Integrand size = 23, antiderivative size = 184 \[ \int \frac {\sqrt [3]{\sin (c+d x)}}{(a+a \sin (c+d x))^2} \, dx=\frac {4 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\sin ^2(c+d x)\right ) \sqrt [3]{\sin (c+d x)}}{9 a^2 d \sqrt {\cos ^2(c+d x)}}-\frac {\cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\sin ^2(c+d x)\right ) \sin ^{\frac {4}{3}}(c+d x)}{36 a^2 d \sqrt {\cos ^2(c+d x)}}-\frac {\cos (c+d x) \sqrt [3]{\sin (c+d x)}}{9 a^2 d (1+\sin (c+d x))}-\frac {\cos (c+d x) \sqrt [3]{\sin (c+d x)}}{3 d (a+a \sin (c+d x))^2} \] Output:
4/9*cos(d*x+c)*hypergeom([1/6, 1/2],[7/6],sin(d*x+c)^2)*sin(d*x+c)^(1/3)/a ^2/d/(cos(d*x+c)^2)^(1/2)-1/36*cos(d*x+c)*hypergeom([1/2, 2/3],[5/3],sin(d *x+c)^2)*sin(d*x+c)^(4/3)/a^2/d/(cos(d*x+c)^2)^(1/2)-1/9*cos(d*x+c)*sin(d* x+c)^(1/3)/a^2/d/(1+sin(d*x+c))-1/3*cos(d*x+c)*sin(d*x+c)^(1/3)/d/(a+a*sin (d*x+c))^2
Time = 0.32 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.66 \[ \int \frac {\sqrt [3]{\sin (c+d x)}}{(a+a \sin (c+d x))^2} \, dx=\frac {\sec ^3(c+d x) \sqrt [3]{\sin (c+d x)} \left (80 \cos ^2(c+d x)^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\sin ^2(c+d x)\right )+27 \cos ^2(c+d x)^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {5}{2},\frac {5}{3},\sin ^2(c+d x)\right ) \sin (c+d x)+4 (-25+5 \cos (2 (c+d x))+27 \sin (c+d x))\right )}{180 a^2 d} \] Input:
Integrate[Sin[c + d*x]^(1/3)/(a + a*Sin[c + d*x])^2,x]
Output:
(Sec[c + d*x]^3*Sin[c + d*x]^(1/3)*(80*(Cos[c + d*x]^2)^(3/2)*Hypergeometr ic2F1[1/6, 1/2, 7/6, Sin[c + d*x]^2] + 27*(Cos[c + d*x]^2)^(3/2)*Hypergeom etric2F1[2/3, 5/2, 5/3, Sin[c + d*x]^2]*Sin[c + d*x] + 4*(-25 + 5*Cos[2*(c + d*x)] + 27*Sin[c + d*x])))/(180*a^2*d)
Time = 0.65 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.05, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3042, 3243, 27, 3042, 3457, 27, 3042, 3227, 3042, 3122}
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.
\(\displaystyle \int \frac {\sqrt [3]{\sin (c+d x)}}{(a \sin (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt [3]{\sin (c+d x)}}{(a \sin (c+d x)+a)^2}dx\) |
\(\Big \downarrow \) 3243 |
\(\displaystyle \frac {\int \frac {2 \sin (c+d x) a+a}{3 \sin ^{\frac {2}{3}}(c+d x) (\sin (c+d x) a+a)}dx}{3 a^2}-\frac {\sqrt [3]{\sin (c+d x)} \cos (c+d x)}{3 d (a \sin (c+d x)+a)^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {2 \sin (c+d x) a+a}{\sin ^{\frac {2}{3}}(c+d x) (\sin (c+d x) a+a)}dx}{9 a^2}-\frac {\sqrt [3]{\sin (c+d x)} \cos (c+d x)}{3 d (a \sin (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {2 \sin (c+d x) a+a}{\sin (c+d x)^{2/3} (\sin (c+d x) a+a)}dx}{9 a^2}-\frac {\sqrt [3]{\sin (c+d x)} \cos (c+d x)}{3 d (a \sin (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3457 |
\(\displaystyle \frac {\frac {\int \frac {4 a^2-a^2 \sin (c+d x)}{3 \sin ^{\frac {2}{3}}(c+d x)}dx}{a^2}-\frac {\sqrt [3]{\sin (c+d x)} \cos (c+d x)}{d (\sin (c+d x)+1)}}{9 a^2}-\frac {\sqrt [3]{\sin (c+d x)} \cos (c+d x)}{3 d (a \sin (c+d x)+a)^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int \frac {4 a^2-a^2 \sin (c+d x)}{\sin ^{\frac {2}{3}}(c+d x)}dx}{3 a^2}-\frac {\sqrt [3]{\sin (c+d x)} \cos (c+d x)}{d (\sin (c+d x)+1)}}{9 a^2}-\frac {\sqrt [3]{\sin (c+d x)} \cos (c+d x)}{3 d (a \sin (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {4 a^2-a^2 \sin (c+d x)}{\sin (c+d x)^{2/3}}dx}{3 a^2}-\frac {\sqrt [3]{\sin (c+d x)} \cos (c+d x)}{d (\sin (c+d x)+1)}}{9 a^2}-\frac {\sqrt [3]{\sin (c+d x)} \cos (c+d x)}{3 d (a \sin (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {\frac {4 a^2 \int \frac {1}{\sin ^{\frac {2}{3}}(c+d x)}dx-a^2 \int \sqrt [3]{\sin (c+d x)}dx}{3 a^2}-\frac {\sqrt [3]{\sin (c+d x)} \cos (c+d x)}{d (\sin (c+d x)+1)}}{9 a^2}-\frac {\sqrt [3]{\sin (c+d x)} \cos (c+d x)}{3 d (a \sin (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {4 a^2 \int \frac {1}{\sin (c+d x)^{2/3}}dx-a^2 \int \sqrt [3]{\sin (c+d x)}dx}{3 a^2}-\frac {\sqrt [3]{\sin (c+d x)} \cos (c+d x)}{d (\sin (c+d x)+1)}}{9 a^2}-\frac {\sqrt [3]{\sin (c+d x)} \cos (c+d x)}{3 d (a \sin (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \frac {\frac {\frac {12 a^2 \sqrt [3]{\sin (c+d x)} \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\sin ^2(c+d x)\right )}{d \sqrt {\cos ^2(c+d x)}}-\frac {3 a^2 \sin ^{\frac {4}{3}}(c+d x) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\sin ^2(c+d x)\right )}{4 d \sqrt {\cos ^2(c+d x)}}}{3 a^2}-\frac {\sqrt [3]{\sin (c+d x)} \cos (c+d x)}{d (\sin (c+d x)+1)}}{9 a^2}-\frac {\sqrt [3]{\sin (c+d x)} \cos (c+d x)}{3 d (a \sin (c+d x)+a)^2}\) |
Input:
Int[Sin[c + d*x]^(1/3)/(a + a*Sin[c + d*x])^2,x]
Output:
-1/3*(Cos[c + d*x]*Sin[c + d*x]^(1/3))/(d*(a + a*Sin[c + d*x])^2) + (-((Co s[c + d*x]*Sin[c + d*x]^(1/3))/(d*(1 + Sin[c + d*x]))) + ((12*a^2*Cos[c + d*x]*Hypergeometric2F1[1/6, 1/2, 7/6, Sin[c + d*x]^2]*Sin[c + d*x]^(1/3))/ (d*Sqrt[Cos[c + d*x]^2]) - (3*a^2*Cos[c + d*x]*Hypergeometric2F1[1/2, 2/3, 5/3, Sin[c + d*x]^2]*Sin[c + d*x]^(4/3))/(4*d*Sqrt[Cos[c + d*x]^2]))/(3*a ^2))/(9*a^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2 F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[2*n]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[e + f*x]*(a + b*Sin[e + f*x])^m* ((c + d*Sin[e + f*x])^n/(a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int [(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 1)*Simp[a*d*n - b*c *(m + 1) - b*d*(m + n + 1)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e , f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && LtQ[0, n, 1] && (IntegersQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c , 0]))
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b *d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ [b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
\[\int \frac {\sin \left (d x +c \right )^{\frac {1}{3}}}{\left (a +a \sin \left (d x +c \right )\right )^{2}}d x\]
Input:
int(sin(d*x+c)^(1/3)/(a+a*sin(d*x+c))^2,x)
Output:
int(sin(d*x+c)^(1/3)/(a+a*sin(d*x+c))^2,x)
\[ \int \frac {\sqrt [3]{\sin (c+d x)}}{(a+a \sin (c+d x))^2} \, dx=\int { \frac {\sin \left (d x + c\right )^{\frac {1}{3}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:
integrate(sin(d*x+c)^(1/3)/(a+a*sin(d*x+c))^2,x, algorithm="fricas")
Output:
integral(-sin(d*x + c)^(1/3)/(a^2*cos(d*x + c)^2 - 2*a^2*sin(d*x + c) - 2* a^2), x)
\[ \int \frac {\sqrt [3]{\sin (c+d x)}}{(a+a \sin (c+d x))^2} \, dx=\frac {\int \frac {\sqrt [3]{\sin {\left (c + d x \right )}}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:
integrate(sin(d*x+c)**(1/3)/(a+a*sin(d*x+c))**2,x)
Output:
Integral(sin(c + d*x)**(1/3)/(sin(c + d*x)**2 + 2*sin(c + d*x) + 1), x)/a* *2
\[ \int \frac {\sqrt [3]{\sin (c+d x)}}{(a+a \sin (c+d x))^2} \, dx=\int { \frac {\sin \left (d x + c\right )^{\frac {1}{3}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:
integrate(sin(d*x+c)^(1/3)/(a+a*sin(d*x+c))^2,x, algorithm="maxima")
Output:
integrate(sin(d*x + c)^(1/3)/(a*sin(d*x + c) + a)^2, x)
\[ \int \frac {\sqrt [3]{\sin (c+d x)}}{(a+a \sin (c+d x))^2} \, dx=\int { \frac {\sin \left (d x + c\right )^{\frac {1}{3}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:
integrate(sin(d*x+c)^(1/3)/(a+a*sin(d*x+c))^2,x, algorithm="giac")
Output:
integrate(sin(d*x + c)^(1/3)/(a*sin(d*x + c) + a)^2, x)
Timed out. \[ \int \frac {\sqrt [3]{\sin (c+d x)}}{(a+a \sin (c+d x))^2} \, dx=\int \frac {{\sin \left (c+d\,x\right )}^{1/3}}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^2} \,d x \] Input:
int(sin(c + d*x)^(1/3)/(a + a*sin(c + d*x))^2,x)
Output:
int(sin(c + d*x)^(1/3)/(a + a*sin(c + d*x))^2, x)
\[ \int \frac {\sqrt [3]{\sin (c+d x)}}{(a+a \sin (c+d x))^2} \, dx=\frac {\int \frac {\sin \left (d x +c \right )^{\frac {1}{3}}}{\sin \left (d x +c \right )^{2}+2 \sin \left (d x +c \right )+1}d x}{a^{2}} \] Input:
int(sin(d*x+c)^(1/3)/(a+a*sin(d*x+c))^2,x)
Output:
int(sin(c + d*x)**(1/3)/(sin(c + d*x)**2 + 2*sin(c + d*x) + 1),x)/a**2