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

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

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2748, 2635, 8, 2633} \[ -\frac {a \sin ^3(c+d x)}{3 d}+\frac {a \sin (c+d x)}{d}+\frac {a \sin (c+d x) \cos (c+d x)}{2 d}+\frac {a x}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 57, normalized size = 1.06 \[ \frac {a (c+d x)}{2 d}-\frac {a \sin ^3(c+d x)}{3 d}+\frac {a \sin (c+d x)}{d}+\frac {a \sin (2 (c+d x))}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.81, size = 42, normalized size = 0.78 \[ \frac {3 \, a d x + {\left (2 \, a \cos \left (d x + c\right )^{2} + 3 \, a \cos \left (d x + c\right ) + 4 \, a\right )} \sin \left (d x + c\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.48, size = 47, normalized size = 0.87 \[ \frac {1}{2} \, a x + \frac {a \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {a \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {3 \, a \sin \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 49, normalized size = 0.91 \[ \frac {\frac {a \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(1/3*a*(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 = 0.30, size = 46, normalized size = 0.85 \[ -\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.38, size = 55, normalized size = 1.02 \[ \frac {a\,x}{2}+\frac {2\,a\,\sin \left (c+d\,x\right )}{3\,d}+\frac {a\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d}+\frac {a\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.45, size = 92, normalized size = 1.70 \[ \begin {cases} \frac {a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {a x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {2 a \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {a \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {a \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\relax (c )} + a\right ) \cos ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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