\(\int \frac {(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx\) [1046]

Optimal result

Integrand size = 35, antiderivative size = 149 \[ \int \frac {(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx=-\frac {2^{-\frac {3}{2}+\frac {p}{2}} \operatorname {AppellF1}\left (\frac {1+p}{2},\frac {5-p}{2},-n,\frac {3+p}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) (g \cos (e+f x))^{1+p} (1+\sin (e+f x))^{-2+\frac {3-p}{2}} (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n}}{a^2 f g (1+p)} \]

[Out]

-2^(-3/2+1/2*p)*AppellF1(1/2+1/2*p,-n,5/2-1/2*p,3/2+1/2*p,d*(1-sin(f*x+e))/(c+d),1/2-1/2*sin(f*x+e))*(g*cos(f*
x+e))^(p+1)*(1+sin(f*x+e))^(-1/2-1/2*p)*(c+d*sin(f*x+e))^n/a^2/f/g/(p+1)/(((c+d*sin(f*x+e))/(c+d))^n)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {2999, 144, 143} \[ \int \frac {(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx=-\frac {g 2^{\frac {p-3}{2}} (1-\sin (e+f x)) (\sin (e+f x)+1)^{\frac {1-p}{2}} (g \cos (e+f x))^{p-1} (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} \operatorname {AppellF1}\left (\frac {p+1}{2},\frac {5-p}{2},-n,\frac {p+3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right )}{a^2 f (p+1)} \]

[In]

Int[((g*Cos[e + f*x])^p*(c + d*Sin[e + f*x])^n)/(a + a*Sin[e + f*x])^2,x]

[Out]

-((2^((-3 + p)/2)*g*AppellF1[(1 + p)/2, (5 - p)/2, -n, (3 + p)/2, (1 - Sin[e + f*x])/2, (d*(1 - Sin[e + f*x]))
/(c + d)]*(g*Cos[e + f*x])^(-1 + p)*(1 - Sin[e + f*x])*(1 + Sin[e + f*x])^((1 - p)/2)*(c + d*Sin[e + f*x])^n)/
(a^2*f*(1 + p)*((c + d*Sin[e + f*x])/(c + d))^n))

Rule 143

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(b*c
- a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !Inte
gerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) &&  !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])

Rule 144

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(e + f*x)^
FracPart[p]/((b/(b*e - a*f))^IntPart[p]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p]), Int[(a + b*x)^m*(c + d*x)^n*
(b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !GtQ[b/(b*e - a*f), 0]

Rule 2999

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

Rubi steps \begin{align*} \text {integral}& = \frac {\left (g (g \cos (e+f x))^{-1+p} (1-\sin (e+f x))^{\frac {1-p}{2}} (1+\sin (e+f x))^{\frac {1-p}{2}}\right ) \text {Subst}\left (\int (1-x)^{\frac {1}{2} (-1+p)} (1+x)^{-2+\frac {1}{2} (-1+p)} (c+d x)^n \, dx,x,\sin (e+f x)\right )}{a^2 f} \\ & = \frac {\left (g (g \cos (e+f x))^{-1+p} (1-\sin (e+f x))^{\frac {1-p}{2}} (1+\sin (e+f x))^{\frac {1-p}{2}} (c+d \sin (e+f x))^n \left (-\frac {c+d \sin (e+f x)}{-c-d}\right )^{-n}\right ) \text {Subst}\left (\int (1-x)^{\frac {1}{2} (-1+p)} (1+x)^{-2+\frac {1}{2} (-1+p)} \left (-\frac {c}{-c-d}-\frac {d x}{-c-d}\right )^n \, dx,x,\sin (e+f x)\right )}{a^2 f} \\ & = -\frac {2^{\frac {1}{2} (-3+p)} g \operatorname {AppellF1}\left (\frac {1+p}{2},\frac {5-p}{2},-n,\frac {3+p}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) (g \cos (e+f x))^{-1+p} (1-\sin (e+f x)) (1+\sin (e+f x))^{\frac {1-p}{2}} (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n}}{a^2 f (1+p)} \\ \end{align*}

Mathematica [F]

\[ \int \frac {(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx=\int \frac {(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx \]

[In]

Integrate[((g*Cos[e + f*x])^p*(c + d*Sin[e + f*x])^n)/(a + a*Sin[e + f*x])^2,x]

[Out]

Integrate[((g*Cos[e + f*x])^p*(c + d*Sin[e + f*x])^n)/(a + a*Sin[e + f*x])^2, x]

Maple [F]

\[\int \frac {\left (g \cos \left (f x +e \right )\right )^{p} \left (c +d \sin \left (f x +e \right )\right )^{n}}{\left (a +a \sin \left (f x +e \right )\right )^{2}}d x\]

[In]

int((g*cos(f*x+e))^p*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^2,x)

[Out]

int((g*cos(f*x+e))^p*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^2,x)

Fricas [F]

\[ \int \frac {(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{p} {\left (d \sin \left (f x + e\right ) + c\right )}^{n}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{2}} \,d x } \]

[In]

integrate((g*cos(f*x+e))^p*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

integral(-(g*cos(f*x + e))^p*(d*sin(f*x + e) + c)^n/(a^2*cos(f*x + e)^2 - 2*a^2*sin(f*x + e) - 2*a^2), x)

Sympy [F]

\[ \int \frac {(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx=\frac {\int \frac {\left (g \cos {\left (e + f x \right )}\right )^{p} \left (c + d \sin {\left (e + f x \right )}\right )^{n}}{\sin ^{2}{\left (e + f x \right )} + 2 \sin {\left (e + f x \right )} + 1}\, dx}{a^{2}} \]

[In]

integrate((g*cos(f*x+e))**p*(c+d*sin(f*x+e))**n/(a+a*sin(f*x+e))**2,x)

[Out]

Integral((g*cos(e + f*x))**p*(c + d*sin(e + f*x))**n/(sin(e + f*x)**2 + 2*sin(e + f*x) + 1), x)/a**2

Maxima [F]

\[ \int \frac {(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{p} {\left (d \sin \left (f x + e\right ) + c\right )}^{n}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{2}} \,d x } \]

[In]

integrate((g*cos(f*x+e))^p*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

integrate((g*cos(f*x + e))^p*(d*sin(f*x + e) + c)^n/(a*sin(f*x + e) + a)^2, x)

Giac [F]

\[ \int \frac {(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{p} {\left (d \sin \left (f x + e\right ) + c\right )}^{n}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{2}} \,d x } \]

[In]

integrate((g*cos(f*x+e))^p*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^2,x, algorithm="giac")

[Out]

integrate((g*cos(f*x + e))^p*(d*sin(f*x + e) + c)^n/(a*sin(f*x + e) + a)^2, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx=\int \frac {{\left (g\,\cos \left (e+f\,x\right )\right )}^p\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^n}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2} \,d x \]

[In]

int(((g*cos(e + f*x))^p*(c + d*sin(e + f*x))^n)/(a + a*sin(e + f*x))^2,x)

[Out]

int(((g*cos(e + f*x))^p*(c + d*sin(e + f*x))^n)/(a + a*sin(e + f*x))^2, x)