Integrand size = 31, antiderivative size = 191 \[ \int (d \sin (e+f x))^n (a+a \sin (e+f x)) (A+B \sin (e+f x)) \, dx=-\frac {a B \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (2+n)}+\frac {a (B (1+n)+A (2+n)) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\sin ^2(e+f x)\right ) (d \sin (e+f x))^{1+n}}{d f (1+n) (2+n) \sqrt {\cos ^2(e+f x)}}+\frac {a (A+B) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+n}{2},\frac {4+n}{2},\sin ^2(e+f x)\right ) (d \sin (e+f x))^{2+n}}{d^2 f (2+n) \sqrt {\cos ^2(e+f x)}} \]
-a*B*cos(f*x+e)*(d*sin(f*x+e))^(1+n)/d/f/(2+n)+a*(B*(1+n)+A*(2+n))*cos(f*x +e)*hypergeom([1/2, 1/2+1/2*n],[3/2+1/2*n],sin(f*x+e)^2)*(d*sin(f*x+e))^(1 +n)/d/f/(1+n)/(2+n)/(cos(f*x+e)^2)^(1/2)+a*(A+B)*cos(f*x+e)*hypergeom([1/2 , 1+1/2*n],[1/2*n+2],sin(f*x+e)^2)*(d*sin(f*x+e))^(2+n)/d^2/f/(2+n)/(cos(f *x+e)^2)^(1/2)
Time = 0.66 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.76 \[ \int (d \sin (e+f x))^n (a+a \sin (e+f x)) (A+B \sin (e+f x)) \, dx=\frac {a \cos (e+f x) \sin (e+f x) (d \sin (e+f x))^n \left ((B (1+n)+A (2+n)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\sin ^2(e+f x)\right )-(1+n) \left (B \sqrt {\cos ^2(e+f x)}-(A+B) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+n}{2},\frac {4+n}{2},\sin ^2(e+f x)\right ) \sin (e+f x)\right )\right )}{f (1+n) (2+n) \sqrt {\cos ^2(e+f x)}} \]
(a*Cos[e + f*x]*Sin[e + f*x]*(d*Sin[e + f*x])^n*((B*(1 + n) + A*(2 + n))*H ypergeometric2F1[1/2, (1 + n)/2, (3 + n)/2, Sin[e + f*x]^2] - (1 + n)*(B*S qrt[Cos[e + f*x]^2] - (A + B)*Hypergeometric2F1[1/2, (2 + n)/2, (4 + n)/2, Sin[e + f*x]^2]*Sin[e + f*x])))/(f*(1 + n)*(2 + n)*Sqrt[Cos[e + f*x]^2])
Time = 0.63 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {3042, 3447, 3042, 3502, 3042, 3227, 3042, 3122}
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.
\(\displaystyle \int (a \sin (e+f x)+a) (A+B \sin (e+f x)) (d \sin (e+f x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \sin (e+f x)+a) (A+B \sin (e+f x)) (d \sin (e+f x))^ndx\) |
\(\Big \downarrow \) 3447 |
\(\displaystyle \int (d \sin (e+f x))^n \left ((a A+a B) \sin (e+f x)+a A+a B \sin ^2(e+f x)\right )dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (d \sin (e+f x))^n \left ((a A+a B) \sin (e+f x)+a A+a B \sin (e+f x)^2\right )dx\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {\int (d \sin (e+f x))^n (a d (B (n+1)+A (n+2))+a (A+B) d (n+2) \sin (e+f x))dx}{d (n+2)}-\frac {a B \cos (e+f x) (d \sin (e+f x))^{n+1}}{d f (n+2)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int (d \sin (e+f x))^n (a d (B (n+1)+A (n+2))+a (A+B) d (n+2) \sin (e+f x))dx}{d (n+2)}-\frac {a B \cos (e+f x) (d \sin (e+f x))^{n+1}}{d f (n+2)}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {a d (A (n+2)+B (n+1)) \int (d \sin (e+f x))^ndx+a (n+2) (A+B) \int (d \sin (e+f x))^{n+1}dx}{d (n+2)}-\frac {a B \cos (e+f x) (d \sin (e+f x))^{n+1}}{d f (n+2)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {a d (A (n+2)+B (n+1)) \int (d \sin (e+f x))^ndx+a (n+2) (A+B) \int (d \sin (e+f x))^{n+1}dx}{d (n+2)}-\frac {a B \cos (e+f x) (d \sin (e+f x))^{n+1}}{d f (n+2)}\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \frac {\frac {a (A (n+2)+B (n+1)) \cos (e+f x) (d \sin (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n+1}{2},\frac {n+3}{2},\sin ^2(e+f x)\right )}{f (n+1) \sqrt {\cos ^2(e+f x)}}+\frac {a (A+B) \cos (e+f x) (d \sin (e+f x))^{n+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n+2}{2},\frac {n+4}{2},\sin ^2(e+f x)\right )}{d f \sqrt {\cos ^2(e+f x)}}}{d (n+2)}-\frac {a B \cos (e+f x) (d \sin (e+f x))^{n+1}}{d f (n+2)}\) |
-((a*B*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + n))/(d*f*(2 + n))) + ((a*(B*(1 + n) + A*(2 + n))*Cos[e + f*x]*Hypergeometric2F1[1/2, (1 + n)/2, (3 + n)/2, Sin[e + f*x]^2]*(d*Sin[e + f*x])^(1 + n))/(f*(1 + n)*Sqrt[Cos[e + f*x]^2] ) + (a*(A + B)*Cos[e + f*x]*Hypergeometric2F1[1/2, (2 + n)/2, (4 + n)/2, S in[e + f*x]^2]*(d*Sin[e + f*x])^(2 + n))/(d*f*Sqrt[Cos[e + f*x]^2]))/(d*(2 + n))
3.1.3.3.1 Defintions of rubi rules used
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2 F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[2*n]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
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]
\[\int \left (d \sin \left (f x +e \right )\right )^{n} \left (a +a \sin \left (f x +e \right )\right ) \left (A +B \sin \left (f x +e \right )\right )d x\]
\[ \int (d \sin (e+f x))^n (a+a \sin (e+f x)) (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )} \left (d \sin \left (f x + e\right )\right )^{n} \,d x } \]
Timed out. \[ \int (d \sin (e+f x))^n (a+a \sin (e+f x)) (A+B \sin (e+f x)) \, dx=\text {Timed out} \]
\[ \int (d \sin (e+f x))^n (a+a \sin (e+f x)) (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )} \left (d \sin \left (f x + e\right )\right )^{n} \,d x } \]
\[ \int (d \sin (e+f x))^n (a+a \sin (e+f x)) (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )} \left (d \sin \left (f x + e\right )\right )^{n} \,d x } \]
Timed out. \[ \int (d \sin (e+f x))^n (a+a \sin (e+f x)) (A+B \sin (e+f x)) \, dx=\int {\left (d\,\sin \left (e+f\,x\right )\right )}^n\,\left (A+B\,\sin \left (e+f\,x\right )\right )\,\left (a+a\,\sin \left (e+f\,x\right )\right ) \,d x \]