\(\int \cos ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx\) [46]

Optimal result

Integrand size = 28, antiderivative size = 126 \[ \int \cos ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {3 a^2 x}{8}+\frac {b^2 x}{8}-\frac {a b \cos ^4(c+d x)}{2 d}+\frac {3 a^2 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {b^2 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {b^2 \cos ^3(c+d x) \sin (c+d x)}{4 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3169, 2715, 8, 2645, 30, 2648} \[ \int \cos ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3 a^2 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {3 a^2 x}{8}-\frac {a b \cos ^4(c+d x)}{2 d}-\frac {b^2 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {b^2 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {b^2 x}{8} \]

[In]

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

[Out]

(3*a^2*x)/8 + (b^2*x)/8 - (a*b*Cos[c + d*x]^4)/(2*d) + (3*a^2*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (b^2*Cos[c +
d*x]*Sin[c + d*x])/(8*d) + (a^2*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) - (b^2*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 2648

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-a)*(b*Cos[e
+ f*x])^(n + 1)*((a*Sin[e + f*x])^(m - 1)/(b*f*(m + n))), x] + Dist[a^2*((m - 1)/(m + n)), Int[(b*Cos[e + f*x
])^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[
2*m, 2*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^2 \cos ^4(c+d x)+2 a b \cos ^3(c+d x) \sin (c+d x)+b^2 \cos ^2(c+d x) \sin ^2(c+d x)\right ) \, dx \\ & = a^2 \int \cos ^4(c+d x) \, dx+(2 a b) \int \cos ^3(c+d x) \sin (c+d x) \, dx+b^2 \int \cos ^2(c+d x) \sin ^2(c+d x) \, dx \\ & = \frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {b^2 \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {1}{4} \left (3 a^2\right ) \int \cos ^2(c+d x) \, dx+\frac {1}{4} b^2 \int \cos ^2(c+d x) \, dx-\frac {(2 a b) \text {Subst}\left (\int x^3 \, dx,x,\cos (c+d x)\right )}{d} \\ & = -\frac {a b \cos ^4(c+d x)}{2 d}+\frac {3 a^2 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {b^2 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {b^2 \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {1}{8} \left (3 a^2\right ) \int 1 \, dx+\frac {1}{8} b^2 \int 1 \, dx \\ & = \frac {3 a^2 x}{8}+\frac {b^2 x}{8}-\frac {a b \cos ^4(c+d x)}{2 d}+\frac {3 a^2 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {b^2 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {b^2 \cos ^3(c+d x) \sin (c+d x)}{4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.96 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.78 \[ \int \cos ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {\left (3 a^2+b^2\right ) (c+d x)}{8 d}-\frac {a b \cos (2 (c+d x))}{4 d}-\frac {a b \cos (4 (c+d x))}{16 d}+\frac {a^2 \sin (2 (c+d x))}{4 d}+\frac {\left (a^2-b^2\right ) \sin (4 (c+d x))}{32 d} \]

[In]

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

[Out]

((3*a^2 + b^2)*(c + d*x))/(8*d) - (a*b*Cos[2*(c + d*x)])/(4*d) - (a*b*Cos[4*(c + d*x)])/(16*d) + (a^2*Sin[2*(c
+ d*x)])/(4*d) + ((a^2 - b^2)*Sin[4*(c + d*x)])/(32*d)

Maple [A] (verified)

