3.293 \(\int \frac{x \tan ^{-1}(a x)^2}{(c+a^2 c x^2)^2} \, dx\)

Optimal. Leaf size=91 \[ \frac{1}{4 a^2 c^2 \left (a^2 x^2+1\right )}-\frac{\tan ^{-1}(a x)^2}{2 a^2 c^2 \left (a^2 x^2+1\right )}+\frac{x \tan ^{-1}(a x)}{2 a c^2 \left (a^2 x^2+1\right )}+\frac{\tan ^{-1}(a x)^2}{4 a^2 c^2} \]

[Out]

1/(4*a^2*c^2*(1 + a^2*x^2)) + (x*ArcTan[a*x])/(2*a*c^2*(1 + a^2*x^2)) + ArcTan[a*x]^2/(4*a^2*c^2) - ArcTan[a*x
]^2/(2*a^2*c^2*(1 + a^2*x^2))

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Rubi [A]  time = 0.069906, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {4930, 4892, 261} \[ \frac{1}{4 a^2 c^2 \left (a^2 x^2+1\right )}-\frac{\tan ^{-1}(a x)^2}{2 a^2 c^2 \left (a^2 x^2+1\right )}+\frac{x \tan ^{-1}(a x)}{2 a c^2 \left (a^2 x^2+1\right )}+\frac{\tan ^{-1}(a x)^2}{4 a^2 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

1/(4*a^2*c^2*(1 + a^2*x^2)) + (x*ArcTan[a*x])/(2*a*c^2*(1 + a^2*x^2)) + ArcTan[a*x]^2/(4*a^2*c^2) - ArcTan[a*x
]^2/(2*a^2*c^2*(1 + a^2*x^2))

Rule 4930

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[((d + e*x^2)^
(q + 1)*(a + b*ArcTan[c*x])^p)/(2*e*(q + 1)), x] - Dist[(b*p)/(2*c*(q + 1)), Int[(d + e*x^2)^q*(a + b*ArcTan[c
*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, q}, x] && EqQ[e, c^2*d] && GtQ[p, 0] && NeQ[q, -1]

Rule 4892

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2)^2, x_Symbol] :> Simp[(x*(a + b*ArcTan[c*x])
^p)/(2*d*(d + e*x^2)), x] + (-Dist[(b*c*p)/2, Int[(x*(a + b*ArcTan[c*x])^(p - 1))/(d + e*x^2)^2, x], x] + Simp
[(a + b*ArcTan[c*x])^(p + 1)/(2*b*c*d^2*(p + 1)), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[p,
0]

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

\begin{align*} \int \frac{x \tan ^{-1}(a x)^2}{\left (c+a^2 c x^2\right )^2} \, dx &=-\frac{\tan ^{-1}(a x)^2}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac{\int \frac{\tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^2} \, dx}{a}\\ &=\frac{x \tan ^{-1}(a x)}{2 a c^2 \left (1+a^2 x^2\right )}+\frac{\tan ^{-1}(a x)^2}{4 a^2 c^2}-\frac{\tan ^{-1}(a x)^2}{2 a^2 c^2 \left (1+a^2 x^2\right )}-\frac{1}{2} \int \frac{x}{\left (c+a^2 c x^2\right )^2} \, dx\\ &=\frac{1}{4 a^2 c^2 \left (1+a^2 x^2\right )}+\frac{x \tan ^{-1}(a x)}{2 a c^2 \left (1+a^2 x^2\right )}+\frac{\tan ^{-1}(a x)^2}{4 a^2 c^2}-\frac{\tan ^{-1}(a x)^2}{2 a^2 c^2 \left (1+a^2 x^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.0314355, size = 47, normalized size = 0.52 \[ \frac{\left (a^2 x^2-1\right ) \tan ^{-1}(a x)^2+2 a x \tan ^{-1}(a x)+1}{4 a^2 c^2 \left (a^2 x^2+1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 + 2*a*x*ArcTan[a*x] + (-1 + a^2*x^2)*ArcTan[a*x]^2)/(4*a^2*c^2*(1 + a^2*x^2))

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Maple [A]  time = 0.032, size = 84, normalized size = 0.9 \begin{align*}{\frac{1}{4\,{a}^{2}{c}^{2} \left ({a}^{2}{x}^{2}+1 \right ) }}+{\frac{x\arctan \left ( ax \right ) }{2\,a{c}^{2} \left ({a}^{2}{x}^{2}+1 \right ) }}+{\frac{ \left ( \arctan \left ( ax \right ) \right ) ^{2}}{4\,{a}^{2}{c}^{2}}}-{\frac{ \left ( \arctan \left ( ax \right ) \right ) ^{2}}{2\,{a}^{2}{c}^{2} \left ({a}^{2}{x}^{2}+1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/a^2/c^2/(a^2*x^2+1)+1/2*x*arctan(a*x)/a/c^2/(a^2*x^2+1)+1/4*arctan(a*x)^2/a^2/c^2-1/2*arctan(a*x)^2/a^2/c^
2/(a^2*x^2+1)

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Maxima [A]  time = 1.52726, size = 140, normalized size = 1.54 \begin{align*} \frac{{\left (\frac{x}{a^{2} c x^{2} + c} + \frac{\arctan \left (a x\right )}{a c}\right )} \arctan \left (a x\right )}{2 \, a c} - \frac{{\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} - 1}{4 \,{\left (a^{4} c x^{2} + a^{2} c\right )} c} - \frac{\arctan \left (a x\right )^{2}}{2 \,{\left (a^{2} c x^{2} + c\right )} a^{2} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(x/(a^2*c*x^2 + c) + arctan(a*x)/(a*c))*arctan(a*x)/(a*c) - 1/4*((a^2*x^2 + 1)*arctan(a*x)^2 - 1)/((a^4*c*
x^2 + a^2*c)*c) - 1/2*arctan(a*x)^2/((a^2*c*x^2 + c)*a^2*c)

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Fricas [A]  time = 2.17174, size = 112, normalized size = 1.23 \begin{align*} \frac{2 \, a x \arctan \left (a x\right ) +{\left (a^{2} x^{2} - 1\right )} \arctan \left (a x\right )^{2} + 1}{4 \,{\left (a^{4} c^{2} x^{2} + a^{2} c^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*a*x*arctan(a*x) + (a^2*x^2 - 1)*arctan(a*x)^2 + 1)/(a^4*c^2*x^2 + a^2*c^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{x \operatorname{atan}^{2}{\left (a x \right )}}{a^{4} x^{4} + 2 a^{2} x^{2} + 1}\, dx}{c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*atan(a*x)**2/(a**2*c*x**2+c)**2,x)

[Out]

Integral(x*atan(a*x)**2/(a**4*x**4 + 2*a**2*x**2 + 1), x)/c**2

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \arctan \left (a x\right )^{2}}{{\left (a^{2} c x^{2} + c\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*arctan(a*x)^2/(a^2*c*x^2 + c)^2, x)