3.1525 \(\int \cos ^2(e+f x) (c+d \sin (e+f x))^n \, dx\)

Optimal. Leaf size=127 \[ \frac{\cos (e+f x) (c+d \sin (e+f x))^{n+1} F_1\left (n+1;-\frac{1}{2},-\frac{1}{2};n+2;\frac{c+d \sin (e+f x)}{c-d},\frac{c+d \sin (e+f x)}{c+d}\right )}{d f (n+1) \sqrt{1-\frac{c+d \sin (e+f x)}{c-d}} \sqrt{1-\frac{c+d \sin (e+f x)}{c+d}}} \]

[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)])

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Rubi [A]  time = 0.0930213, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2704, 138} \[ \frac{\cos (e+f x) (c+d \sin (e+f x))^{n+1} F_1\left (n+1;-\frac{1}{2},-\frac{1}{2};n+2;\frac{c+d \sin (e+f x)}{c-d},\frac{c+d \sin (e+f x)}{c+d}\right )}{d f (n+1) \sqrt{1-\frac{c+d \sin (e+f x)}{c-d}} \sqrt{1-\frac{c+d \sin (e+f x)}{c+d}}} \]

Antiderivative was successfully verified.

[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 2704

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]

Rule 138

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b*(m + 1
)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p), 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])

Rubi steps

\begin{align*} \int \cos ^2(e+f x) (c+d \sin (e+f x))^n \, dx &=\frac{\cos (e+f x) \operatorname{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{F_1\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]  time = 0.310351, size = 0, normalized size = 0. \[ \int \cos ^2(e+f x) (c+d \sin (e+f x))^n \, dx \]

Verification is Not applicable to the result.

[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]

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Maple [F]  time = 0.234, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( fx+e \right ) \right ) ^{2} \left ( c+d\sin \left ( fx+e \right ) \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d \sin \left (f x + e\right ) + c\right )}^{n} \cos \left (f x + e\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (d \sin \left (f x + e\right ) + c\right )}^{n} \cos \left (f x + e\right )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d \sin \left (f x + e\right ) + c\right )}^{n} \cos \left (f x + e\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)