3.1284 \(\int \frac{\tan ^{-1}(x) \log (1+x^2)}{x^5} \, dx\)

Optimal. Leaf size=102 \[ -\frac{1}{4} i \text{PolyLog}(2,-i x)+\frac{1}{4} i \text{PolyLog}(2,i x)+\frac{\log \left (x^2+1\right )}{4 x}-\frac{\log \left (x^2+1\right )}{12 x^3}-\frac{\tan ^{-1}(x)}{4 x^2}-\frac{\log \left (x^2+1\right ) \tan ^{-1}(x)}{4 x^4}+\frac{1}{4} \log \left (x^2+1\right ) \tan ^{-1}(x)-\frac{5}{12 x}-\frac{11}{12} \tan ^{-1}(x) \]

[Out]

-5/(12*x) - (11*ArcTan[x])/12 - ArcTan[x]/(4*x^2) - Log[1 + x^2]/(12*x^3) + Log[1 + x^2]/(4*x) + (ArcTan[x]*Lo
g[1 + x^2])/4 - (ArcTan[x]*Log[1 + x^2])/(4*x^4) - (I/4)*PolyLog[2, (-I)*x] + (I/4)*PolyLog[2, I*x]

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Rubi [A]  time = 0.129238, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {4852, 325, 203, 5021, 453, 4980, 4848, 2391} \[ -\frac{1}{4} i \text{PolyLog}(2,-i x)+\frac{1}{4} i \text{PolyLog}(2,i x)+\frac{\log \left (x^2+1\right )}{4 x}-\frac{\log \left (x^2+1\right )}{12 x^3}-\frac{\tan ^{-1}(x)}{4 x^2}-\frac{\log \left (x^2+1\right ) \tan ^{-1}(x)}{4 x^4}+\frac{1}{4} \log \left (x^2+1\right ) \tan ^{-1}(x)-\frac{5}{12 x}-\frac{11}{12} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(ArcTan[x]*Log[1 + x^2])/x^5,x]

[Out]

-5/(12*x) - (11*ArcTan[x])/12 - ArcTan[x]/(4*x^2) - Log[1 + x^2]/(12*x^3) + Log[1 + x^2]/(4*x) + (ArcTan[x]*Lo
g[1 + x^2])/4 - (ArcTan[x]*Log[1 + x^2])/(4*x^4) - (I/4)*PolyLog[2, (-I)*x] + (I/4)*PolyLog[2, I*x]

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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 5021

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

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 4980

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> With
[{u = ExpandIntegrand[(a + b*ArcTan[c*x])^p, (f*x)^m*(d + e*x^2)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b,
c, d, e, f, m}, x] && IntegerQ[q] && IGtQ[p, 0] && ((EqQ[p, 1] && GtQ[q, 0]) || IntegerQ[m])

Rule 4848

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Dist[(I*b)/2, Int[Log[1 - I*c*x
]/x, x], x] - Dist[(I*b)/2, Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]

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]

