\(\int (a-a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} (d+(c+d) m+(c+(c+d) m) \sin (e+f x)) \, dx\) [351]

Optimal result

Integrand size = 51, antiderivative size = 40 \[ \int (a-a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} (d+(c+d) m+(c+(c+d) m) \sin (e+f x)) \, dx=-\frac {\cos (e+f x) (a-a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m}}{f} \]

[Out]

-cos(f*x+e)*(a-a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(-1-m)/f

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.020, Rules used = {3053} \[ \int (a-a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} (d+(c+d) m+(c+(c+d) m) \sin (e+f x)) \, dx=-\frac {\cos (e+f x) (a-a \sin (e+f x))^m (c+d \sin (e+f x))^{-m-1}}{f} \]

[In]

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

[Out]

-((Cos[e + f*x]*(a - a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(-1 - m))/f)

Rule 3053

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\cos (e+f x) (a-a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m}}{f} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.41 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00 \[ \int (a-a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} (d+(c+d) m+(c+(c+d) m) \sin (e+f x)) \, dx=-\frac {\cos (e+f x) (a-a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m}}{f} \]

[In]

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

[Out]

-((Cos[e + f*x]*(a - a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(-1 - m))/f)

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.42 \[ \int (a-a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} (d+(c+d) m+(c+(c+d) m) \sin (e+f x)) \, dx=-\frac {{\left (d \cos \left (f x + e\right ) \sin \left (f x + e\right ) + c \cos \left (f x + e\right )\right )} {\left (-a \sin \left (f x + e\right ) + a\right )}^{m} {\left (d \sin \left (f x + e\right ) + c\right )}^{-m - 2}}{f} \]

[In]

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

[Out]

-(d*cos(f*x + e)*sin(f*x + e) + c*cos(f*x + e))*(-a*sin(f*x + e) + a)^m*(d*sin(f*x + e) + c)^(-m - 2)/f

Sympy [F(-1)]

Timed out. \[ \int (a-a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} (d+(c+d) m+(c+(c+d) m) \sin (e+f x)) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-1)]

Timed out. \[ \int (a-a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} (d+(c+d) m+(c+(c+d) m) \sin (e+f x)) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 15.29 (sec) , antiderivative size = 99, normalized size of antiderivative = 2.48 \[ \int (a-a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} (d+(c+d) m+(c+(c+d) m) \sin (e+f x)) \, dx=-\frac {{\left (-a\,\left (\sin \left (e+f\,x\right )-1\right )\right )}^m\,\left (d\,\sin \left (2\,e+2\,f\,x\right )-2\,c\,\left (2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )\right )}{f\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^m\,\left (d^2\,\left (2\,{\sin \left (e+f\,x\right )}^2-1\right )+2\,c^2+d^2+4\,c\,d\,\sin \left (e+f\,x\right )\right )} \]

[In]

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

[Out]

-((-a*(sin(e + f*x) - 1))^m*(d*sin(2*e + 2*f*x) - 2*c*(2*sin(e/2 + (f*x)/2)^2 - 1)))/(f*(c + d*sin(e + f*x))^m
*(d^2*(2*sin(e + f*x)^2 - 1) + 2*c^2 + d^2 + 4*c*d*sin(e + f*x)))