\(\int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{4/3}} \, dx\) [110]

Optimal result

Integrand size = 23, antiderivative size = 162 \[ \int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{4/3}} \, dx=\frac {6 \cos (c+d x)}{5 d (a+a \sin (c+d x))^{4/3}}-\frac {3 \cos (c+d x) \sin ^2(c+d x)}{5 d (a+a \sin (c+d x))^{4/3}}+\frac {6 \cos (c+d x)}{5 a d \sqrt [3]{a+a \sin (c+d x)}}-\frac {2 \sqrt [6]{2} \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{a d \sqrt [6]{1+\sin (c+d x)} \sqrt [3]{a+a \sin (c+d x)}} \]

[Out]

6/5*cos(d*x+c)/d/(a+a*sin(d*x+c))^(4/3)-3/5*cos(d*x+c)*sin(d*x+c)^2/d/(a+a*sin(d*x+c))^(4/3)+6/5*cos(d*x+c)/a/
d/(a+a*sin(d*x+c))^(1/3)-2*2^(1/6)*cos(d*x+c)*hypergeom([1/2, 5/6],[3/2],1/2-1/2*sin(d*x+c))/a/d/(1+sin(d*x+c)
)^(1/6)/(a+a*sin(d*x+c))^(1/3)

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 162, 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, 3098, 2830, 2731, 2730} \[ \int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{4/3}} \, dx=-\frac {2 \sqrt [6]{2} \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{a d \sqrt [6]{\sin (c+d x)+1} \sqrt [3]{a \sin (c+d x)+a}}-\frac {3 \sin ^2(c+d x) \cos (c+d x)}{5 d (a \sin (c+d x)+a)^{4/3}}+\frac {6 \cos (c+d x)}{5 a d \sqrt [3]{a \sin (c+d x)+a}}+\frac {6 \cos (c+d x)}{5 d (a \sin (c+d x)+a)^{4/3}} \]

[In]

Int[Sin[c + d*x]^3/(a + a*Sin[c + d*x])^(4/3),x]

[Out]

(6*Cos[c + d*x])/(5*d*(a + a*Sin[c + d*x])^(4/3)) - (3*Cos[c + d*x]*Sin[c + d*x]^2)/(5*d*(a + a*Sin[c + d*x])^
(4/3)) + (6*Cos[c + d*x])/(5*a*d*(a + a*Sin[c + d*x])^(1/3)) - (2*2^(1/6)*Cos[c + d*x]*Hypergeometric2F1[1/2,
5/6, 3/2, (1 - Sin[c + d*x])/2])/(a*d*(1 + Sin[c + d*x])^(1/6)*(a + a*Sin[c + d*x])^(1/3))

