3.1039 \(\int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{2-p} \, dx\)

Optimal. Leaf size=163 \[ \frac{c^3 2^{\frac{5}{2}-\frac{p}{2}} (3 A-B (2-p)) (1-\sin (e+f x))^{\frac{p+1}{2}} (c-c \sin (e+f x))^{-p-1} (g \cos (e+f x))^{p+1} \, _2F_1\left (\frac{p-3}{2},\frac{p+1}{2};\frac{p+3}{2};\frac{1}{2} (\sin (e+f x)+1)\right )}{3 f g (p+1)}-\frac{B (c-c \sin (e+f x))^{2-p} (g \cos (e+f x))^{p+1}}{3 f g} \]

[Out]

(2^(5/2 - p/2)*c^3*(3*A - B*(2 - p))*(g*Cos[e + f*x])^(1 + p)*Hypergeometric2F1[(-3 + p)/2, (1 + p)/2, (3 + p)
/2, (1 + Sin[e + f*x])/2]*(1 - Sin[e + f*x])^((1 + p)/2)*(c - c*Sin[e + f*x])^(-1 - p))/(3*f*g*(1 + p)) - (B*(
g*Cos[e + f*x])^(1 + p)*(c - c*Sin[e + f*x])^(2 - p))/(3*f*g)

________________________________________________________________________________________

Rubi [A]  time = 0.275605, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2860, 2689, 70, 69} \[ \frac{c^3 2^{\frac{5}{2}-\frac{p}{2}} (3 A-B (2-p)) (1-\sin (e+f x))^{\frac{p+1}{2}} (c-c \sin (e+f x))^{-p-1} (g \cos (e+f x))^{p+1} \, _2F_1\left (\frac{p-3}{2},\frac{p+1}{2};\frac{p+3}{2};\frac{1}{2} (\sin (e+f x)+1)\right )}{3 f g (p+1)}-\frac{B (c-c \sin (e+f x))^{2-p} (g \cos (e+f x))^{p+1}}{3 f g} \]

Antiderivative was successfully verified.

[In]

Int[(g*Cos[e + f*x])^p*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(2 - p),x]

[Out]

(2^(5/2 - p/2)*c^3*(3*A - B*(2 - p))*(g*Cos[e + f*x])^(1 + p)*Hypergeometric2F1[(-3 + p)/2, (1 + p)/2, (3 + p)
/2, (1 + Sin[e + f*x])/2]*(1 - Sin[e + f*x])^((1 + p)/2)*(c - c*Sin[e + f*x])^(-1 - p))/(3*f*g*(1 + p)) - (B*(
g*Cos[e + f*x])^(1 + p)*(c - c*Sin[e + f*x])^(2 - p))/(3*f*g)

Rule 2860

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

Rule 2689

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

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*Simp[(b*c)/(b*c - a*d) + (b*d*x)/(b*c -
a*d), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

\begin{align*} \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{2-p} \, dx &=-\frac{B (g \cos (e+f x))^{1+p} (c-c \sin (e+f x))^{2-p}}{3 f g}-\frac{1}{3} (-3 A+B (2-p)) \int (g \cos (e+f x))^p (c-c \sin (e+f x))^{2-p} \, dx\\ &=-\frac{B (g \cos (e+f x))^{1+p} (c-c \sin (e+f x))^{2-p}}{3 f g}-\frac{\left (c^2 (-3 A+B (2-p)) (g \cos (e+f x))^{1+p} (c-c \sin (e+f x))^{\frac{1}{2} (-1-p)} (c+c \sin (e+f x))^{\frac{1}{2} (-1-p)}\right ) \operatorname{Subst}\left (\int (c-c x)^{2+\frac{1}{2} (-1+p)-p} (c+c x)^{\frac{1}{2} (-1+p)} \, dx,x,\sin (e+f x)\right )}{3 f g}\\ &=-\frac{B (g \cos (e+f x))^{1+p} (c-c \sin (e+f x))^{2-p}}{3 f g}-\frac{\left (2^{\frac{3}{2}-\frac{p}{2}} c^4 (-3 A+B (2-p)) (g \cos (e+f x))^{1+p} (c-c \sin (e+f x))^{-\frac{1}{2}+\frac{1}{2} (-1-p)-\frac{p}{2}} \left (\frac{c-c \sin (e+f x)}{c}\right )^{\frac{1}{2}+\frac{p}{2}} (c+c \sin (e+f x))^{\frac{1}{2} (-1-p)}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2}-\frac{x}{2}\right )^{2+\frac{1}{2} (-1+p)-p} (c+c x)^{\frac{1}{2} (-1+p)} \, dx,x,\sin (e+f x)\right )}{3 f g}\\ &=\frac{2^{\frac{5}{2}-\frac{p}{2}} c^3 (3 A-B (2-p)) (g \cos (e+f x))^{1+p} \, _2F_1\left (\frac{1}{2} (-3+p),\frac{1+p}{2};\frac{3+p}{2};\frac{1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac{1+p}{2}} (c-c \sin (e+f x))^{-1-p}}{3 f g (1+p)}-\frac{B (g \cos (e+f x))^{1+p} (c-c \sin (e+f x))^{2-p}}{3 f g}\\ \end{align*}

Mathematica [A]  time = 0.870102, size = 155, normalized size = 0.95 \[ -\frac{c^2 2^{\frac{1}{2} (-p-1)} \cos (e+f x) (c-c \sin (e+f x))^{-p} (g \cos (e+f x))^p \left (8 (3 A+B (p-2)) (1-\sin (e+f x))^{\frac{p+1}{2}} \, _2F_1\left (\frac{p-3}{2},\frac{p+1}{2};\frac{p+3}{2};\frac{1}{2} (\sin (e+f x)+1)\right )+B 2^{\frac{p+1}{2}} (p+1) (\sin (e+f x)-1)^3\right )}{3 f (p+1) (\sin (e+f x)-1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(g*Cos[e + f*x])^p*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(2 - p),x]

[Out]

-(2^((-1 - p)/2)*c^2*Cos[e + f*x]*(g*Cos[e + f*x])^p*(8*(3*A + B*(-2 + p))*Hypergeometric2F1[(-3 + p)/2, (1 +
p)/2, (3 + p)/2, (1 + Sin[e + f*x])/2]*(1 - Sin[e + f*x])^((1 + p)/2) + 2^((1 + p)/2)*B*(1 + p)*(-1 + Sin[e +
f*x])^3))/(3*f*(1 + p)*(-1 + Sin[e + f*x])*(c - c*Sin[e + f*x])^p)

________________________________________________________________________________________

Maple [F]  time = 0.515, 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 ) ^{2-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))^(2-p),x)

[Out]

int((g*cos(f*x+e))^p*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(2-p),x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sin \left (f x + e\right ) + A\right )} \left (g \cos \left (f x + e\right )\right )^{p}{\left (-c \sin \left (f x + e\right ) + c\right )}^{-p + 2}\,{d x} \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))^(2-p),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(g*cos(f*x + e))^p*(-c*sin(f*x + e) + c)^(-p + 2), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (B \sin \left (f x + e\right ) + A\right )} \left (g \cos \left (f x + e\right )\right )^{p}{\left (-c \sin \left (f x + e\right ) + c\right )}^{-p + 2}, x\right ) \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))^(2-p),x, algorithm="fricas")

[Out]

integral((B*sin(f*x + e) + A)*(g*cos(f*x + e))^p*(-c*sin(f*x + e) + c)^(-p + 2), x)

________________________________________________________________________________________

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))**(2-p),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \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))^(2-p),x, algorithm="giac")

[Out]

Exception raised: AttributeError