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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.16, antiderivative size = 140, normalized size of antiderivative = 1.00, 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 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 ^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 = 0.29, size = 150, normalized size = 1.07 \[ \frac {150 a^3 \sin (c+d x)+25 a^3 \sin (3 (c+d x))+3 a^3 \sin (5 (c+d x))-5 \left (9 a^2 b+b^3\right ) \cos (3 (c+d x))-30 b \left (3 a^2+b^2\right ) \cos (c+d x)-9 a^2 b \cos (5 (c+d x))+90 a b^2 \sin (c+d x)-15 a b^2 \sin (3 (c+d x))-9 a b^2 \sin (5 (c+d x))+3 b^3 \cos (5 (c+d x))}{240 d} \]

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)

________________________________________________________________________________________

fricas [A]  time = 0.70, size = 102, normalized size = 0.73 \[ -\frac {5 \, b^{3} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{5} - {\left (3 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + 8 \, a^{3} + 6 \, a b^{2} + {\left (4 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{15 \, d} \]

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

________________________________________________________________________________________

giac [A]  time = 0.48, size = 145, normalized size = 1.04 \[ -\frac {{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac {{\left (9 \, a^{2} b + b^{3}\right )} \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac {{\left (3 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )}{8 \, d} + \frac {{\left (a^{3} - 3 \, a b^{2}\right )} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {{\left (5 \, a^{3} - 3 \, a b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (5 \, a^{3} + 3 \, a b^{2}\right )} \sin \left (d x + c\right )}{8 \, d} \]

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

________________________________________________________________________________________

maple [A]  time = 10.50, size = 125, normalized size = 0.89 \[ \frac {b^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )+3 b^{2} a \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{15}\right )-\frac {3 a^{2} b \left (\cos ^{5}\left (d x +c \right )\right )}{5}+\frac {a^{3} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^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*b^2*a*(-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))

________________________________________________________________________________________

maxima [A]  time = 0.32, size = 107, normalized size = 0.76 \[ -\frac {9 \, a^{2} b \cos \left (d x + c\right )^{5} - {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{3} + 3 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 5 \, \sin \left (d x + c\right )^{3}\right )} a b^{2} - {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} b^{3}}{15 \, d} \]

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

________________________________________________________________________________________

mupad [B]  time = 0.70, size = 147, normalized size = 1.05 \[ \frac {2\,\left (\frac {3\,\sin \left (c+d\,x\right )\,a^3\,{\cos \left (c+d\,x\right )}^4}{2}+2\,\sin \left (c+d\,x\right )\,a^3\,{\cos \left (c+d\,x\right )}^2+4\,\sin \left (c+d\,x\right )\,a^3-\frac {9\,a^2\,b\,{\cos \left (c+d\,x\right )}^5}{2}-\frac {9\,\sin \left (c+d\,x\right )\,a\,b^2\,{\cos \left (c+d\,x\right )}^4}{2}+\frac {3\,\sin \left (c+d\,x\right )\,a\,b^2\,{\cos \left (c+d\,x\right )}^2}{2}+3\,\sin \left (c+d\,x\right )\,a\,b^2+\frac {3\,b^3\,{\cos \left (c+d\,x\right )}^5}{2}-\frac {5\,b^3\,{\cos \left (c+d\,x\right )}^3}{2}\right )}{15\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2*(a*cos(c + d*x) + b*sin(c + d*x))^3,x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 1.93, size = 182, normalized size = 1.30 \[ \begin {cases} \frac {8 a^{3} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {a^{3} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {3 a^{2} b \cos ^{5}{\left (c + d x \right )}}{5 d} + \frac {2 a b^{2} \sin ^{5}{\left (c + d x \right )}}{5 d} + \frac {a b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {b^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {2 b^{3} \cos ^{5}{\left (c + d x \right )}}{15 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\relax (c )} + b \sin {\relax (c )}\right )^{3} \cos ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]

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))

________________________________________________________________________________________