3.1283 \(\int \frac {\tan ^{-1}(x) \log (1+x^2)}{x^4} \, dx\)

Optimal. Leaf size=81 \[ \frac {\text {Li}_2\left (-x^2\right )}{6}+\frac {1}{12} \log ^2\left (x^2+1\right )-\frac {\log \left (x^2+1\right )}{6 x^2}-\frac {1}{2} \log \left (x^2+1\right )-\frac {\log \left (x^2+1\right ) \tan ^{-1}(x)}{3 x^3}+\log (x)-\frac {1}{3} \tan ^{-1}(x)^2-\frac {2 \tan ^{-1}(x)}{3 x} \]

[Out]

-2/3*arctan(x)/x-1/3*arctan(x)^2+ln(x)-1/2*ln(x^2+1)-1/6*ln(x^2+1)/x^2-1/3*arctan(x)*ln(x^2+1)/x^3+1/12*ln(x^2
+1)^2+1/6*polylog(2,-x^2)

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Rubi [A]  time = 0.21, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 15, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.250, Rules used = {4852, 266, 44, 5017, 2475, 2410, 2395, 36, 29, 31, 2391, 2390, 2301, 4918, 4884} \[ \frac {1}{6} \text {PolyLog}\left (2,-x^2\right )+\frac {1}{12} \log ^2\left (x^2+1\right )-\frac {\log \left (x^2+1\right )}{6 x^2}-\frac {1}{2} \log \left (x^2+1\right )-\frac {\log \left (x^2+1\right ) \tan ^{-1}(x)}{3 x^3}+\log (x)-\frac {1}{3} \tan ^{-1}(x)^2-\frac {2 \tan ^{-1}(x)}{3 x} \]

Antiderivative was successfully verified.

[In]

Int[(ArcTan[x]*Log[1 + x^2])/x^4,x]

[Out]

(-2*ArcTan[x])/(3*x) - ArcTan[x]^2/3 + Log[x] - Log[1 + x^2]/2 - Log[1 + x^2]/(6*x^2) - (ArcTan[x]*Log[1 + x^2
])/(3*x^3) + Log[1 + x^2]^2/12 + PolyLog[2, -x^2]/6

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 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 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 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[((f*x)/d)^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rule 2410

Int[(Log[(c_.)*((d_) + (e_.)*(x_))]*(x_)^(m_.))/((f_) + (g_.)*(x_)), x_Symbol] :> Int[ExpandIntegrand[Log[c*(d
+ e*x)], x^m/(f + g*x), x], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[e*f - d*g, 0] && EqQ[c*d, 1] && IntegerQ[m
]

Rule 2475

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.),
x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q,
x], x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && IntegerQ[r] && IntegerQ[s/n] && Intege
rQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0])

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]

Rule 5017

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

Rubi steps

\begin {align*} \int \frac {\tan ^{-1}(x) \log \left (1+x^2\right )}{x^4} \, dx &=-\frac {\tan ^{-1}(x) \log \left (1+x^2\right )}{3 x^3}+\frac {1}{3} \int \frac {\log \left (1+x^2\right )}{x^3 \left (1+x^2\right )} \, dx+\frac {2}{3} \int \frac {\tan ^{-1}(x)}{x^2 \left (1+x^2\right )} \, dx\\ &=-\frac {\tan ^{-1}(x) \log \left (1+x^2\right )}{3 x^3}+\frac {1}{6} \operatorname {Subst}\left (\int \frac {\log (1+x)}{x^2 (1+x)} \, dx,x,x^2\right )+\frac {2}{3} \int \frac {\tan ^{-1}(x)}{x^2} \, dx-\frac {2}{3} \int \frac {\tan ^{-1}(x)}{1+x^2} \, dx\\ &=-\frac {2 \tan ^{-1}(x)}{3 x}-\frac {1}{3} \tan ^{-1}(x)^2-\frac {\tan ^{-1}(x) \log \left (1+x^2\right )}{3 x^3}+\frac {1}{6} \operatorname {Subst}\left (\int \left (\frac {\log (1+x)}{x^2}-\frac {\log (1+x)}{x}+\frac {\log (1+x)}{1+x}\right ) \, dx,x,x^2\right )+\frac {2}{3} \int \frac {1}{x \left (1+x^2\right )} \, dx\\ &=-\frac {2 \tan ^{-1}(x)}{3 x}-\frac {1}{3} \tan ^{-1}(x)^2-\frac {\tan ^{-1}(x) \log \left (1+x^2\right )}{3 x^3}+\frac {1}{6} \operatorname {Subst}\left (\int \frac {\log (1+x)}{x^2} \, dx,x,x^2\right )-\frac {1}{6} \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,x^2\right )+\frac {1}{6} \operatorname {Subst}\left (\int \frac {\log (1+x)}{1+x} \, dx,x,x^2\right )+\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{x (1+x)} \, dx,x,x^2\right )\\ &=-\frac {2 \tan ^{-1}(x)}{3 x}-\frac {1}{3} \tan ^{-1}(x)^2-\frac {\log \left (1+x^2\right )}{6 x^2}-\frac {\tan ^{-1}(x) \log \left (1+x^2\right )}{3 x^3}+\frac {\text {Li}_2\left (-x^2\right )}{6}+\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{x (1+x)} \, dx,x,x^2\right )+\frac {1}{6} \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,1+x^2\right )+\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )-\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{1+x} \, dx,x,x^2\right )\\ &=-\frac {2 \tan ^{-1}(x)}{3 x}-\frac {1}{3} \tan ^{-1}(x)^2+\frac {2 \log (x)}{3}-\frac {1}{3} \log \left (1+x^2\right )-\frac {\log \left (1+x^2\right )}{6 x^2}-\frac {\tan ^{-1}(x) \log \left (1+x^2\right )}{3 x^3}+\frac {1}{12} \log ^2\left (1+x^2\right )+\frac {\text {Li}_2\left (-x^2\right )}{6}+\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )-\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{1+x} \, dx,x,x^2\right )\\ &=-\frac {2 \tan ^{-1}(x)}{3 x}-\frac {1}{3} \tan ^{-1}(x)^2+\log (x)-\frac {1}{2} \log \left (1+x^2\right )-\frac {\log \left (1+x^2\right )}{6 x^2}-\frac {\tan ^{-1}(x) \log \left (1+x^2\right )}{3 x^3}+\frac {1}{12} \log ^2\left (1+x^2\right )+\frac {\text {Li}_2\left (-x^2\right )}{6}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 81, normalized size = 1.00 \[ \frac {\text {Li}_2\left (-x^2\right )}{6}+\frac {1}{12} \log ^2\left (x^2+1\right )-\frac {\log \left (x^2+1\right )}{6 x^2}-\frac {1}{2} \log \left (x^2+1\right )-\frac {\log \left (x^2+1\right ) \tan ^{-1}(x)}{3 x^3}+\log (x)-\frac {1}{3} \tan ^{-1}(x)^2-\frac {2 \tan ^{-1}(x)}{3 x} \]

