\(\int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx\) [47]

Optimal result

Integrand size = 26, antiderivative size = 67 \[ \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=-\frac {2 a b \cos ^3(c+d x)}{3 d}+\frac {a^2 \sin (c+d x)}{d}-\frac {a^2 \sin ^3(c+d x)}{3 d}+\frac {b^2 \sin ^3(c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3169, 2713, 2645, 30, 2644} \[ \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=-\frac {a^2 \sin ^3(c+d x)}{3 d}+\frac {a^2 \sin (c+d x)}{d}-\frac {2 a b \cos ^3(c+d x)}{3 d}+\frac {b^2 \sin ^3(c+d x)}{3 d} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00 \[ \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=-\frac {2 a b \cos ^3(c+d x)}{3 d}+\frac {a^2 \sin (c+d x)}{d}-\frac {a^2 \sin ^3(c+d x)}{3 d}+\frac {b^2 \sin ^3(c+d x)}{3 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.78

method	result	size
derivativedivides	\(\frac {\frac {a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}-\frac {2 a b \cos \left (d x +c \right )^{3}}{3}+\frac {b^{2} \sin \left (d x +c \right )^{3}}{3}}{d}\)	\(52\)
default	\(\frac {\frac {a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}-\frac {2 a b \cos \left (d x +c \right )^{3}}{3}+\frac {b^{2} \sin \left (d x +c \right )^{3}}{3}}{d}\)	\(52\)
parts	\(\frac {a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3 d}+\frac {b^{2} \sin \left (d x +c \right )^{3}}{3 d}-\frac {2 a b \cos \left (d x +c \right )^{3}}{3 d}\)	\(57\)
risch	\(-\frac {a b \cos \left (d x +c \right )}{2 d}+\frac {3 a^{2} \sin \left (d x +c \right )}{4 d}+\frac {b^{2} \sin \left (d x +c \right )}{4 d}-\frac {a b \cos \left (3 d x +3 c \right )}{6 d}+\frac {\sin \left (3 d x +3 c \right ) a^{2}}{12 d}-\frac {\sin \left (3 d x +3 c \right ) b^{2}}{12 d}\)	\(93\)
parallelrisch	\(\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} a^{2}-4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a b +\frac {4 \left (a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {4 a b}{3}}{d \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}\)	\(93\)
norman	\(\frac {-\frac {4 a b}{3 d}+\frac {2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {4 \left (a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}-\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}\)	\(104\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.79 \[ \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=-\frac {2 \, a b \cos \left (d x + c\right )^{3} - {\left ({\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, a^{2} + b^{2}\right )} \sin \left (d x + c\right )}{3 \, d} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.27 \[ \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\begin {cases} \frac {2 a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {2 a b \cos ^{3}{\left (c + d x \right )}}{3 d} + \frac {b^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\left (c \right )} + b \sin {\left (c \right )}\right )^{2} \cos {\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.78 \[ \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=-\frac {2 \, a b \cos \left (d x + c\right )^{3} - b^{2} \sin \left (d x + c\right )^{3} + {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{2}}{3 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.09 \[ \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=-\frac {a b \cos \left (3 \, d x + 3 \, c\right )}{6 \, d} - \frac {a b \cos \left (d x + c\right )}{2 \, d} + \frac {{\left (a^{2} - b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {{\left (3 \, a^{2} + b^{2}\right )} \sin \left (d x + c\right )}{4 \, d} \]

[In]

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

[Out]

-1/6*a*b*cos(3*d*x + 3*c)/d - 1/2*a*b*cos(d*x + c)/d + 1/12*(a^2 - b^2)*sin(3*d*x + 3*c)/d + 1/4*(3*a^2 + b^2)
*sin(d*x + c)/d

Mupad [B] (verification not implemented)

Time = 21.75 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.15 \[ \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {2\,\left (\frac {\sin \left (c+d\,x\right )\,a^2\,{\cos \left (c+d\,x\right )}^2}{2}+\sin \left (c+d\,x\right )\,a^2-a\,b\,{\cos \left (c+d\,x\right )}^3-\frac {\sin \left (c+d\,x\right )\,b^2\,{\cos \left (c+d\,x\right )}^2}{2}+\frac {\sin \left (c+d\,x\right )\,b^2}{2}\right )}{3\,d} \]

[In]

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

[Out]

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