Integrand size = 23, antiderivative size = 161 \[ \int \frac {\sin ^3(c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx=-\frac {99 \cos (c+d x)}{80 d \sqrt [3]{a+a \sin (c+d x)}}-\frac {3 \cos (c+d x) \sin ^2(c+d x)}{8 d \sqrt [3]{a+a \sin (c+d x)}}+\frac {37 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{40\ 2^{5/6} d \sqrt [6]{1+\sin (c+d x)} \sqrt [3]{a+a \sin (c+d x)}}+\frac {3 \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{40 a d} \]
-99/80*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/3)-3/8*cos(d*x+c)*sin(d*x+c)^2/d/( a+a*sin(d*x+c))^(1/3)+37/80*cos(d*x+c)*hypergeom([1/2, 5/6],[3/2],1/2-1/2* sin(d*x+c))*2^(1/6)/d/(1+sin(d*x+c))^(1/6)/(a+a*sin(d*x+c))^(1/3)+3/40*cos (d*x+c)*(a+a*sin(d*x+c))^(2/3)/a/d
Time = 0.66 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.68 \[ \int \frac {\sin ^3(c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx=\frac {3 \cos (c+d x) \left (-37 \sqrt {2} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\sin ^2\left (\frac {1}{4} (2 c+\pi +2 d x)\right )\right )+\sqrt {1-\sin (c+d x)} (-36+5 \cos (2 (c+d x))+2 \sin (c+d x))\right )}{80 d \sqrt {1-\sin (c+d x)} \sqrt [3]{a (1+\sin (c+d x))}} \]
(3*Cos[c + d*x]*(-37*Sqrt[2]*Hypergeometric2F1[1/6, 1/2, 7/6, Sin[(2*c + P i + 2*d*x)/4]^2] + Sqrt[1 - Sin[c + d*x]]*(-36 + 5*Cos[2*(c + d*x)] + 2*Si n[c + d*x])))/(80*d*Sqrt[1 - Sin[c + d*x]]*(a*(1 + Sin[c + d*x]))^(1/3))
Time = 0.85 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.11, 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.
| \(\displaystyle \int \frac {\sin ^3(c+d x)}{\sqrt [3]{a \sin (c+d x)+a}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^3}{\sqrt [3]{a \sin (c+d x)+a}}dx\) |
| \(\Big \downarrow \) 3262 |
| \(\displaystyle \frac {3 \int \frac {\sin (c+d x) (6 a-a \sin (c+d x))}{3 \sqrt [3]{\sin (c+d x) a+a}}dx}{8 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x)}{8 d \sqrt [3]{a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sin (c+d x) (6 a-a \sin (c+d x))}{\sqrt [3]{\sin (c+d x) a+a}}dx}{8 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x)}{8 d \sqrt [3]{a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sin (c+d x) (6 a-a \sin (c+d x))}{\sqrt [3]{\sin (c+d x) a+a}}dx}{8 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x)}{8 d \sqrt [3]{a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {\int \frac {6 a \sin (c+d x)-a \sin ^2(c+d x)}{\sqrt [3]{\sin (c+d x) a+a}}dx}{8 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x)}{8 d \sqrt [3]{a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {6 a \sin (c+d x)-a \sin (c+d x)^2}{\sqrt [3]{\sin (c+d x) a+a}}dx}{8 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x)}{8 d \sqrt [3]{a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {\frac {3 \int -\frac {2 a^2-33 a^2 \sin (c+d x)}{3 \sqrt [3]{\sin (c+d x) a+a}}dx}{5 a}+\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{5 d}}{8 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x)}{8 d \sqrt [3]{a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{5 d}-\frac {\int \frac {2 a^2-33 a^2 \sin (c+d x)}{\sqrt [3]{\sin (c+d x) a+a}}dx}{5 a}}{8 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x)}{8 d \sqrt [3]{a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{5 d}-\frac {\int \frac {2 a^2-33 a^2 \sin (c+d x)}{\sqrt [3]{\sin (c+d x) a+a}}dx}{5 a}}{8 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x)}{8 d \sqrt [3]{a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3230 |
