Integrand size = 23, antiderivative size = 161 \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{2/3} \, dx=-\frac {63 \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{220 d}-\frac {3 \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{2/3}}{11 d}-\frac {67 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right ) (a+a \sin (c+d x))^{2/3}}{55\ 2^{5/6} d (1+\sin (c+d x))^{7/6}}-\frac {3 \cos (c+d x) (a+a \sin (c+d x))^{5/3}}{44 a d} \]
-63/220*cos(d*x+c)*(a+a*sin(d*x+c))^(2/3)/d-3/11*cos(d*x+c)*sin(d*x+c)^2*( a+a*sin(d*x+c))^(2/3)/d-67/110*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)-3/44 *cos(d*x+c)*(a+a*sin(d*x+c))^(5/3)/a/d
Time = 0.81 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.99 \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{2/3} \, dx=\frac {3 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a (1+\sin (c+d x)))^{2/3} \left (67 \sqrt {2} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\sin ^2\left (\frac {1}{4} (2 c+\pi +2 d x)\right )\right )+\sqrt {1-\sin (c+d x)} (-144+25 \cos (2 (c+d x))-92 \sin (c+d x)+10 \sin (3 (c+d x)))\right )}{440 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {1-\sin (c+d x)}} \]
(3*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(a*(1 + Sin[c + d*x]))^(2/3)*(67* 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]]*(-144 + 25*Cos[2*(c + d*x)] - 92*Sin[c + d*x] + 10*Si n[3*(c + d*x)])))/(440*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*Sqrt[1 - Si n[c + d*x]])
Time = 0.86 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.12, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.609, Rules used = {3042, 3262, 27, 3042, 3447, 3042, 3502, 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.
| \(\displaystyle \int \sin ^3(c+d x) (a \sin (c+d x)+a)^{2/3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (c+d x)^3 (a \sin (c+d x)+a)^{2/3}dx\) |
| \(\Big \downarrow \) 3262 |
| \(\displaystyle \frac {3 \int \frac {2}{3} \sin (c+d x) (\sin (c+d x) a+a)^{2/3} (\sin (c+d x) a+3 a)dx}{11 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{11 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 \int \sin (c+d x) (\sin (c+d x) a+a)^{2/3} (\sin (c+d x) a+3 a)dx}{11 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{11 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \int \sin (c+d x) (\sin (c+d x) a+a)^{2/3} (\sin (c+d x) a+3 a)dx}{11 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{11 d}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {2 \int (\sin (c+d x) a+a)^{2/3} \left (a \sin ^2(c+d x)+3 a \sin (c+d x)\right )dx}{11 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{11 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \int (\sin (c+d x) a+a)^{2/3} \left (a \sin (c+d x)^2+3 a \sin (c+d x)\right )dx}{11 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{11 d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {2 \left (\frac {3 \int \frac {1}{3} (\sin (c+d x) a+a)^{2/3} \left (21 \sin (c+d x) a^2+5 a^2\right )dx}{8 a}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{5/3}}{8 d}\right )}{11 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{11 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 \left (\frac {\int (\sin (c+d x) a+a)^{2/3} \left (21 \sin (c+d x) a^2+5 a^2\right )dx}{8 a}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{5/3}}{8 d}\right )}{11 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{11 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \left (\frac {\int (\sin (c+d x) a+a)^{2/3} \left (21 \sin (c+d x) a^2+5 a^2\right )dx}{8 a}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{5/3}}{8 d}\right )}{11 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{11 d}\) |
| \(\Big \downarrow \) 3230 |
| \(\displaystyle \frac {2 \left (\frac {\frac {67}{5} a^2 \int (\sin (c+d x) a+a)^{2/3}dx-\frac {63 a^2 \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{5 d}}{8 a}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{5/3}}{8 d}\right )}{11 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{11 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \left (\frac {\frac {67}{5} a^2 \int (\sin (c+d x) a+a)^{2/3}dx-\frac {63 a^2 \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{5 d}}{8 a}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{5/3}}{8 d}\right )}{11 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{11 d}\) |
| \(\Big \downarrow \) 3131 |
| \(\displaystyle \frac {2 \left (\frac {\frac {67 a^2 (a \sin (c+d x)+a)^{2/3} \int (\sin (c+d x)+1)^{2/3}dx}{5 (\sin (c+d x)+1)^{2/3}}-\frac {63 a^2 \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{5 d}}{8 a}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{5/3}}{8 d}\right )}{11 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{11 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \left (\frac {\frac {67 a^2 (a \sin (c+d x)+a)^{2/3} \int (\sin (c+d x)+1)^{2/3}dx}{5 (\sin (c+d x)+1)^{2/3}}-\frac {63 a^2 \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{5 d}}{8 a}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{5/3}}{8 d}\right )}{11 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{11 d}\) |
| \(\Big \downarrow \) 3130 |
| \(\displaystyle \frac {2 \left (\frac {-\frac {134 \sqrt [6]{2} a^2 \cos (c+d x) (a \sin (c+d x)+a)^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{5 d (\sin (c+d x)+1)^{7/6}}-\frac {63 a^2 \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{5 d}}{8 a}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{5/3}}{8 d}\right )}{11 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{11 d}\) |
(-3*Cos[c + d*x]*Sin[c + d*x]^2*(a + a*Sin[c + d*x])^(2/3))/(11*d) + (2*(( -3*Cos[c + d*x]*(a + a*Sin[c + d*x])^(5/3))/(8*d) + ((-63*a^2*Cos[c + d*x] *(a + a*Sin[c + d*x])^(2/3))/(5*d) - (134*2^(1/6)*a^2*Cos[c + d*x]*Hyperge ometric2F1[-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)))/(11*a)
3.1.92.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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]
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)]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*(a + b*Sin[e + f*x]) ^m*((c + d*Sin[e + f*x])^(n - 1)/(f*(m + n))), x] + Simp[1/(b*(m + n)) In t[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 2)*Simp[d*(a*c*m + b*d*( n - 1)) + b*c^2*(m + n) + d*(a*d*m + b*c*(m + 2*n - 1))*Sin[e + f*x], x], x ], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 1] && IntegerQ[n]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[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*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
\[\int \left (\sin ^{3}\left (d x +c \right )\right ) \left (a +a \sin \left (d x +c \right )\right )^{\frac {2}{3}}d x\]
\[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{2/3} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \sin \left (d x + c\right )^{3} \,d x } \]
Timed out. \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{2/3} \, dx=\text {Timed out} \]
\[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{2/3} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \sin \left (d x + c\right )^{3} \,d x } \]
\[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{2/3} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \sin \left (d x + c\right )^{3} \,d x } \]
Timed out. \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{2/3} \, dx=\int {\sin \left (c+d\,x\right )}^3\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{2/3} \,d x \]