Integrand size = 29, antiderivative size = 115 \[ \int (-3-5 \sin (e+f x))^{-1-m} (3+3 \sin (e+f x))^m \, dx=\frac {\cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,1-m,\frac {3+5 \sin (e+f x)}{4 (1+\sin (e+f x))}\right ) (-3-5 \sin (e+f x))^{-m} \sqrt {\frac {1-\sin (e+f x)}{1+\sin (e+f x)}} (3+3 \sin (e+f x))^m}{4 f m (1-\sin (e+f x))} \]
1/4*cos(f*x+e)*hypergeom([1/2, -m],[1-m],1/4*(3+5*sin(f*x+e))/(1+sin(f*x+e )))*(a+a*sin(f*x+e))^m*((1-sin(f*x+e))/(1+sin(f*x+e)))^(1/2)/f/m/((-3-5*si n(f*x+e))^m)/(1-sin(f*x+e))
Time = 9.27 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.73 \[ \int (-3-5 \sin (e+f x))^{-1-m} (3+3 \sin (e+f x))^m \, dx=-\frac {3^m \operatorname {Hypergeometric2F1}\left (1+m,1+2 m,2 (1+m),\frac {4 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )}{\cos \left (\frac {1}{2} (e+f x)\right )+3 \sin \left (\frac {1}{2} (e+f x)\right )}\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (-\frac {3 \cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )}{\cos \left (\frac {1}{2} (e+f x)\right )+3 \sin \left (\frac {1}{2} (e+f x)\right )}\right )^m (-3-5 \sin (e+f x))^{-m} (1+\sin (e+f x))^m}{f (1+2 m) \left (\cos \left (\frac {1}{2} (e+f x)\right )+3 \sin \left (\frac {1}{2} (e+f x)\right )\right )} \]
-((3^m*Hypergeometric2F1[1 + m, 1 + 2*m, 2*(1 + m), (4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))/(Cos[(e + f*x)/2] + 3*Sin[(e + f*x)/2])]*(Cos[(e + f*x )/2] + Sin[(e + f*x)/2])*(-((3*Cos[(e + f*x)/2] + Sin[(e + f*x)/2])/(Cos[( e + f*x)/2] + 3*Sin[(e + f*x)/2])))^m*(1 + Sin[e + f*x])^m)/(f*(1 + 2*m)*( Cos[(e + f*x)/2] + 3*Sin[(e + f*x)/2])*(-3 - 5*Sin[e + f*x])^m))
Time = 0.29 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3042, 3267, 142}
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.
\(\displaystyle \int (-5 \sin (e+f x)-3)^{-m-1} (a \sin (e+f x)+a)^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (-5 \sin (e+f x)-3)^{-m-1} (a \sin (e+f x)+a)^mdx\) |
\(\Big \downarrow \) 3267 |
\(\displaystyle \frac {a^2 \cos (e+f x) \int \frac {(-5 \sin (e+f x)-3)^{-m-1} (\sin (e+f x) a+a)^{m-\frac {1}{2}}}{\sqrt {a-a \sin (e+f x)}}d\sin (e+f x)}{f \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 142 |
\(\displaystyle \frac {a \sqrt {\frac {1-\sin (e+f x)}{\sin (e+f x)+1}} \cos (e+f x) (-5 \sin (e+f x)-3)^{-m} (a \sin (e+f x)+a)^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,1-m,\frac {5 \sin (e+f x)+3}{4 (\sin (e+f x)+1)}\right )}{4 f m (a-a \sin (e+f x))}\) |
(a*Cos[e + f*x]*Hypergeometric2F1[1/2, -m, 1 - m, (3 + 5*Sin[e + f*x])/(4* (1 + Sin[e + f*x]))]*Sqrt[(1 - Sin[e + f*x])/(1 + Sin[e + f*x])]*(a + a*Si n[e + f*x])^m)/(4*f*m*(-3 - 5*Sin[e + f*x])^m*(a - a*Sin[e + f*x]))
3.7.52.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((b*e - a*f)*(m + 1)))*Hypergeometric2F1[m + 1, -n, m + 2, (-(d*e - c*f))*((a + b*x)/((b*c - a*d)*(e + f*x)))])/((b*e - a*f)*((c + d*x)/((b*c - a*d)*(e + f *x))))^n, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[m + n + p + 2, 0] && !IntegerQ[n]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^2*(Cos[e + f*x]/(f*Sqrt[a + b*Sin[e + f*x]]*Sqrt[a - b*Sin[e + f*x]])) Subst[Int[(a + b*x)^(m - 1/2)*((c + d* x)^n/Sqrt[a - b*x]), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m , n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !IntegerQ[m]
\[\int \left (-3-5 \sin \left (f x +e \right )\right )^{-1-m} \left (a +a \sin \left (f x +e \right )\right )^{m}d x\]
\[ \int (-3-5 \sin (e+f x))^{-1-m} (3+3 \sin (e+f x))^m \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-5 \, \sin \left (f x + e\right ) - 3\right )}^{-m - 1} \,d x } \]
\[ \int (-3-5 \sin (e+f x))^{-1-m} (3+3 \sin (e+f x))^m \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (- 5 \sin {\left (e + f x \right )} - 3\right )^{- m - 1}\, dx \]
\[ \int (-3-5 \sin (e+f x))^{-1-m} (3+3 \sin (e+f x))^m \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-5 \, \sin \left (f x + e\right ) - 3\right )}^{-m - 1} \,d x } \]
\[ \int (-3-5 \sin (e+f x))^{-1-m} (3+3 \sin (e+f x))^m \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-5 \, \sin \left (f x + e\right ) - 3\right )}^{-m - 1} \,d x } \]
Timed out. \[ \int (-3-5 \sin (e+f x))^{-1-m} (3+3 \sin (e+f x))^m \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (-5\,\sin \left (e+f\,x\right )-3\right )}^{m+1}} \,d x \]