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

Optimal result

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

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
 

Mathematica [A] (verified)

Time = 0.91 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.20 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{2/3} \, dx=\frac {3 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a (1+\sin (c+d x)))^{2/3} \left (19 \sqrt {2} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\sin ^2\left (\frac {1}{4} (2 c+\pi +2 d x)\right )\right )+\sqrt {1-\sin (c+d x)} (5 \cos (2 (c+d x))-14 (2+\sin (c+d x)))\right )}{80 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {1-\sin (c+d x)}} \] Input:

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

Output:

(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] + Sq 
rt[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]])
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 3238, 27, 3042, 3230, 3042, 3131, 3042, 3130}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3238

\(\displaystyle \frac {3 \int \frac {1}{3} (5 a-3 a \sin (c+d x)) (\sin (c+d x) a+a)^{2/3}dx}{8 a}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{5/3}}{8 a d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int (5 a-3 a \sin (c+d x)) (\sin (c+d x) a+a)^{2/3}dx}{8 a}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{5/3}}{8 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int (5 a-3 a \sin (c+d x)) (\sin (c+d x) a+a)^{2/3}dx}{8 a}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{5/3}}{8 a d}\)

\(\Big \downarrow \) 3230

\(\displaystyle \frac {\frac {19}{5} a \int (\sin (c+d x) a+a)^{2/3}dx+\frac {9 a \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{5 d}}{8 a}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{5/3}}{8 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {19}{5} a \int (\sin (c+d x) a+a)^{2/3}dx+\frac {9 a \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{5 d}}{8 a}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{5/3}}{8 a d}\)

\(\Big \downarrow \) 3131

\(\displaystyle \frac {\frac {19 a (a \sin (c+d x)+a)^{2/3} \int (\sin (c+d x)+1)^{2/3}dx}{5 (\sin (c+d x)+1)^{2/3}}+\frac {9 a \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{5 d}}{8 a}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{5/3}}{8 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {19 a (a \sin (c+d x)+a)^{2/3} \int (\sin (c+d x)+1)^{2/3}dx}{5 (\sin (c+d x)+1)^{2/3}}+\frac {9 a \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{5 d}}{8 a}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{5/3}}{8 a d}\)

\(\Big \downarrow \) 3130

\(\displaystyle \frac {\frac {9 a \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{5 d}-\frac {38 \sqrt [6]{2} a \cos (c+d x) (a \sin (c+d x)+a)^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{5 d (\sin (c+d x)+1)^{7/6}}}{8 a}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{5/3}}{8 a d}\)

Input:

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

Output:

(-3*Cos[c + d*x]*(a + a*Sin[c + d*x])^(5/3))/(8*a*d) + ((9*a*Cos[c + d*x]* 
(a + a*Sin[c + d*x])^(2/3))/(5*d) - (38*2^(1/6)*a*Cos[c + d*x]*Hypergeomet 
ric2F1[-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)))/(8*a)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3130

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]]))*Hypergeome 
tric2F1[1/2, 1/2 - n, 3/2, (1/2)*(1 - b*(Sin[c + d*x]/a))], x] /; FreeQ[{a, 
 b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]
 

rule 3131

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a^IntPar 
t[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 3230

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] + Simp[(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 3238

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] + Simp[1/(b*(m + 2))   Int[(a + b*Sin[e + f*x])^m*(b*(m + 1) - a*Si 
n[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  ! 
LtQ[m, -2^(-1)]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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