3.1034 \(\int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-3-p} \, dx\)

Optimal. Leaf size=168 \[ \frac{(2 A-B (p+3)) (c-c \sin (e+f x))^{-p-1} (g \cos (e+f x))^{p+1}}{c^2 f g (p+1) (p+3) (p+5)}+\frac{(A+B) (c-c \sin (e+f x))^{-p-3} (g \cos (e+f x))^{p+1}}{f g (p+5)}+\frac{(2 A-B (p+3)) (c-c \sin (e+f x))^{-p-2} (g \cos (e+f x))^{p+1}}{c f g (p+3) (p+5)} \]

[Out]

((A + B)*(g*Cos[e + f*x])^(1 + p)*(c - c*Sin[e + f*x])^(-3 - p))/(f*g*(5 + p)) + ((2*A - B*(3 + p))*(g*Cos[e +
f*x])^(1 + p)*(c - c*Sin[e + f*x])^(-2 - p))/(c*f*g*(3 + p)*(5 + p)) + ((2*A - B*(3 + p))*(g*Cos[e + f*x])^(1
+ p)*(c - c*Sin[e + f*x])^(-1 - p))/(c^2*f*g*(1 + p)*(3 + p)*(5 + p))

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Rubi [A]  time = 0.306062, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.079, Rules used = {2859, 2672, 2671} \[ \frac{(2 A-B (p+3)) (c-c \sin (e+f x))^{-p-1} (g \cos (e+f x))^{p+1}}{c^2 f g (p+1) (p+3) (p+5)}+\frac{(A+B) (c-c \sin (e+f x))^{-p-3} (g \cos (e+f x))^{p+1}}{f g (p+5)}+\frac{(2 A-B (p+3)) (c-c \sin (e+f x))^{-p-2} (g \cos (e+f x))^{p+1}}{c f g (p+3) (p+5)} \]

Antiderivative was successfully verified.

[In]

Int[(g*Cos[e + f*x])^p*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(-3 - p),x]

[Out]

((A + B)*(g*Cos[e + f*x])^(1 + p)*(c - c*Sin[e + f*x])^(-3 - p))/(f*g*(5 + p)) + ((2*A - B*(3 + p))*(g*Cos[e +
f*x])^(1 + p)*(c - c*Sin[e + f*x])^(-2 - p))/(c*f*g*(3 + p)*(5 + p)) + ((2*A - B*(3 + p))*(g*Cos[e + f*x])^(1
+ p)*(c - c*Sin[e + f*x])^(-1 - p))/(c^2*f*g*(1 + p)*(3 + p)*(5 + p))

Rule 2859

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

Rule 2672

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

Rule 2671

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

Rubi steps

\begin{align*} \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-3-p} \, dx &=\frac{(A+B) (g \cos (e+f x))^{1+p} (c-c \sin (e+f x))^{-3-p}}{f g (5+p)}+\frac{(2 A-B (3+p)) \int (g \cos (e+f x))^p (c-c \sin (e+f x))^{-2-p} \, dx}{c (5+p)}\\ &=\frac{(A+B) (g \cos (e+f x))^{1+p} (c-c \sin (e+f x))^{-3-p}}{f g (5+p)}+\frac{(2 A-B (3+p)) (g \cos (e+f x))^{1+p} (c-c \sin (e+f x))^{-2-p}}{c f g (3+p) (5+p)}+\frac{(2 A-B (3+p)) \int (g \cos (e+f x))^p (c-c \sin (e+f x))^{-1-p} \, dx}{c^2 (3+p) (5+p)}\\ &=\frac{(A+B) (g \cos (e+f x))^{1+p} (c-c \sin (e+f x))^{-3-p}}{f g (5+p)}+\frac{(2 A-B (3+p)) (g \cos (e+f x))^{1+p} (c-c \sin (e+f x))^{-2-p}}{c f g (3+p) (5+p)}+\frac{(2 A-B (3+p)) (g \cos (e+f x))^{1+p} (c-c \sin (e+f x))^{-1-p}}{c^2 f g (1+p) (3+p) (5+p)}\\ \end{align*}

Mathematica [A]  time = 0.261668, size = 119, normalized size = 0.71 \[ -\frac{\cos (e+f x) (c-c \sin (e+f x))^{-p} (g \cos (e+f x))^p \left ((2 A-B (p+3)) \sin ^2(e+f x)+(p+3) (B (p+3)-2 A) \sin (e+f x)+A \left (p^2+6 p+7\right )-B (p+3)\right )}{c^3 f (p+1) (p+3) (p+5) (\sin (e+f x)-1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(g*Cos[e + f*x])^p*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(-3 - p),x]

[Out]

-((Cos[e + f*x]*(g*Cos[e + f*x])^p*(-(B*(3 + p)) + A*(7 + 6*p + p^2) + (3 + p)*(-2*A + B*(3 + p))*Sin[e + f*x]
+ (2*A - B*(3 + p))*Sin[e + f*x]^2))/(c^3*f*(1 + p)*(3 + p)*(5 + p)*(-1 + Sin[e + f*x])^3*(c - c*Sin[e + f*x]
)^p))

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Maple [F]  time = 0.632, size = 0, normalized size = 0. \begin{align*} \int \left ( g\cos \left ( fx+e \right ) \right ) ^{p} \left ( A+B\sin \left ( fx+e \right ) \right ) \left ( c-c\sin \left ( fx+e \right ) \right ) ^{-3-p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*cos(f*x+e))^p*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(-3-p),x)

[Out]

int((g*cos(f*x+e))^p*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(-3-p),x)

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Maxima [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))^p*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(-3-p),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.43571, size = 317, normalized size = 1.89 \begin{align*} \frac{{\left ({\left (B p - 2 \, A + 3 \, B\right )} \cos \left (f x + e\right )^{3} +{\left (B p^{2} - 2 \,{\left (A - 3 \, B\right )} p - 6 \, A + 9 \, B\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) +{\left (A p^{2} + 2 \,{\left (3 \, A - B\right )} p + 9 \, A - 6 \, B\right )} \cos \left (f x + e\right )\right )} \left (g \cos \left (f x + e\right )\right )^{p}{\left (-c \sin \left (f x + e\right ) + c\right )}^{-p - 3}}{f p^{3} + 9 \, f p^{2} + 23 \, f p + 15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*cos(f*x+e))^p*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(-3-p),x, algorithm="fricas")

[Out]

((B*p - 2*A + 3*B)*cos(f*x + e)^3 + (B*p^2 - 2*(A - 3*B)*p - 6*A + 9*B)*cos(f*x + e)*sin(f*x + e) + (A*p^2 + 2
*(3*A - B)*p + 9*A - 6*B)*cos(f*x + e))*(g*cos(f*x + e))^p*(-c*sin(f*x + e) + c)^(-p - 3)/(f*p^3 + 9*f*p^2 + 2
3*f*p + 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((g*cos(f*x+e))**p*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(-3-p),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*cos(f*x+e))^p*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(-3-p),x, algorithm="giac")

[Out]

sage2