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

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

[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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Rubi [A]  time = 0.07, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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 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 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]

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]

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*}

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Mathematica [A]  time = 0.03, size = 49, normalized size = 1.00 \[ -\frac {\tan ^{-1}(a x)^2}{2 a^3 c}+\frac {x \tan ^{-1}(a x)}{a^2 c}-\frac {\log \left (a^2 x^2+1\right )}{2 a^3 c} \]

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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fricas [A]  time = 0.46, size = 37, normalized size = 0.76 \[ \frac {2 \, a x \arctan \left (a x\right ) - \arctan \left (a x\right )^{2} - \log \left (a^{2} x^{2} + 1\right )}{2 \, a^{3} c} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

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]

sage0*x

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maple [A]  time = 0.03, size = 46, normalized size = 0.94 \[ \frac {x \arctan \left (a x \right )}{a^{2} c}-\frac {\arctan \left (a x \right )^{2}}{2 a^{3} c}-\frac {\ln \left (a^{2} x^{2}+1\right )}{2 a^{3} c} \]

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 = 0.42, size = 54, normalized size = 1.10 \[ {\left (\frac {x}{a^{2} c} - \frac {\arctan \left (a x\right )}{a^{3} c}\right )} \arctan \left (a x\right ) + \frac {\arctan \left (a x\right )^{2} - \log \left (a^{2} x^{2} + 1\right )}{2 \, a^{3} c} \]

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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mupad [B]  time = 0.16, size = 33, normalized size = 0.67 \[ -\frac {{\mathrm {atan}\left (a\,x\right )}^2-2\,a\,x\,\mathrm {atan}\left (a\,x\right )+\ln \left (a^2\,x^2+1\right )}{2\,a^3\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.05, size = 75, normalized size = 1.53 \[ \begin {cases} \frac {x \operatorname {atan}{\left (a x \right )}}{a^{2} c} - \frac {\log {\left (x^{2} + \frac {1}{a^{2}} \right )}}{2 a^{3} c} - \frac {\operatorname {atan}^{2}{\left (a x \right )}}{2 a^{3} c} & \text {for}\: c \neq 0 \\\tilde {\infty } \left (\frac {x^{3} \operatorname {atan}{\left (a x \right )}}{3} - \frac {x^{2}}{6 a} + \frac {\log {\left (a^{2} x^{2} + 1 \right )}}{6 a^{3}}\right ) & \text {otherwise} \end {cases} \]

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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