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

Optimal. Leaf size=123 \[ -\frac{5 a \sqrt [6]{\sin (e+f x)+1} \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{7}{6};\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{3 \sqrt [6]{2} f (a \sin (e+f x)+a)^{2/3}}-\frac{3 \sec (e+f x) (a \sin (e+f x)+a)^{4/3}}{a f}+\frac{7 \sec (e+f x) \sqrt [3]{a \sin (e+f x)+a}}{f} \]

[Out]

(-5*a*Cos[e + f*x]*Hypergeometric2F1[1/2, 7/6, 3/2, (1 - Sin[e + f*x])/2]*(1 + Sin[e + f*x])^(1/6))/(3*2^(1/6)
*f*(a + a*Sin[e + f*x])^(2/3)) + (7*Sec[e + f*x]*(a + a*Sin[e + f*x])^(1/3))/f - (3*Sec[e + f*x]*(a + a*Sin[e
+ f*x])^(4/3))/(a*f)

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Rubi [A]  time = 0.19337, antiderivative size = 123, 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 = {2713, 2855, 2652, 2651} \[ -\frac{5 a \sqrt [6]{\sin (e+f x)+1} \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{7}{6};\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{3 \sqrt [6]{2} f (a \sin (e+f x)+a)^{2/3}}-\frac{3 \sec (e+f x) (a \sin (e+f x)+a)^{4/3}}{a f}+\frac{7 \sec (e+f x) \sqrt [3]{a \sin (e+f x)+a}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*a*Cos[e + f*x]*Hypergeometric2F1[1/2, 7/6, 3/2, (1 - Sin[e + f*x])/2]*(1 + Sin[e + f*x])^(1/6))/(3*2^(1/6)
*f*(a + a*Sin[e + f*x])^(2/3)) + (7*Sec[e + f*x]*(a + a*Sin[e + f*x])^(1/3))/f - (3*Sec[e + f*x]*(a + a*Sin[e
+ f*x])^(4/3))/(a*f)

Rule 2713

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^2, x_Symbol] :> -Simp[(a + b*Sin[e + f
*x])^(m + 1)/(b*f*m*Cos[e + f*x]), x] + Dist[1/(b*m), Int[((a + b*Sin[e + f*x])^m*(b*(m + 1) + a*Sin[e + f*x])
)/Cos[e + f*x]^2, x], x] /; FreeQ[{a, b, e, f, m}, 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 \sqrt [3]{a+a \sin (e+f x)} \tan ^2(e+f x) \, dx &=-\frac{3 \sec (e+f x) (a+a \sin (e+f x))^{4/3}}{a f}+\frac{3 \int \sec ^2(e+f x) \sqrt [3]{a+a \sin (e+f x)} \left (\frac{4 a}{3}+a \sin (e+f x)\right ) \, dx}{a}\\ &=\frac{7 \sec (e+f x) \sqrt [3]{a+a \sin (e+f x)}}{f}-\frac{3 \sec (e+f x) (a+a \sin (e+f x))^{4/3}}{a f}+\frac{1}{3} (5 a) \int \frac{1}{(a+a \sin (e+f x))^{2/3}} \, dx\\ &=\frac{7 \sec (e+f x) \sqrt [3]{a+a \sin (e+f x)}}{f}-\frac{3 \sec (e+f x) (a+a \sin (e+f x))^{4/3}}{a f}+\frac{\left (5 a (1+\sin (e+f x))^{2/3}\right ) \int \frac{1}{(1+\sin (e+f x))^{2/3}} \, dx}{3 (a+a \sin (e+f x))^{2/3}}\\ &=-\frac{5 a \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{7}{6};\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right ) \sqrt [6]{1+\sin (e+f x)}}{3 \sqrt [6]{2} f (a+a \sin (e+f x))^{2/3}}+\frac{7 \sec (e+f x) \sqrt [3]{a+a \sin (e+f x)}}{f}-\frac{3 \sec (e+f x) (a+a \sin (e+f x))^{4/3}}{a f}\\ \end{align*}

Mathematica [C]  time = 2.746, size = 290, normalized size = 2.36 \[ \frac{\sqrt [3]{a (\sin (e+f x)+1)} \left (-3 (-2 \tan (e+f x)+\sec (e+f x)+5)+\frac{\left (\frac{3}{2}+\frac{3 i}{2}\right ) (-1)^{3/4} e^{-i (e+f x)} \left (2 \left (1+i e^{-i (e+f x)}\right )^{2/3} \left (1+e^{2 i (e+f x)}\right ) \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{11}{6};\sin ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right )\right )-5 i \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{5}{3};-i e^{-i (e+f x)}\right ) \sqrt{2-2 \sin (e+f x)}-20 e^{i (e+f x)} \, _2F_1\left (-\frac{1}{3},\frac{1}{3};\frac{2}{3};-i e^{-i (e+f x)}\right ) \sqrt{\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right )}\right )}{\sqrt{2} \left (1+i e^{-i (e+f x)}\right )^{2/3} \sqrt{i e^{-i (e+f x)} \left (e^{i (e+f x)}-i\right )^2}}\right )}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*(1 + Sin[e + f*x]))^(1/3)*(((3/2 + (3*I)/2)*(-1)^(3/4)*(-20*E^(I*(e + f*x))*Sqrt[Cos[(2*e + Pi + 2*f*x)/4]
^2]*Hypergeometric2F1[-1/3, 1/3, 2/3, (-I)/E^(I*(e + f*x))] + 2*(1 + I/E^(I*(e + f*x)))^(2/3)*(1 + E^((2*I)*(e
+ f*x)))*Hypergeometric2F1[1/2, 5/6, 11/6, Sin[(2*e + Pi + 2*f*x)/4]^2] - (5*I)*Hypergeometric2F1[1/3, 2/3, 5
/3, (-I)/E^(I*(e + f*x))]*Sqrt[2 - 2*Sin[e + f*x]]))/(Sqrt[2]*E^(I*(e + f*x))*(1 + I/E^(I*(e + f*x)))^(2/3)*Sq
rt[(I*(-I + E^(I*(e + f*x)))^2)/E^(I*(e + f*x))]) - 3*(5 + Sec[e + f*x] - 2*Tan[e + f*x])))/(3*f)

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Maple [F]  time = 0.104, size = 0, normalized size = 0. \begin{align*} \int \sqrt [3]{a+a\sin \left ( fx+e \right ) } \left ( \tan \left ( fx+e \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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