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

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

[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)

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Rubi [A]  time = 0.181317, antiderivative size = 165, normalized size of antiderivative = 1., 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 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]

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 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]]

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*}

Mathematica [A]  time = 0.379479, size = 146, normalized size = 0.88 \[ \frac{30 \left (6 a^2 b^2+5 a^4+b^4\right ) \sin (c+d x)+5 \left (-6 a^2 b^2+5 a^4-3 b^4\right ) \sin (3 (c+d x))+3 \left (-6 a^2 b^2+a^4+b^4\right ) \sin (5 (c+d x))-120 a b \left (a^2+b^2\right ) \cos (c+d x)-20 a b \left (3 a^2+b^2\right ) \cos (3 (c+d x))-12 a b \left (a^2-b^2\right ) \cos (5 (c+d x))}{240 d} \]

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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Maple [A]  time = 0.069, size = 142, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({\frac{{b}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{5}}+4\,a{b}^{3} \left ( -1/5\, \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}-2/15\, \left ( \cos \left ( dx+c \right ) \right ) ^{3} \right ) +6\,{a}^{2}{b}^{2} \left ( -1/5\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}+1/15\, \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) \right ) -{\frac{4\,{a}^{3}b \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{5}}+{\frac{{a}^{4}\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 ) } \right ) } \end{align*}

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 = 1.15388, size = 166, normalized size = 1.01 \begin{align*} -\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} \end{align*}

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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Fricas [A]  time = 0.517038, size = 277, normalized size = 1.68 \begin{align*} -\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} \end{align*}

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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Sympy [A]  time = 2.45826, size = 206, normalized size = 1.25 \begin{align*} \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{\left (c \right )} + b \sin{\left (c \right )}\right )^{4} \cos{\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)*(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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Giac [A]  time = 1.19437, size = 223, normalized size = 1.35 \begin{align*} -\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} \end{align*}

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