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

Optimal. Leaf size=67 \[ -\frac{a^2 \sin ^3(c+d x)}{3 d}+\frac{a^2 \sin (c+d x)}{d}-\frac{2 a b \cos ^3(c+d x)}{3 d}+\frac{b^2 \sin ^3(c+d x)}{3 d} \]

[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)

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Rubi [A]  time = 0.0908333, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {3090, 2633, 2565, 30, 2564} \[ -\frac{a^2 \sin ^3(c+d x)}{3 d}+\frac{a^2 \sin (c+d x)}{d}-\frac{2 a b \cos ^3(c+d x)}{3 d}+\frac{b^2 \sin ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[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 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 2633

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 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 2564

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])

Rubi steps

\begin{align*} \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx &=\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 \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d}-\frac{(2 a b) \operatorname{Subst}\left (\int x^2 \, dx,x,\cos (c+d x)\right )}{d}+\frac{b^2 \operatorname{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]  time = 0.388145, size = 64, normalized size = 0.96 \[ \frac{\sin (c+d x) \left (\left (a^2-b^2\right ) \cos (2 (c+d x))+5 a^2+b^2\right )-3 a b \cos (c+d x)-a b \cos (3 (c+d x))}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a*b*Cos[c + d*x] - a*b*Cos[3*(c + d*x)] + (5*a^2 + b^2 + (a^2 - b^2)*Cos[2*(c + d*x)])*Sin[c + d*x])/(6*d)

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Maple [A]  time = 0.051, size = 52, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ({\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3}}-{\frac{2\,ab \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{3}}+{\frac{{a}^{2} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.124, size = 70, normalized size = 1.04 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 0.472204, size = 120, normalized size = 1.79 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [A]  time = 0.545085, size = 85, normalized size = 1.27 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Giac [A]  time = 1.15405, size = 99, normalized size = 1.48 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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