Time = 0.70 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.67

method	result	size
parallelrisch	\(\frac {\left (a^{2}-b^{2}\right ) \sin \left (4 d x +4 c \right )+12 a^{2} x d +4 b^{2} d x +8 \sin \left (2 d x +2 c \right ) a^{2}-8 a b \cos \left (2 d x +2 c \right )-2 a b \cos \left (4 d x +4 c \right )+10 a b}{32 d}\)	\(84\)
derivativedivides	\(\frac {a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )-\frac {a b \cos \left (d x +c \right )^{4}}{2}+b^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}{4}+\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )}{d}\)	\(97\)
default	\(\frac {a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )-\frac {a b \cos \left (d x +c \right )^{4}}{2}+b^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}{4}+\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )}{d}\)	\(97\)
risch	\(\frac {3 a^{2} x}{8}+\frac {x \,b^{2}}{8}-\frac {a b \cos \left (4 d x +4 c \right )}{16 d}+\frac {\sin \left (4 d x +4 c \right ) a^{2}}{32 d}-\frac {\sin \left (4 d x +4 c \right ) b^{2}}{32 d}-\frac {a b \cos \left (2 d x +2 c \right )}{4 d}+\frac {\sin \left (2 d x +2 c \right ) a^{2}}{4 d}\)	\(97\)
parts	\(\frac {a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}+\frac {b^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}{4}+\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )}{d}-\frac {a b \cos \left (d x +c \right )^{4}}{2 d}\)	\(102\)
norman	\(\frac {\left (\frac {3 a^{2}}{8}+\frac {b^{2}}{8}\right ) x +\left (\frac {3 a^{2}}{2}+\frac {b^{2}}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (\frac {3 a^{2}}{2}+\frac {b^{2}}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (\frac {3 a^{2}}{8}+\frac {b^{2}}{8}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (\frac {9 a^{2}}{4}+\frac {3 b^{2}}{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\frac {\left (3 a^{2}-7 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 d}+\frac {\left (3 a^{2}-7 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 d}+\frac {\left (5 a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {\left (5 a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}+\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}+\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}\)	\(269\)

[In]

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

[Out]

1/32*((a^2-b^2)*sin(4*d*x+4*c)+12*a^2*x*d+4*b^2*d*x+8*sin(2*d*x+2*c)*a^2-8*a*b*cos(2*d*x+2*c)-2*a*b*cos(4*d*x+
4*c)+10*a*b)/d

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.60 \[ \int \cos ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=-\frac {4 \, a b \cos \left (d x + c\right )^{4} - {\left (3 \, a^{2} + b^{2}\right )} d x - {\left (2 \, {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, d} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 238 vs. \(2 (116) = 232\).

Time = 0.20 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.89 \[ \int \cos ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\begin {cases} \frac {3 a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac {a b \cos ^{4}{\left (c + d x \right )}}{2 d} + \frac {b^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {b^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {b^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {b^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\left (c \right )} + b \sin {\left (c \right )}\right )^{2} \cos ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.60 \[ \int \cos ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=-\frac {16 \, a b \cos \left (d x + c\right )^{4} - {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} - {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} b^{2}}{32 \, d} \]

[In]

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

[Out]

-1/32*(16*a*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^2 - (4*d*x + 4*c - si
n(4*d*x + 4*c))*b^2)/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.67 \[ \int \cos ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {1}{8} \, {\left (3 \, a^{2} + b^{2}\right )} x - \frac {a b \cos \left (4 \, d x + 4 \, c\right )}{16 \, d} - \frac {a b \cos \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {a^{2} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (a^{2} - b^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} \]

[In]

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

[Out]

1/8*(3*a^2 + b^2)*x - 1/16*a*b*cos(4*d*x + 4*c)/d - 1/4*a*b*cos(2*d*x + 2*c)/d + 1/4*a^2*sin(2*d*x + 2*c)/d +
1/32*(a^2 - b^2)*sin(4*d*x + 4*c)/d

Mupad [B] (verification not implemented)

Time = 20.90 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.71 \[ \int \cos ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {4\,a^2\,\sin \left (2\,c+2\,d\,x\right )+\frac {a^2\,\sin \left (4\,c+4\,d\,x\right )}{2}-\frac {b^2\,\sin \left (4\,c+4\,d\,x\right )}{2}+2\,a\,b\,{\sin \left (2\,c+2\,d\,x\right )}^2+8\,a\,b\,{\sin \left (c+d\,x\right )}^2+6\,a^2\,d\,x+2\,b^2\,d\,x}{16\,d} \]

[In]

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

[Out]

(4*a^2*sin(2*c + 2*d*x) + (a^2*sin(4*c + 4*d*x))/2 - (b^2*sin(4*c + 4*d*x))/2 + 2*a*b*sin(2*c + 2*d*x)^2 + 8*a
*b*sin(c + d*x)^2 + 6*a^2*d*x + 2*b^2*d*x)/(16*d)