\(\int \sin ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx\) [99]

Optimal result

Integrand size = 23, antiderivative size = 127 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=-\frac {37\ 2^{5/6} a \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right ) \sqrt [3]{a+a \sin (c+d x)}}{35 d (1+\sin (c+d x))^{5/6}}+\frac {9 \cos (c+d x) (a+a \sin (c+d x))^{4/3}}{70 d}-\frac {3 \cos (c+d x) (a+a \sin (c+d x))^{7/3}}{10 a d} \] Output:

-37/35*2^(5/6)*a*cos(d*x+c)*hypergeom([-5/6, 1/2],[3/2],1/2-1/2*sin(d*x+c) 
)*(a+a*sin(d*x+c))^(1/3)/d/(1+sin(d*x+c))^(5/6)+9/70*cos(d*x+c)*(a+a*sin(d 
*x+c))^(4/3)/d-3/10*cos(d*x+c)*(a+a*sin(d*x+c))^(7/3)/a/d
 

Mathematica [A] (warning: unable to verify)

Time = 4.06 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.97 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\frac {(a (1+\sin (c+d x)))^{4/3} \left (-\frac {3 (-185+60 \cos (c+d x)-7 \cos (3 (c+d x))+22 \sin (2 (c+d x)))}{20 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {37 \left (-2 \cos \left (\frac {1}{4} (2 c+\pi +2 d x)\right ) \, _2F_1\left (-\frac {1}{2},-\frac {1}{6};\frac {5}{6};\sin ^2\left (\frac {1}{4} (2 c+\pi +2 d x)\right )\right )+\sqrt {\cos ^2\left (\frac {1}{4} (2 c+\pi +2 d x)\right )} \left (2 \cos \left (\frac {1}{4} (2 c+\pi +2 d x)\right )+3 \sin \left (\frac {1}{4} (2 c+\pi +2 d x)\right )\right )\right )}{2\ 2^{2/3} \sqrt {\cos ^2\left (\frac {1}{4} (2 c+\pi +2 d x)\right )} \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^{8/3} \sqrt [3]{\sin \left (\frac {1}{4} (2 c+\pi +2 d x)\right )}}\right )}{14 d} \] Input:

Integrate[Sin[c + d*x]^2*(a + a*Sin[c + d*x])^(4/3),x]
 

Output:

((a*(1 + Sin[c + d*x]))^(4/3)*((-3*(-185 + 60*Cos[c + d*x] - 7*Cos[3*(c + 
d*x)] + 22*Sin[2*(c + d*x)]))/(20*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) 
 - (37*(-2*Cos[(2*c + Pi + 2*d*x)/4]*HypergeometricPFQ[{-1/2, -1/6}, {5/6} 
, Sin[(2*c + Pi + 2*d*x)/4]^2] + Sqrt[Cos[(2*c + Pi + 2*d*x)/4]^2]*(2*Cos[ 
(2*c + Pi + 2*d*x)/4] + 3*Sin[(2*c + Pi + 2*d*x)/4])))/(2*2^(2/3)*Sqrt[Cos 
[(2*c + Pi + 2*d*x)/4]^2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^(8/3)*Sin[ 
(2*c + Pi + 2*d*x)/4]^(1/3))))/(14*d)
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.09, 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.

\(\displaystyle \int \sin ^2(c+d x) (a \sin (c+d x)+a)^{4/3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (c+d x)^2 (a \sin (c+d x)+a)^{4/3}dx\)

\(\Big \downarrow \) 3238

\(\displaystyle \frac {3 \int \frac {1}{3} (7 a-3 a \sin (c+d x)) (\sin (c+d x) a+a)^{4/3}dx}{10 a}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{7/3}}{10 a d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int (7 a-3 a \sin (c+d x)) (\sin (c+d x) a+a)^{4/3}dx}{10 a}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{7/3}}{10 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int (7 a-3 a \sin (c+d x)) (\sin (c+d x) a+a)^{4/3}dx}{10 a}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{7/3}}{10 a d}\)

\(\Big \downarrow \) 3230

\(\displaystyle \frac {\frac {37}{7} a \int (\sin (c+d x) a+a)^{4/3}dx+\frac {9 a \cos (c+d x) (a \sin (c+d x)+a)^{4/3}}{7 d}}{10 a}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{7/3}}{10 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {37}{7} a \int (\sin (c+d x) a+a)^{4/3}dx+\frac {9 a \cos (c+d x) (a \sin (c+d x)+a)^{4/3}}{7 d}}{10 a}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{7/3}}{10 a d}\)

