3.419 \(\int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m} \, dx\)

Optimal. Leaf size=164 \[ \frac{2 \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1}}{c^2 f (2 m+5) \left (4 m^2+8 m+3\right )}+\frac{2 \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-2}}{c f \left (4 m^2+16 m+15\right )}+\frac{\cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-3}}{f (2 m+5)} \]

[Out]

(Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-3 - m))/(f*(5 + 2*m)) + (2*Cos[e + f*x]*(a + a*Sin
[e + f*x])^m*(c - c*Sin[e + f*x])^(-2 - m))/(c*f*(15 + 16*m + 4*m^2)) + (2*Cos[e + f*x]*(a + a*Sin[e + f*x])^m
*(c - c*Sin[e + f*x])^(-1 - m))/(c^2*f*(5 + 2*m)*(3 + 8*m + 4*m^2))

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Rubi [A]  time = 0.223402, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {2743, 2742} \[ \frac{2 \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1}}{c^2 f (2 m+5) \left (4 m^2+8 m+3\right )}+\frac{2 \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-2}}{c f \left (4 m^2+16 m+15\right )}+\frac{\cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-3}}{f (2 m+5)} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-3 - m),x]

[Out]

(Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-3 - m))/(f*(5 + 2*m)) + (2*Cos[e + f*x]*(a + a*Sin
[e + f*x])^m*(c - c*Sin[e + f*x])^(-2 - m))/(c*f*(15 + 16*m + 4*m^2)) + (2*Cos[e + f*x]*(a + a*Sin[e + f*x])^m
*(c - c*Sin[e + f*x])^(-1 - m))/(c^2*f*(5 + 2*m)*(3 + 8*m + 4*m^2))

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])

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)]

Rubi steps

\begin{align*} \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m} \, dx &=\frac{\cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m}}{f (5+2 m)}+\frac{2 \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \, dx}{c (5+2 m)}\\ &=\frac{\cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m}}{f (5+2 m)}+\frac{2 \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m}}{c f \left (15+16 m+4 m^2\right )}+\frac{2 \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m} \, dx}{c^2 (3+2 m) (5+2 m)}\\ &=\frac{\cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m}}{f (5+2 m)}+\frac{2 \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m}}{c f \left (15+16 m+4 m^2\right )}+\frac{2 \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{c^2 f (1+2 m) (3+2 m) (5+2 m)}\\ \end{align*}

Mathematica [A]  time = 8.60579, size = 174, normalized size = 1.06 \[ \frac{2^{-m-2} \cos \left (\frac{1}{2} \left (-e-f x+\frac{\pi }{2}\right )\right ) \sin ^{-2 m-5}\left (\frac{1}{2} \left (-e-f x+\frac{\pi }{2}\right )\right ) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-3} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^{-2 (-m-3)} \left (-2 (2 m+3) \sin (e+f x)+\cos \left (2 \left (-e-f x+\frac{\pi }{2}\right )\right )+4 \left (m^2+3 m+2\right )\right )}{f (2 m+1) (2 m+3) (2 m+5)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-3 - m),x]

[Out]

(2^(-2 - m)*Cos[(-e + Pi/2 - f*x)/2]*Sin[(-e + Pi/2 - f*x)/2]^(-5 - 2*m)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e +
f*x])^(-3 - m)*(4*(2 + 3*m + m^2) + Cos[2*(-e + Pi/2 - f*x)] - 2*(3 + 2*m)*Sin[e + f*x]))/(f*(1 + 2*m)*(3 + 2
*m)*(5 + 2*m)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^(2*(-3 - m)))

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Maple [F]  time = 0.342, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( c-c\sin \left ( fx+e \right ) \right ) ^{-3-m}\, dx \end{align*}

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))^(-3-m),x)

[Out]

int((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-3-m),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (f x + e\right ) + a\right )}^{m}{\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 3}\,{d x} \end{align*}

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))^(-3-m),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(-m - 3), x)

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Fricas [A]  time = 1.08885, size = 251, normalized size = 1.53 \begin{align*} -\frac{{\left (2 \, \cos \left (f x + e\right )^{3} + 2 \,{\left (2 \, m + 3\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) -{\left (4 \, m^{2} + 12 \, m + 9\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 - 3}}{8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m + 15 \, f} \end{align*}

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))^(-3-m),x, algorithm="fricas")

[Out]

-(2*cos(f*x + e)^3 + 2*(2*m + 3)*cos(f*x + e)*sin(f*x + e) - (4*m^2 + 12*m + 9)*cos(f*x + e))*(a*sin(f*x + e)
+ a)^m*(-c*sin(f*x + e) + c)^(-m - 3)/(8*f*m^3 + 36*f*m^2 + 46*f*m + 15*f)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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))**(-3-m),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (f x + e\right ) + a\right )}^{m}{\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 3}\,{d x} \end{align*}

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))^(-3-m),x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(-m - 3), x)