Antiderivative was successfully verified.

[In]

Integrate[(ArcTan[x]*Log[1 + x^2])/x^4,x]

[Out]

(-2*ArcTan[x])/(3*x) - ArcTan[x]^2/3 + Log[x] - Log[1 + x^2]/2 - Log[1 + x^2]/(6*x^2) - (ArcTan[x]*Log[1 + x^2
])/(3*x^3) + Log[1 + x^2]^2/12 + PolyLog[2, -x^2]/6

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\arctan \relax (x) \log \left (x^{2} + 1\right )}{x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(x)*log(x^2+1)/x^4,x, algorithm="fricas")

[Out]

integral(arctan(x)*log(x^2 + 1)/x^4, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\arctan \relax (x) \log \left (x^{2} + 1\right )}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(x)*log(x^2+1)/x^4,x, algorithm="giac")

[Out]

integrate(arctan(x)*log(x^2 + 1)/x^4, x)

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maple [F]  time = 4.93, size = 0, normalized size = 0.00 \[ \int \frac {\arctan \relax (x ) \ln \left (x^{2}+1\right )}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(x)*ln(x^2+1)/x^4,x)

[Out]

int(arctan(x)*ln(x^2+1)/x^4,x)

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maxima [A]  time = 0.42, size = 95, normalized size = 1.17 \[ -\frac {1}{3} \, {\left (\frac {2}{x} + \frac {\log \left (x^{2} + 1\right )}{x^{3}} + 2 \, \arctan \relax (x)\right )} \arctan \relax (x) + \frac {4 \, x^{2} \arctan \relax (x)^{2} + x^{2} \log \left (x^{2} + 1\right )^{2} - 2 \, x^{2} {\rm Li}_2\left (x^{2} + 1\right ) + 12 \, x^{2} \log \relax (x) - 2 \, {\left (x^{2} \log \left (-x^{2}\right ) + 3 \, x^{2} + 1\right )} \log \left (x^{2} + 1\right )}{12 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(x)*log(x^2+1)/x^4,x, algorithm="maxima")

[Out]

-1/3*(2/x + log(x^2 + 1)/x^3 + 2*arctan(x))*arctan(x) + 1/12*(4*x^2*arctan(x)^2 + x^2*log(x^2 + 1)^2 - 2*x^2*d
ilog(x^2 + 1) + 12*x^2*log(x) - 2*(x^2*log(-x^2) + 3*x^2 + 1)*log(x^2 + 1))/x^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\ln \left (x^2+1\right )\,\mathrm {atan}\relax (x)}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x^2 + 1)*atan(x))/x^4,x)

[Out]

int((log(x^2 + 1)*atan(x))/x^4, x)

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sympy [C]  time = 28.09, size = 97, normalized size = 1.20 \[ \frac {2 \log {\relax (x )}}{3} + \frac {\log {\left (2 x^{2} \right )}}{6} + \frac {\log {\left (x^{2} + 1 \right )}^{2}}{12} - \frac {\log {\left (x^{2} + 1 \right )}}{3} - \frac {\log {\left (2 x^{2} + 2 \right )}}{6} - \frac {\operatorname {atan}^{2}{\relax (x )}}{3} + \frac {\operatorname {Li}_{2}\left (x^{2} e^{i \pi }\right )}{6} - \frac {2 \operatorname {atan}{\relax (x )}}{3 x} - \frac {\log {\left (x^{2} + 1 \right )}}{6 x^{2}} - \frac {\log {\left (x^{2} + 1 \right )} \operatorname {atan}{\relax (x )}}{3 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atan(x)*ln(x**2+1)/x**4,x)

[Out]

2*log(x)/3 + log(2*x**2)/6 + log(x**2 + 1)**2/12 - log(x**2 + 1)/3 - log(2*x**2 + 2)/6 - atan(x)**2/3 + polylo
g(2, x**2*exp_polar(I*pi))/6 - 2*atan(x)/(3*x) - log(x**2 + 1)/(6*x**2) - log(x**2 + 1)*atan(x)/(3*x**3)

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