3.105 \(\int \frac{\sin ^2(c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx\)

Optimal. Leaf size=126 \[ -\frac{7 \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{5\ 2^{5/6} d \sqrt [6]{\sin (c+d x)+1} \sqrt [3]{a \sin (c+d x)+a}}-\frac{3 \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{5 a d}+\frac{9 \cos (c+d x)}{10 d \sqrt [3]{a \sin (c+d x)+a}} \]

[Out]

(9*Cos[c + d*x])/(10*d*(a + a*Sin[c + d*x])^(1/3)) - (7*Cos[c + d*x]*Hypergeometric2F1[1/2, 5/6, 3/2, (1 - Sin
[c + d*x])/2])/(5*2^(5/6)*d*(1 + Sin[c + d*x])^(1/6)*(a + a*Sin[c + d*x])^(1/3)) - (3*Cos[c + d*x]*(a + a*Sin[
c + d*x])^(2/3))/(5*a*d)

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Rubi [A]  time = 0.131674, 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 = {2759, 2751, 2652, 2651} \[ -\frac{7 \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{5\ 2^{5/6} d \sqrt [6]{\sin (c+d x)+1} \sqrt [3]{a \sin (c+d x)+a}}-\frac{3 \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{5 a d}+\frac{9 \cos (c+d x)}{10 d \sqrt [3]{a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^2/(a + a*Sin[c + d*x])^(1/3),x]

[Out]

(9*Cos[c + d*x])/(10*d*(a + a*Sin[c + d*x])^(1/3)) - (7*Cos[c + d*x]*Hypergeometric2F1[1/2, 5/6, 3/2, (1 - Sin
[c + d*x])/2])/(5*2^(5/6)*d*(1 + Sin[c + d*x])^(1/6)*(a + a*Sin[c + d*x])^(1/3)) - (3*Cos[c + d*x]*(a + a*Sin[
c + d*x])^(2/3))/(5*a*d)

Rule 2759

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

Rule 2751

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*Sin
[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m,
-2^(-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{\sin ^2(c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx &=-\frac{3 \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{5 a d}+\frac{3 \int \frac{\frac{2 a}{3}-a \sin (c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx}{5 a}\\ &=\frac{9 \cos (c+d x)}{10 d \sqrt [3]{a+a \sin (c+d x)}}-\frac{3 \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{5 a d}+\frac{7}{10} \int \frac{1}{\sqrt [3]{a+a \sin (c+d x)}} \, dx\\ &=\frac{9 \cos (c+d x)}{10 d \sqrt [3]{a+a \sin (c+d x)}}-\frac{3 \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{5 a d}+\frac{\left (7 \sqrt [3]{1+\sin (c+d x)}\right ) \int \frac{1}{\sqrt [3]{1+\sin (c+d x)}} \, dx}{10 \sqrt [3]{a+a \sin (c+d x)}}\\ &=\frac{9 \cos (c+d x)}{10 d \sqrt [3]{a+a \sin (c+d x)}}-\frac{7 \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{5\ 2^{5/6} d \sqrt [6]{1+\sin (c+d x)} \sqrt [3]{a+a \sin (c+d x)}}-\frac{3 \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{5 a d}\\ \end{align*}

Mathematica [A]  time = 0.296551, size = 95, normalized size = 0.75 \[ -\frac{3 \cos (c+d x) \left (\sqrt{2-2 \sin (c+d x)} (2 \sin (c+d x)-1)-14 \, _2F_1\left (\frac{1}{6},\frac{1}{2};\frac{7}{6};\sin ^2\left (\frac{1}{4} (2 c+2 d x+\pi )\right )\right )\right )}{10 d \sqrt{2-2 \sin (c+d x)} \sqrt [3]{a (\sin (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^2/(a + a*Sin[c + d*x])^(1/3),x]

[Out]

(-3*Cos[c + d*x]*(-14*Hypergeometric2F1[1/6, 1/2, 7/6, Sin[(2*c + Pi + 2*d*x)/4]^2] + Sqrt[2 - 2*Sin[c + d*x]]
*(-1 + 2*Sin[c + d*x])))/(10*d*Sqrt[2 - 2*Sin[c + d*x]]*(a*(1 + Sin[c + d*x]))^(1/3))

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Maple [F]  time = 0.162, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \sin \left ( dx+c \right ) \right ) ^{2}{\frac{1}{\sqrt [3]{a+a\sin \left ( dx+c \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^2/(a+a*sin(d*x+c))^(1/3),x)

[Out]

int(sin(d*x+c)^2/(a+a*sin(d*x+c))^(1/3),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^2/(a+a*sin(d*x+c))^(1/3),x, algorithm="maxima")

[Out]

integrate(sin(d*x + c)^2/(a*sin(d*x + c) + a)^(1/3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^2/(a+a*sin(d*x+c))^(1/3),x, algorithm="fricas")

[Out]

integral(-(cos(d*x + c)^2 - 1)/(a*sin(d*x + c) + a)^(1/3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)**2/(a+a*sin(d*x+c))**(1/3),x)

[Out]

Integral(sin(c + d*x)**2/(a*(sin(c + d*x) + 1))**(1/3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^2/(a+a*sin(d*x+c))^(1/3),x, algorithm="giac")

[Out]

integrate(sin(d*x + c)^2/(a*sin(d*x + c) + a)^(1/3), x)