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 3098

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_
.)*(x_)]^2), x_Symbol] :> Simp[(A*b - a*B + b*C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(a*f*(2*m + 1))), x] + D
ist[1/(a^2*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b*B - a*C) + b*C*(2*m + 1)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {3 \cos (c+d x) \sin ^2(c+d x)}{5 d (a+a \sin (c+d x))^{4/3}}+\frac {3 \int \frac {\sin (c+d x) \left (2 a-\frac {4}{3} a \sin (c+d x)\right )}{(a+a \sin (c+d x))^{4/3}} \, dx}{5 a} \\ & = -\frac {3 \cos (c+d x) \sin ^2(c+d x)}{5 d (a+a \sin (c+d x))^{4/3}}+\frac {3 \int \frac {2 a \sin (c+d x)-\frac {4}{3} a \sin ^2(c+d x)}{(a+a \sin (c+d x))^{4/3}} \, dx}{5 a} \\ & = \frac {6 \cos (c+d x)}{5 d (a+a \sin (c+d x))^{4/3}}-\frac {3 \cos (c+d x) \sin ^2(c+d x)}{5 d (a+a \sin (c+d x))^{4/3}}-\frac {9 \int \frac {-\frac {40 a^2}{9}+\frac {20}{9} a^2 \sin (c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx}{25 a^3} \\ & = \frac {6 \cos (c+d x)}{5 d (a+a \sin (c+d x))^{4/3}}-\frac {3 \cos (c+d x) \sin ^2(c+d x)}{5 d (a+a \sin (c+d x))^{4/3}}+\frac {6 \cos (c+d x)}{5 a d \sqrt [3]{a+a \sin (c+d x)}}+\frac {2 \int \frac {1}{\sqrt [3]{a+a \sin (c+d x)}} \, dx}{a} \\ & = \frac {6 \cos (c+d x)}{5 d (a+a \sin (c+d x))^{4/3}}-\frac {3 \cos (c+d x) \sin ^2(c+d x)}{5 d (a+a \sin (c+d x))^{4/3}}+\frac {6 \cos (c+d x)}{5 a d \sqrt [3]{a+a \sin (c+d x)}}+\frac {\left (2 \sqrt [3]{1+\sin (c+d x)}\right ) \int \frac {1}{\sqrt [3]{1+\sin (c+d x)}} \, dx}{a \sqrt [3]{a+a \sin (c+d x)}} \\ & = \frac {6 \cos (c+d x)}{5 d (a+a \sin (c+d x))^{4/3}}-\frac {3 \cos (c+d x) \sin ^2(c+d x)}{5 d (a+a \sin (c+d x))^{4/3}}+\frac {6 \cos (c+d x)}{5 a d \sqrt [3]{a+a \sin (c+d x)}}-\frac {2 \sqrt [6]{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 )}{a d \sqrt [6]{1+\sin (c+d x)} \sqrt [3]{a+a \sin (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.71 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.72 \[ \int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{4/3}} \, dx=\frac {3 \cos (c+d x) \left (20 \sqrt {2} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\sin ^2\left (\frac {1}{4} (2 c+\pi +2 d x)\right )\right ) (1+\sin (c+d x))+\sqrt {1-\sin (c+d x)} (7+\cos (2 (c+d x))+4 \sin (c+d x))\right )}{10 d \sqrt {1-\sin (c+d x)} (a (1+\sin (c+d x)))^{4/3}} \]

[In]

Integrate[Sin[c + d*x]^3/(a + a*Sin[c + d*x])^(4/3),x]

[Out]

(3*Cos[c + d*x]*(20*Sqrt[2]*Hypergeometric2F1[1/6, 1/2, 7/6, Sin[(2*c + Pi + 2*d*x)/4]^2]*(1 + Sin[c + d*x]) +
Sqrt[1 - Sin[c + d*x]]*(7 + Cos[2*(c + d*x)] + 4*Sin[c + d*x])))/(10*d*Sqrt[1 - Sin[c + d*x]]*(a*(1 + Sin[c +
d*x]))^(4/3))

Maple [F]

\[\int \frac {\sin ^{3}\left (d x +c \right )}{\left (a +a \sin \left (d x +c \right )\right )^{\frac {4}{3}}}d x\]

[In]

int(sin(d*x+c)^3/(a+a*sin(d*x+c))^(4/3),x)

[Out]

int(sin(d*x+c)^3/(a+a*sin(d*x+c))^(4/3),x)

Fricas [F]

\[ \int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{4/3}} \, dx=\int { \frac {\sin \left (d x + c\right )^{3}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {4}{3}}} \,d x } \]

[In]

integrate(sin(d*x+c)^3/(a+a*sin(d*x+c))^(4/3),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{4/3}} \, dx=\text {Timed out} \]

[In]

integrate(sin(d*x+c)**3/(a+a*sin(d*x+c))**(4/3),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{4/3}} \, dx=\int { \frac {\sin \left (d x + c\right )^{3}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {4}{3}}} \,d x } \]

[In]

integrate(sin(d*x+c)^3/(a+a*sin(d*x+c))^(4/3),x, algorithm="maxima")

[Out]

integrate(sin(d*x + c)^3/(a*sin(d*x + c) + a)^(4/3), x)

Giac [F]

\[ \int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{4/3}} \, dx=\int { \frac {\sin \left (d x + c\right )^{3}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {4}{3}}} \,d x } \]

[In]

integrate(sin(d*x+c)^3/(a+a*sin(d*x+c))^(4/3),x, algorithm="giac")

[Out]

integrate(sin(d*x + c)^3/(a*sin(d*x + c) + a)^(4/3), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{4/3}} \, dx=\int \frac {{\sin \left (c+d\,x\right )}^3}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{4/3}} \,d x \]

[In]

int(sin(c + d*x)^3/(a + a*sin(c + d*x))^(4/3),x)

[Out]

int(sin(c + d*x)^3/(a + a*sin(c + d*x))^(4/3), x)