\(\int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-p} \, dx\) [1037]

Optimal result

Integrand size = 36, antiderivative size = 147 \[ \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-p} \, dx=\frac {2^{\frac {1}{2}-\frac {p}{2}} c (A+B p) (g \cos (e+f x))^{1+p} \operatorname {Hypergeometric2F1}\left (\frac {1+p}{2},\frac {1+p}{2},\frac {3+p}{2},\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {1+p}{2}} (c-c \sin (e+f x))^{-1-p}}{f g (1+p)}-\frac {B (g \cos (e+f x))^{1+p} (c-c \sin (e+f x))^{-p}}{f g} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2939, 2768, 72, 71} \[ \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-p} \, dx=\frac {c 2^{\frac {1}{2}-\frac {p}{2}} (A+B p) (1-\sin (e+f x))^{\frac {p+1}{2}} (c-c \sin (e+f x))^{-p-1} (g \cos (e+f x))^{p+1} \operatorname {Hypergeometric2F1}\left (\frac {p+1}{2},\frac {p+1}{2},\frac {p+3}{2},\frac {1}{2} (\sin (e+f x)+1)\right )}{f g (p+1)}-\frac {B (c-c \sin (e+f x))^{-p} (g \cos (e+f x))^{p+1}}{f g} \]

[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 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], 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 72

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 2768

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 2939

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] /; F
reeQ[{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*} \text {integral}& = -\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 ) \text {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 ) \text {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} \operatorname {Hypergeometric2F1}\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*}

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.98 \[ \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-p} \, dx=-\frac {2^{\frac {1}{2} (-1-p)} \cos (e+f x) (g \cos (e+f x))^p \left (2 (A+B p) \operatorname {Hypergeometric2F1}\left (\frac {1+p}{2},\frac {1+p}{2},\frac {3+p}{2},\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {1+p}{2}}+2^{\frac {1+p}{2}} B (1+p) (-1+\sin (e+f x))\right ) (c-c \sin (e+f x))^{-p}}{f (1+p) (-1+\sin (e+f x))} \]

[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))

Maple [F]

\[\int \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{-p} \left (g \cos \left (f x +e \right )\right )^{p} \left (A +B \sin \left (f x +e \right )\right )d x\]

[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)

Fricas [F]

\[ \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-p} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} \left (g \cos \left (f x + e\right )\right )^{p}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{p}} \,d x } \]

[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)

Sympy [F(-1)]

Timed out. \[ \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-p} \, dx=\text {Timed out} \]

[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

Maxima [F]

\[ \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-p} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} \left (g \cos \left (f x + e\right )\right )^{p}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{p}} \,d x } \]

[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)

Giac [F]

\[ \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-p} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} \left (g \cos \left (f x + e\right )\right )^{p}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{p}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-p} \, dx=\int \frac {{\left (g\,\cos \left (e+f\,x\right )\right )}^p\,\left (A+B\,\sin \left (e+f\,x\right )\right )}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^p} \,d x \]

[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)