3.2.16 \(\int \sqrt [3]{a+a \sin (e+f x)} \tan ^2(e+f x) \, dx\) [116]

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

[Out]

-5/6*a*cos(f*x+e)*hypergeom([1/2, 7/6],[3/2],1/2-1/2*sin(f*x+e))*(1+sin(f*x+e))^(1/6)*2^(5/6)/f/(a+a*sin(f*x+e
))^(2/3)+7*sec(f*x+e)*(a+a*sin(f*x+e))^(1/3)/f-3*sec(f*x+e)*(a+a*sin(f*x+e))^(4/3)/a/f

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Rubi [A]
time = 0.15, antiderivative size = 123, normalized size of antiderivative = 1.00, 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 = {2792, 2934, 2731, 2730} \begin {gather*} -\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} \end {gather*}

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 2730

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2^(n + 1/2))*a^(n - 1/2)*b*(Cos[c + d*x]/
(d*Sqrt[a + b*Sin[c + d*x]]))*Hypergeometric2F1[1/2, 1/2 - n, 3/2, (1/2)*(1 - b*(Sin[c + d*x]/a))], x] /; Free
Q[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]

Rule 2731

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

Rule 2792

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 2934

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]

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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 9.37, size = 290, normalized size = 2.36 \begin {gather*} \frac {\sqrt [3]{a (1+\sin (e+f x))} \left (\frac {\left (\frac {3}{2}+\frac {3 i}{2}\right ) (-1)^{3/4} e^{-i (e+f x)} \left (-20 e^{i (e+f x)} \sqrt {\cos ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right )} \, _2F_1\left (-\frac {1}{3},\frac {1}{3};\frac {2}{3};-i e^{-i (e+f x)}\right )+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+\pi +2 f x)\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)}\right )}{\sqrt {2} \left (1+i e^{-i (e+f x)}\right )^{2/3} \sqrt {i e^{-i (e+f x)} \left (-i+e^{i (e+f x)}\right )^2}}-3 (5+\sec (e+f x)-2 \tan (e+f x))\right )}{3 f} \end {gather*}

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 = 0.14, size = 0, normalized size = 0.00 \[\int \left (a +a \sin \left (f x +e \right )\right )^{\frac {1}{3}} \left (\tan ^{2}\left (f x +e \right )\right )\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt [3]{a \left (\sin {\left (e + f x \right )} + 1\right )} \tan ^{2}{\left (e + f x \right )}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {tan}\left (e+f\,x\right )}^2\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{1/3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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