\(\int \frac {\sqrt [3]{\sin (c+d x)}}{(a+a \sin (c+d x))^2} \, dx\) [91]

Optimal result

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
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

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)
 

Rubi [A] (verified)

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)
 

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 3122

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]
 

rule 3227

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]
 

rule 3243

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]))
 

rule 3457

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])
 

Maple [F]

\[\int \frac {\sin \left (d x +c \right )^{\frac {1}{3}}}{\left (a +a \sin \left (d x +c \right )\right )^{2}}d x\]

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)
 

Fricas [F]

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

Sympy [F]

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

Maxima [F]

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

Giac [F]

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

Mupad [F(-1)]

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)
 

Reduce [F]

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