Integrand size = 14, antiderivative size = 66 \[ \int \frac {1}{(a+a \sin (c+d x))^{2/3}} \, dx=-\frac {\cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right ) \sqrt [6]{1+\sin (c+d x)}}{\sqrt [6]{2} d (a+a \sin (c+d x))^{2/3}} \] Output:
-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)
Leaf count is larger than twice the leaf count of optimal. \(215\) vs. \(2(66)=132\).
Time = 1.13 (sec) , antiderivative size = 215, normalized size of antiderivative = 3.26 \[ \int \frac {1}{(a+a \sin (c+d x))^{2/3}} \, dx=\frac {2 \left (-3 \cos \left (\frac {1}{2} (c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^{4/3} \left (-2 \cos \left (\frac {1}{4} (2 c+\pi +2 d x)\right ) \, _2F_1\left (-\frac {1}{2},-\frac {1}{6};\frac {5}{6};\sin ^2\left (\frac {1}{4} (2 c+\pi +2 d x)\right )\right )+\sqrt {\cos ^2\left (\frac {1}{4} (2 c+\pi +2 d x)\right )} \left (2 \cos \left (\frac {1}{4} (2 c+\pi +2 d x)\right )+3 \sin \left (\frac {1}{4} (2 c+\pi +2 d x)\right )\right )\right )}{\sqrt [6]{2} \sqrt {1-\sin (c+d x)} \sqrt [3]{\sin \left (\frac {1}{4} (2 c+\pi +2 d x)\right )}}\right )}{d (a (1+\sin (c+d x)))^{2/3}} \] Input:
Integrate[(a + a*Sin[c + d*x])^(-2/3),x]
Output:
(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]*Hypergeom etricPFQ[{-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))
Time = 0.26 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {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.
\(\displaystyle \int \frac {1}{(a \sin (c+d x)+a)^{2/3}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a \sin (c+d x)+a)^{2/3}}dx\) |
\(\Big \downarrow \) 3131 |
\(\displaystyle \frac {(\sin (c+d x)+1)^{2/3} \int \frac {1}{(\sin (c+d x)+1)^{2/3}}dx}{(a \sin (c+d x)+a)^{2/3}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(\sin (c+d x)+1)^{2/3} \int \frac {1}{(\sin (c+d x)+1)^{2/3}}dx}{(a \sin (c+d x)+a)^{2/3}}\) |
\(\Big \downarrow \) 3130 |
\(\displaystyle -\frac {\sqrt [6]{\sin (c+d x)+1} \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{\sqrt [6]{2} d (a \sin (c+d x)+a)^{2/3}}\) |
Input:
Int[(a + a*Sin[c + d*x])^(-2/3),x]
Output:
-((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)))
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]
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]
\[\int \frac {1}{\left (a +a \sin \left (d x +c \right )\right )^{\frac {2}{3}}}d x\]
Input:
int(1/(a+a*sin(d*x+c))^(2/3),x)
Output:
int(1/(a+a*sin(d*x+c))^(2/3),x)
\[ \int \frac {1}{(a+a \sin (c+d x))^{2/3}} \, dx=\int { \frac {1}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {2}{3}}} \,d x } \] Input:
integrate(1/(a+a*sin(d*x+c))^(2/3),x, algorithm="fricas")
Output:
integral((a*sin(d*x + c) + a)^(-2/3), x)
\[ \int \frac {1}{(a+a \sin (c+d x))^{2/3}} \, dx=\int \frac {1}{\left (a \sin {\left (c + d x \right )} + a\right )^{\frac {2}{3}}}\, dx \] Input:
integrate(1/(a+a*sin(d*x+c))**(2/3),x)
Output:
Integral((a*sin(c + d*x) + a)**(-2/3), x)
\[ \int \frac {1}{(a+a \sin (c+d x))^{2/3}} \, dx=\int { \frac {1}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {2}{3}}} \,d x } \] Input:
integrate(1/(a+a*sin(d*x+c))^(2/3),x, algorithm="maxima")
Output:
integrate((a*sin(d*x + c) + a)^(-2/3), x)
\[ \int \frac {1}{(a+a \sin (c+d x))^{2/3}} \, dx=\int { \frac {1}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {2}{3}}} \,d x } \] Input:
integrate(1/(a+a*sin(d*x+c))^(2/3),x, algorithm="giac")
Output:
integrate((a*sin(d*x + c) + a)^(-2/3), x)
Timed out. \[ \int \frac {1}{(a+a \sin (c+d x))^{2/3}} \, dx=\int \frac {1}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{2/3}} \,d x \] Input:
int(1/(a + a*sin(c + d*x))^(2/3),x)
Output:
int(1/(a + a*sin(c + d*x))^(2/3), x)
\[ \int \frac {1}{(a+a \sin (c+d x))^{2/3}} \, dx=\frac {\int \frac {1}{\left (\sin \left (d x +c \right )+1\right )^{\frac {2}{3}}}d x}{a^{\frac {2}{3}}} \] Input:
int(1/(a+a*sin(d*x+c))^(2/3),x)
Output:
int(1/(sin(c + d*x) + 1)**(2/3),x)/a**(2/3)