\(\int \cos ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx\) [32]

Optimal result

Integrand size = 26, antiderivative size = 65 \[ \int \cos ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\frac {3 a x}{8}-\frac {b \cos ^4(c+d x)}{4 d}+\frac {3 a \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a \cos ^3(c+d x) \sin (c+d x)}{4 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3169, 2715, 8, 2645, 30} \[ \int \cos ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\frac {a \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3 a \sin (c+d x) \cos (c+d x)}{8 d}+\frac {3 a x}{8}-\frac {b \cos ^4(c+d x)}{4 d} \]

[In]

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

[Out]

(3*a*x)/8 - (b*Cos[c + d*x]^4)/(4*d) + (3*a*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a*Cos[c + d*x]^3*Sin[c + d*x])
/(4*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 2715

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] && Integ
erQ[2*n]

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.95 \[ \int \cos ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\frac {3 a (c+d x)}{8 d}-\frac {b \cos ^4(c+d x)}{4 d}+\frac {a \sin (2 (c+d x))}{4 d}+\frac {a \sin (4 (c+d x))}{32 d} \]

[In]

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

[Out]

(3*a*(c + d*x))/(8*d) - (b*Cos[c + d*x]^4)/(4*d) + (a*Sin[2*(c + d*x)])/(4*d) + (a*Sin[4*(c + d*x)])/(32*d)

Maple [A] (verified)

Time = 0.65 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.80

method	result	size
derivativedivides	\(\frac {a \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )-\frac {\cos \left (d x +c \right )^{4} b}{4}}{d}\)	\(52\)
default	\(\frac {a \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )-\frac {\cos \left (d x +c \right )^{4} b}{4}}{d}\)	\(52\)
parts	\(\frac {a \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}-\frac {b \cos \left (d x +c \right )^{4}}{4 d}\)	\(54\)
parallelrisch	\(\frac {12 a x d -b \cos \left (4 d x +4 c \right )+a \sin \left (4 d x +4 c \right )+8 a \sin \left (2 d x +2 c \right )-4 b \cos \left (2 d x +2 c \right )+5 b}{32 d}\)	\(62\)
risch	\(\frac {3 a x}{8}-\frac {b \cos \left (4 d x +4 c \right )}{32 d}+\frac {a \sin \left (4 d x +4 c \right )}{32 d}-\frac {b \cos \left (2 d x +2 c \right )}{8 d}+\frac {a \sin \left (2 d x +2 c \right )}{4 d}\)	\(66\)
norman	\(\frac {\frac {3 a x}{8}+\frac {5 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {3 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 d}+\frac {3 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 d}-\frac {5 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}+\frac {3 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}+\frac {9 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4}+\frac {3 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2}+\frac {3 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8}+\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}+\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}\)	\(182\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.78 \[ \int \cos ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=-\frac {2 \, b \cos \left (d x + c\right )^{4} - 3 \, a d x - {\left (2 \, a \cos \left (d x + c\right )^{3} + 3 \, a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, d} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 128 vs. \(2 (60) = 120\).

Time = 0.18 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.97 \[ \int \cos ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\begin {cases} \frac {3 a x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 a x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 a x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 a \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 a \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac {b \cos ^{4}{\left (c + d x \right )}}{4 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\left (c \right )} + b \sin {\left (c \right )}\right ) \cos ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.74 \[ \int \cos ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=-\frac {8 \, b \cos \left (d x + c\right )^{4} - {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a}{32 \, d} \]

[In]

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

[Out]

-1/32*(8*b*cos(d*x + c)^4 - (12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a)/d

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00 \[ \int \cos ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\frac {3}{8} \, a x - \frac {b \cos \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac {b \cos \left (2 \, d x + 2 \, c\right )}{8 \, d} + \frac {a \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {a \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} \]

[In]

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

[Out]

3/8*a*x - 1/32*b*cos(4*d*x + 4*c)/d - 1/8*b*cos(2*d*x + 2*c)/d + 1/32*a*sin(4*d*x + 4*c)/d + 1/4*a*sin(2*d*x +
2*c)/d

Mupad [B] (verification not implemented)

Time = 24.89 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.65 \[ \int \cos ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\frac {3\,a\,x}{8}+\frac {-\frac {5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}-\frac {3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {5\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^4} \]

[In]

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

[Out]

(3*a*x)/8 + ((5*a*tan(c/2 + (d*x)/2))/4 - (3*a*tan(c/2 + (d*x)/2)^3)/4 + (3*a*tan(c/2 + (d*x)/2)^5)/4 - (5*a*t
an(c/2 + (d*x)/2)^7)/4 + 2*b*tan(c/2 + (d*x)/2)^2 + 2*b*tan(c/2 + (d*x)/2)^6)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^4)