\(\int \frac {\cos (e+f x) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx\) [918]

Optimal result

Integrand size = 31, antiderivative size = 60 \[ \int \frac {\cos (e+f x) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx=\frac {d \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,\frac {c+d \sin (e+f x)}{c-d}\right ) (c+d \sin (e+f x))^{1+n}}{a^2 (c-d)^2 f (1+n)} \]

[Out]

d*hypergeom([2, 1+n],[2+n],(c+d*sin(f*x+e))/(c-d))*(c+d*sin(f*x+e))^(1+n)/a^2/(c-d)^2/f/(1+n)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {2912, 70} \[ \int \frac {\cos (e+f x) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx=\frac {d (c+d \sin (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (2,n+1,n+2,\frac {c+d \sin (e+f x)}{c-d}\right )}{a^2 f (n+1) (c-d)^2} \]

[In]

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

[Out]

(d*Hypergeometric2F1[2, 1 + n, 2 + n, (c + d*Sin[e + f*x])/(c - d)]*(c + d*Sin[e + f*x])^(1 + n))/(a^2*(c - d)
^2*f*(1 + n))

Rule 70

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

Rule 2912

Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)
])^(n_.), x_Symbol] :> Dist[1/(b*f), Subst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (c+\frac {d x}{a}\right )^n}{(a+x)^2} \, dx,x,a \sin (e+f x)\right )}{a f} \\ & = \frac {d \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,\frac {c+d \sin (e+f x)}{c-d}\right ) (c+d \sin (e+f x))^{1+n}}{a^2 (c-d)^2 f (1+n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.02 \[ \int \frac {\cos (e+f x) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx=\frac {d \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,-\frac {c+d \sin (e+f x)}{-c+d}\right ) (c+d \sin (e+f x))^{1+n}}{a^2 (-c+d)^2 f (1+n)} \]

[In]

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

[Out]

(d*Hypergeometric2F1[2, 1 + n, 2 + n, -((c + d*Sin[e + f*x])/(-c + d))]*(c + d*Sin[e + f*x])^(1 + n))/(a^2*(-c
+ d)^2*f*(1 + n))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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