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

Optimal. Leaf size=44 \[ -\frac{a \sin ^3(c+d x)}{3 d}+\frac{a \sin (c+d x)}{d}-\frac{b \cos ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.0646631, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3090, 2633, 2565, 30} \[ -\frac{a \sin ^3(c+d x)}{3 d}+\frac{a \sin (c+d x)}{d}-\frac{b \cos ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

\begin{align*} \int \cos ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx &=\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 \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d}-\frac{b \operatorname{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]  time = 0.0116904, size = 44, normalized size = 1. \[ -\frac{a \sin ^3(c+d x)}{3 d}+\frac{a \sin (c+d x)}{d}-\frac{b \cos ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.037, size = 36, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ({\frac{a \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}-{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}b}{3}} \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)),x)

[Out]

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

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Maxima [A]  time = 1.18483, size = 47, normalized size = 1.07 \begin{align*} -\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} \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)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.47403, size = 90, normalized size = 2.05 \begin{align*} -\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} \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)),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

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Sympy [A]  time = 0.513386, size = 63, normalized size = 1.43 \begin{align*} \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} \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)),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))

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Giac [A]  time = 1.13003, size = 74, normalized size = 1.68 \begin{align*} -\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} \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)),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