Integrand size = 21, antiderivative size = 99 \[ \int \frac {\sin (c+d x)}{(a+a \sin (c+d x))^{4/3}} \, dx=\frac {3 \cos (c+d x)}{5 d (a+a \sin (c+d x))^{4/3}}-\frac {4 \sqrt [6]{2} \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{5 a d \sqrt [6]{1+\sin (c+d x)} \sqrt [3]{a+a \sin (c+d x)}} \]
3/5*cos(d*x+c)/d/(a+a*sin(d*x+c))^(4/3)-4/5*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)
Time = 0.36 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.31 \[ \int \frac {\sin (c+d x)}{(a+a \sin (c+d x))^{4/3}} \, dx=\frac {3 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sqrt {2-2 \sin (c+d x)}+8 \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\sin ^2\left (\frac {1}{4} (2 c+\pi +2 d x)\right )\right ) (1+\sin (c+d x))\right )}{5 d \sqrt {2-2 \sin (c+d x)} (a (1+\sin (c+d x)))^{4/3}} \]
(3*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x) /2])*(Sqrt[2 - 2*Sin[c + d*x]] + 8*Hypergeometric2F1[1/6, 1/2, 7/6, Sin[(2 *c + Pi + 2*d*x)/4]^2]*(1 + Sin[c + d*x])))/(5*d*Sqrt[2 - 2*Sin[c + d*x]]* (a*(1 + Sin[c + d*x]))^(4/3))
Time = 0.36 (sec) , antiderivative size = 99, 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, 3229, 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.
\(\displaystyle \int \frac {\sin (c+d x)}{(a \sin (c+d x)+a)^{4/3}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)}{(a \sin (c+d x)+a)^{4/3}}dx\) |
\(\Big \downarrow \) 3229 |
\(\displaystyle \frac {4 \int \frac {1}{\sqrt [3]{\sin (c+d x) a+a}}dx}{5 a}+\frac {3 \cos (c+d x)}{5 d (a \sin (c+d x)+a)^{4/3}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4 \int \frac {1}{\sqrt [3]{\sin (c+d x) a+a}}dx}{5 a}+\frac {3 \cos (c+d x)}{5 d (a \sin (c+d x)+a)^{4/3}}\) |
\(\Big \downarrow \) 3131 |
\(\displaystyle \frac {4 \sqrt [3]{\sin (c+d x)+1} \int \frac {1}{\sqrt [3]{\sin (c+d x)+1}}dx}{5 a \sqrt [3]{a \sin (c+d x)+a}}+\frac {3 \cos (c+d x)}{5 d (a \sin (c+d x)+a)^{4/3}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4 \sqrt [3]{\sin (c+d x)+1} \int \frac {1}{\sqrt [3]{\sin (c+d x)+1}}dx}{5 a \sqrt [3]{a \sin (c+d x)+a}}+\frac {3 \cos (c+d x)}{5 d (a \sin (c+d x)+a)^{4/3}}\) |
\(\Big \downarrow \) 3130 |
\(\displaystyle \frac {3 \cos (c+d x)}{5 d (a \sin (c+d x)+a)^{4/3}}-\frac {4 \sqrt [6]{2} \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{5 a d \sqrt [6]{\sin (c+d x)+1} \sqrt [3]{a \sin (c+d x)+a}}\) |
(3*Cos[c + d*x])/(5*d*(a + a*Sin[c + d*x])^(4/3)) - (4*2^(1/6)*Cos[c + d*x ]*Hypergeometric2F1[1/2, 5/6, 3/2, (1 - Sin[c + d*x])/2])/(5*a*d*(1 + Sin[ c + d*x])^(1/6)*(a + a*Sin[c + d*x])^(1/3))
3.2.12.3.1 Defintions of rubi rules used
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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*((a + b*Sin[e + f* x])^m/(a*f*(2*m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)) I nt[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
\[\int \frac {\sin \left (d x +c \right )}{\left (a +a \sin \left (d x +c \right )\right )^{\frac {4}{3}}}d x\]
\[ \int \frac {\sin (c+d x)}{(a+a \sin (c+d x))^{4/3}} \, dx=\int { \frac {\sin \left (d x + c\right )}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {4}{3}}} \,d x } \]
integral(-(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)
\[ \int \frac {\sin (c+d x)}{(a+a \sin (c+d x))^{4/3}} \, dx=\int \frac {\sin {\left (c + d x \right )}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {4}{3}}}\, dx \]
\[ \int \frac {\sin (c+d x)}{(a+a \sin (c+d x))^{4/3}} \, dx=\int { \frac {\sin \left (d x + c\right )}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {4}{3}}} \,d x } \]
\[ \int \frac {\sin (c+d x)}{(a+a \sin (c+d x))^{4/3}} \, dx=\int { \frac {\sin \left (d x + c\right )}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {4}{3}}} \,d x } \]
Timed out. \[ \int \frac {\sin (c+d x)}{(a+a \sin (c+d x))^{4/3}} \, dx=\int \frac {\sin \left (c+d\,x\right )}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{4/3}} \,d x \]