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

Optimal. Leaf size=126 \[ \frac{a^2 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{3 a^2 \sin (c+d x) \cos (c+d x)}{8 d}+\frac{3 a^2 x}{8}-\frac{a b \cos ^4(c+d x)}{2 d}-\frac{b^2 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{b^2 \sin (c+d x) \cos (c+d x)}{8 d}+\frac{b^2 x}{8} \]

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.134596, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3090, 2635, 8, 2565, 30, 2568} \[ \frac{a^2 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{3 a^2 \sin (c+d x) \cos (c+d x)}{8 d}+\frac{3 a^2 x}{8}-\frac{a b \cos ^4(c+d x)}{2 d}-\frac{b^2 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{b^2 \sin (c+d x) \cos (c+d x)}{8 d}+\frac{b^2 x}{8} \]

Antiderivative was successfully verified.

[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 3090

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]

Rule 2635

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] && Integer
Q[2*n]

Rule 8

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

Rule 2565

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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2568

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]

Rubi steps

\begin{align*} \int \cos ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx &=\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) \operatorname{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]  time = 0.225099, size = 98, normalized size = 0.78 \[ \frac{\left (3 a^2+b^2\right ) (c+d x)}{8 d}+\frac{\left (a^2-b^2\right ) \sin (4 (c+d x))}{32 d}+\frac{a^2 \sin (2 (c+d x))}{4 d}-\frac{a b \cos (2 (c+d x))}{4 d}-\frac{a b \cos (4 (c+d x))}{16 d} \]

Antiderivative was successfully verified.

[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]  time = 0.058, size = 97, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ({b}^{2} \left ( -{\frac{\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4}}+{\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{8}}+{\frac{dx}{8}}+{\frac{c}{8}} \right ) -{\frac{ab \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{2}}+{a}^{2} \left ({\frac{\sin \left ( dx+c \right ) }{4} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.12365, size = 101, normalized size = 0.8 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

Fricas [A]  time = 0.482353, size = 170, normalized size = 1.35 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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 [A]  time = 1.36889, size = 260, normalized size = 2.06 \begin{align*} \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 \sin ^{4}{\left (c + d x \right )}}{2 d} + \frac{a b \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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*sin(c + d*x)**
4/(2*d) + a*b*sin(c + d*x)**2*cos(c + d*x)**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*sin(c))**2*cos(c)**2, True))

________________________________________________________________________________________

Giac [A]  time = 1.17283, size = 115, normalized size = 0.91 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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