3.46 \(\int \frac {\cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx\)

Optimal. Leaf size=43 \[ \frac {\sin (c+d x)}{a d}+\frac {\sin (c+d x)}{a d (\cos (c+d x)+1)}-\frac {x}{a} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2746, 12, 2735, 2648} \[ \frac {\sin (c+d x)}{a d}+\frac {\sin (c+d x)}{a d (\cos (c+d x)+1)}-\frac {x}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/a) + Sin[c + d*x]/(a*d) + Sin[c + d*x]/(a*d*(1 + Cos[c + d*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2746

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b^2
*Cos[e + f*x])/(d*f), x] + Dist[1/d, Int[Simp[a^2*d - b*(b*c - 2*a*d)*Sin[e + f*x], x]/(c + d*Sin[e + f*x]), x
], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rubi steps

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

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Mathematica [B]  time = 0.20, size = 89, normalized size = 2.07 \[ \frac {\sec \left (\frac {c}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (c+\frac {d x}{2}\right )+\sin \left (c+\frac {3 d x}{2}\right )+\sin \left (2 c+\frac {3 d x}{2}\right )-2 d x \cos \left (c+\frac {d x}{2}\right )+5 \sin \left (\frac {d x}{2}\right )-2 d x \cos \left (\frac {d x}{2}\right )\right )}{4 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]*(-2*d*x*Cos[(d*x)/2] - 2*d*x*Cos[c + (d*x)/2] + 5*Sin[(d*x)/2] + Sin[c + (d*x)/2] +
Sin[c + (3*d*x)/2] + Sin[2*c + (3*d*x)/2]))/(4*a*d)

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fricas [A]  time = 1.00, size = 46, normalized size = 1.07 \[ -\frac {d x \cos \left (d x + c\right ) + d x - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right )}{a d \cos \left (d x + c\right ) + a d} \]

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]

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

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giac [A]  time = 0.45, size = 58, normalized size = 1.35 \[ -\frac {\frac {d x + c}{a} - \frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a} - \frac {2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a}}{d} \]

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]

-((d*x + c)/a - tan(1/2*d*x + 1/2*c)/a - 2*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 + 1)*a))/d

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maple [A]  time = 0.05, size = 68, normalized size = 1.58 \[ \frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d} \]

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/a/d*tan(1/2*d*x+1/2*c)+2/a/d*tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x+1/2*c)^2)-2/a/d*arctan(tan(1/2*d*x+1/2*c))

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maxima [B]  time = 2.25, size = 92, normalized size = 2.14 \[ -\frac {\frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {2 \, \sin \left (d x + c\right )}{{\left (a + \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}}{d} \]

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]

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

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mupad [B]  time = 0.40, size = 66, normalized size = 1.53 \[ \frac {2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (-c-d\,x\right )\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a\,d\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )} \]

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]

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

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sympy [A]  time = 1.41, size = 129, normalized size = 3.00 \[ \begin {cases} - \frac {d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} - \frac {d x}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} + \frac {\tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} + \frac {3 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{2}{\relax (c )}}{a \cos {\relax (c )} + a} & \text {otherwise} \end {cases} \]

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((-d*x*tan(c/2 + d*x/2)**2/(a*d*tan(c/2 + d*x/2)**2 + a*d) - d*x/(a*d*tan(c/2 + d*x/2)**2 + a*d) + ta
n(c/2 + d*x/2)**3/(a*d*tan(c/2 + d*x/2)**2 + a*d) + 3*tan(c/2 + d*x/2)/(a*d*tan(c/2 + d*x/2)**2 + a*d), Ne(d,
0)), (x*cos(c)**2/(a*cos(c) + a), True))

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