tan[e + f*x])^(5/6))

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 \frac{(a+i a \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx &=\frac{\left (\sqrt [6]{a-i a \tan (e+f x)} \sqrt [6]{a+i a \tan (e+f x)}\right ) \int \frac{(a+i a \tan (e+f x))^{11/6}}{\sqrt [6]{a-i a \tan (e+f x)}} \, dx}{\sqrt [3]{d \sec (e+f x)}}\\ &=\frac{\left (a^2 \sqrt [6]{a-i a \tan (e+f x)} \sqrt [6]{a+i a \tan (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{(a+i a x)^{5/6}}{(a-i a x)^{7/6}} \, dx,x,\tan (e+f x)\right )}{f \sqrt [3]{d \sec (e+f x)}}\\ &=\frac{\left (2^{5/6} a^2 \sqrt [6]{a-i a \tan (e+f x)} (a+i a \tan (e+f x))\right ) \operatorname{Subst}\left (\int \frac{\left (\frac{1}{2}+\frac{i x}{2}\right )^{5/6}}{(a-i a x)^{7/6}} \, dx,x,\tan (e+f x)\right )}{f \sqrt [3]{d \sec (e+f x)} \left (\frac{a+i a \tan (e+f x)}{a}\right )^{5/6}}\\ &=-\frac{6 i 2^{5/6} \, _2F_1\left (-\frac{5}{6},-\frac{1}{6};\frac{5}{6};\frac{1}{2} (1-i \tan (e+f x))\right ) \left (a^2+i a^2 \tan (e+f x)\right )}{f \sqrt [3]{d \sec (e+f x)} (1+i \tan (e+f x))^{5/6}}\\ \end{align*}

Mathematica [A]  time = 1.11866, size = 132, normalized size = 1.59 \[ -\frac{3 i a^2 e^{2 i (e+f x)} \left (\left (1+e^{2 i (e+f x)}\right ) \text{Hypergeometric2F1}\left (\frac{2}{3},\frac{5}{6},\frac{11}{6},-e^{2 i (e+f x)}\right )-\sqrt [3]{1+e^{2 i (e+f x)}}\right )}{\sqrt [3]{2} f \left (1+e^{2 i (e+f x)}\right )^{4/3} \sqrt [3]{\frac{d e^{i (e+f x)}}{1+e^{2 i (e+f x)}}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*I)*a^2*E^((2*I)*(e + f*x))*(-(1 + E^((2*I)*(e + f*x)))^(1/3) + (1 + E^((2*I)*(e + f*x)))*Hypergeometric2F
1[2/3, 5/6, 11/6, -E^((2*I)*(e + f*x))]))/(2^(1/3)*((d*E^(I*(e + f*x)))/(1 + E^((2*I)*(e + f*x))))^(1/3)*(1 +
E^((2*I)*(e + f*x)))^(4/3)*f)

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Maple [F]  time = 0.129, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+ia\tan \left ( fx+e \right ) \right ) ^{2}{\frac{1}{\sqrt [3]{d\sec \left ( fx+e \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{2^{\frac{2}{3}}{\left (-12 i \, a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 3 i \, a^{2} e^{\left (i \, f x + i \, e\right )} - 15 i \, a^{2}\right )} \left (\frac{d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac{2}{3}} e^{\left (\frac{2}{3} i \, f x + \frac{2}{3} i \, e\right )} + 2 \,{\left (d f e^{\left (i \, f x + i \, e\right )} - d f\right )}{\rm integral}\left (\frac{2^{\frac{2}{3}}{\left (-5 i \, a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 5 i \, a^{2} e^{\left (i \, f x + i \, e\right )} - 5 i \, a^{2}\right )} \left (\frac{d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac{2}{3}} e^{\left (\frac{2}{3} i \, f x + \frac{2}{3} i \, e\right )}}{d f e^{\left (3 i \, f x + 3 i \, e\right )} - 2 \, d f e^{\left (2 i \, f x + 2 i \, e\right )} + d f e^{\left (i \, f x + i \, e\right )}}, x\right )}{2 \,{\left (d f e^{\left (i \, f x + i \, e\right )} - d f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2^(2/3)*(-12*I*a^2*e^(2*I*f*x + 2*I*e) - 3*I*a^2*e^(I*f*x + I*e) - 15*I*a^2)*(d/(e^(2*I*f*x + 2*I*e) + 1)
)^(2/3)*e^(2/3*I*f*x + 2/3*I*e) + 2*(d*f*e^(I*f*x + I*e) - d*f)*integral(2^(2/3)*(-5*I*a^2*e^(2*I*f*x + 2*I*e)
- 5*I*a^2*e^(I*f*x + I*e) - 5*I*a^2)*(d/(e^(2*I*f*x + 2*I*e) + 1))^(2/3)*e^(2/3*I*f*x + 2/3*I*e)/(d*f*e^(3*I*
f*x + 3*I*e) - 2*d*f*e^(2*I*f*x + 2*I*e) + d*f*e^(I*f*x + I*e)), x))/(d*f*e^(I*f*x + I*e) - d*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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