| \(\displaystyle \frac {\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{5 d}-\frac {\frac {37}{2} a^2 \int \frac {1}{\sqrt [3]{\sin (c+d x) a+a}}dx+\frac {99 a^2 \cos (c+d x)}{2 d \sqrt [3]{a \sin (c+d x)+a}}}{5 a}}{8 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x)}{8 d \sqrt [3]{a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{5 d}-\frac {\frac {37}{2} a^2 \int \frac {1}{\sqrt [3]{\sin (c+d x) a+a}}dx+\frac {99 a^2 \cos (c+d x)}{2 d \sqrt [3]{a \sin (c+d x)+a}}}{5 a}}{8 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x)}{8 d \sqrt [3]{a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3131 |
| \(\displaystyle \frac {\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{5 d}-\frac {\frac {37 a^2 \sqrt [3]{\sin (c+d x)+1} \int \frac {1}{\sqrt [3]{\sin (c+d x)+1}}dx}{2 \sqrt [3]{a \sin (c+d x)+a}}+\frac {99 a^2 \cos (c+d x)}{2 d \sqrt [3]{a \sin (c+d x)+a}}}{5 a}}{8 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x)}{8 d \sqrt [3]{a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{5 d}-\frac {\frac {37 a^2 \sqrt [3]{\sin (c+d x)+1} \int \frac {1}{\sqrt [3]{\sin (c+d x)+1}}dx}{2 \sqrt [3]{a \sin (c+d x)+a}}+\frac {99 a^2 \cos (c+d x)}{2 d \sqrt [3]{a \sin (c+d x)+a}}}{5 a}}{8 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x)}{8 d \sqrt [3]{a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3130 |
| \(\displaystyle \frac {\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{5 d}-\frac {\frac {99 a^2 \cos (c+d x)}{2 d \sqrt [3]{a \sin (c+d x)+a}}-\frac {37 a^2 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{2^{5/6} d \sqrt [6]{\sin (c+d x)+1} \sqrt [3]{a \sin (c+d x)+a}}}{5 a}}{8 a}-\frac {3 \sin ^2(c+d x) \cos (c+d x)}{8 d \sqrt [3]{a \sin (c+d x)+a}}\) |
(-3*Cos[c + d*x]*Sin[c + d*x]^2)/(8*d*(a + a*Sin[c + d*x])^(1/3)) + ((3*Co s[c + d*x]*(a + a*Sin[c + d*x])^(2/3))/(5*d) - ((99*a^2*Cos[c + d*x])/(2*d *(a + a*Sin[c + d*x])^(1/3)) - (37*a^2*Cos[c + d*x]*Hypergeometric2F1[1/2, 5/6, 3/2, (1 - Sin[c + d*x])/2])/(2^(5/6)*d*(1 + Sin[c + d*x])^(1/6)*(a + a*Sin[c + d*x])^(1/3)))/(5*a))/(8*a)
3.2.4.3.1 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]
\[\int \frac {\sin ^{3}\left (d x +c \right )}{\left (a +a \sin \left (d x +c \right )\right )^{\frac {1}{3}}}d x\]
\[ \int \frac {\sin ^3(c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx=\int { \frac {\sin \left (d x + c\right )^{3}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \]
Timed out. \[ \int \frac {\sin ^3(c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx=\text {Timed out} \]
\[ \int \frac {\sin ^3(c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx=\int { \frac {\sin \left (d x + c\right )^{3}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \]
\[ \int \frac {\sin ^3(c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx=\int { \frac {\sin \left (d x + c\right )^{3}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \]
Timed out. \[ \int \frac {\sin ^3(c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx=\int \frac {{\sin \left (c+d\,x\right )}^3}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{1/3}} \,d x \]