3.637 \(\int \cos ^4(c+d x) (a+b \sin (c+d x))^m \, dx\)

Optimal. Leaf size=129 \[ \frac{\cos ^3(c+d x) (a+b \sin (c+d x))^{m+1} F_1\left (m+1;-\frac{3}{2},-\frac{3}{2};m+2;\frac{a+b \sin (c+d x)}{a-b},\frac{a+b \sin (c+d x)}{a+b}\right )}{b d (m+1) \left (1-\frac{a+b \sin (c+d x)}{a-b}\right )^{3/2} \left (1-\frac{a+b \sin (c+d x)}{a+b}\right )^{3/2}} \]

[Out]

(AppellF1[1 + m, -3/2, -3/2, 2 + m, (a + b*Sin[c + d*x])/(a - b), (a + b*Sin[c + d*x])/(a + b)]*Cos[c + d*x]^3
*(a + b*Sin[c + d*x])^(1 + m))/(b*d*(1 + m)*(1 - (a + b*Sin[c + d*x])/(a - b))^(3/2)*(1 - (a + b*Sin[c + d*x])
/(a + b))^(3/2))

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Rubi [A]  time = 0.0940427, antiderivative size = 129, 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} \[ \frac{\cos ^3(c+d x) (a+b \sin (c+d x))^{m+1} F_1\left (m+1;-\frac{3}{2},-\frac{3}{2};m+2;\frac{a+b \sin (c+d x)}{a-b},\frac{a+b \sin (c+d x)}{a+b}\right )}{b d (m+1) \left (1-\frac{a+b \sin (c+d x)}{a-b}\right )^{3/2} \left (1-\frac{a+b \sin (c+d x)}{a+b}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + b*Sin[c + d*x])^m,x]

[Out]

(AppellF1[1 + m, -3/2, -3/2, 2 + m, (a + b*Sin[c + d*x])/(a - b), (a + b*Sin[c + d*x])/(a + b)]*Cos[c + d*x]^3
*(a + b*Sin[c + d*x])^(1 + m))/(b*d*(1 + m)*(1 - (a + b*Sin[c + d*x])/(a - b))^(3/2)*(1 - (a + b*Sin[c + d*x])
/(a + b))^(3/2))

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 ^4(c+d x) (a+b \sin (c+d x))^m \, dx &=\frac{\cos ^3(c+d x) \operatorname{Subst}\left (\int (a+b x)^m \left (-\frac{b}{a-b}-\frac{b x}{a-b}\right )^{3/2} \left (\frac{b}{a+b}-\frac{b x}{a+b}\right )^{3/2} \, dx,x,\sin (c+d x)\right )}{d \left (1-\frac{a+b \sin (c+d x)}{a-b}\right )^{3/2} \left (1-\frac{a+b \sin (c+d x)}{a+b}\right )^{3/2}}\\ &=\frac{F_1\left (1+m;-\frac{3}{2},-\frac{3}{2};2+m;\frac{a+b \sin (c+d x)}{a-b},\frac{a+b \sin (c+d x)}{a+b}\right ) \cos ^3(c+d x) (a+b \sin (c+d x))^{1+m}}{b d (1+m) \left (1-\frac{a+b \sin (c+d x)}{a-b}\right )^{3/2} \left (1-\frac{a+b \sin (c+d x)}{a+b}\right )^{3/2}}\\ \end{align*}

Mathematica [F]  time = 4.11705, size = 0, normalized size = 0. \[ \int \cos ^4(c+d x) (a+b \sin (c+d x))^m \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Cos[c + d*x]^4*(a + b*Sin[c + d*x])^m,x]

[Out]

Integrate[Cos[c + d*x]^4*(a + b*Sin[c + d*x])^m, x]

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Maple [F]  time = 0.174, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( dx+c \right ) \right ) ^{4} \left ( a+b\sin \left ( dx+c \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(a+b*sin(d*x+c))^m,x)

[Out]

int(cos(d*x+c)^4*(a+b*sin(d*x+c))^m,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (d x + c\right ) + a\right )}^{m} \cos \left (d x + c\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*sin(d*x+c))^m,x, algorithm="maxima")

[Out]

integrate((b*sin(d*x + c) + a)^m*cos(d*x + c)^4, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \sin \left (d x + c\right ) + a\right )}^{m} \cos \left (d x + c\right )^{4}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*sin(d*x+c))^m,x, algorithm="fricas")

[Out]

integral((b*sin(d*x + c) + a)^m*cos(d*x + c)^4, 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(d*x+c)**4*(a+b*sin(d*x+c))**m,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (d x + c\right ) + a\right )}^{m} \cos \left (d x + c\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*sin(d*x+c))^m,x, algorithm="giac")

[Out]

integrate((b*sin(d*x + c) + a)^m*cos(d*x + c)^4, x)