\(\Big \downarrow \) 3131

\(\displaystyle \frac {\frac {37 a^2 \sqrt [3]{a \sin (c+d x)+a} \int (\sin (c+d x)+1)^{4/3}dx}{7 \sqrt [3]{\sin (c+d x)+1}}+\frac {9 a \cos (c+d x) (a \sin (c+d x)+a)^{4/3}}{7 d}}{10 a}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{7/3}}{10 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {37 a^2 \sqrt [3]{a \sin (c+d x)+a} \int (\sin (c+d x)+1)^{4/3}dx}{7 \sqrt [3]{\sin (c+d x)+1}}+\frac {9 a \cos (c+d x) (a \sin (c+d x)+a)^{4/3}}{7 d}}{10 a}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{7/3}}{10 a d}\)

\(\Big \downarrow \) 3130

\(\displaystyle \frac {\frac {9 a \cos (c+d x) (a \sin (c+d x)+a)^{4/3}}{7 d}-\frac {74\ 2^{5/6} a^2 \cos (c+d x) \sqrt [3]{a \sin (c+d x)+a} \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{7 d (\sin (c+d x)+1)^{5/6}}}{10 a}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{7/3}}{10 a d}\)

Input:

Int[Sin[c + d*x]^2*(a + a*Sin[c + d*x])^(4/3),x]
 

Output:

(-3*Cos[c + d*x]*(a + a*Sin[c + d*x])^(7/3))/(10*a*d) + ((-74*2^(5/6)*a^2* 
Cos[c + d*x]*Hypergeometric2F1[-5/6, 1/2, 3/2, (1 - Sin[c + d*x])/2]*(a + 
a*Sin[c + d*x])^(1/3))/(7*d*(1 + Sin[c + d*x])^(5/6)) + (9*a*Cos[c + d*x]* 
(a + a*Sin[c + d*x])^(4/3))/(7*d))/(10*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]

\[\int \sin \left (d x +c \right )^{2} \left (a +a \sin \left (d x +c \right )\right )^{\frac {4}{3}}d x\]

Input:

int(sin(d*x+c)^2*(a+a*sin(d*x+c))^(4/3),x)
 

Output:

int(sin(d*x+c)^2*(a+a*sin(d*x+c))^(4/3),x)
 

Fricas [F]

\[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {4}{3}} \sin \left (d x + c\right )^{2} \,d x } \] Input:

integrate(sin(d*x+c)^2*(a+a*sin(d*x+c))^(4/3),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {4}{3}} \sin ^{2}{\left (c + d x \right )}\, dx \] Input:

integrate(sin(d*x+c)**2*(a+a*sin(d*x+c))**(4/3),x)
 

Output:

Integral((a*(sin(c + d*x) + 1))**(4/3)*sin(c + d*x)**2, x)
 

Maxima [F]

\[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {4}{3}} \sin \left (d x + c\right )^{2} \,d x } \] Input:

integrate(sin(d*x+c)^2*(a+a*sin(d*x+c))^(4/3),x, algorithm="maxima")
 

Output:

integrate((a*sin(d*x + c) + a)^(4/3)*sin(d*x + c)^2, x)
 

Giac [F]

\[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {4}{3}} \sin \left (d x + c\right )^{2} \,d x } \] Input:

integrate(sin(d*x+c)^2*(a+a*sin(d*x+c))^(4/3),x, algorithm="giac")
 

Output:

integrate((a*sin(d*x + c) + a)^(4/3)*sin(d*x + c)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\int {\sin \left (c+d\,x\right )}^2\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{4/3} \,d x \] Input:

int(sin(c + d*x)^2*(a + a*sin(c + d*x))^(4/3),x)
 

Output:

int(sin(c + d*x)^2*(a + a*sin(c + d*x))^(4/3), x)
 

Reduce [F]

\[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=a^{\frac {4}{3}} \left (\int \left (\sin \left (d x +c \right )+1\right )^{\frac {1}{3}} \sin \left (d x +c \right )^{3}d x +\int \left (\sin \left (d x +c \right )+1\right )^{\frac {1}{3}} \sin \left (d x +c \right )^{2}d x \right ) \] Input:

int(sin(d*x+c)^2*(a+a*sin(d*x+c))^(4/3),x)
                                                                                    
                                                                                    
 

Output:

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