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

Optimal. Leaf size=49 \[ -\frac{\log \left (a^2 x^2+1\right )}{2 a^3 c}-\frac{\tan ^{-1}(a x)^2}{2 a^3 c}+\frac{x \tan ^{-1}(a x)}{a^2 c} \]

[Out]

(x*ArcTan[a*x])/(a^2*c) - ArcTan[a*x]^2/(2*a^3*c) - Log[1 + a^2*x^2]/(2*a^3*c)

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Rubi [A]  time = 0.072521, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4916, 4846, 260, 4884} \[ -\frac{\log \left (a^2 x^2+1\right )}{2 a^3 c}-\frac{\tan ^{-1}(a x)^2}{2 a^3 c}+\frac{x \tan ^{-1}(a x)}{a^2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*ArcTan[a*x])/(a^2*c) - ArcTan[a*x]^2/(2*a^3*c) - Log[1 + a^2*x^2]/(2*a^3*c)

Rule 4916

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

Rule 4846

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

Rule 260

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

Rule 4884

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

Rubi steps

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

Mathematica [A]  time = 0.0278609, size = 49, normalized size = 1. \[ -\frac{\log \left (a^2 x^2+1\right )}{2 a^3 c}-\frac{\tan ^{-1}(a x)^2}{2 a^3 c}+\frac{x \tan ^{-1}(a x)}{a^2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*ArcTan[a*x])/(a^2*c) - ArcTan[a*x]^2/(2*a^3*c) - Log[1 + a^2*x^2]/(2*a^3*c)

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Maple [A]  time = 0.03, size = 46, normalized size = 0.9 \begin{align*}{\frac{x\arctan \left ( ax \right ) }{{a}^{2}c}}-{\frac{ \left ( \arctan \left ( ax \right ) \right ) ^{2}}{2\,{a}^{3}c}}-{\frac{\ln \left ({a}^{2}{x}^{2}+1 \right ) }{2\,{a}^{3}c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*arctan(a*x)/a^2/c-1/2*arctan(a*x)^2/a^3/c-1/2*ln(a^2*x^2+1)/a^3/c

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Maxima [A]  time = 1.67045, size = 73, normalized size = 1.49 \begin{align*}{\left (\frac{x}{a^{2} c} - \frac{\arctan \left (a x\right )}{a^{3} c}\right )} \arctan \left (a x\right ) + \frac{\arctan \left (a x\right )^{2} - \log \left (a^{2} x^{2} + 1\right )}{2 \, a^{3} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x/(a^2*c) - arctan(a*x)/(a^3*c))*arctan(a*x) + 1/2*(arctan(a*x)^2 - log(a^2*x^2 + 1))/(a^3*c)

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Fricas [A]  time = 1.70957, size = 92, normalized size = 1.88 \begin{align*} \frac{2 \, a x \arctan \left (a x\right ) - \arctan \left (a x\right )^{2} - \log \left (a^{2} x^{2} + 1\right )}{2 \, a^{3} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*a*x*arctan(a*x) - arctan(a*x)^2 - log(a^2*x^2 + 1))/(a^3*c)

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Sympy [A]  time = 1.41703, size = 75, normalized size = 1.53 \begin{align*} \begin{cases} \frac{x \operatorname{atan}{\left (a x \right )}}{a^{2} c} - \frac{\log{\left (x^{2} + \frac{1}{a^{2}} \right )}}{2 a^{3} c} - \frac{\operatorname{atan}^{2}{\left (a x \right )}}{2 a^{3} c} & \text{for}\: c \neq 0 \\\tilde{\infty } \left (\frac{x^{3} \operatorname{atan}{\left (a x \right )}}{3} - \frac{x^{2}}{6 a} + \frac{\log{\left (a^{2} x^{2} + 1 \right )}}{6 a^{3}}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x*atan(a*x)/(a**2*c) - log(x**2 + a**(-2))/(2*a**3*c) - atan(a*x)**2/(2*a**3*c), Ne(c, 0)), (zoo*(x
**3*atan(a*x)/3 - x**2/(6*a) + log(a**2*x**2 + 1)/(6*a**3)), True))

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Giac [A]  time = 1.15817, size = 50, normalized size = 1.02 \begin{align*} \frac{2 \, a x \arctan \left (a x\right ) - \arctan \left (a x\right )^{2} - \log \left (a^{2} x^{2} + 1\right )}{2 \, a^{3} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*a*x*arctan(a*x) - arctan(a*x)^2 - log(a^2*x^2 + 1))/(a^3*c)