3.59 \(\int \cos ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=216 \[ -\frac{a^2 b \cos ^6(c+d x)}{2 d}+\frac{a^3 \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac{5 a^3 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{5 a^3 \sin (c+d x) \cos (c+d x)}{16 d}+\frac{5 a^3 x}{16}-\frac{a b^2 \sin (c+d x) \cos ^5(c+d x)}{2 d}+\frac{a b^2 \sin (c+d x) \cos ^3(c+d x)}{8 d}+\frac{3 a b^2 \sin (c+d x) \cos (c+d x)}{16 d}+\frac{3}{16} a b^2 x-\frac{b^3 \sin ^6(c+d x)}{6 d}+\frac{b^3 \sin ^4(c+d x)}{4 d} \]

[Out]

(5*a^3*x)/16 + (3*a*b^2*x)/16 - (a^2*b*Cos[c + d*x]^6)/(2*d) + (5*a^3*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (3*a
*b^2*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (5*a^3*Cos[c + d*x]^3*Sin[c + d*x])/(24*d) + (a*b^2*Cos[c + d*x]^3*Si
n[c + d*x])/(8*d) + (a^3*Cos[c + d*x]^5*Sin[c + d*x])/(6*d) - (a*b^2*Cos[c + d*x]^5*Sin[c + d*x])/(2*d) + (b^3
*Sin[c + d*x]^4)/(4*d) - (b^3*Sin[c + d*x]^6)/(6*d)

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Rubi [A]  time = 0.212641, antiderivative size = 216, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 8, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3090, 2635, 8, 2565, 30, 2568, 2564, 14} \[ -\frac{a^2 b \cos ^6(c+d x)}{2 d}+\frac{a^3 \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac{5 a^3 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{5 a^3 \sin (c+d x) \cos (c+d x)}{16 d}+\frac{5 a^3 x}{16}-\frac{a b^2 \sin (c+d x) \cos ^5(c+d x)}{2 d}+\frac{a b^2 \sin (c+d x) \cos ^3(c+d x)}{8 d}+\frac{3 a b^2 \sin (c+d x) \cos (c+d x)}{16 d}+\frac{3}{16} a b^2 x-\frac{b^3 \sin ^6(c+d x)}{6 d}+\frac{b^3 \sin ^4(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]

[Out]

(5*a^3*x)/16 + (3*a*b^2*x)/16 - (a^2*b*Cos[c + d*x]^6)/(2*d) + (5*a^3*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (3*a
*b^2*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (5*a^3*Cos[c + d*x]^3*Sin[c + d*x])/(24*d) + (a*b^2*Cos[c + d*x]^3*Si
n[c + d*x])/(8*d) + (a^3*Cos[c + d*x]^5*Sin[c + d*x])/(6*d) - (a*b^2*Cos[c + d*x]^5*Sin[c + d*x])/(2*d) + (b^3
*Sin[c + d*x]^4)/(4*d) - (b^3*Sin[c + d*x]^6)/(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]

Rule 2568

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*(b*Cos[e
+ f*x])^(n + 1)*(a*Sin[e + f*x])^(m - 1))/(b*f*(m + n)), x] + Dist[(a^2*(m - 1))/(m + n), Int[(b*Cos[e + f*x])
^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*
m, 2*n]

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

Mathematica [A]  time = 0.30174, size = 171, normalized size = 0.79 \[ \frac{a \left (5 a^2+3 b^2\right ) (c+d x)}{16 d}+\frac{3 a \left (5 a^2+b^2\right ) \sin (2 (c+d x))}{64 d}+\frac{3 a \left (a^2-b^2\right ) \sin (4 (c+d x))}{64 d}+\frac{a \left (a^2-3 b^2\right ) \sin (6 (c+d x))}{192 d}-\frac{3 b \left (5 a^2+b^2\right ) \cos (2 (c+d x))}{64 d}-\frac{b \left (3 a^2-b^2\right ) \cos (6 (c+d x))}{192 d}-\frac{3 a^2 b \cos (4 (c+d x))}{32 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]

[Out]

(a*(5*a^2 + 3*b^2)*(c + d*x))/(16*d) - (3*b*(5*a^2 + b^2)*Cos[2*(c + d*x)])/(64*d) - (3*a^2*b*Cos[4*(c + d*x)]
)/(32*d) - (b*(3*a^2 - b^2)*Cos[6*(c + d*x)])/(192*d) + (3*a*(5*a^2 + b^2)*Sin[2*(c + d*x)])/(64*d) + (3*a*(a^
2 - b^2)*Sin[4*(c + d*x)])/(64*d) + (a*(a^2 - 3*b^2)*Sin[6*(c + d*x)])/(192*d)

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Maple [A]  time = 0.078, size = 155, normalized size = 0.7 \begin{align*}{\frac{1}{d} \left ({b}^{3} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{6}}-{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{12}} \right ) +3\,a{b}^{2} \left ( -1/6\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{5}+1/24\, \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+3/2\,\cos \left ( dx+c \right ) \right ) \sin \left ( dx+c \right ) +1/16\,dx+c/16 \right ) -{\frac{{a}^{2}b \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{2}}+{a}^{3} \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 ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a*cos(d*x+c)+b*sin(d*x+c))^3,x)

