3.187 \(\int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2+n} \, dx\)

Optimal. Leaf size=204 \[ \frac{2 (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n (g \cos (e+f x))^{-m-n}}{c^2 f g (m-n) (m-n+2) (m-n+4)}+\frac{(a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n-2} (g \cos (e+f x))^{-m-n}}{f g (m-n+4)}+\frac{2 (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n-1} (g \cos (e+f x))^{-m-n}}{c f g (m-n+2) (m-n+4)} \]

[Out]

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

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Rubi [A]  time = 0.665382, antiderivative size = 204, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.044, Rules used = {2849, 2848} \[ \frac{2 (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n (g \cos (e+f x))^{-m-n}}{c^2 f g (m-n) (m-n+2) (m-n+4)}+\frac{(a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n-2} (g \cos (e+f x))^{-m-n}}{f g (m-n+4)}+\frac{2 (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n-1} (g \cos (e+f x))^{-m-n}}{c f g (m-n+2) (m-n+4)} \]

Antiderivative was successfully verified.

[In]

Int[(g*Cos[e + f*x])^(-1 - m - n)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-2 + n),x]

[Out]

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

Rule 2849

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) +
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^
n)/(a*f*g*(2*m + p + 1)), x] + Dist[(m + n + p + 1)/(a*(2*m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f
*x])^(m + 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && EqQ[b*c + a*d, 0] &
& EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + n + p + 1], 0] && NeQ[2*m + p + 1, 0] && (SumSimplerQ[m, 1] ||  !SumS
implerQ[n, 1])

Rule 2848

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

Rubi steps

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

Mathematica [A]  time = 30.0394, size = 183, normalized size = 0.9 \[ \frac{2^{n-2} \cos (e+f x) \sin ^{2 n-4}\left (\frac{1}{2} \left (-e-f x+\frac{\pi }{2}\right )\right ) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n-2} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^{-2 (n-2)} (g \cos (e+f x))^{-m-n-1} \left (-2 (m-n+2) \sin (e+f x)+\cos \left (2 \left (-e-f x+\frac{\pi }{2}\right )\right )+m^2-2 m n+4 m+n^2-4 n+3\right )}{f (m-n) (m-n+2) (m-n+4)} \]

Antiderivative was successfully verified.

[In]

Integrate[(g*Cos[e + f*x])^(-1 - m - n)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-2 + n),x]

[Out]

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

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Maple [F]  time = 0.636, size = 0, normalized size = 0. \begin{align*} \int \left ( g\cos \left ( fx+e \right ) \right ) ^{-1-m-n} \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( c-c\sin \left ( fx+e \right ) \right ) ^{-2+n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*cos(f*x+e))^(-1-m-n)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-2+n),x)

[Out]

int((g*cos(f*x+e))^(-1-m-n)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-2+n),x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*cos(f*x+e))^(-1-m-n)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-2+n),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.90677, size = 460, normalized size = 2.25 \begin{align*} -\frac{{\left (2 \, \cos \left (f x + e\right )^{3} + 2 \,{\left (m - n + 2\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) -{\left (m^{2} - 2 \,{\left (m + 2\right )} n + n^{2} + 4 \, m + 4\right )} \cos \left (f x + e\right )\right )} \left (g \cos \left (f x + e\right )\right )^{-m - n - 1}{\left (a \sin \left (f x + e\right ) + a\right )}^{m} e^{\left (2 \,{\left (n - 2\right )} \log \left (g \cos \left (f x + e\right )\right ) -{\left (n - 2\right )} \log \left (a \sin \left (f x + e\right ) + a\right ) +{\left (n - 2\right )} \log \left (\frac{a c}{g^{2}}\right )\right )}}{f m^{3} - f n^{3} + 6 \, f m^{2} + 3 \,{\left (f m + 2 \, f\right )} n^{2} + 8 \, f m -{\left (3 \, f m^{2} + 12 \, f m + 8 \, f\right )} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*cos(f*x+e))^(-1-m-n)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-2+n),x, algorithm="fricas")

[Out]

-(2*cos(f*x + e)^3 + 2*(m - n + 2)*cos(f*x + e)*sin(f*x + e) - (m^2 - 2*(m + 2)*n + n^2 + 4*m + 4)*cos(f*x + e
))*(g*cos(f*x + e))^(-m - n - 1)*(a*sin(f*x + e) + a)^m*e^(2*(n - 2)*log(g*cos(f*x + e)) - (n - 2)*log(a*sin(f
*x + e) + a) + (n - 2)*log(a*c/g^2))/(f*m^3 - f*n^3 + 6*f*m^2 + 3*(f*m + 2*f)*n^2 + 8*f*m - (3*f*m^2 + 12*f*m
+ 8*f)*n)

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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((g*cos(f*x+e))**(-1-m-n)*(a+a*sin(f*x+e))**m*(c-c*sin(f*x+e))**(-2+n),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (g \cos \left (f x + e\right )\right )^{-m - n - 1}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}{\left (-c \sin \left (f x + e\right ) + c\right )}^{n - 2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*cos(f*x+e))^(-1-m-n)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-2+n),x, algorithm="giac")

[Out]

integrate((g*cos(f*x + e))^(-m - n - 1)*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(n - 2), x)