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

Optimal. Leaf size=52 \[ -\frac{a \log \left (a^2 x^2+1\right )}{2 c}+\frac{a \log (x)}{c}-\frac{a \tan ^{-1}(a x)^2}{2 c}-\frac{\tan ^{-1}(a x)}{c x} \]

[Out]

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

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Rubi [A]  time = 0.0865378, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {4918, 4852, 266, 36, 29, 31, 4884} \[ -\frac{a \log \left (a^2 x^2+1\right )}{2 c}+\frac{a \log (x)}{c}-\frac{a \tan ^{-1}(a x)^2}{2 c}-\frac{\tan ^{-1}(a x)}{c x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4918

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

Rule 4852

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
n[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^
2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 266

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

Rule 36

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

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

Mathematica [A]  time = 0.0077087, size = 52, normalized size = 1. \[ -\frac{a \log \left (a^2 x^2+1\right )}{2 c}+\frac{a \log (x)}{c}-\frac{a \tan ^{-1}(a x)^2}{2 c}-\frac{\tan ^{-1}(a x)}{c x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.036, size = 51, normalized size = 1. \begin{align*} -{\frac{a \left ( \arctan \left ( ax \right ) \right ) ^{2}}{2\,c}}-{\frac{\arctan \left ( ax \right ) }{cx}}-{\frac{a\ln \left ({a}^{2}{x}^{2}+1 \right ) }{2\,c}}+{\frac{a\ln \left ( ax \right ) }{c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.60181, size = 72, normalized size = 1.38 \begin{align*} -{\left (\frac{a \arctan \left (a x\right )}{c} + \frac{1}{c x}\right )} \arctan \left (a x\right ) + \frac{{\left (\arctan \left (a x\right )^{2} - \log \left (a^{2} x^{2} + 1\right ) + 2 \, \log \left (x\right )\right )} a}{2 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.69865, size = 116, normalized size = 2.23 \begin{align*} -\frac{a x \arctan \left (a x\right )^{2} + a x \log \left (a^{2} x^{2} + 1\right ) - 2 \, a x \log \left (x\right ) + 2 \, \arctan \left (a x\right )}{2 \, c x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.0772, size = 68, normalized size = 1.31 \begin{align*} \begin{cases} \frac{a \log{\left (x \right )}}{c} - \frac{a \log{\left (x^{2} + \frac{1}{a^{2}} \right )}}{2 c} - \frac{a \operatorname{atan}^{2}{\left (a x \right )}}{2 c} - \frac{\operatorname{atan}{\left (a x \right )}}{c x} & \text{for}\: c \neq 0 \\\tilde{\infty } \left (a \log{\left (x \right )} - \frac{a \log{\left (a^{2} x^{2} + 1 \right )}}{2} - \frac{\operatorname{atan}{\left (a x \right )}}{x}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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