[Out]

1/d*(b^3*(-1/6*sin(d*x+c)^2*cos(d*x+c)^4-1/12*cos(d*x+c)^4)+3*a*b^2*(-1/6*sin(d*x+c)*cos(d*x+c)^5+1/24*(cos(d*
x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+1/16*d*x+1/16*c)-1/2*a^2*b*cos(d*x+c)^6+a^3*(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))

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Maxima [A]  time = 1.22564, size = 177, normalized size = 0.82 \begin{align*} -\frac{96 \, a^{2} 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^{3} - 3 \,{\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 12 \, d x + 12 \, c - 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a b^{2} + 16 \,{\left (2 \, \sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4}\right )} b^{3}}{192 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/192*(96*a^2*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^3 - 3*(4*sin(2*d*x + 2*c)^3 + 12*d*x + 12*c - 3*sin(4*d*x + 4*c))*a*b^2 + 16*(2*sin(d*x + c)^6 - 3*sin
(d*x + c)^4)*b^3)/d

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Fricas [A]  time = 0.515885, size = 292, normalized size = 1.35 \begin{align*} -\frac{12 \, b^{3} \cos \left (d x + c\right )^{4} + 8 \,{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{6} - 3 \,{\left (5 \, a^{3} + 3 \, a b^{2}\right )} d x -{\left (8 \,{\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} + 2 \,{\left (5 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (5 \, a^{3} + 3 \, a b^{2}\right )} \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)^3*(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 6.82375, size = 454, normalized size = 2.1 \begin{align*} \begin{cases} \frac{5 a^{3} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac{15 a^{3} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac{15 a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac{5 a^{3} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac{5 a^{3} \sin ^{5}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{16 d} + \frac{5 a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac{11 a^{3} \sin{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} + \frac{a^{2} b \sin ^{6}{\left (c + d x \right )}}{2 d} + \frac{3 a^{2} b \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2 d} + \frac{3 a^{2} b \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{2 d} + \frac{3 a b^{2} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac{9 a b^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac{9 a b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac{3 a b^{2} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac{3 a b^{2} \sin ^{5}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{16 d} + \frac{a b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{2 d} - \frac{3 a b^{2} \sin{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} + \frac{b^{3} \sin ^{6}{\left (c + d x \right )}}{12 d} + \frac{b^{3} \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4 d} & \text{for}\: d \neq 0 \\x \left (a \cos{\left (c \right )} + b \sin{\left (c \right )}\right )^{3} \cos ^{3}{\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)**3*(a*cos(d*x+c)+b*sin(d*x+c))**3,x)

[Out]

Piecewise((5*a**3*x*sin(c + d*x)**6/16 + 15*a**3*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 15*a**3*x*sin(c + d*x)
**2*cos(c + d*x)**4/16 + 5*a**3*x*cos(c + d*x)**6/16 + 5*a**3*sin(c + d*x)**5*cos(c + d*x)/(16*d) + 5*a**3*sin
(c + d*x)**3*cos(c + d*x)**3/(6*d) + 11*a**3*sin(c + d*x)*cos(c + d*x)**5/(16*d) + a**2*b*sin(c + d*x)**6/(2*d
) + 3*a**2*b*sin(c + d*x)**4*cos(c + d*x)**2/(2*d) + 3*a**2*b*sin(c + d*x)**2*cos(c + d*x)**4/(2*d) + 3*a*b**2
*x*sin(c + d*x)**6/16 + 9*a*b**2*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 9*a*b**2*x*sin(c + d*x)**2*cos(c + d*x
)**4/16 + 3*a*b**2*x*cos(c + d*x)**6/16 + 3*a*b**2*sin(c + d*x)**5*cos(c + d*x)/(16*d) + a*b**2*sin(c + d*x)**
3*cos(c + d*x)**3/(2*d) - 3*a*b**2*sin(c + d*x)*cos(c + d*x)**5/(16*d) + b**3*sin(c + d*x)**6/(12*d) + b**3*si
n(c + d*x)**4*cos(c + d*x)**2/(4*d), Ne(d, 0)), (x*(a*cos(c) + b*sin(c))**3*cos(c)**3, True))

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Giac [A]  time = 1.18411, size = 212, normalized size = 0.98 \begin{align*} -\frac{3 \, a^{2} b \cos \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac{1}{16} \,{\left (5 \, a^{3} + 3 \, a b^{2}\right )} x - \frac{{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac{3 \,{\left (5 \, a^{2} b + b^{3}\right )} \cos \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac{{\left (a^{3} - 3 \, a b^{2}\right )} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac{3 \,{\left (a^{3} - a b^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac{3 \,{\left (5 \, a^{3} + a b^{2}\right )} \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)^3*(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

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