Integrand size = 50, antiderivative size = 232 \[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=-\frac {2^{-\frac {1}{2}-m} C \cos ^3(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (3+2 m),\frac {1}{2} (3+2 m),\frac {1}{2} (5+2 m),\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {1}{2}+m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m}}{f (3+2 m)}+\frac {(A+B+C) \cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-2-m}}{2 a f (3+2 m)}+\frac {(A-B+C) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{2 c f (1+2 m)} \]
-2^(-1/2-m)*C*cos(f*x+e)^3*hypergeom([3/2+m, 3/2+m],[5/2+m],1/2+1/2*sin(f* x+e))*(1-sin(f*x+e))^(1/2+m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-2-m)/f/ (3+2*m)+1/2*(A+B+C)*cos(f*x+e)*(a+a*sin(f*x+e))^(1+m)*(c-c*sin(f*x+e))^(-2 -m)/a/f/(3+2*m)+1/2*(A-B+C)*cos(f*x+e)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e)) ^(-1-m)/c/f/(1+2*m)
Time = 1.16 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.84 \[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=-\frac {\sec (e+f x) (1+\sin (e+f x))^{-m} (a (1+\sin (e+f x)))^m (c-c \sin (e+f x))^{-m} \left (2^{\frac {3}{2}+m} C (3+2 m) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2}-m,-\frac {1}{2}-m,\frac {1}{2}-m,\frac {1}{2} (1-\sin (e+f x))\right ) (-1+\sin (e+f x)) \sqrt {1+\sin (e+f x)}+(1+\sin (e+f x))^{1+m} (2 A-B+2 C+2 A m+2 C m-(A+C-2 B (1+m)) \sin (e+f x))\right )}{c^2 f (1+2 m) (3+2 m) (-1+\sin (e+f x))} \]
Integrate[(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-2 - m)*(A + B*Sin[ e + f*x] + C*Sin[e + f*x]^2),x]
-((Sec[e + f*x]*(a*(1 + Sin[e + f*x]))^m*(2^(3/2 + m)*C*(3 + 2*m)*Hypergeo metric2F1[-1/2 - m, -1/2 - m, 1/2 - m, (1 - Sin[e + f*x])/2]*(-1 + Sin[e + f*x])*Sqrt[1 + Sin[e + f*x]] + (1 + Sin[e + f*x])^(1 + m)*(2*A - B + 2*C + 2*A*m + 2*C*m - (A + C - 2*B*(1 + m))*Sin[e + f*x])))/(c^2*f*(1 + 2*m)*( 3 + 2*m)*(-1 + Sin[e + f*x])*(1 + Sin[e + f*x])^m*(c - c*Sin[e + f*x])^m))
Time = 1.28 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.29, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3514, 3042, 3451, 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.
| \(\displaystyle \int (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-2} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-2} \left (A+B \sin (e+f x)+C \sin (e+f x)^2\right )dx\) |
| \(\Big \downarrow \) 3514 |
| \(\displaystyle \frac {\int (\sin (e+f x) a+a)^{m+1} (c-c \sin (e+f x))^{-m-2} \left ((A+B-C) (2 m+1) c^2+2 C (2 m+1) \sin (e+f x) c^2\right )dx}{2 a c^2 (2 m+1)}+\frac {(A-B+C) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1}}{2 c f (2 m+1)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int (\sin (e+f x) a+a)^{m+1} (c-c \sin (e+f x))^{-m-2} \left ((A+B-C) (2 m+1) c^2+2 C (2 m+1) \sin (e+f x) c^2\right )dx}{2 a c^2 (2 m+1)}+\frac {(A-B+C) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1}}{2 c f (2 m+1)}\) |
| \(\Big \downarrow \) 3451 |
| \(\displaystyle \frac {\frac {c^2 (2 m+1) (A+B+C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-2}}{f (2 m+3)}-2 c C (2 m+1) \int (\sin (e+f x) a+a)^{m+1} (c-c \sin (e+f x))^{-m-1}dx}{2 a c^2 (2 m+1)}+\frac {(A-B+C) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1}}{2 c f (2 m+1)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {c^2 (2 m+1) (A+B+C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-2}}{f (2 m+3)}-2 c C (2 m+1) \int (\sin (e+f x) a+a)^{m+1} (c-c \sin (e+f x))^{-m-1}dx}{2 a c^2 (2 m+1)}+\frac {(A-B+C) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1}}{2 c f (2 m+1)}\) |
| \(\Big \downarrow \) 3224 |
| \(\displaystyle \frac {\frac {c^2 (2 m+1) (A+B+C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-2}}{f (2 m+3)}-2 a c^2 C (2 m+1) \cos ^{-2 m}(e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m \int \cos ^{2 (m+1)}(e+f x) (c-c \sin (e+f x))^{-2 (m+1)}dx}{2 a c^2 (2 m+1)}+\frac {(A-B+C) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1}}{2 c f (2 m+1)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {c^2 (2 m+1) (A+B+C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-2}}{f (2 m+3)}-2 a c^2 C (2 m+1) \cos ^{-2 m}(e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m \int \cos (e+f x)^{2 (m+1)} (c-c \sin (e+f x))^{-2 (m+1)}dx}{2 a c^2 (2 m+1)}+\frac {(A-B+C) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1}}{2 c f (2 m+1)}\) |
| \(\Big \downarrow \) 3168 |
