\(\int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-2-m} \, dx\) [213]

Optimal result

Integrand size = 40, antiderivative size = 114 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-2-m} \, dx=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m}}{f (3+2 m)}+\frac {(A-2 B (1+m)) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{c f (1+2 m) (3+2 m)} \]

[Out]

(A+B)*cos(f*x+e)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-2-m)/f/(3+2*m)+(A-2*B*(1+m))*cos(f*x+e)*(a+a*sin(f*x+e)
)^m*(c-c*sin(f*x+e))^(-1-m)/c/f/(4*m^2+8*m+3)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {3051, 2821} \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-2-m} \, dx=\frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-2}}{f (2 m+3)}+\frac {(A-2 B (m+1)) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1}}{c f (2 m+1) (2 m+3)} \]

[In]

Int[(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(-2 - m),x]

[Out]

((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-2 - m))/(f*(3 + 2*m)) + ((A - 2*B*(1 + m))
*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-1 - m))/(c*f*(1 + 2*m)*(3 + 2*m))

Rule 2821

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[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] /; FreeQ[{a, b, c, d, e, f
, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] && NeQ[m, -2^(-1)]

Rule 3051

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[(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] + Dist[(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]

Rubi steps \begin{align*} \text {integral}& = \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m}}{f (3+2 m)}+\frac {(A-2 B (1+m)) \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m} \, dx}{c (3+2 m)} \\ & = \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m}}{f (3+2 m)}+\frac {(A-2 B (1+m)) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{c f (1+2 m) (3+2 m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.81 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-2-m} \, dx=\frac {\sec (e+f x) (a (1+\sin (e+f x)))^{1+m} (c-c \sin (e+f x))^{-m} (B-2 A (1+m)+(A-2 B (1+m)) \sin (e+f x))}{a c^2 f (1+2 m) (3+2 m) (-1+\sin (e+f x))} \]

[In]

Integrate[(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(-2 - m),x]

[Out]

(Sec[e + f*x]*(a*(1 + Sin[e + f*x]))^(1 + m)*(B - 2*A*(1 + m) + (A - 2*B*(1 + m))*Sin[e + f*x]))/(a*c^2*f*(1 +
2*m)*(3 + 2*m)*(-1 + Sin[e + f*x])*(c - c*Sin[e + f*x])^m)

Maple [F]

\[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right ) \left (c -c \sin \left (f x +e \right )\right )^{-2-m}d x\]

[In]

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(-2-m),x)

[Out]

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(-2-m),x)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.78 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-2-m} \, dx=\frac {{\left ({\left (2 \, B m - A + 2 \, B\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (2 \, A m + 2 \, A - B\right )} \cos \left (f x + e\right )\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 2}}{4 \, f m^{2} + 8 \, f m + 3 \, f} \]

[In]

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(-2-m),x, algorithm="fricas")

[Out]

((2*B*m - A + 2*B)*cos(f*x + e)*sin(f*x + e) + (2*A*m + 2*A - B)*cos(f*x + e))*(a*sin(f*x + e) + a)^m*(-c*sin(
f*x + e) + c)^(-m - 2)/(4*f*m^2 + 8*f*m + 3*f)

Sympy [F]

\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-2-m} \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{- m - 2} \left (A + B \sin {\left (e + f x \right )}\right )\, dx \]

[In]

integrate((a+a*sin(f*x+e))**m*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(-2-m),x)

[Out]

Integral((a*(sin(e + f*x) + 1))**m*(-c*(sin(e + f*x) - 1))**(-m - 2)*(A + B*sin(e + f*x)), x)

Maxima [F]

\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-2-m} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 2} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(-2-m),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(-m - 2), x)

Giac [F]

\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-2-m} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 2} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(-2-m),x, algorithm="giac")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(-m - 2), x)

Mupad [B] (verification not implemented)

Time = 14.18 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.18 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-2-m} \, dx=-\frac {{\left (a\,\left (\sin \left (e+f\,x\right )+1\right )\right )}^m\,\left (4\,A\,\cos \left (e+f\,x\right )-2\,B\,\cos \left (e+f\,x\right )-A\,\sin \left (2\,e+2\,f\,x\right )+2\,B\,\sin \left (2\,e+2\,f\,x\right )+4\,A\,m\,\cos \left (e+f\,x\right )+2\,B\,m\,\sin \left (2\,e+2\,f\,x\right )\right )}{c^2\,f\,{\left (-c\,\left (\sin \left (e+f\,x\right )-1\right )\right )}^m\,\left (4\,m^2+8\,m+3\right )\,\left (4\,\sin \left (e+f\,x\right )+\cos \left (2\,e+2\,f\,x\right )-3\right )} \]

[In]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^m)/(c - c*sin(e + f*x))^(m + 2),x)

[Out]

-((a*(sin(e + f*x) + 1))^m*(4*A*cos(e + f*x) - 2*B*cos(e + f*x) - A*sin(2*e + 2*f*x) + 2*B*sin(2*e + 2*f*x) +
4*A*m*cos(e + f*x) + 2*B*m*sin(2*e + 2*f*x)))/(c^2*f*(-c*(sin(e + f*x) - 1))^m*(8*m + 4*m^2 + 3)*(4*sin(e + f*
x) + cos(2*e + 2*f*x) - 3))