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]

5/16*a*x-1/6*b*cos(d*x+c)^6/d+5/16*a*cos(d*x+c)*sin(d*x+c)/d+5/24*a*cos(d*x+c)^3*sin(d*x+c)/d+1/6*a*cos(d*x+c)
^5*sin(d*x+c)/d

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 87, normalized size of antiderivative = 1.00, 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 8

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

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 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 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 ^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 = 0.12, 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)

________________________________________________________________________________________

fricas [A]  time = 0.60, size = 62, normalized size = 0.71 \[ -\frac {8 \, b \cos \left (d x + c\right )^{6} - 15 \, a d x - {\left (8 \, a \cos \left (d x + c\right )^{5} + 10 \, a \cos \left (d x + c\right )^{3} + 15 \, a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d} \]

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

________________________________________________________________________________________

giac [A]  time = 0.41, size = 95, normalized size = 1.09 \[ \frac {5}{16} \, a x - \frac {b \cos \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac {b \cos \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac {5 \, b \cos \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {a \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {3 \, a \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {15 \, a \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \]

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

________________________________________________________________________________________

maple [A]  time = 1.09, size = 62, normalized size = 0.71 \[ \frac {a \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )-\frac {b \left (\cos ^{6}\left (d x +c \right )\right )}{6}}{d} \]

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)

________________________________________________________________________________________

maxima [A]  time = 0.33, size = 62, normalized size = 0.71 \[ -\frac {32 \, b \cos \left (d x + c\right )^{6} + {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a}{192 \, d} \]

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

________________________________________________________________________________________

mupad [B]  time = 4.20, size = 149, normalized size = 1.71 \[ \frac {5\,a\,x}{16}+\frac {-\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{8}+2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\frac {5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{24}-\frac {15\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {20\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+\frac {15\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}-\frac {5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24}+2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {11\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*a*x)/16 + ((11*a*tan(c/2 + (d*x)/2))/8 - (5*a*tan(c/2 + (d*x)/2)^3)/24 + (15*a*tan(c/2 + (d*x)/2)^5)/4 - (1
5*a*tan(c/2 + (d*x)/2)^7)/4 + (5*a*tan(c/2 + (d*x)/2)^9)/24 - (11*a*tan(c/2 + (d*x)/2)^11)/8 + 2*b*tan(c/2 + (
d*x)/2)^2 + (20*b*tan(c/2 + (d*x)/2)^6)/3 + 2*b*tan(c/2 + (d*x)/2)^10)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^6)

________________________________________________________________________________________

sympy [A]  time = 3.10, size = 175, normalized size = 2.01 \[ \begin {cases} \frac {5 a x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {15 a x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {15 a x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {5 a x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {5 a \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {5 a \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac {11 a \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac {b \cos ^{6}{\left (c + d x \right )}}{6 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\relax (c )} + b \sin {\relax (c )}\right ) \cos ^{5}{\relax (c )} & \text {otherwise} \end {cases} \]

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*cos(c + d*x)**6/(6*d), Ne(d, 0)), (x*(a*cos(c)
+ b*sin(c))*cos(c)**5, True))

________________________________________________________________________________________