Integrand size = 46, antiderivative size = 269 \[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\frac {2^{\frac {1}{2}+n} c ((1+m+n) (C (1-m+n)+A (2+m+n))+(m-n) (C+2 C m+B (2+m+n))) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (1+2 m),\frac {1}{2} (1-2 n),\frac {1}{2} (3+2 m),\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {1}{2}-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1+n}}{f (1+2 m) (1+m+n) (2+m+n)}-\frac {(C+2 C m+B (2+m+n)) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n}{f (1+m+n) (2+m+n)}+\frac {C \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1+n}}{c f (2+m+n)} \]
2^(1/2+n)*c*((1+m+n)*(C*(1-m+n)+A*(2+m+n))+(m-n)*(C+2*C*m+B*(2+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)/(2+m +n)-(C+2*C*m+B*(2+m+n))*cos(f*x+e)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n/f /(1+m+n)/(2+m+n)+C*cos(f*x+e)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(1+n)/c/ f/(2+m+n)
\[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx \]
Integrate[(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^n*(A + B*Sin[e + f*x ] + C*Sin[e + f*x]^2),x]
Integrate[(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^n*(A + B*Sin[e + f*x ] + C*Sin[e + f*x]^2), x]
Time = 1.20 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.18, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.239, Rules used = {3042, 3518, 25, 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.
| \(\displaystyle \int (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n \left (A+B \sin (e+f x)+C \sin (e+f x)^2\right )dx\) |
| \(\Big \downarrow \) 3518 |
| \(\displaystyle \frac {C \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n+1}}{c f (m+n+2)}-\frac {\int -(\sin (e+f x) a+a)^m (c-c \sin (e+f x))^n (a c (C (-m+n+1)+A (m+n+2))+a c (2 m C+C+B (m+n+2)) \sin (e+f x))dx}{a c (m+n+2)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int (\sin (e+f x) a+a)^m (c-c \sin (e+f x))^n (a c (C (-m+n+1)+A (m+n+2))+a c (2 m C+C+B (m+n+2)) \sin (e+f x))dx}{a c (m+n+2)}+\frac {C \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n+1}}{c f (m+n+2)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int (\sin (e+f x) a+a)^m (c-c \sin (e+f x))^n (a c (C (-m+n+1)+A (m+n+2))+a c (2 m C+C+B (m+n+2)) \sin (e+f x))dx}{a c (m+n+2)}+\frac {C \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n+1}}{c f (m+n+2)}\) |
| \(\Big \downarrow \) 3452 |
| \(\displaystyle \frac {\frac {a c ((m+n+1) (A (m+n+2)+C (-m+n+1))+(m-n) (B (m+n+2)+2 C m+C)) \int (\sin (e+f x) a+a)^m (c-c \sin (e+f x))^ndx}{m+n+1}-\frac {a c (B (m+n+2)+2 C m+C) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n}{f (m+n+1)}}{a c (m+n+2)}+\frac {C \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n+1}}{c f (m+n+2)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a c ((m+n+1) (A (m+n+2)+C (-m+n+1))+(m-n) (B (m+n+2)+2 C m+C)) \int (\sin (e+f x) a+a)^m (c-c \sin (e+f x))^ndx}{m+n+1}-\frac {a c (B (m+n+2)+2 C m+C) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n}{f (m+n+1)}}{a c (m+n+2)}+\frac {C \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n+1}}{c f (m+n+2)}\) |
| \(\Big \downarrow \) 3224 |
| \(\displaystyle \frac {\frac {a c ((m+n+1) (A (m+n+2)+C (-m+n+1))+(m-n) (B (m+n+2)+2 C m+C)) \cos ^{-2 m}(e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m \int \cos ^{2 m}(e+f x) (c-c \sin (e+f x))^{n-m}dx}{m+n+1}-\frac {a c (B (m+n+2)+2 C m+C) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n}{f (m+n+1)}}{a c (m+n+2)}+\frac {C \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n+1}}{c f (m+n+2)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a c ((m+n+1) (A (m+n+2)+C (-m+n+1))+(m-n) (B (m+n+2)+2 C m+C)) \cos ^{-2 m}(e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m \int \cos (e+f x)^{2 m} (c-c \sin (e+f x))^{n-m}dx}{m+n+1}-\frac {a c (B (m+n+2)+2 C m+C) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n}{f (m+n+1)}}{a c (m+n+2)}+\frac {C \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n+1}}{c f (m+n+2)}\) |
| \(\Big \downarrow \) 3168 |
