\(\int \frac {\sin ^3(c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx\) [104]

Optimal result

Integrand size = 23, antiderivative size = 161 \[ \int \frac {\sin ^3(c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx=-\frac {99 \cos (c+d x)}{80 d \sqrt [3]{a+a \sin (c+d x)}}-\frac {3 \cos (c+d x) \sin ^2(c+d x)}{8 d \sqrt [3]{a+a \sin (c+d x)}}+\frac {37 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{40\ 2^{5/6} d \sqrt [6]{1+\sin (c+d x)} \sqrt [3]{a+a \sin (c+d x)}}+\frac {3 \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{40 a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2862, 3047, 3102, 2830, 2731, 2730} \[ \int \frac {\sin ^3(c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx=\frac {37 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{40\ 2^{5/6} d \sqrt [6]{\sin (c+d x)+1} \sqrt [3]{a \sin (c+d x)+a}}-\frac {3 \sin ^2(c+d x) \cos (c+d x)}{8 d \sqrt [3]{a \sin (c+d x)+a}}+\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{40 a d}-\frac {99 \cos (c+d x)}{80 d \sqrt [3]{a \sin (c+d x)+a}} \]

[In]

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

[Out]

(-99*Cos[c + d*x])/(80*d*(a + a*Sin[c + d*x])^(1/3)) - (3*Cos[c + d*x]*Sin[c + d*x]^2)/(8*d*(a + a*Sin[c + d*x
])^(1/3)) + (37*Cos[c + d*x]*Hypergeometric2F1[1/2, 5/6, 3/2, (1 - Sin[c + d*x])/2])/(40*2^(5/6)*d*(1 + Sin[c
+ d*x])^(1/6)*(a + a*Sin[c + d*x])^(1/3)) + (3*Cos[c + d*x]*(a + a*Sin[c + d*x])^(2/3))/(40*a*d)

Rule 2730

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]]))*Hypergeometric2F1[1/2, 1/2 - n, 3/2, (1/2)*(1 - b*(Sin[c + d*x]/a))], x] /; Free
Q[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]

Rule 2731

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[a^IntPart[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 2830

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] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*S
in[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)]

Rule 2862

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n - 1)/(f*(m + n))), x] + Dist[1/(b*(m + n))
, Int[(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] && E
qQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 1] && IntegerQ[n]

Rule 3047

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]

Rule 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[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]

Rubi steps \begin{align*} \text {integral}& = -\frac {3 \cos (c+d x) \sin ^2(c+d x)}{8 d \sqrt [3]{a+a \sin (c+d x)}}+\frac {3 \int \frac {\sin (c+d x) \left (2 a-\frac {1}{3} a \sin (c+d x)\right )}{\sqrt [3]{a+a \sin (c+d x)}} \, dx}{8 a} \\ & = -\frac {3 \cos (c+d x) \sin ^2(c+d x)}{8 d \sqrt [3]{a+a \sin (c+d x)}}+\frac {3 \int \frac {2 a \sin (c+d x)-\frac {1}{3} a \sin ^2(c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx}{8 a} \\ & = -\frac {3 \cos (c+d x) \sin ^2(c+d x)}{8 d \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}+\frac {9 \int \frac {-\frac {2 a^2}{9}+\frac {11}{3} a^2 \sin (c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx}{40 a^2} \\ & = -\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 {3 \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{40 a d}-\frac {37}{80} \int \frac {1}{\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 {3 \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{40 a d}-\frac {\left (37 \sqrt [3]{1+\sin (c+d x)}\right ) \int \frac {1}{\sqrt [3]{1+\sin (c+d x)}} \, dx}{80 \sqrt [3]{a+a \sin (c+d x)}} \\ & = -\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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.68 \[ \int \frac {\sin ^3(c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx=\frac {3 \cos (c+d x) \left (-37 \sqrt {2} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\sin ^2\left (\frac {1}{4} (2 c+\pi +2 d x)\right )\right )+\sqrt {1-\sin (c+d x)} (-36+5 \cos (2 (c+d x))+2 \sin (c+d x))\right )}{80 d \sqrt {1-\sin (c+d x)} \sqrt [3]{a (1+\sin (c+d x))}} \]

[In]

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

[Out]

(3*Cos[c + d*x]*(-37*Sqrt[2]*Hypergeometric2F1[1/6, 1/2, 7/6, Sin[(2*c + Pi + 2*d*x)/4]^2] + Sqrt[1 - Sin[c +
d*x]]*(-36 + 5*Cos[2*(c + d*x)] + 2*Sin[c + d*x])))/(80*d*Sqrt[1 - Sin[c + d*x]]*(a*(1 + Sin[c + d*x]))^(1/3))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^3(c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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