3.94 \(\int \sin (c+d x) (a+a \sin (c+d x))^{2/3} \, dx\)

Optimal. Leaf size=96 \[ -\frac{4 \sqrt [6]{2} \cos (c+d x) (a \sin (c+d x)+a)^{2/3} \, _2F_1\left (-\frac{1}{6},\frac{1}{2};\frac{3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{5 d (\sin (c+d x)+1)^{7/6}}-\frac{3 \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.0682446, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2751, 2652, 2651} \[ -\frac{4 \sqrt [6]{2} \cos (c+d x) (a \sin (c+d x)+a)^{2/3} \, _2F_1\left (-\frac{1}{6},\frac{1}{2};\frac{3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{5 d (\sin (c+d x)+1)^{7/6}}-\frac{3 \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 \sin (c+d x) (a+a \sin (c+d x))^{2/3} \, dx &=-\frac{3 \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{5 d}+\frac{2}{5} \int (a+a \sin (c+d x))^{2/3} \, dx\\ &=-\frac{3 \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{5 d}+\frac{\left (2 (a+a \sin (c+d x))^{2/3}\right ) \int (1+\sin (c+d x))^{2/3} \, dx}{5 (1+\sin (c+d x))^{2/3}}\\ &=-\frac{3 \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{5 d}-\frac{4 \sqrt [6]{2} \cos (c+d x) \, _2F_1\left (-\frac{1}{6},\frac{1}{2};\frac{3}{2};\frac{1}{2} (1-\sin (c+d x))\right ) (a+a \sin (c+d x))^{2/3}}{5 d (1+\sin (c+d x))^{7/6}}\\ \end{align*}

Mathematica [A]  time = 0.230909, size = 138, normalized size = 1.44 \[ -\frac{3 (a (\sin (c+d x)+1))^{2/3} \left (\sqrt{1-\sin (c+d x)} (\sin (c+d x)+2)-\sqrt{2} \, _2F_1\left (\frac{1}{6},\frac{1}{2};\frac{7}{6};\sin ^2\left (\frac{1}{4} (2 c+2 d x+\pi )\right )\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{5 d \sqrt{1-\sin (c+d x)} \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.075, size = 0, normalized size = 0. \begin{align*} \int \sin \left ( dx+c \right ) \left ( a+a\sin \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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