Integrand size = 36, antiderivative size = 174 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^n \, dx=\frac {2^{\frac {1}{2}+n} c (B (m-n)+A (1+m+n)) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (1+2 m),\frac {1}{2} (1-2 n),\frac {1}{2} (3+2 m),\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {1}{2}-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1+n}}{f (1+2 m) (1+m+n)}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n}{f (1+m+n)} \]
2^(1/2+n)*c*(B*(m-n)+A*(1+m+n))*cos(f*x+e)*hypergeom([1/2-n, 1/2+m],[3/2+m ],1/2+1/2*sin(f*x+e))*(1-sin(f*x+e))^(1/2-n)*(a+a*sin(f*x+e))^m*(c-c*sin(f *x+e))^(-1+n)/f/(1+2*m)/(1+m+n)-B*cos(f*x+e)*(a+a*sin(f*x+e))^m*(c-c*sin(f *x+e))^n/f/(1+m+n)
\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^n \, dx=\int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^n \, dx \]
Time = 0.73 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.21, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3042, 3452, 3042, 3224, 3042, 3168, 80, 79}
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.
\(\displaystyle \int (a \sin (e+f x)+a)^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \sin (e+f x)+a)^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^ndx\) |
\(\Big \downarrow \) 3452 |
\(\displaystyle \left (A+\frac {B (m-n)}{m+n+1}\right ) \int (\sin (e+f x) a+a)^m (c-c \sin (e+f x))^ndx-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n}{f (m+n+1)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \left (A+\frac {B (m-n)}{m+n+1}\right ) \int (\sin (e+f x) a+a)^m (c-c \sin (e+f x))^ndx-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n}{f (m+n+1)}\) |
\(\Big \downarrow \) 3224 |
\(\displaystyle \left (A+\frac {B (m-n)}{m+n+1}\right ) \cos ^{-2 m}(e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m \int \cos ^{2 m}(e+f x) (c-c \sin (e+f x))^{n-m}dx-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n}{f (m+n+1)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \left (A+\frac {B (m-n)}{m+n+1}\right ) \cos ^{-2 m}(e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m \int \cos (e+f x)^{2 m} (c-c \sin (e+f x))^{n-m}dx-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n}{f (m+n+1)}\) |
\(\Big \downarrow \) 3168 |
\(\displaystyle \frac {c^2 \left (A+\frac {B (m-n)}{m+n+1}\right ) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{\frac {1}{2} (-2 m-1)+m} (c \sin (e+f x)+c)^{\frac {1}{2} (-2 m-1)} \int (c-c \sin (e+f x))^{\frac {1}{2} (2 n-1)} (\sin (e+f x) c+c)^{\frac {1}{2} (2 m-1)}d\sin (e+f x)}{f}-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n}{f (m+n+1)}\) |
\(\Big \downarrow \) 80 |
\(\displaystyle \frac {c^2 2^{n-\frac {1}{2}} \left (A+\frac {B (m-n)}{m+n+1}\right ) \cos (e+f x) (1-\sin (e+f x))^{\frac {1}{2}-n} (a \sin (e+f x)+a)^m (c \sin (e+f x)+c)^{\frac {1}{2} (-2 m-1)} (c-c \sin (e+f x))^{\frac {1}{2} (-2 m-1)+m+n-\frac {1}{2}} \int \left (\frac {1}{2}-\frac {1}{2} \sin (e+f x)\right )^{\frac {1}{2} (2 n-1)} (\sin (e+f x) c+c)^{\frac {1}{2} (2 m-1)}d\sin (e+f x)}{f}-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n}{f (m+n+1)}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle \frac {c 2^{n+\frac {1}{2}} \left (A+\frac {B (m-n)}{m+n+1}\right ) \cos (e+f x) (1-\sin (e+f x))^{\frac {1}{2}-n} (a \sin (e+f x)+a)^m (c \sin (e+f x)+c)^{\frac {1}{2} (-2 m-1)+\frac {1}{2} (2 m+1)} (c-c \sin (e+f x))^{\frac {1}{2} (-2 m-1)+m+n-\frac {1}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (2 m+1),\frac {1}{2} (1-2 n),\frac {1}{2} (2 m+3),\frac {1}{2} (\sin (e+f x)+1)\right )}{f (2 m+1)}-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n}{f (m+n+1)}\) |
-((B*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^n)/(f*(1 + m + n))) + (2^(1/2 + n)*c*(A + (B*(m - n))/(1 + m + n))*Cos[e + f*x]*Hyperg eometric2F1[(1 + 2*m)/2, (1 - 2*n)/2, (3 + 2*m)/2, (1 + Sin[e + f*x])/2]*( 1 - Sin[e + f*x])^(1/2 - n)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(- 1/2 + (-1 - 2*m)/2 + m + n)*(c + c*Sin[e + f*x])^((-1 - 2*m)/2 + (1 + 2*m) /2))/(f*(1 + 2*m))
3.2.95.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 , m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c + d*x)^FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d))) ^FracPart[n]) Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d) ), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !Integ erQ[n] && (RationalQ[m] || !SimplerQ[n + 1, m + 1])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.), x_Symbol] :> Simp[a^2*((g*Cos[e + f*x])^(p + 1)/(f*g*(a + b*Sin [e + f*x])^((p + 1)/2)*(a - b*Sin[e + f*x])^((p + 1)/2))) Subst[Int[(a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; Fre eQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[m]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[a^IntPart[m]*c^IntPart[m]*(a + b*Sin[e + f*x])^FracPart[m]*((c + d*Sin[e + f*x])^FracPart[m]/Cos[e + f*x]^(2*FracP art[m])) Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; F reeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && (FractionQ[m] || !FractionQ[n])
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*(m + n + 1))), x] - Simp[(B*c*(m - n) - A*d*(m + n + 1))/(d*(m + n + 1)) Int[( a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && NeQ[m + n + 1, 0]
\[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right ) \left (c -c \sin \left (f x +e \right )\right )^{n}d x\]
\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^n \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \]
\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^n \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{n} \left (A + B \sin {\left (e + f x \right )}\right )\, dx \]
\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^n \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \]
\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^n \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \]
Timed out. \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^n \, dx=\int \left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^n \,d x \]