\(\int \cos ^2(c+d x) (a+b \cos (c+d x)) \, dx\) [408]

Optimal result

Integrand size = 19, antiderivative size = 54 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x)) \, dx=\frac {a x}{2}+\frac {b \sin (c+d x)}{d}+\frac {a \cos (c+d x) \sin (c+d x)}{2 d}-\frac {b \sin ^3(c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {2827, 2715, 8, 2713} \[ \int \cos ^2(c+d x) (a+b \cos (c+d x)) \, dx=\frac {a \sin (c+d x) \cos (c+d x)}{2 d}+\frac {a x}{2}-\frac {b \sin ^3(c+d x)}{3 d}+\frac {b \sin (c+d x)}{d} \]

[In]

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

[Out]

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

Rule 8

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

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 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 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.06 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x)) \, dx=\frac {a (c+d x)}{2 d}+\frac {b \sin (c+d x)}{d}-\frac {b \sin ^3(c+d x)}{3 d}+\frac {a \sin (2 (c+d x))}{4 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.81 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.81

method	result	size
parallelrisch	\(\frac {6 a x d +b \sin \left (3 d x +3 c \right )+3 \sin \left (2 d x +2 c \right ) a +9 b \sin \left (d x +c \right )}{12 d}\)	\(44\)
risch	\(\frac {a x}{2}+\frac {3 b \sin \left (d x +c \right )}{4 d}+\frac {b \sin \left (3 d x +3 c \right )}{12 d}+\frac {a \sin \left (2 d x +2 c \right )}{4 d}\)	\(48\)
derivativedivides	\(\frac {\frac {b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(49\)
default	\(\frac {\frac {b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(49\)
parts	\(\frac {a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}\)	\(51\)
norman	\(\frac {\frac {\left (a +2 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a x}{2}+\frac {3 a x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {3 a x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {a x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {4 b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {\left (a -2 b \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}\)	\(123\)

[In]

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

[Out]

1/12*(6*a*x*d+b*sin(3*d*x+3*c)+3*sin(2*d*x+2*c)*a+9*b*sin(d*x+c))/d

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.78 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x)) \, dx=\frac {3 \, a d x + {\left (2 \, b \cos \left (d x + c\right )^{2} + 3 \, a \cos \left (d x + c\right ) + 4 \, b\right )} \sin \left (d x + c\right )}{6 \, d} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.70 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x)) \, dx=\begin {cases} \frac {a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {a x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {a \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 b \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {b \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \cos {\left (c \right )}\right ) \cos ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**2*(a+b*cos(d*x+c)),x)

[Out]

Piecewise((a*x*sin(c + d*x)**2/2 + a*x*cos(c + d*x)**2/2 + a*sin(c + d*x)*cos(c + d*x)/(2*d) + 2*b*sin(c + d*x
)**3/(3*d) + b*sin(c + d*x)*cos(c + d*x)**2/d, Ne(d, 0)), (x*(a + b*cos(c))*cos(c)**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x)) \, dx=\frac {3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} b}{12 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.87 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x)) \, dx=\frac {1}{2} \, a x + \frac {b \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {a \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {3 \, b \sin \left (d x + c\right )}{4 \, d} \]

[In]

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

[Out]

1/2*a*x + 1/12*b*sin(3*d*x + 3*c)/d + 1/4*a*sin(2*d*x + 2*c)/d + 3/4*b*sin(d*x + c)/d

Mupad [B] (verification not implemented)

Time = 14.05 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.02 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x)) \, dx=\frac {a\,x}{2}+\frac {2\,b\,\sin \left (c+d\,x\right )}{3\,d}+\frac {a\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d}+\frac {b\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d} \]

[In]

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

[Out]

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