\(\int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx\) [34]

Optimal result

Integrand size = 24, antiderivative size = 43 \[ \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\frac {a x}{2}+\frac {a \cos (c+d x) \sin (c+d x)}{2 d}+\frac {b \sin ^2(c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {3169, 2715, 8, 2644, 30} \[ \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\frac {a \sin (c+d x) \cos (c+d x)}{2 d}+\frac {a x}{2}+\frac {b \sin ^2(c+d x)}{2 d} \]

[In]

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

[Out]

(a*x)/2 + (a*Cos[c + d*x]*Sin[c + d*x])/(2*d) + (b*Sin[c + d*x]^2)/(2*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 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 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 ^2(c+d x)+b \cos (c+d x) \sin (c+d x)\right ) \, dx \\ & = a \int \cos ^2(c+d x) \, dx+b \int \cos (c+d x) \sin (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}(\int x \, dx,x,\sin (c+d x))}{d} \\ & = \frac {a x}{2}+\frac {a \cos (c+d x) \sin (c+d x)}{2 d}+\frac {b \sin ^2(c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.07 \[ \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\frac {a (c+d x)}{2 d}-\frac {b \cos ^2(c+d x)}{2 d}+\frac {a \sin (2 (c+d x))}{4 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84

method	result	size
risch	\(\frac {a x}{2}-\frac {b \cos \left (2 d x +2 c \right )}{4 d}+\frac {a \sin \left (2 d x +2 c \right )}{4 d}\)	\(36\)
parallelrisch	\(\frac {2 a x d -b \cos \left (2 d x +2 c \right )+a \sin \left (2 d x +2 c \right )+b}{4 d}\)	\(36\)
derivativedivides	\(\frac {a \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-\frac {\cos \left (d x +c \right )^{2} b}{2}}{d}\)	\(41\)
default	\(\frac {a \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-\frac {\cos \left (d x +c \right )^{2} b}{2}}{d}\)	\(41\)
parts	\(\frac {a \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {b \sin \left (d x +c \right )^{2}}{2 d}\)	\(43\)
norman	\(\frac {\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\frac {a x}{2}-\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}+\frac {a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2}+\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}\)	\(99\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\frac {a d x - b \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right ) \sin \left (d x + c\right )}{2 \, d} \]

[In]

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

[Out]

1/2*(a*d*x - b*cos(d*x + c)^2 + 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. 73 vs. \(2 (36) = 72\).

Time = 0.10 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.70 \[ \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\begin {cases} \frac {a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {a x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {a \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} - \frac {b \cos ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\left (c \right )} + b \sin {\left (c \right )}\right ) \cos {\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.86 \[ \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=-\frac {2 \, b \cos \left (d x + c\right )^{2} - {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a}{4 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\frac {1}{2} \, a x - \frac {b \cos \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {a \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 21.44 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\frac {a\,x}{2}-\frac {b\,\cos \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \]

[In]

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

[Out]

(a*x)/2 - (b*cos(2*c + 2*d*x))/(4*d) + (a*sin(2*c + 2*d*x))/(4*d)