\(\int \frac {1}{(a+a \sin (c+d x))^{2/3}} \, dx\) [12]

Optimal result

Integrand size = 14, antiderivative size = 66 \[ \int \frac {1}{(a+a \sin (c+d x))^{2/3}} \, dx=-\frac {\cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right ) \sqrt [6]{1+\sin (c+d x)}}{\sqrt [6]{2} d (a+a \sin (c+d x))^{2/3}} \]

[Out]

-1/2*cos(d*x+c)*hypergeom([1/2, 7/6],[3/2],1/2-1/2*sin(d*x+c))*(1+sin(d*x+c))^(1/6)*2^(5/6)/d/(a+a*sin(d*x+c))
^(2/3)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2731, 2730} \[ \int \frac {1}{(a+a \sin (c+d x))^{2/3}} \, dx=-\frac {\sqrt [6]{\sin (c+d x)+1} \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{\sqrt [6]{2} d (a \sin (c+d x)+a)^{2/3}} \]

[In]

Int[(a + a*Sin[c + d*x])^(-2/3),x]

[Out]

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

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]

Rubi steps \begin{align*} \text {integral}& = \frac {(1+\sin (c+d x))^{2/3} \int \frac {1}{(1+\sin (c+d x))^{2/3}} \, dx}{(a+a \sin (c+d x))^{2/3}} \\ & = -\frac {\cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right ) \sqrt [6]{1+\sin (c+d x)}}{\sqrt [6]{2} d (a+a \sin (c+d x))^{2/3}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(215\) vs. \(2(66)=132\).

Time = 0.65 (sec) , antiderivative size = 215, normalized size of antiderivative = 3.26 \[ \int \frac {1}{(a+a \sin (c+d x))^{2/3}} \, dx=\frac {2 \left (-3 \cos \left (\frac {1}{2} (c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^{4/3} \left (-2 \cos \left (\frac {1}{4} (2 c+\pi +2 d x)\right ) \, _2F_1\left (-\frac {1}{2},-\frac {1}{6};\frac {5}{6};\sin ^2\left (\frac {1}{4} (2 c+\pi +2 d x)\right )\right )+\sqrt {\cos ^2\left (\frac {1}{4} (2 c+\pi +2 d x)\right )} \left (2 \cos \left (\frac {1}{4} (2 c+\pi +2 d x)\right )+3 \sin \left (\frac {1}{4} (2 c+\pi +2 d x)\right )\right )\right )}{\sqrt [6]{2} \sqrt {1-\sin (c+d x)} \sqrt [3]{\sin \left (\frac {1}{4} (2 c+\pi +2 d x)\right )}}\right )}{d (a (1+\sin (c+d x)))^{2/3}} \]

[In]

Integrate[(a + a*Sin[c + d*x])^(-2/3),x]

[Out]

(2*(-3*Cos[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + ((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^(4/3)*(
-2*Cos[(2*c + Pi + 2*d*x)/4]*HypergeometricPFQ[{-1/2, -1/6}, {5/6}, Sin[(2*c + Pi + 2*d*x)/4]^2] + Sqrt[Cos[(2
*c + Pi + 2*d*x)/4]^2]*(2*Cos[(2*c + Pi + 2*d*x)/4] + 3*Sin[(2*c + Pi + 2*d*x)/4])))/(2^(1/6)*Sqrt[1 - Sin[c +
d*x]]*Sin[(2*c + Pi + 2*d*x)/4]^(1/3))))/(d*(a*(1 + Sin[c + d*x]))^(2/3))

Maple [F]

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

[In]

int(1/(a+a*sin(d*x+c))^(2/3),x)

[Out]

int(1/(a+a*sin(d*x+c))^(2/3),x)

Fricas [F]

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

[In]

integrate(1/(a+a*sin(d*x+c))^(2/3),x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

integrate(1/(a+a*sin(d*x+c))**(2/3),x)

[Out]

Integral((a*sin(c + d*x) + a)**(-2/3), x)

Maxima [F]

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

[In]

integrate(1/(a+a*sin(d*x+c))^(2/3),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate(1/(a+a*sin(d*x+c))^(2/3),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

int(1/(a + a*sin(c + d*x))^(2/3),x)

[Out]

int(1/(a + a*sin(c + d*x))^(2/3), x)