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]

1/3*2^(5/2-1/2*p)*c^3*(3*A-B*(2-p))*(g*cos(f*x+e))^(1+p)*hypergeom([1/2+1/2*p, -3/2+1/2*p],[3/2+1/2*p],1/2+1/2
*sin(f*x+e))*(1-sin(f*x+e))^(1/2+1/2*p)*(c-c*sin(f*x+e))^(-1-p)/f/g/(1+p)-1/3*B*(g*cos(f*x+e))^(1+p)*(c-c*sin(
f*x+e))^(2-p)/f/g

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Rubi [A]  time = 0.28, antiderivative size = 163, normalized size of antiderivative = 1.00, 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 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]))

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 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 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]

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*}

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Mathematica [A]  time = 0.91, 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]

-1/3*(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))/(f*(1 + p)*(-1 + Sin[e + f*x])*(c - c*Sin[e + f*x])^p)

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fricas [F]  time = 0.74, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B \sin \left (f x + e\right ) + A\right )} \left (g \cos \left (f x + e\right )\right )^{p} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-p + 2}, x\right ) \]

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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sin \left (f x + e\right ) + A\right )} \left (g \cos \left (f x + e\right )\right )^{p} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-p + 2}\,{d x} \]

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]

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

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maple [F]  time = 4.14, size = 0, normalized size = 0.00 \[ \int \left (g \cos \left (f x +e \right )\right )^{p} \left (A +B \sin \left (f x +e \right )\right ) \left (c -c \sin \left (f x +e \right )\right )^{2-p}\, dx \]

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)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sin \left (f x + e\right ) + A\right )} \left (g \cos \left (f x + e\right )\right )^{p} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-p + 2}\,{d x} \]

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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (g\,\cos \left (e+f\,x\right )\right )}^p\,\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{2-p} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

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