3.60 \(\int \cos ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=140 \[ -\frac{3 a^2 b \cos ^5(c+d x)}{5 d}+\frac{a^3 \sin ^5(c+d x)}{5 d}-\frac{2 a^3 \sin ^3(c+d x)}{3 d}+\frac{a^3 \sin (c+d x)}{d}-\frac{3 a b^2 \sin ^5(c+d x)}{5 d}+\frac{a b^2 \sin ^3(c+d x)}{d}+\frac{b^3 \cos ^5(c+d x)}{5 d}-\frac{b^3 \cos ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.163265, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3090, 2633, 2565, 30, 2564, 14} \[ -\frac{3 a^2 b \cos ^5(c+d x)}{5 d}+\frac{a^3 \sin ^5(c+d x)}{5 d}-\frac{2 a^3 \sin ^3(c+d x)}{3 d}+\frac{a^3 \sin (c+d x)}{d}-\frac{3 a b^2 \sin ^5(c+d x)}{5 d}+\frac{a b^2 \sin ^3(c+d x)}{d}+\frac{b^3 \cos ^5(c+d x)}{5 d}-\frac{b^3 \cos ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]

[Out]

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

Mathematica [A]  time = 0.28896, size = 150, normalized size = 1.07 \[ \frac{-30 b \left (3 a^2+b^2\right ) \cos (c+d x)-5 \left (9 a^2 b+b^3\right ) \cos (3 (c+d x))-9 a^2 b \cos (5 (c+d x))+150 a^3 \sin (c+d x)+25 a^3 \sin (3 (c+d x))+3 a^3 \sin (5 (c+d x))+90 a b^2 \sin (c+d x)-15 a b^2 \sin (3 (c+d x))-9 a b^2 \sin (5 (c+d x))+3 b^3 \cos (5 (c+d x))}{240 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]

[Out]

(-30*b*(3*a^2 + b^2)*Cos[c + d*x] - 5*(9*a^2*b + b^3)*Cos[3*(c + d*x)] - 9*a^2*b*Cos[5*(c + d*x)] + 3*b^3*Cos[
5*(c + d*x)] + 150*a^3*Sin[c + d*x] + 90*a*b^2*Sin[c + d*x] + 25*a^3*Sin[3*(c + d*x)] - 15*a*b^2*Sin[3*(c + d*
x)] + 3*a^3*Sin[5*(c + d*x)] - 9*a*b^2*Sin[5*(c + d*x)])/(240*d)

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Maple [A]  time = 0.07, size = 125, normalized size = 0.9 \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 ) ^{3}}{5}}-{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{15}} \right ) +3\,a{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{3\,{a}^{2}b \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{5}}+{\frac{{a}^{3}\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)^2*(a*cos(d*x+c)+b*sin(d*x+c))^3,x)

[Out]

1/d*(b^3*(-1/5*sin(d*x+c)^2*cos(d*x+c)^3-2/15*cos(d*x+c)^3)+3*a*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))-3/5*a^2*b*cos(d*x+c)^5+1/5*a^3*(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 = 1.23966, size = 144, normalized size = 1.03 \begin{align*} -\frac{9 \, a^{2} b \cos \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^{3} + 3 \,{\left (3 \, \sin \left (d x + c\right )^{5} - 5 \, \sin \left (d x + c\right )^{3}\right )} a b^{2} -{\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} b^{3}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/15*(9*a^2*b*cos(d*x + c)^5 - (3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*a^3 + 3*(3*sin(d*x +
c)^5 - 5*sin(d*x + c)^3)*a*b^2 - (3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*b^3)/d

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Fricas [A]  time = 0.494217, size = 230, normalized size = 1.64 \begin{align*} -\frac{5 \, b^{3} \cos \left (d x + c\right )^{3} + 3 \,{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{5} -{\left (3 \,{\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + 8 \, a^{3} + 6 \, a b^{2} +{\left (4 \, a^{3} + 3 \, a b^{2}\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)^2*(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 2.78219, size = 182, normalized size = 1.3 \begin{align*} \begin{cases} \frac{8 a^{3} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{4 a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac{a^{3} \sin{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac{3 a^{2} b \cos ^{5}{\left (c + d x \right )}}{5 d} + \frac{2 a b^{2} \sin ^{5}{\left (c + d x \right )}}{5 d} + \frac{a b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac{b^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac{2 b^{3} \cos ^{5}{\left (c + d x \right )}}{15 d} & \text{for}\: d \neq 0 \\x \left (a \cos{\left (c \right )} + b \sin{\left (c \right )}\right )^{3} \cos ^{2}{\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)**2*(a*cos(d*x+c)+b*sin(d*x+c))**3,x)

[Out]

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

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Giac [A]  time = 1.1482, size = 196, normalized size = 1.4 \begin{align*} -\frac{{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac{{\left (9 \, a^{2} b + b^{3}\right )} \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac{{\left (3 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )}{8 \, d} + \frac{{\left (a^{3} - 3 \, a b^{2}\right )} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac{{\left (5 \, a^{3} - 3 \, a b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac{{\left (5 \, a^{3} + 3 \, a b^{2}\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)^2*(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/80*(3*a^2*b - b^3)*cos(5*d*x + 5*c)/d - 1/48*(9*a^2*b + b^3)*cos(3*d*x + 3*c)/d - 1/8*(3*a^2*b + b^3)*cos(d
*x + c)/d + 1/80*(a^3 - 3*a*b^2)*sin(5*d*x + 5*c)/d + 1/48*(5*a^3 - 3*a*b^2)*sin(3*d*x + 3*c)/d + 1/8*(5*a^3 +
3*a*b^2)*sin(d*x + c)/d