| \(\displaystyle \frac {\frac {c^2 (2 m+1) (A+B+C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-2}}{f (2 m+3)}-\frac {2 a c^4 C (2 m+1) \cos ^3(e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{\frac {1}{2} (-2 m-3)+m} (c \sin (e+f x)+c)^{\frac {1}{2} (-2 m-3)} \int (c-c \sin (e+f x))^{\frac {1}{2} (-2 m-3)} (\sin (e+f x) c+c)^{\frac {1}{2} (2 m+1)}d\sin (e+f x)}{f}}{2 a c^2 (2 m+1)}+\frac {(A-B+C) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1}}{2 c f (2 m+1)}\) |
| \(\Big \downarrow \) 80 |
| \(\displaystyle \frac {\frac {c^2 (2 m+1) (A+B+C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-2}}{f (2 m+3)}-\frac {a c^3 C 2^{-m-\frac {1}{2}} (2 m+1) \cos ^3(e+f x) (1-\sin (e+f x))^{m+\frac {1}{2}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{\frac {1}{2} (-2 m-3)-\frac {1}{2}} (c \sin (e+f x)+c)^{\frac {1}{2} (-2 m-3)} \int \left (\frac {1}{2}-\frac {1}{2} \sin (e+f x)\right )^{\frac {1}{2} (-2 m-3)} (\sin (e+f x) c+c)^{\frac {1}{2} (2 m+1)}d\sin (e+f x)}{f}}{2 a c^2 (2 m+1)}+\frac {(A-B+C) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1}}{2 c f (2 m+1)}\) |
| \(\Big \downarrow \) 79 |
| \(\displaystyle \frac {\frac {c^2 (2 m+1) (A+B+C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^{-m-2}}{f (2 m+3)}-\frac {a c^2 C 2^{\frac {1}{2}-m} (2 m+1) \cos ^3(e+f x) (1-\sin (e+f x))^{m+\frac {1}{2}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{\frac {1}{2} (-2 m-3)-\frac {1}{2}} (c \sin (e+f x)+c)^{\frac {1}{2} (-2 m-3)+\frac {1}{2} (2 m+3)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (2 m+3),\frac {1}{2} (2 m+3),\frac {1}{2} (2 m+5),\frac {1}{2} (\sin (e+f x)+1)\right )}{f (2 m+3)}}{2 a c^2 (2 m+1)}+\frac {(A-B+C) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1}}{2 c f (2 m+1)}\) |
Int[(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-2 - m)*(A + B*Sin[e + f* x] + C*Sin[e + f*x]^2),x]
((A - B + C)*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-1 - m))/(2*c*f*(1 + 2*m)) + ((c^2*(A + B + C)*(1 + 2*m)*Cos[e + f*x]*(a + a* Sin[e + f*x])^(1 + m)*(c - c*Sin[e + f*x])^(-2 - m))/(f*(3 + 2*m)) - (2^(1 /2 - m)*a*c^2*C*(1 + 2*m)*Cos[e + f*x]^3*Hypergeometric2F1[(3 + 2*m)/2, (3 + 2*m)/2, (5 + 2*m)/2, (1 + Sin[e + f*x])/2]*(1 - Sin[e + f*x])^(1/2 + m) *(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-1/2 + (-3 - 2*m)/2)*(c + c* Sin[e + f*x])^((-3 - 2*m)/2 + (3 + 2*m)/2))/(f*(3 + 2*m)))/(2*a*c^2*(1 + 2 *m))
3.1.24.3.1 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[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/( a*f*(2*m + 1))), x] + Simp[(a*B*(m - n) + A*b*(m + n + 1))/(a*b*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(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)] || (ILtQ[m + n, 0] && !SumSimplerQ[n, 1])) && NeQ[2* m + 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[(a*A - b*B + a*C)*Cos[e + f*x]*(a + b* Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(2*b*c*f*(2*m + 1))), x] - Si mp[1/(2*b*c*d*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(c^2*(m + 1) + d^2*(2*m + n + 2)) - B*c*d*(m - n - 1) - C*(c ^2*m - d^2*(n + 1)) + d*((A*c + B*d)*(m + n + 2) - c*C*(3*m - n))*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)] || (EqQ[m + n + 2, 0] && NeQ[2*m + 1, 0]))
\[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{-2-m} \left (A +B \sin \left (f x +e \right )+C \left (\sin ^{2}\left (f x +e \right )\right )\right )d x\]
\[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\int { {\left (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 2} \,d x } \]
integrate((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-2-m)*(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)^(-m - 2), x)
\[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{- m - 2} \left (A + B \sin {\left (e + f x \right )} + C \sin ^{2}{\left (e + f x \right )}\right )\, dx \]
Integral((a*(sin(e + f*x) + 1))**m*(-c*(sin(e + f*x) - 1))**(-m - 2)*(A + B*sin(e + f*x) + C*sin(e + f*x)**2), x)
\[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\int { {\left (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 2} \,d x } \]
integrate((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-2-m)*(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)^(-m - 2), x)
\[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\int { {\left (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 2} \,d x } \]
integrate((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-2-m)*(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)^(-m - 2), x)
Timed out. \[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (C\,{\sin \left (e+f\,x\right )}^2+B\,\sin \left (e+f\,x\right )+A\right )}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{m+2}} \,d x \]
int(((a + a*sin(e + f*x))^m*(A + B*sin(e + f*x) + C*sin(e + f*x)^2))/(c - c*sin(e + f*x))^(m + 2),x)