3.30 \(\int \cos ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx\)

Optimal. Leaf size=87 \[ \frac{a \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac{5 a \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{5 a \sin (c+d x) \cos (c+d x)}{16 d}+\frac{5 a x}{16}-\frac{b \cos ^6(c+d x)}{6 d} \]

[Out]

(5*a*x)/16 - (b*Cos[c + d*x]^6)/(6*d) + (5*a*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (5*a*Cos[c + d*x]^3*Sin[c + d
*x])/(24*d) + (a*Cos[c + d*x]^5*Sin[c + d*x])/(6*d)

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Rubi [A]  time = 0.0907608, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {3090, 2635, 8, 2565, 30} \[ \frac{a \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac{5 a \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{5 a \sin (c+d x) \cos (c+d x)}{16 d}+\frac{5 a x}{16}-\frac{b \cos ^6(c+d x)}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a*Cos[c + d*x] + b*Sin[c + d*x]),x]

[Out]

(5*a*x)/16 - (b*Cos[c + d*x]^6)/(6*d) + (5*a*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (5*a*Cos[c + d*x]^3*Sin[c + d
*x])/(24*d) + (a*Cos[c + d*x]^5*Sin[c + d*x])/(6*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 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx &=\int \left (a \cos ^6(c+d x)+b \cos ^5(c+d x) \sin (c+d x)\right ) \, dx\\ &=a \int \cos ^6(c+d x) \, dx+b \int \cos ^5(c+d x) \sin (c+d x) \, dx\\ &=\frac{a \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac{1}{6} (5 a) \int \cos ^4(c+d x) \, dx-\frac{b \operatorname{Subst}\left (\int x^5 \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{b \cos ^6(c+d x)}{6 d}+\frac{5 a \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac{a \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac{1}{8} (5 a) \int \cos ^2(c+d x) \, dx\\ &=-\frac{b \cos ^6(c+d x)}{6 d}+\frac{5 a \cos (c+d x) \sin (c+d x)}{16 d}+\frac{5 a \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac{a \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac{1}{16} (5 a) \int 1 \, dx\\ &=\frac{5 a x}{16}-\frac{b \cos ^6(c+d x)}{6 d}+\frac{5 a \cos (c+d x) \sin (c+d x)}{16 d}+\frac{5 a \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac{a \cos ^5(c+d x) \sin (c+d x)}{6 d}\\ \end{align*}

Mathematica [A]  time = 0.104173, size = 57, normalized size = 0.66 \[ \frac{a (45 \sin (2 (c+d x))+9 \sin (4 (c+d x))+\sin (6 (c+d x))+60 c+60 d x)-32 b \cos ^6(c+d x)}{192 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a*Cos[c + d*x] + b*Sin[c + d*x]),x]

[Out]

(-32*b*Cos[c + d*x]^6 + a*(60*c + 60*d*x + 45*Sin[2*(c + d*x)] + 9*Sin[4*(c + d*x)] + Sin[6*(c + d*x)]))/(192*
d)

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Maple [A]  time = 0.042, size = 62, normalized size = 0.7 \begin{align*}{\frac{1}{d} \left ( a \left ({\frac{\sin \left ( dx+c \right ) }{6} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{5}+{\frac{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4}}+{\frac{15\,\cos \left ( dx+c \right ) }{8}} \right ) }+{\frac{5\,dx}{16}}+{\frac{5\,c}{16}} \right ) -{\frac{b \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{6}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a*cos(d*x+c)+b*sin(d*x+c)),x)

[Out]

1/d*(a*(1/6*(cos(d*x+c)^5+5/4*cos(d*x+c)^3+15/8*cos(d*x+c))*sin(d*x+c)+5/16*d*x+5/16*c)-1/6*b*cos(d*x+c)^6)

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Maxima [A]  time = 1.22386, size = 84, normalized size = 0.97 \begin{align*} -\frac{32 \, b \cos \left (d x + c\right )^{6} +{\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a}{192 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a*cos(d*x+c)+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/192*(32*b*cos(d*x + c)^6 + (4*sin(2*d*x + 2*c)^3 - 60*d*x - 60*c - 9*sin(4*d*x + 4*c) - 48*sin(2*d*x + 2*c)
)*a)/d

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Fricas [A]  time = 0.494747, size = 161, normalized size = 1.85 \begin{align*} -\frac{8 \, b \cos \left (d x + c\right )^{6} - 15 \, a d x -{\left (8 \, a \cos \left (d x + c\right )^{5} + 10 \, a \cos \left (d x + c\right )^{3} + 15 \, a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a*cos(d*x+c)+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/48*(8*b*cos(d*x + c)^6 - 15*a*d*x - (8*a*cos(d*x + c)^5 + 10*a*cos(d*x + c)^3 + 15*a*cos(d*x + c))*sin(d*x
+ c))/d

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Sympy [A]  time = 4.50227, size = 219, normalized size = 2.52 \begin{align*} \begin{cases} \frac{5 a x \sin ^{6}{\left (c + d x \right )}}{16} + \frac{15 a x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac{15 a x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac{5 a x \cos ^{6}{\left (c + d x \right )}}{16} + \frac{5 a \sin ^{5}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{16 d} + \frac{5 a \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac{11 a \sin{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} + \frac{b \sin ^{6}{\left (c + d x \right )}}{6 d} + \frac{b \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2 d} + \frac{b \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{2 d} & \text{for}\: d \neq 0 \\x \left (a \cos{\left (c \right )} + b \sin{\left (c \right )}\right ) \cos ^{5}{\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)**5*(a*cos(d*x+c)+b*sin(d*x+c)),x)

[Out]

Piecewise((5*a*x*sin(c + d*x)**6/16 + 15*a*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 15*a*x*sin(c + d*x)**2*cos(c
+ d*x)**4/16 + 5*a*x*cos(c + d*x)**6/16 + 5*a*sin(c + d*x)**5*cos(c + d*x)/(16*d) + 5*a*sin(c + d*x)**3*cos(c
+ d*x)**3/(6*d) + 11*a*sin(c + d*x)*cos(c + d*x)**5/(16*d) + b*sin(c + d*x)**6/(6*d) + b*sin(c + d*x)**4*cos(
c + d*x)**2/(2*d) + b*sin(c + d*x)**2*cos(c + d*x)**4/(2*d), Ne(d, 0)), (x*(a*cos(c) + b*sin(c))*cos(c)**5, Tr
ue))

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Giac [A]  time = 1.09659, size = 128, normalized size = 1.47 \begin{align*} \frac{5}{16} \, a x - \frac{b \cos \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac{b \cos \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac{5 \, b \cos \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac{a \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac{3 \, a \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac{15 \, a \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a*cos(d*x+c)+b*sin(d*x+c)),x, algorithm="giac")

[Out]

5/16*a*x - 1/192*b*cos(6*d*x + 6*c)/d - 1/32*b*cos(4*d*x + 4*c)/d - 5/64*b*cos(2*d*x + 2*c)/d + 1/192*a*sin(6*
d*x + 6*c)/d + 3/64*a*sin(4*d*x + 4*c)/d + 15/64*a*sin(2*d*x + 2*c)/d