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

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Rubi [A]  time = 0.09, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, 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 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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.01, size = 52, normalized size = 1.00 \[ -\frac {a \log \left (a^2 x^2+1\right )}{2 c}+\frac {a \log (x)}{c}-\frac {a \tan ^{-1}(a x)^2}{2 c}-\frac {\tan ^{-1}(a x)}{c x} \]

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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fricas [A]  time = 0.68, size = 43, normalized size = 0.83 \[ -\frac {a x \arctan \left (a x\right )^{2} + a x \log \left (a^{2} x^{2} + 1\right ) - 2 \, a x \log \relax (x) + 2 \, \arctan \left (a x\right )}{2 \, c x} \]

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

[Out]

sage0*x

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maple [A]  time = 0.04, size = 51, normalized size = 0.98 \[ -\frac {\arctan \left (a x \right )}{c x}-\frac {a \arctan \left (a x \right )^{2}}{2 c}+\frac {a \ln \left (a x \right )}{c}-\frac {a \ln \left (a^{2} x^{2}+1\right )}{2 c} \]

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]

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

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maxima [A]  time = 0.44, size = 53, normalized size = 1.02 \[ -{\left (\frac {a \arctan \left (a x\right )}{c} + \frac {1}{c x}\right )} \arctan \left (a x\right ) + \frac {{\left (\arctan \left (a x\right )^{2} - \log \left (a^{2} x^{2} + 1\right ) + 2 \, \log \relax (x)\right )} a}{2 \, c} \]

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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mupad [B]  time = 0.45, size = 48, normalized size = 0.92 \[ \frac {a\,\ln \relax (x)}{c}-\frac {a\,\ln \left (a^2\,x^2+1\right )}{2\,c}-\frac {a\,{\mathrm {atan}\left (a\,x\right )}^2}{2\,c}-\frac {\mathrm {atan}\left (a\,x\right )}{c\,x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.47, size = 68, normalized size = 1.31 \[ \begin {cases} \frac {a \log {\relax (x )}}{c} - \frac {a \log {\left (x^{2} + \frac {1}{a^{2}} \right )}}{2 c} - \frac {a \operatorname {atan}^{2}{\left (a x \right )}}{2 c} - \frac {\operatorname {atan}{\left (a x \right )}}{c x} & \text {for}\: c \neq 0 \\\tilde {\infty } \left (a \log {\relax (x )} - \frac {a \log {\left (a^{2} x^{2} + 1 \right )}}{2} - \frac {\operatorname {atan}{\left (a x \right )}}{x}\right ) & \text {otherwise} \end {cases} \]

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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