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

Optimal. Leaf size=125 \[ \frac{(a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n-1} (g \cos (e+f x))^{-m-n}}{f g (m-n+2)}+\frac{(a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n (g \cos (e+f x))^{-m-n}}{c f g (m-n) (m-n+2)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.410686, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.044, Rules used = {2849, 2848} \[ \frac{(a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n-1} (g \cos (e+f x))^{-m-n}}{f g (m-n+2)}+\frac{(a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n (g \cos (e+f x))^{-m-n}}{c f g (m-n) (m-n+2)} \]

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])^(-1 + n),x]

[Out]

((g*Cos[e + f*x])^(-m - n)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-1 + n))/(f*g*(2 + m - n)) + ((g*Cos[e
+ f*x])^(-m - n)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^n)/(c*f*g*(m - n)*(2 + 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))^{-1+n} \, dx &=\frac{(g \cos (e+f x))^{-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1+n}}{f g (2+m-n)}+\frac{\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+m-n)}\\ &=\frac{(g \cos (e+f x))^{-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1+n}}{f g (2+m-n)}+\frac{(g \cos (e+f x))^{-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n}{c f g (m-n) (2+m-n)}\\ \end{align*}

Mathematica [A]  time = 26.008, size = 132, normalized size = 1.06 \[ -\frac{2^{n-1} \cos ^{2 (n-1)}\left (\frac{1}{4} (2 e+2 f x+\pi )\right ) (a (\sin (e+f x)+1))^m (c-c \sin (e+f x))^{n-1} (\sin (e+f x)-m+n-1) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^{2-2 n} (g \cos (e+f x))^{-m-n}}{f g (m-n) (m-n+2)} \]

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])^(-1 + n),x]

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.53, 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 ) ^{-1+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))^(-1+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))^(-1+n),x)

________________________________________________________________________________________

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))^(-1+n),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

________________________________________________________________________________________

Fricas [A]  time = 1.84344, size = 324, normalized size = 2.59 \begin{align*} \frac{{\left ({\left (m - n + 1\right )} \cos \left (f x + e\right ) - \cos \left (f x + e\right ) \sin \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 - 1\right )} \log \left (g \cos \left (f x + e\right )\right ) -{\left (n - 1\right )} \log \left (a \sin \left (f x + e\right ) + a\right ) +{\left (n - 1\right )} \log \left (\frac{a c}{g^{2}}\right )\right )}}{f m^{2} + f n^{2} + 2 \, f m - 2 \,{\left (f m + 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))^(-1+n),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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))**(-1+n),x)

[Out]

Timed out

________________________________________________________________________________________

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 - 1}\,{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))^(-1+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 - 1), x)