\(\int \cos ^2(e+f x) (c+d \sin (e+f x))^n \, dx\) [1525]

Optimal result

Integrand size = 21, antiderivative size = 127 \[ \int \cos ^2(e+f x) (c+d \sin (e+f x))^n \, dx=\frac {\operatorname {AppellF1}\left (1+n,-\frac {1}{2},-\frac {1}{2},2+n,\frac {c+d \sin (e+f x)}{c-d},\frac {c+d \sin (e+f x)}{c+d}\right ) \cos (e+f x) (c+d \sin (e+f x))^{1+n}}{d f (1+n) \sqrt {1-\frac {c+d \sin (e+f x)}{c-d}} \sqrt {1-\frac {c+d \sin (e+f x)}{c+d}}} \]

[Out]

AppellF1(1+n,-1/2,-1/2,2+n,(c+d*sin(f*x+e))/(c-d),(c+d*sin(f*x+e))/(c+d))*cos(f*x+e)*(c+d*sin(f*x+e))^(1+n)/d/
f/(1+n)/(1+(-c-d*sin(f*x+e))/(c-d))^(1/2)/(1+(-c-d*sin(f*x+e))/(c+d))^(1/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2783, 143} \[ \int \cos ^2(e+f x) (c+d \sin (e+f x))^n \, dx=\frac {\cos (e+f x) (c+d \sin (e+f x))^{n+1} \operatorname {AppellF1}\left (n+1,-\frac {1}{2},-\frac {1}{2},n+2,\frac {c+d \sin (e+f x)}{c-d},\frac {c+d \sin (e+f x)}{c+d}\right )}{d f (n+1) \sqrt {1-\frac {c+d \sin (e+f x)}{c-d}} \sqrt {1-\frac {c+d \sin (e+f x)}{c+d}}} \]

[In]

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

[Out]

(AppellF1[1 + n, -1/2, -1/2, 2 + n, (c + d*Sin[e + f*x])/(c - d), (c + d*Sin[e + f*x])/(c + d)]*Cos[e + f*x]*(
c + d*Sin[e + f*x])^(1 + n))/(d*f*(1 + n)*Sqrt[1 - (c + d*Sin[e + f*x])/(c - d)]*Sqrt[1 - (c + d*Sin[e + f*x])
/(c + d)])

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 2783

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[g*((g*
Cos[e + f*x])^(p - 1)/(f*(1 - (a + b*Sin[e + f*x])/(a - b))^((p - 1)/2)*(1 - (a + b*Sin[e + f*x])/(a + b))^((p
- 1)/2))), Subst[Int[(-b/(a - b) - b*(x/(a - b)))^((p - 1)/2)*(b/(a + b) - b*(x/(a + b)))^((p - 1)/2)*(a + b*
x)^m, x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && NeQ[a^2 - b^2, 0] &&  !IGtQ[m, 0]

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

Mathematica [F]

\[ \int \cos ^2(e+f x) (c+d \sin (e+f x))^n \, dx=\int \cos ^2(e+f x) (c+d \sin (e+f x))^n \, dx \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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