\(\int \frac {\sin (c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx\) [106]

Optimal result

Integrand size = 21, antiderivative size = 93 \[ \int \frac {\sin (c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx=-\frac {3 \cos (c+d x)}{2 d \sqrt [3]{a+a \sin (c+d x)}}+\frac {\cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{2^{5/6} d \sqrt [6]{1+\sin (c+d x)} \sqrt [3]{a+a \sin (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.90 \[ \int \frac {\sin (c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx=-\frac {3 \cos (c+d x) \left (2 \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\sin ^2\left (\frac {1}{4} (2 c+\pi +2 d x)\right )\right )+\sqrt {2-2 \sin (c+d x)}\right )}{2 d \sqrt {2-2 \sin (c+d x)} \sqrt [3]{a (1+\sin (c+d x))}} \] Input:

Integrate[Sin[c + d*x]/(a + a*Sin[c + d*x])^(1/3),x]
 

Output:

(-3*Cos[c + d*x]*(2*Hypergeometric2F1[1/6, 1/2, 7/6, Sin[(2*c + Pi + 2*d*x 
)/4]^2] + Sqrt[2 - 2*Sin[c + d*x]]))/(2*d*Sqrt[2 - 2*Sin[c + d*x]]*(a*(1 + 
 Sin[c + d*x]))^(1/3))
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (c+d x)}{\sqrt [3]{a \sin (c+d x)+a}}dx\)

\(\Big \downarrow \) 3230

\(\displaystyle -\frac {1}{2} \int \frac {1}{\sqrt [3]{\sin (c+d x) a+a}}dx-\frac {3 \cos (c+d x)}{2 d \sqrt [3]{a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {1}{2} \int \frac {1}{\sqrt [3]{\sin (c+d x) a+a}}dx-\frac {3 \cos (c+d x)}{2 d \sqrt [3]{a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3131

\(\displaystyle -\frac {\sqrt [3]{\sin (c+d x)+1} \int \frac {1}{\sqrt [3]{\sin (c+d x)+1}}dx}{2 \sqrt [3]{a \sin (c+d x)+a}}-\frac {3 \cos (c+d x)}{2 d \sqrt [3]{a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\sqrt [3]{\sin (c+d x)+1} \int \frac {1}{\sqrt [3]{\sin (c+d x)+1}}dx}{2 \sqrt [3]{a \sin (c+d x)+a}}-\frac {3 \cos (c+d x)}{2 d \sqrt [3]{a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3130

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

Input:

Int[Sin[c + d*x]/(a + a*Sin[c + d*x])^(1/3),x]
 

Output:

(-3*Cos[c + d*x])/(2*d*(a + a*Sin[c + d*x])^(1/3)) + (Cos[c + d*x]*Hyperge 
ometric2F1[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))
 

Defintions of rubi rules used

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)]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

integrate(sin(d*x+c)/(a+a*sin(d*x+c))^(1/3),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(sin(d*x+c)/(a+a*sin(d*x+c))^(1/3),x, algorithm="maxima")
 

Output:

integrate(sin(d*x + c)/(a*sin(d*x + c) + a)^(1/3), x)
 

Giac [F]

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

integrate(sin(d*x+c)/(a+a*sin(d*x+c))^(1/3),x, algorithm="giac")
 

Output:

integrate(sin(d*x + c)/(a*sin(d*x + c) + a)^(1/3), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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