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

Optimal. Leaf size=57 \[ -\frac{5 \sin (c+d x)}{3 a^2 d (\cos (c+d x)+1)}+\frac{x}{a^2}+\frac{\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[Out]

x/a^2 - (5*Sin[c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])) + Sin[c + d*x]/(3*d*(a + a*Cos[c + d*x])^2)

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Rubi [A]  time = 0.0845487, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2758, 2735, 2648} \[ -\frac{5 \sin (c+d x)}{3 a^2 d (\cos (c+d x)+1)}+\frac{x}{a^2}+\frac{\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/a^2 - (5*Sin[c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])) + Sin[c + d*x]/(3*d*(a + a*Cos[c + d*x])^2)

Rule 2758

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

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 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]

Rubi steps

\begin{align*} \int \frac{\cos ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx &=\frac{\sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \frac{-2 a+3 a \cos (c+d x)}{a+a \cos (c+d x)} \, dx}{3 a^2}\\ &=\frac{x}{a^2}+\frac{\sin (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac{5 \int \frac{1}{a+a \cos (c+d x)} \, dx}{3 a}\\ &=\frac{x}{a^2}+\frac{\sin (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac{5 \sin (c+d x)}{3 d \left (a^2+a^2 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.228174, size = 105, normalized size = 1.84 \[ \frac{2 \cos \left (\frac{1}{2} (c+d x)\right ) \left (6 d x \cos ^3\left (\frac{1}{2} (c+d x)\right )+\tan \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right )+\sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right )-10 \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \cos ^2\left (\frac{1}{2} (c+d x)\right )\right )}{3 a^2 d (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[(c + d*x)/2]*(6*d*x*Cos[(c + d*x)/2]^3 + Sec[c/2]*Sin[(d*x)/2] - 10*Cos[(c + d*x)/2]^2*Sec[c/2]*Sin[(d*
x)/2] + Cos[(c + d*x)/2]*Tan[c/2]))/(3*a^2*d*(1 + Cos[c + d*x])^2)

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Maple [A]  time = 0.04, size = 56, normalized size = 1. \begin{align*}{\frac{1}{6\,{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{3}{2\,{a}^{2}d}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{{a}^{2}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/d/a^2*tan(1/2*d*x+1/2*c)^3-3/2/d/a^2*tan(1/2*d*x+1/2*c)+2/d/a^2*arctan(tan(1/2*d*x+1/2*c))

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Maxima [A]  time = 1.74071, size = 97, normalized size = 1.7 \begin{align*} -\frac{\frac{\frac{9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{12 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*((9*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 12*arctan(sin(d*x + c)/(
cos(d*x + c) + 1))/a^2)/d

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Fricas [A]  time = 1.80739, size = 198, normalized size = 3.47 \begin{align*} \frac{3 \, d x \cos \left (d x + c\right )^{2} + 6 \, d x \cos \left (d x + c\right ) + 3 \, d x -{\left (5 \, \cos \left (d x + c\right ) + 4\right )} \sin \left (d x + c\right )}{3 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*d*x*cos(d*x + c)^2 + 6*d*x*cos(d*x + c) + 3*d*x - (5*cos(d*x + c) + 4)*sin(d*x + c))/(a^2*d*cos(d*x + c
)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

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Sympy [A]  time = 3.12045, size = 56, normalized size = 0.98 \begin{align*} \begin{cases} \frac{x}{a^{2}} + \frac{\tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{6 a^{2} d} - \frac{3 \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{2 a^{2} d} & \text{for}\: d \neq 0 \\\frac{x \cos ^{2}{\left (c \right )}}{\left (a \cos{\left (c \right )} + a\right )^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x/a**2 + tan(c/2 + d*x/2)**3/(6*a**2*d) - 3*tan(c/2 + d*x/2)/(2*a**2*d), Ne(d, 0)), (x*cos(c)**2/(a
*cos(c) + a)**2, True))

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Giac [A]  time = 1.40093, size = 68, normalized size = 1.19 \begin{align*} \frac{\frac{6 \,{\left (d x + c\right )}}{a^{2}} + \frac{a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 9 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{6}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*(d*x + c)/a^2 + (a^4*tan(1/2*d*x + 1/2*c)^3 - 9*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d