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

Optimal. Leaf size=60 \[ \frac{a \sin ^5(c+d x)}{5 d}-\frac{2 a \sin ^3(c+d x)}{3 d}+\frac{a \sin (c+d x)}{d}-\frac{b \cos ^5(c+d x)}{5 d} \]

[Out]

-(b*Cos[c + d*x]^5)/(5*d) + (a*Sin[c + d*x])/d - (2*a*Sin[c + d*x]^3)/(3*d) + (a*Sin[c + d*x]^5)/(5*d)

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Rubi [A]  time = 0.0700506, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3090, 2633, 2565, 30} \[ \frac{a \sin ^5(c+d x)}{5 d}-\frac{2 a \sin ^3(c+d x)}{3 d}+\frac{a \sin (c+d x)}{d}-\frac{b \cos ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*Cos[c + d*x]^5)/(5*d) + (a*Sin[c + d*x])/d - (2*a*Sin[c + d*x]^3)/(3*d) + (a*Sin[c + d*x]^5)/(5*d)

Rule 3090

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Sym
bol] :> Int[ExpandTrig[cos[c + d*x]^m*(a*cos[c + d*x] + b*sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d}, x] &&
IntegerQ[m] && IGtQ[n, 0]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \cos ^4(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx &=\int \left (a \cos ^5(c+d x)+b \cos ^4(c+d x) \sin (c+d x)\right ) \, dx\\ &=a \int \cos ^5(c+d x) \, dx+b \int \cos ^4(c+d x) \sin (c+d x) \, dx\\ &=-\frac{a \operatorname{Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,-\sin (c+d x)\right )}{d}-\frac{b \operatorname{Subst}\left (\int x^4 \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{b \cos ^5(c+d x)}{5 d}+\frac{a \sin (c+d x)}{d}-\frac{2 a \sin ^3(c+d x)}{3 d}+\frac{a \sin ^5(c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.0156501, size = 60, normalized size = 1. \[ \frac{a \sin ^5(c+d x)}{5 d}-\frac{2 a \sin ^3(c+d x)}{3 d}+\frac{a \sin (c+d x)}{d}-\frac{b \cos ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*Cos[c + d*x]^5)/(5*d) + (a*Sin[c + d*x])/d - (2*a*Sin[c + d*x]^3)/(3*d) + (a*Sin[c + d*x]^5)/(5*d)

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Maple [A]  time = 0.04, size = 46, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ({\frac{a\sin \left ( dx+c \right ) }{5} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) }-{\frac{b \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{5}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(1/5*a*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)-1/5*b*cos(d*x+c)^5)

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Maxima [A]  time = 1.25828, size = 66, normalized size = 1.1 \begin{align*} -\frac{3 \, b \cos \left (d x + c\right )^{5} -{\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(3*b*cos(d*x + c)^5 - (3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*a)/d

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Fricas [A]  time = 0.478467, size = 126, normalized size = 2.1 \begin{align*} -\frac{3 \, b \cos \left (d x + c\right )^{5} -{\left (3 \, a \cos \left (d x + c\right )^{4} + 4 \, a \cos \left (d x + c\right )^{2} + 8 \, a\right )} \sin \left (d x + c\right )}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(3*b*cos(d*x + c)^5 - (3*a*cos(d*x + c)^4 + 4*a*cos(d*x + c)^2 + 8*a)*sin(d*x + c))/d

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Sympy [A]  time = 2.23337, size = 87, normalized size = 1.45 \begin{align*} \begin{cases} \frac{8 a \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{4 a \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac{a \sin{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac{b \cos ^{5}{\left (c + d x \right )}}{5 d} & \text{for}\: d \neq 0 \\x \left (a \cos{\left (c \right )} + b \sin{\left (c \right )}\right ) \cos ^{4}{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((8*a*sin(c + d*x)**5/(15*d) + 4*a*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + a*sin(c + d*x)*cos(c + d*x
)**4/d - b*cos(c + d*x)**5/(5*d), Ne(d, 0)), (x*(a*cos(c) + b*sin(c))*cos(c)**4, True))

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Giac [A]  time = 1.12746, size = 115, normalized size = 1.92 \begin{align*} -\frac{b \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac{b \cos \left (3 \, d x + 3 \, c\right )}{16 \, d} - \frac{b \cos \left (d x + c\right )}{8 \, d} + \frac{a \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac{5 \, a \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac{5 \, a \sin \left (d x + c\right )}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/80*b*cos(5*d*x + 5*c)/d - 1/16*b*cos(3*d*x + 3*c)/d - 1/8*b*cos(d*x + c)/d + 1/80*a*sin(5*d*x + 5*c)/d + 5/
48*a*sin(3*d*x + 3*c)/d + 5/8*a*sin(d*x + c)/d