3.1037 \(\int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-p} \, dx\)

Optimal. Leaf size=147 \[ \frac {c 2^{\frac {1}{2}-\frac {p}{2}} (A+B p) (1-\sin (e+f x))^{\frac {p+1}{2}} (c-c \sin (e+f x))^{-p-1} (g \cos (e+f x))^{p+1} \, _2F_1\left (\frac {p+1}{2},\frac {p+1}{2};\frac {p+3}{2};\frac {1}{2} (\sin (e+f x)+1)\right )}{f g (p+1)}-\frac {B (c-c \sin (e+f x))^{-p} (g \cos (e+f x))^{p+1}}{f g} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2860, 2689, 70, 69} \[ \frac {c 2^{\frac {1}{2}-\frac {p}{2}} (A+B p) (1-\sin (e+f x))^{\frac {p+1}{2}} (c-c \sin (e+f x))^{-p-1} (g \cos (e+f x))^{p+1} \, _2F_1\left (\frac {p+1}{2},\frac {p+1}{2};\frac {p+3}{2};\frac {1}{2} (\sin (e+f x)+1)\right )}{f g (p+1)}-\frac {B (c-c \sin (e+f x))^{-p} (g \cos (e+f x))^{p+1}}{f g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.56, size = 144, normalized size = 0.98 \[ -\frac {2^{\frac {1}{2} (-p-1)} \cos (e+f x) (c-c \sin (e+f x))^{-p} (g \cos (e+f x))^p \left (2 (A+B p) (1-\sin (e+f x))^{\frac {p+1}{2}} \, _2F_1\left (\frac {p+1}{2},\frac {p+1}{2};\frac {p+3}{2};\frac {1}{2} (\sin (e+f x)+1)\right )+B 2^{\frac {p+1}{2}} (p+1) (\sin (e+f x)-1)\right )}{f (p+1) (\sin (e+f x)-1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B \sin \left (f x + e\right ) + A\right )} \left (g \cos \left (f x + e\right )\right )^{p}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{p}}, x\right ) \]

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))^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, x)

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

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))^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, x)

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maple [F]  time = 4.82, size = 0, normalized size = 0.00 \[ \int \left (g \cos \left (f x +e \right )\right )^{p} \left (A +B \sin \left (f x +e \right )\right ) \left (c -c \sin \left (f x +e \right )\right )^{-p}\, dx \]

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))^p),x)

[Out]

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

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

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))^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, x)

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

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))^p,x)

[Out]

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

[Out]

Timed out

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