3.420 \(\int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \, dx\)

Optimal. Leaf size=101 \[ \frac {\cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1}}{c f \left (4 m^2+8 m+3\right )}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-2}}{f (2 m+3)} \]

[Out]

cos(f*x+e)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-2-m)/f/(3+2*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]  time = 0.13, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2743, 2742} \[ \frac {\cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1}}{c f \left (4 m^2+8 m+3\right )}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-2}}{f (2 m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-2 - m))/(f*(3 + 2*m)) + (Cos[e + f*x]*(a + a*Sin[e
+ f*x])^m*(c - c*Sin[e + f*x])^(-1 - m))/(c*f*(3 + 8*m + 4*m^2))

Rule 2742

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 2743

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] + Dist[(m + n + 1)/(a*(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, m, n}, x]
&& EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + n + 1], 0] && NeQ[m, -2^(-1)] && (SumSimplerQ[m
, 1] ||  !SumSimplerQ[n, 1])

Rubi steps

\begin {align*} \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \, dx &=\frac {\cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m}}{f (3+2 m)}+\frac {\int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m} \, dx}{c (3+2 m)}\\ &=\frac {\cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m}}{f (3+2 m)}+\frac {\cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{c f \left (3+8 m+4 m^2\right )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 3.04, size = 136, normalized size = 1.35 \[ -\frac {2^{-m} \cos \left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right ) \sin ^{-2 m-3}\left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right ) (\sin (e+f x)-2 (m+1)) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-2} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^{-2 (-m-2)}}{f \left (8 m^2+16 m+6\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Cos[(-e + Pi/2 - f*x)/2]*Sin[(-e + Pi/2 - f*x)/2]^(-3 - 2*m)*(-2*(1 + m) + Sin[e + f*x])*(a + a*Sin[e + f*x
])^m*(c - c*Sin[e + f*x])^(-2 - m))/(2^m*f*(6 + 16*m + 8*m^2)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^(2*(-2 - m
))))

________________________________________________________________________________________

fricas [A]  time = 0.46, size = 72, normalized size = 0.71 \[ \frac {{\left (2 \, {\left (m + 1\right )} \cos \left (f x + e\right ) - \cos \left (f x + e\right ) \sin \left (f x + e\right )\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 2}}{4 \, f m^{2} + 8 \, f m + 3 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(m + 1)*cos(f*x + e) - cos(f*x + e)*sin(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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.79, size = 0, normalized size = 0.00 \[ \int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{-2-m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.88, size = 111, normalized size = 1.10 \[ -\frac {{\left (a\,\left (\sin \left (e+f\,x\right )+1\right )\right )}^m\,\left (\sin \left (2\,e+2\,f\,x\right )+8\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+4\,m\,\left (2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )-4\right )}{c^2\,f\,{\left (-c\,\left (\sin \left (e+f\,x\right )-1\right )\right )}^m\,\left (4\,m^2+8\,m+3\right )\,\left (2\,{\sin \left (e+f\,x\right )}^2-4\,\sin \left (e+f\,x\right )+2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________