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

Optimal. Leaf size=126 \[ -\frac{19 \cos (c+d x) (a \sin (c+d x)+a)^{2/3} \, _2F_1\left (-\frac{1}{6},\frac{1}{2};\frac{3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{10\ 2^{5/6} d (\sin (c+d x)+1)^{7/6}}-\frac{3 \cos (c+d x) (a \sin (c+d x)+a)^{5/3}}{8 a d}+\frac{9 \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{40 d} \]

[Out]

(9*Cos[c + d*x]*(a + a*Sin[c + d*x])^(2/3))/(40*d) - (19*Cos[c + d*x]*Hypergeometric2F1[-1/6, 1/2, 3/2, (1 - S
in[c + d*x])/2]*(a + a*Sin[c + d*x])^(2/3))/(10*2^(5/6)*d*(1 + Sin[c + d*x])^(7/6)) - (3*Cos[c + d*x]*(a + a*S
in[c + d*x])^(5/3))/(8*a*d)

________________________________________________________________________________________

Rubi [A]  time = 0.145496, 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} \[ -\frac{19 \cos (c+d x) (a \sin (c+d x)+a)^{2/3} \, _2F_1\left (-\frac{1}{6},\frac{1}{2};\frac{3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{10\ 2^{5/6} d (\sin (c+d x)+1)^{7/6}}-\frac{3 \cos (c+d x) (a \sin (c+d x)+a)^{5/3}}{8 a d}+\frac{9 \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{40 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(9*Cos[c + d*x]*(a + a*Sin[c + d*x])^(2/3))/(40*d) - (19*Cos[c + d*x]*Hypergeometric2F1[-1/6, 1/2, 3/2, (1 - S
in[c + d*x])/2]*(a + a*Sin[c + d*x])^(2/3))/(10*2^(5/6)*d*(1 + Sin[c + d*x])^(7/6)) - (3*Cos[c + d*x]*(a + a*S
in[c + d*x])^(5/3))/(8*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 \sin ^2(c+d x) (a+a \sin (c+d x))^{2/3} \, dx &=-\frac{3 \cos (c+d x) (a+a \sin (c+d x))^{5/3}}{8 a d}+\frac{3 \int \left (\frac{5 a}{3}-a \sin (c+d x)\right ) (a+a \sin (c+d x))^{2/3} \, dx}{8 a}\\ &=\frac{9 \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{40 d}-\frac{3 \cos (c+d x) (a+a \sin (c+d x))^{5/3}}{8 a d}+\frac{19}{40} \int (a+a \sin (c+d x))^{2/3} \, dx\\ &=\frac{9 \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{40 d}-\frac{3 \cos (c+d x) (a+a \sin (c+d x))^{5/3}}{8 a d}+\frac{\left (19 (a+a \sin (c+d x))^{2/3}\right ) \int (1+\sin (c+d x))^{2/3} \, dx}{40 (1+\sin (c+d x))^{2/3}}\\ &=\frac{9 \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{40 d}-\frac{19 \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}}{10\ 2^{5/6} d (1+\sin (c+d x))^{7/6}}-\frac{3 \cos (c+d x) (a+a \sin (c+d x))^{5/3}}{8 a d}\\ \end{align*}

Mathematica [A]  time = 0.438797, size = 151, normalized size = 1.2 \[ \frac{3 (a (\sin (c+d x)+1))^{2/3} \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (19 \sqrt{2} \, _2F_1\left (\frac{1}{6},\frac{1}{2};\frac{7}{6};\sin ^2\left (\frac{1}{4} (2 c+2 d x+\pi )\right )\right )+\sqrt{1-\sin (c+d x)} (5 \cos (2 (c+d x))-14 (\sin (c+d x)+2))\right )}{80 d \sqrt{1-\sin (c+d x)} \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^2*(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)*(19*Sqrt[2]*Hypergeometric2F1[1/6, 1/2,
7/6, Sin[(2*c + Pi + 2*d*x)/4]^2] + Sqrt[1 - Sin[c + d*x]]*(5*Cos[2*(c + d*x)] - 14*(2 + Sin[c + d*x]))))/(80*
d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*Sqrt[1 - Sin[c + d*x]])

________________________________________________________________________________________

Maple [F]  time = 0.197, size = 0, normalized size = 0. \begin{align*} \int \left ( \sin \left ( dx+c \right ) \right ) ^{2} \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)^2*(a+a*sin(d*x+c))^(2/3),x)

[Out]

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

________________________________________________________________________________________

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 )^{2}\,{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))^(2/3),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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 ^{2}{\left (c + d x \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))**(2/3),x)

[Out]

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

________________________________________________________________________________________

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 )^{2}\,{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))^(2/3),x, algorithm="giac")

[Out]

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