\(\int \sin ^2(c+d x) (a+a \sin (c+d x))^{2/3} \, dx\) [93]

Optimal result

Integrand size = 23, antiderivative size = 126 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{2/3} \, dx=\frac {9 \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{40 d}-\frac {19 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right ) (a+a \sin (c+d x))^{2/3}}{10\ 2^{5/6} d (1+\sin (c+d x))^{7/6}}-\frac {3 \cos (c+d x) (a+a \sin (c+d x))^{5/3}}{8 a d} \]

[Out]

9/40*cos(d*x+c)*(a+a*sin(d*x+c))^(2/3)/d-19/20*cos(d*x+c)*hypergeom([-1/6, 1/2],[3/2],1/2-1/2*sin(d*x+c))*(a+a
*sin(d*x+c))^(2/3)*2^(1/6)/d/(1+sin(d*x+c))^(7/6)-3/8*cos(d*x+c)*(a+a*sin(d*x+c))^(5/3)/a/d

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2838, 2830, 2731, 2730} \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{2/3} \, dx=-\frac {19 \cos (c+d x) (a \sin (c+d x)+a)^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{10\ 2^{5/6} d (\sin (c+d x)+1)^{7/6}}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{5/3}}{8 a d}+\frac {9 \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{40 d} \]

[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 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 2830

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*S
in[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 2838

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)]

Rubi steps \begin{align*} \text {integral}& = -\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) \operatorname {Hypergeometric2F1}\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] (verified)

Time = 0.86 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.20 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{2/3} \, dx=\frac {3 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a (1+\sin (c+d x)))^{2/3} \left (19 \sqrt {2} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\sin ^2\left (\frac {1}{4} (2 c+\pi +2 d x)\right )\right )+\sqrt {1-\sin (c+d x)} (5 \cos (2 (c+d x))-14 (2+\sin (c+d x)))\right )}{80 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {1-\sin (c+d x)}} \]

[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]

\[\int \left (\sin ^{2}\left (d x +c \right )\right ) \left (a +a \sin \left (d x +c \right )\right )^{\frac {2}{3}}d x\]

[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)

Fricas [F]

\[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{2/3} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \sin \left (d x + c\right )^{2} \,d x } \]

[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]

\[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{2/3} \, dx=\int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {2}{3}} \sin ^{2}{\left (c + d x \right )}\, dx \]

[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)

Maxima [F]

\[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{2/3} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \sin \left (d x + c\right )^{2} \,d x } \]

[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)

Giac [F]

\[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{2/3} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \sin \left (d x + c\right )^{2} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{2/3} \, dx=\int {\sin \left (c+d\,x\right )}^2\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{2/3} \,d x \]

[In]

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

[Out]

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