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]

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

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Rubi [A]  time = 0.07, antiderivative size = 60, normalized size of antiderivative = 1.00, 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 30

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

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 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 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]

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*}

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Mathematica [A]  time = 0.02, size = 60, normalized size = 1.00 \[ \frac {a \sin ^5(c+d x)}{5 d}-\frac {2 a \sin ^3(c+d x)}{3 d}+\frac {a \sin (c+d x)}{d}-\frac {b \cos ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*(b*Cos[c + d*x]^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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fricas [A]  time = 0.58, size = 51, normalized size = 0.85 \[ -\frac {3 \, b \cos \left (d x + c\right )^{5} - {\left (3 \, a \cos \left (d x + c\right )^{4} + 4 \, a \cos \left (d x + c\right )^{2} + 8 \, a\right )} \sin \left (d x + c\right )}{15 \, d} \]

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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giac [A]  time = 0.93, size = 85, normalized size = 1.42 \[ -\frac {b \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac {b \cos \left (3 \, d x + 3 \, c\right )}{16 \, d} - \frac {b \cos \left (d x + c\right )}{8 \, d} + \frac {a \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {5 \, a \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {5 \, a \sin \left (d x + c\right )}{8 \, d} \]

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

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maple [A]  time = 1.05, size = 46, normalized size = 0.77 \[ \frac {\frac {a \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}-\frac {b \left (\cos ^{5}\left (d x +c \right )\right )}{5}}{d} \]

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 = 0.32, size = 49, normalized size = 0.82 \[ -\frac {3 \, b \cos \left (d x + c\right )^{5} - {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a}{15 \, d} \]

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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mupad [B]  time = 0.46, size = 67, normalized size = 1.12 \[ \frac {8\,a\,\sin \left (c+d\,x\right )}{15\,d}-\frac {b\,{\cos \left (c+d\,x\right )}^5}{5\,d}+\frac {4\,a\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{15\,d}+\frac {a\,{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )}{5\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.61, size = 87, normalized size = 1.45 \[ \begin {cases} \frac {8 a \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 a \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {a \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {b \cos ^{5}{\left (c + d x \right )}}{5 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\relax (c )} + b \sin {\relax (c )}\right ) \cos ^{4}{\relax (c )} & \text {otherwise} \end {cases} \]

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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