\(\int \cos ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx\) [60]

Optimal result

Integrand size = 28, antiderivative size = 140 \[ \int \cos ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=-\frac {b^3 \cos ^3(c+d x)}{3 d}-\frac {3 a^2 b \cos ^5(c+d x)}{5 d}+\frac {b^3 \cos ^5(c+d x)}{5 d}+\frac {a^3 \sin (c+d x)}{d}-\frac {2 a^3 \sin ^3(c+d x)}{3 d}+\frac {a b^2 \sin ^3(c+d x)}{d}+\frac {a^3 \sin ^5(c+d x)}{5 d}-\frac {3 a b^2 \sin ^5(c+d x)}{5 d} \]

[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] (verified)

Time = 0.19 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3169, 2713, 2645, 30, 2644, 14} \[ \int \cos ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=\frac {a^3 \sin ^5(c+d x)}{5 d}-\frac {2 a^3 \sin ^3(c+d x)}{3 d}+\frac {a^3 \sin (c+d x)}{d}-\frac {3 a^2 b \cos ^5(c+d x)}{5 d}-\frac {3 a b^2 \sin ^5(c+d x)}{5 d}+\frac {a b^2 \sin ^3(c+d x)}{d}+\frac {b^3 \cos ^5(c+d x)}{5 d}-\frac {b^3 \cos ^3(c+d x)}{3 d} \]

[In]

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

[Out]

-1/3*(b^3*Cos[c + d*x]^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 2644

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 2645

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 2713

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 3169

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*} \text {integral}& = \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 \text {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 ) \text {Subst}\left (\int x^4 \, dx,x,\cos (c+d x)\right )}{d}+\frac {\left (3 a b^2\right ) \text {Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\sin (c+d x)\right )}{d}-\frac {b^3 \text {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 ) \text {Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\sin (c+d x)\right )}{d}-\frac {b^3 \text {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] (verified)

Time = 1.22 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.86 \[ \int \cos ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=\frac {-9 a^2 b \cos ^5(c+d x)+15 a^3 \sin (c+d x)-5 a \left (2 a^2-3 b^2\right ) \sin ^3(c+d x)+3 a \left (a^2-3 b^2\right ) \sin ^5(c+d x)+b^3 \cos (c+d x) \left (-2+\frac {2}{\sqrt {\cos ^2(c+d x)}}-\sin ^2(c+d x)+3 \sin ^4(c+d x)\right )}{15 d} \]

[In]

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

[Out]

(-9*a^2*b*Cos[c + d*x]^5 + 15*a^3*Sin[c + d*x] - 5*a*(2*a^2 - 3*b^2)*Sin[c + d*x]^3 + 3*a*(a^2 - 3*b^2)*Sin[c
+ d*x]^5 + b^3*Cos[c + d*x]*(-2 + 2/Sqrt[Cos[c + d*x]^2] - Sin[c + d*x]^2 + 3*Sin[c + d*x]^4))/(15*d)

Maple [A] (verified)

