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

Optimal. Leaf size=165 \[ \frac {a^4 \sin ^5(c+d x)}{5 d}-\frac {2 a^4 \sin ^3(c+d x)}{3 d}+\frac {a^4 \sin (c+d x)}{d}-\frac {4 a^3 b \cos ^5(c+d x)}{5 d}-\frac {6 a^2 b^2 \sin ^5(c+d x)}{5 d}+\frac {2 a^2 b^2 \sin ^3(c+d x)}{d}+\frac {4 a b^3 \cos ^5(c+d x)}{5 d}-\frac {4 a b^3 \cos ^3(c+d x)}{3 d}+\frac {b^4 \sin ^5(c+d x)}{5 d} \]

[Out]

-4/3*a*b^3*cos(d*x+c)^3/d-4/5*a^3*b*cos(d*x+c)^5/d+4/5*a*b^3*cos(d*x+c)^5/d+a^4*sin(d*x+c)/d-2/3*a^4*sin(d*x+c
)^3/d+2*a^2*b^2*sin(d*x+c)^3/d+1/5*a^4*sin(d*x+c)^5/d-6/5*a^2*b^2*sin(d*x+c)^5/d+1/5*b^4*sin(d*x+c)^5/d

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Rubi [A]  time = 0.18, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3090, 2633, 2565, 30, 2564, 14} \[ -\frac {6 a^2 b^2 \sin ^5(c+d x)}{5 d}+\frac {2 a^2 b^2 \sin ^3(c+d x)}{d}-\frac {4 a^3 b \cos ^5(c+d x)}{5 d}+\frac {a^4 \sin ^5(c+d x)}{5 d}-\frac {2 a^4 \sin ^3(c+d x)}{3 d}+\frac {a^4 \sin (c+d x)}{d}+\frac {4 a b^3 \cos ^5(c+d x)}{5 d}-\frac {4 a b^3 \cos ^3(c+d x)}{3 d}+\frac {b^4 \sin ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*a*b^3*Cos[c + d*x]^3)/(3*d) - (4*a^3*b*Cos[c + d*x]^5)/(5*d) + (4*a*b^3*Cos[c + d*x]^5)/(5*d) + (a^4*Sin[c
+ d*x])/d - (2*a^4*Sin[c + d*x]^3)/(3*d) + (2*a^2*b^2*Sin[c + d*x]^3)/d + (a^4*Sin[c + d*x]^5)/(5*d) - (6*a^2
*b^2*Sin[c + d*x]^5)/(5*d) + (b^4*Sin[c + d*x]^5)/(5*d)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

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

Rule 2564

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

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

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Mathematica [A]  time = 0.43, size = 146, normalized size = 0.88 \[ \frac {-120 a b \left (a^2+b^2\right ) \cos (c+d x)-20 a b \left (3 a^2+b^2\right ) \cos (3 (c+d x))-12 a b \left (a^2-b^2\right ) \cos (5 (c+d x))+30 \left (5 a^4+6 a^2 b^2+b^4\right ) \sin (c+d x)+5 \left (5 a^4-6 a^2 b^2-3 b^4\right ) \sin (3 (c+d x))+3 \left (a^4-6 a^2 b^2+b^4\right ) \sin (5 (c+d x))}{240 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-120*a*b*(a^2 + b^2)*Cos[c + d*x] - 20*a*b*(3*a^2 + b^2)*Cos[3*(c + d*x)] - 12*a*b*(a^2 - b^2)*Cos[5*(c + d*x
)] + 30*(5*a^4 + 6*a^2*b^2 + b^4)*Sin[c + d*x] + 5*(5*a^4 - 6*a^2*b^2 - 3*b^4)*Sin[3*(c + d*x)] + 3*(a^4 - 6*a
^2*b^2 + b^4)*Sin[5*(c + d*x)])/(240*d)

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fricas [A]  time = 0.78, size = 123, normalized size = 0.75 \[ -\frac {20 \, a b^{3} \cos \left (d x + c\right )^{3} + 12 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{5} - {\left (3 \, {\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} + 8 \, a^{4} + 12 \, a^{2} b^{2} + 3 \, b^{4} + 2 \, {\left (2 \, a^{4} + 3 \, a^{2} b^{2} - 3 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(20*a*b^3*cos(d*x + c)^3 + 12*(a^3*b - a*b^3)*cos(d*x + c)^5 - (3*(a^4 - 6*a^2*b^2 + b^4)*cos(d*x + c)^4
+ 8*a^4 + 12*a^2*b^2 + 3*b^4 + 2*(2*a^4 + 3*a^2*b^2 - 3*b^4)*cos(d*x + c)^2)*sin(d*x + c))/d

