Integrand size = 29, antiderivative size = 117 \[ \int (-3-4 \sin (e+f x))^{-1-m} (3+3 \sin (e+f x))^m \, dx=\frac {\cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,1-m,\frac {2 (3+4 \sin (e+f x))}{7 (1+\sin (e+f x))}\right ) (-3-4 \sin (e+f x))^{-m} \sqrt {\frac {1-\sin (e+f x)}{1+\sin (e+f x)}} (3+3 \sin (e+f x))^m}{\sqrt {7} f m (1-\sin (e+f x))} \]
1/7*cos(f*x+e)*hypergeom([1/2, -m],[1-m],2/7*(3+4*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/((-3-4*si n(f*x+e))^m)/(1-sin(f*x+e))*7^(1/2)
Result contains complex when optimal does not.
Time = 9.52 (sec) , antiderivative size = 274, normalized size of antiderivative = 2.34 \[ \int (-3-4 \sin (e+f x))^{-1-m} (3+3 \sin (e+f x))^m \, dx=-\frac {3^m \operatorname {Hypergeometric2F1}\left (1+m,1+2 m,2 (1+m),-\frac {8 \sqrt {7} (\cos (e+f x)+i (1+\sin (e+f x)))}{\left (-i+\sqrt {7}\right ) \left (-3 i+\sqrt {7}-4 \cos (e+f x)-4 i \sin (e+f x)\right )}\right ) (-3-4 \sin (e+f x))^{-1-m} \left (3 i+\sqrt {7}+4 \cos (e+f x)+4 i \sin (e+f x)\right ) \left (\frac {\left (i+\sqrt {7}\right ) \left (3 i+\sqrt {7}+4 \cos (e+f x)+4 i \sin (e+f x)\right )}{\left (-i+\sqrt {7}\right ) \left (-3 i+\sqrt {7}-4 \cos (e+f x)-4 i \sin (e+f x)\right )}\right )^m (1+\sin (e+f x))^m (i \cos (e+f x)+\sin (e+f x)) (\cos (e+f x)+i (1+\sin (e+f x)))}{\left (-i+\sqrt {7}\right ) f (1+2 m)} \]
-((3^m*Hypergeometric2F1[1 + m, 1 + 2*m, 2*(1 + m), (-8*Sqrt[7]*(Cos[e + f *x] + I*(1 + Sin[e + f*x])))/((-I + Sqrt[7])*(-3*I + Sqrt[7] - 4*Cos[e + f *x] - (4*I)*Sin[e + f*x]))]*(-3 - 4*Sin[e + f*x])^(-1 - m)*(3*I + Sqrt[7] + 4*Cos[e + f*x] + (4*I)*Sin[e + f*x])*(((I + Sqrt[7])*(3*I + Sqrt[7] + 4* Cos[e + f*x] + (4*I)*Sin[e + f*x]))/((-I + Sqrt[7])*(-3*I + Sqrt[7] - 4*Co s[e + f*x] - (4*I)*Sin[e + f*x])))^m*(1 + Sin[e + f*x])^m*(I*Cos[e + f*x] + Sin[e + f*x])*(Cos[e + f*x] + I*(1 + Sin[e + f*x])))/((-I + Sqrt[7])*f*( 1 + 2*m)))
Time = 0.28 (sec) , antiderivative size = 119, 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.
\(\displaystyle \int (-4 \sin (e+f x)-3)^{-m-1} (a \sin (e+f x)+a)^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (-4 \sin (e+f x)-3)^{-m-1} (a \sin (e+f x)+a)^mdx\) |
\(\Big \downarrow \) 3267 |
\(\displaystyle \frac {a^2 \cos (e+f x) \int \frac {(-4 \sin (e+f x)-3)^{-m-1} (\sin (e+f x) a+a)^{m-\frac {1}{2}}}{\sqrt {a-a \sin (e+f x)}}d\sin (e+f x)}{f \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 142 |
\(\displaystyle \frac {a \sqrt {\frac {1-\sin (e+f x)}{\sin (e+f x)+1}} \cos (e+f x) (-4 \sin (e+f x)-3)^{-m} (a \sin (e+f x)+a)^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,1-m,\frac {2 (4 \sin (e+f x)+3)}{7 (\sin (e+f x)+1)}\right )}{\sqrt {7} f m (a-a \sin (e+f x))}\) |
(a*Cos[e + f*x]*Hypergeometric2F1[1/2, -m, 1 - m, (2*(3 + 4*Sin[e + f*x])) /(7*(1 + Sin[e + f*x]))]*Sqrt[(1 - Sin[e + f*x])/(1 + Sin[e + f*x])]*(a + a*Sin[e + f*x])^m)/(Sqrt[7]*f*m*(-3 - 4*Sin[e + f*x])^m*(a - a*Sin[e + f*x ]))
3.7.51.3.1 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]
\[\int \left (-3-4 \sin \left (f x +e \right )\right )^{-1-m} \left (a +a \sin \left (f x +e \right )\right )^{m}d x\]
\[ \int (-3-4 \sin (e+f x))^{-1-m} (3+3 \sin (e+f x))^m \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-4 \, \sin \left (f x + e\right ) - 3\right )}^{-m - 1} \,d x } \]
\[ \int (-3-4 \sin (e+f x))^{-1-m} (3+3 \sin (e+f x))^m \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (- 4 \sin {\left (e + f x \right )} - 3\right )^{- m - 1}\, dx \]
\[ \int (-3-4 \sin (e+f x))^{-1-m} (3+3 \sin (e+f x))^m \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-4 \, \sin \left (f x + e\right ) - 3\right )}^{-m - 1} \,d x } \]
\[ \int (-3-4 \sin (e+f x))^{-1-m} (3+3 \sin (e+f x))^m \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-4 \, \sin \left (f x + e\right ) - 3\right )}^{-m - 1} \,d x } \]
Timed out. \[ \int (-3-4 \sin (e+f x))^{-1-m} (3+3 \sin (e+f x))^m \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (-4\,\sin \left (e+f\,x\right )-3\right )}^{m+1}} \,d x \]