3.120 \(\int \frac{\tan ^2(e+f x)}{\sqrt [3]{a+a \sin (e+f x)}} \, dx\)

Optimal. Leaf size=126 \[ \frac{11 \sqrt [6]{2} \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{15 f \sqrt [6]{\sin (e+f x)+1} \sqrt [3]{a \sin (e+f x)+a}}+\frac{4 \sec (e+f x) (a \sin (e+f x)+a)^{2/3}}{5 a f}-\frac{3 \sec (e+f x)}{5 f \sqrt [3]{a \sin (e+f x)+a}} \]

[Out]

(-3*Sec[e + f*x])/(5*f*(a + a*Sin[e + f*x])^(1/3)) + (11*2^(1/6)*Cos[e + f*x]*Hypergeometric2F1[1/2, 5/6, 3/2,
(1 - Sin[e + f*x])/2])/(15*f*(1 + Sin[e + f*x])^(1/6)*(a + a*Sin[e + f*x])^(1/3)) + (4*Sec[e + f*x]*(a + a*Si
n[e + f*x])^(2/3))/(5*a*f)

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Rubi [A]  time = 0.216494, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2712, 2855, 2652, 2651} \[ \frac{11 \sqrt [6]{2} \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{15 f \sqrt [6]{\sin (e+f x)+1} \sqrt [3]{a \sin (e+f x)+a}}+\frac{4 \sec (e+f x) (a \sin (e+f x)+a)^{2/3}}{5 a f}-\frac{3 \sec (e+f x)}{5 f \sqrt [3]{a \sin (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Sec[e + f*x])/(5*f*(a + a*Sin[e + f*x])^(1/3)) + (11*2^(1/6)*Cos[e + f*x]*Hypergeometric2F1[1/2, 5/6, 3/2,
(1 - Sin[e + f*x])/2])/(15*f*(1 + Sin[e + f*x])^(1/6)*(a + a*Sin[e + f*x])^(1/3)) + (4*Sec[e + f*x]*(a + a*Si
n[e + f*x])^(2/3))/(5*a*f)

Rule 2712

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^2, x_Symbol] :> Simp[(b*(a + b*Sin[e +
f*x])^m)/(a*f*(2*m - 1)*Cos[e + f*x]), x] - Dist[1/(a^2*(2*m - 1)), Int[((a + b*Sin[e + f*x])^(m + 1)*(a*m -
b*(2*m - 1)*Sin[e + f*x]))/Cos[e + f*x]^2, x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ
[m] && LtQ[m, 0]

Rule 2855

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[((b*c + a*d)*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(p +
1)), x] + Dist[(b*(a*d*m + b*c*(m + p + 1)))/(a*g^2*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x]
)^(m - 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, -1] && LtQ[p, -1]

Rule 2652

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(a^IntPart[n]*(a + b*Sin[c + d*x])^FracPart
[n])/(1 + (b*Sin[c + d*x])/a)^FracPart[n], Int[(1 + (b*Sin[c + d*x])/a)^n, x], x] /; FreeQ[{a, b, c, d, n}, x]
&& EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] &&  !GtQ[a, 0]

Rule 2651

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(2^(n + 1/2)*a^(n - 1/2)*b*Cos[c + d*x]*Hy
pergeometric2F1[1/2, 1/2 - n, 3/2, (1*(1 - (b*Sin[c + d*x])/a))/2])/(d*Sqrt[a + b*Sin[c + d*x]]), x] /; FreeQ[
{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]

Rubi steps

\begin{align*} \int \frac{\tan ^2(e+f x)}{\sqrt [3]{a+a \sin (e+f x)}} \, dx &=-\frac{3 \sec (e+f x)}{5 f \sqrt [3]{a+a \sin (e+f x)}}+\frac{3 \int \sec ^2(e+f x) (a+a \sin (e+f x))^{2/3} \left (-\frac{a}{3}+\frac{5}{3} a \sin (e+f x)\right ) \, dx}{5 a^2}\\ &=-\frac{3 \sec (e+f x)}{5 f \sqrt [3]{a+a \sin (e+f x)}}+\frac{4 \sec (e+f x) (a+a \sin (e+f x))^{2/3}}{5 a f}-\frac{11}{15} \int \frac{1}{\sqrt [3]{a+a \sin (e+f x)}} \, dx\\ &=-\frac{3 \sec (e+f x)}{5 f \sqrt [3]{a+a \sin (e+f x)}}+\frac{4 \sec (e+f x) (a+a \sin (e+f x))^{2/3}}{5 a f}-\frac{\left (11 \sqrt [3]{1+\sin (e+f x)}\right ) \int \frac{1}{\sqrt [3]{1+\sin (e+f x)}} \, dx}{15 \sqrt [3]{a+a \sin (e+f x)}}\\ &=-\frac{3 \sec (e+f x)}{5 f \sqrt [3]{a+a \sin (e+f x)}}+\frac{11 \sqrt [6]{2} \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{15 f \sqrt [6]{1+\sin (e+f x)} \sqrt [3]{a+a \sin (e+f x)}}+\frac{4 \sec (e+f x) (a+a \sin (e+f x))^{2/3}}{5 a f}\\ \end{align*}

Mathematica [A]  time = 0.513245, size = 100, normalized size = 0.79 \[ \frac{\sqrt{2-2 \sin (e+f x)} (4 \tan (e+f x)+\sec (e+f x))-22 \cos (e+f x) \, _2F_1\left (\frac{1}{6},\frac{1}{2};\frac{7}{6};\sin ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right )\right )}{5 f \sqrt{2-2 \sin (e+f x)} \sqrt [3]{a (\sin (e+f x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-22*Cos[e + f*x]*Hypergeometric2F1[1/6, 1/2, 7/6, Sin[(2*e + Pi + 2*f*x)/4]^2] + Sqrt[2 - 2*Sin[e + f*x]]*(Se
c[e + f*x] + 4*Tan[e + f*x]))/(5*f*Sqrt[2 - 2*Sin[e + f*x]]*(a*(1 + Sin[e + f*x]))^(1/3))

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Maple [F]  time = 0.105, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \tan \left ( fx+e \right ) \right ) ^{2}{\frac{1}{\sqrt [3]{a+a\sin \left ( fx+e \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(tan(f*x + e)^2/(a*sin(f*x + e) + a)^(1/3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)**2/(a+a*sin(f*x+e))**(1/3),x)

[Out]

Integral(tan(e + f*x)**2/(a*(sin(e + f*x) + 1))**(1/3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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