| \(\displaystyle \frac {\frac {a c^3 \cos (e+f x) ((m+n+1) (A (m+n+2)+C (-m+n+1))+(m-n) (B (m+n+2)+2 C m+C)) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{\frac {1}{2} (-2 m-1)+m} (c \sin (e+f x)+c)^{\frac {1}{2} (-2 m-1)} \int (c-c \sin (e+f x))^{\frac {1}{2} (2 n-1)} (\sin (e+f x) c+c)^{\frac {1}{2} (2 m-1)}d\sin (e+f x)}{f (m+n+1)}-\frac {a c (B (m+n+2)+2 C m+C) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n}{f (m+n+1)}}{a c (m+n+2)}+\frac {C \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n+1}}{c f (m+n+2)}\) |
| \(\Big \downarrow \) 80 |
| \(\displaystyle \frac {\frac {a c^3 2^{n-\frac {1}{2}} \cos (e+f x) ((m+n+1) (A (m+n+2)+C (-m+n+1))+(m-n) (B (m+n+2)+2 C m+C)) (1-\sin (e+f x))^{\frac {1}{2}-n} (a \sin (e+f x)+a)^m (c \sin (e+f x)+c)^{\frac {1}{2} (-2 m-1)} (c-c \sin (e+f x))^{\frac {1}{2} (-2 m-1)+m+n-\frac {1}{2}} \int \left (\frac {1}{2}-\frac {1}{2} \sin (e+f x)\right )^{\frac {1}{2} (2 n-1)} (\sin (e+f x) c+c)^{\frac {1}{2} (2 m-1)}d\sin (e+f x)}{f (m+n+1)}-\frac {a c (B (m+n+2)+2 C m+C) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n}{f (m+n+1)}}{a c (m+n+2)}+\frac {C \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n+1}}{c f (m+n+2)}\) |
| \(\Big \downarrow \) 79 |
| \(\displaystyle \frac {\frac {a c^2 2^{n+\frac {1}{2}} \cos (e+f x) ((m+n+1) (A (m+n+2)+C (-m+n+1))+(m-n) (B (m+n+2)+2 C m+C)) (1-\sin (e+f x))^{\frac {1}{2}-n} (a \sin (e+f x)+a)^m (c \sin (e+f x)+c)^{\frac {1}{2} (-2 m-1)+\frac {1}{2} (2 m+1)} (c-c \sin (e+f x))^{\frac {1}{2} (-2 m-1)+m+n-\frac {1}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (2 m+1),\frac {1}{2} (1-2 n),\frac {1}{2} (2 m+3),\frac {1}{2} (\sin (e+f x)+1)\right )}{f (2 m+1) (m+n+1)}-\frac {a c (B (m+n+2)+2 C m+C) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n}{f (m+n+1)}}{a c (m+n+2)}+\frac {C \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n+1}}{c f (m+n+2)}\) |
(C*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(1 + n))/(c*f* (2 + m + n)) + (-((a*c*(C + 2*C*m + B*(2 + m + n))*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)*a*c^2 *((1 + m + n)*(C*(1 - m + n) + A*(2 + m + n)) + (m - n)*(C + 2*C*m + B*(2 + m + n)))*Cos[e + f*x]*Hypergeometric2F1[(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)*(1 + m + n)))/(a*c*(2 + m + n))
3.1.17.3.1 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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_. ) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x] )^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(b*d*(m + n + 2)) Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n*Simp[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1)) + (b*B*d*(m + n + 2) - b*c*C*(2*m + 1))* Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && NeQ[m + n + 2, 0]
\[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{n} \left (A +B \sin \left (f x +e \right )+C \left (\sin ^{2}\left (f x +e \right )\right )\right )d x\]
\[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\int { {\left (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \]
integrate((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n*(A+B*sin(f*x+e)+C*sin(f*x+ e)^2),x, algorithm="fricas")
integral(-(C*cos(f*x + e)^2 - B*sin(f*x + e) - A - C)*(a*sin(f*x + e) + a) ^m*(-c*sin(f*x + e) + c)^n, x)
Timed out. \[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\text {Timed out} \]
\[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\int { {\left (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \]
integrate((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n*(A+B*sin(f*x+e)+C*sin(f*x+ e)^2),x, algorithm="maxima")
integrate((C*sin(f*x + e)^2 + B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m*( -c*sin(f*x + e) + c)^n, x)
\[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\int { {\left (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \]
integrate((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n*(A+B*sin(f*x+e)+C*sin(f*x+ e)^2),x, algorithm="giac")
integrate((C*sin(f*x + e)^2 + B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m*( -c*sin(f*x + e) + c)^n, x)
Timed out. \[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^n\,\left (C\,{\sin \left (e+f\,x\right )}^2+B\,\sin \left (e+f\,x\right )+A\right ) \,d x \]