Integrand size = 23, antiderivative size = 127 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=-\frac {37\ 2^{5/6} a \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right ) \sqrt [3]{a+a \sin (c+d x)}}{35 d (1+\sin (c+d x))^{5/6}}+\frac {9 \cos (c+d x) (a+a \sin (c+d x))^{4/3}}{70 d}-\frac {3 \cos (c+d x) (a+a \sin (c+d x))^{7/3}}{10 a d} \]
-37/35*2^(5/6)*a*cos(d*x+c)*hypergeom([-5/6, 1/2],[3/2],1/2-1/2*sin(d*x+c) )*(a+a*sin(d*x+c))^(1/3)/d/(1+sin(d*x+c))^(5/6)+9/70*cos(d*x+c)*(a+a*sin(d *x+c))^(4/3)/d-3/10*cos(d*x+c)*(a+a*sin(d*x+c))^(7/3)/a/d
Time = 3.64 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.97 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\frac {(a (1+\sin (c+d x)))^{4/3} \left (-\frac {3 (-185+60 \cos (c+d x)-7 \cos (3 (c+d x))+22 \sin (2 (c+d x)))}{20 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {37 \left (-2 \cos \left (\frac {1}{4} (2 c+\pi +2 d x)\right ) \, _2F_1\left (-\frac {1}{2},-\frac {1}{6};\frac {5}{6};\sin ^2\left (\frac {1}{4} (2 c+\pi +2 d x)\right )\right )+\sqrt {\cos ^2\left (\frac {1}{4} (2 c+\pi +2 d x)\right )} \left (2 \cos \left (\frac {1}{4} (2 c+\pi +2 d x)\right )+3 \sin \left (\frac {1}{4} (2 c+\pi +2 d x)\right )\right )\right )}{2\ 2^{2/3} \sqrt {\cos ^2\left (\frac {1}{4} (2 c+\pi +2 d x)\right )} \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^{8/3} \sqrt [3]{\sin \left (\frac {1}{4} (2 c+\pi +2 d x)\right )}}\right )}{14 d} \]
((a*(1 + Sin[c + d*x]))^(4/3)*((-3*(-185 + 60*Cos[c + d*x] - 7*Cos[3*(c + d*x)] + 22*Sin[2*(c + d*x)]))/(20*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) - (37*(-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*2^(2/3)*Sqrt[Cos [(2*c + Pi + 2*d*x)/4]^2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^(8/3)*Sin[ (2*c + Pi + 2*d*x)/4]^(1/3))))/(14*d)
Time = 0.53 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.09, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 3238, 27, 3042, 3230, 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.
\(\displaystyle \int \sin ^2(c+d x) (a \sin (c+d x)+a)^{4/3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (c+d x)^2 (a \sin (c+d x)+a)^{4/3}dx\) |
\(\Big \downarrow \) 3238 |
\(\displaystyle \frac {3 \int \frac {1}{3} (7 a-3 a \sin (c+d x)) (\sin (c+d x) a+a)^{4/3}dx}{10 a}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{7/3}}{10 a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int (7 a-3 a \sin (c+d x)) (\sin (c+d x) a+a)^{4/3}dx}{10 a}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{7/3}}{10 a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int (7 a-3 a \sin (c+d x)) (\sin (c+d x) a+a)^{4/3}dx}{10 a}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{7/3}}{10 a d}\) |
\(\Big \downarrow \) 3230 |
\(\displaystyle \frac {\frac {37}{7} a \int (\sin (c+d x) a+a)^{4/3}dx+\frac {9 a \cos (c+d x) (a \sin (c+d x)+a)^{4/3}}{7 d}}{10 a}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{7/3}}{10 a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {37}{7} a \int (\sin (c+d x) a+a)^{4/3}dx+\frac {9 a \cos (c+d x) (a \sin (c+d x)+a)^{4/3}}{7 d}}{10 a}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{7/3}}{10 a d}\) |
\(\Big \downarrow \) 3131 |
\(\displaystyle \frac {\frac {37 a^2 \sqrt [3]{a \sin (c+d x)+a} \int (\sin (c+d x)+1)^{4/3}dx}{7 \sqrt [3]{\sin (c+d x)+1}}+\frac {9 a \cos (c+d x) (a \sin (c+d x)+a)^{4/3}}{7 d}}{10 a}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{7/3}}{10 a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {37 a^2 \sqrt [3]{a \sin (c+d x)+a} \int (\sin (c+d x)+1)^{4/3}dx}{7 \sqrt [3]{\sin (c+d x)+1}}+\frac {9 a \cos (c+d x) (a \sin (c+d x)+a)^{4/3}}{7 d}}{10 a}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{7/3}}{10 a d}\) |
\(\Big \downarrow \) 3130 |
\(\displaystyle \frac {\frac {9 a \cos (c+d x) (a \sin (c+d x)+a)^{4/3}}{7 d}-\frac {74\ 2^{5/6} a^2 \cos (c+d x) \sqrt [3]{a \sin (c+d x)+a} \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{7 d (\sin (c+d x)+1)^{5/6}}}{10 a}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{7/3}}{10 a d}\) |
(-3*Cos[c + d*x]*(a + a*Sin[c + d*x])^(7/3))/(10*a*d) + ((-74*2^(5/6)*a^2* Cos[c + d*x]*Hypergeometric2F1[-5/6, 1/2, 3/2, (1 - Sin[c + d*x])/2]*(a + a*Sin[c + d*x])^(1/3))/(7*d*(1 + Sin[c + d*x])^(5/6)) + (9*a*Cos[c + d*x]* (a + a*Sin[c + d*x])^(4/3))/(7*d))/(10*a)
3.1.99.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[(-d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/( f*(m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(b*(m + 1)) Int[(a + b*Sin[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)]
Int[sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-Cos[e + f*x])*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2 ))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*(b*(m + 1) - a*Si n[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] && ! LtQ[m, -2^(-1)]
\[\int \left (\sin ^{2}\left (d x +c \right )\right ) \left (a +a \sin \left (d x +c \right )\right )^{\frac {4}{3}}d x\]
\[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {4}{3}} \sin \left (d x + c\right )^{2} \,d x } \]
integral(-(a*cos(d*x + c)^2 + (a*cos(d*x + c)^2 - a)*sin(d*x + c) - a)*(a* sin(d*x + c) + a)^(1/3), x)
\[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {4}{3}} \sin ^{2}{\left (c + d x \right )}\, dx \]
\[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {4}{3}} \sin \left (d x + c\right )^{2} \,d x } \]
\[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {4}{3}} \sin \left (d x + c\right )^{2} \,d x } \]
Timed out. \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{4/3} \, dx=\int {\sin \left (c+d\,x\right )}^2\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{4/3} \,d x \]