Integrand size = 33, antiderivative size = 199 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=\frac {(B d-(B c+A d) (2+m)) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m) (2+m)}-\frac {2^{\frac {1}{2}+m} \left (A (2+m) (c+c m+d m)+B \left (c m (2+m)+d \left (1+m+m^2\right )\right )\right ) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^m}{f (1+m) (2+m)}-\frac {B d \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m)} \]
(B*d-(A*d+B*c)*(2+m))*cos(f*x+e)*(a+a*sin(f*x+e))^m/f/(1+m)/(2+m)-2^(1/2+m )*(A*(2+m)*(c*m+d*m+c)+B*(c*m*(2+m)+d*(m^2+m+1)))*cos(f*x+e)*hypergeom([1/ 2, 1/2-m],[3/2],1/2-1/2*sin(f*x+e))*(1+sin(f*x+e))^(-1/2-m)*(a+a*sin(f*x+e ))^m/f/(m^2+3*m+2)-B*d*cos(f*x+e)*(a+a*sin(f*x+e))^(1+m)/a/f/(2+m)
Result contains complex when optimal does not.
Time = 22.75 (sec) , antiderivative size = 317, normalized size of antiderivative = 1.59 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=-\frac {(a (1+\sin (e+f x)))^m (\cos (e+f x)+i (1+\sin (e+f x))) \left (-\frac {2 (2 A c+B d) \operatorname {Hypergeometric2F1}(1,1+m,1-m,i \cos (e+f x)-\sin (e+f x))}{m}-\frac {2 i (B c+A d) \operatorname {Hypergeometric2F1}(1,m,-m,i \cos (e+f x)-\sin (e+f x)) (\cos (e+f x)-i \sin (e+f x))}{1+m}+\frac {2 i (B c+A d) \operatorname {Hypergeometric2F1}(1,2+m,2-m,i \cos (e+f x)-\sin (e+f x)) (\cos (e+f x)+i \sin (e+f x))}{-1+m}+\frac {B d \operatorname {Hypergeometric2F1}(1,-1+m,-1-m,i \cos (e+f x)-\sin (e+f x)) (\cos (2 (e+f x))-i \sin (2 (e+f x)))}{2+m}+\frac {B d \operatorname {Hypergeometric2F1}(1,3+m,3-m,i \cos (e+f x)-\sin (e+f x)) (\cos (2 (e+f x))+i \sin (2 (e+f x)))}{-2+m}\right )}{4 f} \]
-1/4*((a*(1 + Sin[e + f*x]))^m*(Cos[e + f*x] + I*(1 + Sin[e + f*x]))*((-2* (2*A*c + B*d)*Hypergeometric2F1[1, 1 + m, 1 - m, I*Cos[e + f*x] - Sin[e + f*x]])/m - ((2*I)*(B*c + A*d)*Hypergeometric2F1[1, m, -m, I*Cos[e + f*x] - Sin[e + f*x]]*(Cos[e + f*x] - I*Sin[e + f*x]))/(1 + m) + ((2*I)*(B*c + A* d)*Hypergeometric2F1[1, 2 + m, 2 - m, I*Cos[e + f*x] - Sin[e + f*x]]*(Cos[ e + f*x] + I*Sin[e + f*x]))/(-1 + m) + (B*d*Hypergeometric2F1[1, -1 + m, - 1 - m, I*Cos[e + f*x] - Sin[e + f*x]]*(Cos[2*(e + f*x)] - I*Sin[2*(e + f*x )]))/(2 + m) + (B*d*Hypergeometric2F1[1, 3 + m, 3 - m, I*Cos[e + f*x] - Si n[e + f*x]]*(Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)]))/(-2 + m)))/f
Time = 0.86 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {3042, 3447, 3042, 3502, 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.
