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

Optimal. Leaf size=54 \[ \frac{a \sin (c+d x) \cos (c+d x)}{2 d}+\frac{a x}{2}-\frac{b \sin ^3(c+d x)}{3 d}+\frac{b \sin (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.047293, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {2748, 2635, 8, 2633} \[ \frac{a \sin (c+d x) \cos (c+d x)}{2 d}+\frac{a x}{2}-\frac{b \sin ^3(c+d x)}{3 d}+\frac{b \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

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 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]

Rubi steps

\begin{align*} \int \cos ^2(c+d x) (a+b \cos (c+d x)) \, dx &=a \int \cos ^2(c+d x) \, dx+b \int \cos ^3(c+d x) \, dx\\ &=\frac{a \cos (c+d x) \sin (c+d x)}{2 d}+\frac{1}{2} a \int 1 \, dx-\frac{b \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=\frac{a x}{2}+\frac{b \sin (c+d x)}{d}+\frac{a \cos (c+d x) \sin (c+d x)}{2 d}-\frac{b \sin ^3(c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.064984, size = 57, normalized size = 1.06 \[ \frac{a (c+d x)}{2 d}+\frac{a \sin (2 (c+d x))}{4 d}-\frac{b \sin ^3(c+d x)}{3 d}+\frac{b \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.033, size = 49, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({\frac{b \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+a \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.945717, size = 62, normalized size = 1.15 \begin{align*} \frac{3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a - 4 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} b}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.90571, size = 105, normalized size = 1.94 \begin{align*} \frac{3 \, a d x +{\left (2 \, b \cos \left (d x + c\right )^{2} + 3 \, a \cos \left (d x + c\right ) + 4 \, b\right )} \sin \left (d x + c\right )}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.601227, size = 92, normalized size = 1.7 \begin{align*} \begin{cases} \frac{a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{a x \cos ^{2}{\left (c + d x \right )}}{2} + \frac{a \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} + \frac{2 b \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{b \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (a + b \cos{\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+b*cos(d*x+c)),x)

[Out]

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

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Giac [A]  time = 1.30865, size = 63, normalized size = 1.17 \begin{align*} \frac{1}{2} \, a x + \frac{b \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac{a \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac{3 \, b \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+b*cos(d*x+c)),x, algorithm="giac")

[Out]

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