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giac [A]  time = 2.94, size = 165, normalized size = 1.00 \[ -\frac {{\left (a^{3} b - a b^{3}\right )} \cos \left (5 \, d x + 5 \, c\right )}{20 \, d} - \frac {{\left (3 \, a^{3} b + a b^{3}\right )} \cos \left (3 \, d x + 3 \, c\right )}{12 \, d} - \frac {{\left (a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )}{2 \, d} + \frac {{\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {{\left (5 \, a^{4} - 6 \, a^{2} b^{2} - 3 \, b^{4}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (5 \, a^{4} + 6 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20*(a^3*b - a*b^3)*cos(5*d*x + 5*c)/d - 1/12*(3*a^3*b + a*b^3)*cos(3*d*x + 3*c)/d - 1/2*(a^3*b + a*b^3)*cos
(d*x + c)/d + 1/80*(a^4 - 6*a^2*b^2 + b^4)*sin(5*d*x + 5*c)/d + 1/48*(5*a^4 - 6*a^2*b^2 - 3*b^4)*sin(3*d*x + 3
*c)/d + 1/8*(5*a^4 + 6*a^2*b^2 + b^4)*sin(d*x + c)/d

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maple [A]  time = 39.38, size = 142, normalized size = 0.86 \[ \frac {\frac {b^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{5}+4 a \,b^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )+6 a^{2} b^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{15}\right )-\frac {4 a^{3} b \left (\cos ^{5}\left (d x +c \right )\right )}{5}+\frac {a^{4} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(1/5*b^4*sin(d*x+c)^5+4*a*b^3*(-1/5*sin(d*x+c)^2*cos(d*x+c)^3-2/15*cos(d*x+c)^3)+6*a^2*b^2*(-1/5*sin(d*x+c
)*cos(d*x+c)^4+1/15*(2+cos(d*x+c)^2)*sin(d*x+c))-4/5*a^3*b*cos(d*x+c)^5+1/5*a^4*(8/3+cos(d*x+c)^4+4/3*cos(d*x+
c)^2)*sin(d*x+c))

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maxima [A]  time = 0.33, size = 123, normalized size = 0.75 \[ -\frac {12 \, a^{3} b \cos \left (d x + c\right )^{5} - 3 \, b^{4} \sin \left (d x + c\right )^{5} - {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{4} + 6 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 5 \, \sin \left (d x + c\right )^{3}\right )} a^{2} b^{2} - 4 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a b^{3}}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(12*a^3*b*cos(d*x + c)^5 - 3*b^4*sin(d*x + c)^5 - (3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c
))*a^4 + 6*(3*sin(d*x + c)^5 - 5*sin(d*x + c)^3)*a^2*b^2 - 4*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*a*b^3)/d

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mupad [B]  time = 0.83, size = 204, normalized size = 1.24 \[ \frac {2\,\left (\frac {3\,\sin \left (c+d\,x\right )\,a^4\,{\cos \left (c+d\,x\right )}^4}{2}+2\,\sin \left (c+d\,x\right )\,a^4\,{\cos \left (c+d\,x\right )}^2+4\,\sin \left (c+d\,x\right )\,a^4-6\,a^3\,b\,{\cos \left (c+d\,x\right )}^5-9\,\sin \left (c+d\,x\right )\,a^2\,b^2\,{\cos \left (c+d\,x\right )}^4+3\,\sin \left (c+d\,x\right )\,a^2\,b^2\,{\cos \left (c+d\,x\right )}^2+6\,\sin \left (c+d\,x\right )\,a^2\,b^2+6\,a\,b^3\,{\cos \left (c+d\,x\right )}^5-10\,a\,b^3\,{\cos \left (c+d\,x\right )}^3+\frac {3\,\sin \left (c+d\,x\right )\,b^4\,{\cos \left (c+d\,x\right )}^4}{2}-3\,\sin \left (c+d\,x\right )\,b^4\,{\cos \left (c+d\,x\right )}^2+\frac {3\,\sin \left (c+d\,x\right )\,b^4}{2}\right )}{15\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(4*a^4*sin(c + d*x) + (3*b^4*sin(c + d*x))/2 - 10*a*b^3*cos(c + d*x)^3 + 6*a*b^3*cos(c + d*x)^5 - 6*a^3*b*c
os(c + d*x)^5 + 2*a^4*cos(c + d*x)^2*sin(c + d*x) + (3*a^4*cos(c + d*x)^4*sin(c + d*x))/2 + 6*a^2*b^2*sin(c +
d*x) - 3*b^4*cos(c + d*x)^2*sin(c + d*x) + (3*b^4*cos(c + d*x)^4*sin(c + d*x))/2 + 3*a^2*b^2*cos(c + d*x)^2*si
n(c + d*x) - 9*a^2*b^2*cos(c + d*x)^4*sin(c + d*x)))/(15*d)

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sympy [A]  time = 1.99, size = 206, normalized size = 1.25 \[ \begin {cases} \frac {8 a^{4} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 a^{4} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {a^{4} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {4 a^{3} b \cos ^{5}{\left (c + d x \right )}}{5 d} + \frac {4 a^{2} b^{2} \sin ^{5}{\left (c + d x \right )}}{5 d} + \frac {2 a^{2} b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {4 a b^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {8 a b^{3} \cos ^{5}{\left (c + d x \right )}}{15 d} + \frac {b^{4} \sin ^{5}{\left (c + d x \right )}}{5 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\relax (c )} + b \sin {\relax (c )}\right )^{4} \cos {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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