| \(\displaystyle \int (a \sin (e+f x)+a)^m (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a \sin (e+f x)+a)^m (A+B \sin (e+f x)) (c+d \sin (e+f x))dx\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \int (a \sin (e+f x)+a)^m \left ((A d+B c) \sin (e+f x)+A c+B d \sin ^2(e+f x)\right )dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a \sin (e+f x)+a)^m \left ((A d+B c) \sin (e+f x)+A c+B d \sin (e+f x)^2\right )dx\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {\int (\sin (e+f x) a+a)^m (a (B d (m+1)+A c (m+2))-a (B d-(B c+A d) (m+2)) \sin (e+f x))dx}{a (m+2)}-\frac {B d \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (m+2)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int (\sin (e+f x) a+a)^m (a (B d (m+1)+A c (m+2))-a (B d-(B c+A d) (m+2)) \sin (e+f x))dx}{a (m+2)}-\frac {B d \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (m+2)}\) |
| \(\Big \downarrow \) 3230 |
| \(\displaystyle \frac {\frac {a \left (A (m+2) (c m+c+d m)+B c m (m+2)+B d \left (m^2+m+1\right )\right ) \int (\sin (e+f x) a+a)^mdx}{m+1}+\frac {a \cos (e+f x) (B d-(m+2) (A d+B c)) (a \sin (e+f x)+a)^m}{f (m+1)}}{a (m+2)}-\frac {B d \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (m+2)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a \left (A (m+2) (c m+c+d m)+B c m (m+2)+B d \left (m^2+m+1\right )\right ) \int (\sin (e+f x) a+a)^mdx}{m+1}+\frac {a \cos (e+f x) (B d-(m+2) (A d+B c)) (a \sin (e+f x)+a)^m}{f (m+1)}}{a (m+2)}-\frac {B d \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (m+2)}\) |
| \(\Big \downarrow \) 3131 |
| \(\displaystyle \frac {\frac {a \left (A (m+2) (c m+c+d m)+B c m (m+2)+B d \left (m^2+m+1\right )\right ) (\sin (e+f x)+1)^{-m} (a \sin (e+f x)+a)^m \int (\sin (e+f x)+1)^mdx}{m+1}+\frac {a \cos (e+f x) (B d-(m+2) (A d+B c)) (a \sin (e+f x)+a)^m}{f (m+1)}}{a (m+2)}-\frac {B d \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (m+2)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a \left (A (m+2) (c m+c+d m)+B c m (m+2)+B d \left (m^2+m+1\right )\right ) (\sin (e+f x)+1)^{-m} (a \sin (e+f x)+a)^m \int (\sin (e+f x)+1)^mdx}{m+1}+\frac {a \cos (e+f x) (B d-(m+2) (A d+B c)) (a \sin (e+f x)+a)^m}{f (m+1)}}{a (m+2)}-\frac {B d \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (m+2)}\) |
| \(\Big \downarrow \) 3130 |
| \(\displaystyle \frac {\frac {a \cos (e+f x) (B d-(m+2) (A d+B c)) (a \sin (e+f x)+a)^m}{f (m+1)}-\frac {a 2^{m+\frac {1}{2}} \cos (e+f x) \left (A (m+2) (c m+c+d m)+B c m (m+2)+B d \left (m^2+m+1\right )\right ) (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a \sin (e+f x)+a)^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x))\right )}{f (m+1)}}{a (m+2)}-\frac {B d \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (m+2)}\) |
-((B*d*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m))/(a*f*(2 + m))) + ((a*(B* d - (B*c + A*d)*(2 + m))*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(f*(1 + m)) - (2^(1/2 + m)*a*(B*c*m*(2 + m) + A*(2 + m)*(c + c*m + d*m) + B*d*(1 + m + m^2))*Cos[e + f*x]*Hypergeometric2F1[1/2, 1/2 - m, 3/2, (1 - Sin[e + f*x] )/2]*(1 + Sin[e + f*x])^(-1/2 - m)*(a + a*Sin[e + f*x])^m)/(f*(1 + m)))/(a *(2 + m))
3.4.37.3.1 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[(-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[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[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*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
\[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right ) \left (c +d \sin \left (f x +e \right )\right )d x\]
\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (d \sin \left (f x + e\right ) + c\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
integral(-(B*d*cos(f*x + e)^2 - A*c - B*d - (B*c + A*d)*sin(f*x + e))*(a*s in(f*x + e) + a)^m, x)
\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (A + B \sin {\left (e + f x \right )}\right ) \left (c + d \sin {\left (e + f x \right )}\right )\, dx \]
\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (d \sin \left (f x + e\right ) + c\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (d \sin \left (f x + e\right ) + c\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
Timed out. \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=\int \left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (c+d\,\sin \left (e+f\,x\right )\right ) \,d x \]