\(\int \cos ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx\) [33]

Optimal result

Integrand size = 26, antiderivative size = 44 \[ \int \cos ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=-\frac {b \cos ^3(c+d x)}{3 d}+\frac {a \sin (c+d x)}{d}-\frac {a \sin ^3(c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3169, 2713, 2645, 30} \[ \int \cos ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=-\frac {a \sin ^3(c+d x)}{3 d}+\frac {a \sin (c+d x)}{d}-\frac {b \cos ^3(c+d x)}{3 d} \]

[In]

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

[Out]

-1/3*(b*Cos[c + d*x]^3)/d + (a*Sin[c + d*x])/d - (a*Sin[c + d*x]^3)/(3*d)

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00 \[ \int \cos ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=-\frac {b \cos ^3(c+d x)}{3 d}+\frac {a \sin (c+d x)}{d}-\frac {a \sin ^3(c+d x)}{3 d} \]

[In]

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

[Out]

-1/3*(b*Cos[c + d*x]^3)/d + (a*Sin[c + d*x])/d - (a*Sin[c + d*x]^3)/(3*d)

Maple [A] (verified)

Time = 0.61 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.82

method	result	size
derivativedivides	\(\frac {\frac {a \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}-\frac {\cos \left (d x +c \right )^{3} b}{3}}{d}\)	\(36\)
default	\(\frac {\frac {a \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}-\frac {\cos \left (d x +c \right )^{3} b}{3}}{d}\)	\(36\)
parts	\(\frac {a \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3 d}-\frac {b \cos \left (d x +c \right )^{3}}{3 d}\)	\(38\)
risch	\(-\frac {b \cos \left (d x +c \right )}{4 d}+\frac {3 a \sin \left (d x +c \right )}{4 d}-\frac {b \cos \left (3 d x +3 c \right )}{12 d}+\frac {a \sin \left (3 d x +3 c \right )}{12 d}\)	\(56\)
parallelrisch	\(\frac {6 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +4 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+6 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{3 d \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}\)	\(79\)
norman	\(\frac {-\frac {2 b}{3 d}+\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {4 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}\)	\(90\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.86 \[ \int \cos ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=-\frac {b \cos \left (d x + c\right )^{3} - {\left (a \cos \left (d x + c\right )^{2} + 2 \, a\right )} \sin \left (d x + c\right )}{3 \, d} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.43 \[ \int \cos ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\begin {cases} \frac {2 a \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {a \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {b \cos ^{3}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\left (c \right )} + b \sin {\left (c \right )}\right ) \cos ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.80 \[ \int \cos ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=-\frac {b \cos \left (d x + c\right )^{3} + {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a}{3 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.25 \[ \int \cos ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=-\frac {b \cos \left (3 \, d x + 3 \, c\right )}{12 \, d} - \frac {b \cos \left (d x + c\right )}{4 \, d} + \frac {a \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {3 \, a \sin \left (d x + c\right )}{4 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 21.31 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.07 \[ \int \cos ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\frac {2\,a\,\sin \left (c+d\,x\right )}{3\,d}-\frac {b\,{\cos \left (c+d\,x\right )}^3}{3\,d}+\frac {a\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d} \]

[In]

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

[Out]

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