Time = 1.11 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.79

method	result	size
parts	\(\frac {a^{3} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5 d}+\frac {b^{3} \left (\frac {\cos \left (d x +c \right )^{5}}{5}-\frac {\cos \left (d x +c \right )^{3}}{3}\right )}{d}+\frac {3 a \,b^{2} \left (-\frac {\sin \left (d x +c \right )^{5}}{5}+\frac {\sin \left (d x +c \right )^{3}}{3}\right )}{d}-\frac {3 a^{2} b \cos \left (d x +c \right )^{5}}{5 d}\)	\(111\)
derivativedivides	\(\frac {\frac {a^{3} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}-\frac {3 a^{2} b \cos \left (d x +c \right )^{5}}{5}+3 a \,b^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}{5}+\frac {\left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{15}\right )+b^{3} \left (-\frac {\sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{3}}{5}-\frac {2 \cos \left (d x +c \right )^{3}}{15}\right )}{d}\)	\(125\)
default	\(\frac {\frac {a^{3} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}-\frac {3 a^{2} b \cos \left (d x +c \right )^{5}}{5}+3 a \,b^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}{5}+\frac {\left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{15}\right )+b^{3} \left (-\frac {\sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{3}}{5}-\frac {2 \cos \left (d x +c \right )^{3}}{15}\right )}{d}\)	\(125\)
risch	\(-\frac {3 a^{2} b \cos \left (d x +c \right )}{8 d}-\frac {b^{3} \cos \left (d x +c \right )}{8 d}+\frac {5 a^{3} \sin \left (d x +c \right )}{8 d}+\frac {3 a \,b^{2} \sin \left (d x +c \right )}{8 d}-\frac {3 b \cos \left (5 d x +5 c \right ) a^{2}}{80 d}+\frac {b^{3} \cos \left (5 d x +5 c \right )}{80 d}+\frac {a^{3} \sin \left (5 d x +5 c \right )}{80 d}-\frac {3 a \sin \left (5 d x +5 c \right ) b^{2}}{80 d}-\frac {3 b \cos \left (3 d x +3 c \right ) a^{2}}{16 d}-\frac {b^{3} \cos \left (3 d x +3 c \right )}{48 d}+\frac {5 a^{3} \sin \left (3 d x +3 c \right )}{48 d}-\frac {a \sin \left (3 d x +3 c \right ) b^{2}}{16 d}\)	\(200\)
parallelrisch	\(\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9} a^{3}-6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8} a^{2} b +\frac {8 \left (a^{3}+3 a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3}-4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} b^{3}+\frac {4 \left (29 a^{3}-12 a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15}+\frac {4 \left (-9 a^{2} b +b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3}+\frac {8 \left (a^{3}+3 a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b^{3}}{3}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3}-\frac {6 a^{2} b}{5}-\frac {4 b^{3}}{15}}{d \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}\)	\(206\)
norman	\(\frac {-\frac {18 a^{2} b +4 b^{3}}{15 d}+\frac {2 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}-\frac {4 b^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}-\frac {4 b^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3 d}-\frac {6 a^{2} b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}-\frac {2 \left (18 a^{2} b -2 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3 d}+\frac {8 a \left (a^{2}+3 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {8 a \left (a^{2}+3 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}+\frac {4 a \left (29 a^{2}-12 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}\)	\(237\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.73 \[ \int \cos ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=-\frac {5 \, b^{3} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{5} - {\left (3 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + 8 \, a^{3} + 6 \, a b^{2} + {\left (4 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{15 \, d} \]

[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

Sympy [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.30 \[ \int \cos ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=\begin {cases} \frac {8 a^{3} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {a^{3} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {3 a^{2} b \cos ^{5}{\left (c + d x \right )}}{5 d} + \frac {2 a b^{2} \sin ^{5}{\left (c + d x \right )}}{5 d} + \frac {a b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {b^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {2 b^{3} \cos ^{5}{\left (c + d x \right )}}{15 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\left (c \right )} + b \sin {\left (c \right )}\right )^{3} \cos ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.76 \[ \int \cos ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=-\frac {9 \, a^{2} b \cos \left (d x + c\right )^{5} - {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{3} + 3 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 5 \, \sin \left (d x + c\right )^{3}\right )} a b^{2} - {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} b^{3}}{15 \, d} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.04 \[ \int \cos ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=-\frac {{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac {{\left (9 \, a^{2} b + b^{3}\right )} \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac {{\left (3 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )}{8 \, d} + \frac {{\left (a^{3} - 3 \, a b^{2}\right )} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {{\left (5 \, a^{3} - 3 \, a b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (5 \, a^{3} + 3 \, a b^{2}\right )} \sin \left (d x + c\right )}{8 \, d} \]

[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

Mupad [B] (verification not implemented)

Time = 22.84 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.05 \[ \int \cos ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=\frac {2\,\left (\frac {3\,\sin \left (c+d\,x\right )\,a^3\,{\cos \left (c+d\,x\right )}^4}{2}+2\,\sin \left (c+d\,x\right )\,a^3\,{\cos \left (c+d\,x\right )}^2+4\,\sin \left (c+d\,x\right )\,a^3-\frac {9\,a^2\,b\,{\cos \left (c+d\,x\right )}^5}{2}-\frac {9\,\sin \left (c+d\,x\right )\,a\,b^2\,{\cos \left (c+d\,x\right )}^4}{2}+\frac {3\,\sin \left (c+d\,x\right )\,a\,b^2\,{\cos \left (c+d\,x\right )}^2}{2}+3\,\sin \left (c+d\,x\right )\,a\,b^2+\frac {3\,b^3\,{\cos \left (c+d\,x\right )}^5}{2}-\frac {5\,b^3\,{\cos \left (c+d\,x\right )}^3}{2}\right )}{15\,d} \]

[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)