Rubi steps

\begin{align*} \int \frac{\tan ^{-1}(x) \log \left (1+x^2\right )}{x^5} \, dx &=-\frac{\log \left (1+x^2\right )}{12 x^3}+\frac{\log \left (1+x^2\right )}{4 x}+\frac{1}{4} \tan ^{-1}(x) \log \left (1+x^2\right )-\frac{\tan ^{-1}(x) \log \left (1+x^2\right )}{4 x^4}-2 \int \left (\frac{-1+3 x^2}{12 x^2 \left (1+x^2\right )}+\frac{\left (-1+x^2\right ) \tan ^{-1}(x)}{4 x^3}\right ) \, dx\\ &=-\frac{\log \left (1+x^2\right )}{12 x^3}+\frac{\log \left (1+x^2\right )}{4 x}+\frac{1}{4} \tan ^{-1}(x) \log \left (1+x^2\right )-\frac{\tan ^{-1}(x) \log \left (1+x^2\right )}{4 x^4}-\frac{1}{6} \int \frac{-1+3 x^2}{x^2 \left (1+x^2\right )} \, dx-\frac{1}{2} \int \frac{\left (-1+x^2\right ) \tan ^{-1}(x)}{x^3} \, dx\\ &=-\frac{1}{6 x}-\frac{\log \left (1+x^2\right )}{12 x^3}+\frac{\log \left (1+x^2\right )}{4 x}+\frac{1}{4} \tan ^{-1}(x) \log \left (1+x^2\right )-\frac{\tan ^{-1}(x) \log \left (1+x^2\right )}{4 x^4}-\frac{1}{2} \int \left (-\frac{\tan ^{-1}(x)}{x^3}+\frac{\tan ^{-1}(x)}{x}\right ) \, dx-\frac{2}{3} \int \frac{1}{1+x^2} \, dx\\ &=-\frac{1}{6 x}-\frac{2}{3} \tan ^{-1}(x)-\frac{\log \left (1+x^2\right )}{12 x^3}+\frac{\log \left (1+x^2\right )}{4 x}+\frac{1}{4} \tan ^{-1}(x) \log \left (1+x^2\right )-\frac{\tan ^{-1}(x) \log \left (1+x^2\right )}{4 x^4}+\frac{1}{2} \int \frac{\tan ^{-1}(x)}{x^3} \, dx-\frac{1}{2} \int \frac{\tan ^{-1}(x)}{x} \, dx\\ &=-\frac{1}{6 x}-\frac{2}{3} \tan ^{-1}(x)-\frac{\tan ^{-1}(x)}{4 x^2}-\frac{\log \left (1+x^2\right )}{12 x^3}+\frac{\log \left (1+x^2\right )}{4 x}+\frac{1}{4} \tan ^{-1}(x) \log \left (1+x^2\right )-\frac{\tan ^{-1}(x) \log \left (1+x^2\right )}{4 x^4}-\frac{1}{4} i \int \frac{\log (1-i x)}{x} \, dx+\frac{1}{4} i \int \frac{\log (1+i x)}{x} \, dx+\frac{1}{4} \int \frac{1}{x^2 \left (1+x^2\right )} \, dx\\ &=-\frac{5}{12 x}-\frac{2}{3} \tan ^{-1}(x)-\frac{\tan ^{-1}(x)}{4 x^2}-\frac{\log \left (1+x^2\right )}{12 x^3}+\frac{\log \left (1+x^2\right )}{4 x}+\frac{1}{4} \tan ^{-1}(x) \log \left (1+x^2\right )-\frac{\tan ^{-1}(x) \log \left (1+x^2\right )}{4 x^4}-\frac{1}{4} i \text{Li}_2(-i x)+\frac{1}{4} i \text{Li}_2(i x)-\frac{1}{4} \int \frac{1}{1+x^2} \, dx\\ &=-\frac{5}{12 x}-\frac{11}{12} \tan ^{-1}(x)-\frac{\tan ^{-1}(x)}{4 x^2}-\frac{\log \left (1+x^2\right )}{12 x^3}+\frac{\log \left (1+x^2\right )}{4 x}+\frac{1}{4} \tan ^{-1}(x) \log \left (1+x^2\right )-\frac{\tan ^{-1}(x) \log \left (1+x^2\right )}{4 x^4}-\frac{1}{4} i \text{Li}_2(-i x)+\frac{1}{4} i \text{Li}_2(i x)\\ \end{align*}

Mathematica [A]  time = 0.0350324, size = 98, normalized size = 0.96 \[ -\frac{1}{4} i (\text{PolyLog}(2,-i x)-\text{PolyLog}(2,i x))+\frac{1}{2} \left (\frac{1}{2} \left (-\frac{1}{x}-\tan ^{-1}(x)\right )-\frac{\tan ^{-1}(x)}{2 x^2}\right )+\frac{\log \left (x^2+1\right ) \left (3 x^3+3 x^4 \tan ^{-1}(x)-x-3 \tan ^{-1}(x)\right )}{12 x^4}-\frac{1}{6 x}-\frac{2}{3} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Integrate[(ArcTan[x]*Log[1 + x^2])/x^5,x]

[Out]

-1/(6*x) - (2*ArcTan[x])/3 + ((-x^(-1) - ArcTan[x])/2 - ArcTan[x]/(2*x^2))/2 + ((-x + 3*x^3 - 3*ArcTan[x] + 3*
x^4*ArcTan[x])*Log[1 + x^2])/(12*x^4) - (I/4)*(PolyLog[2, (-I)*x] - PolyLog[2, I*x])

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Maple [F]  time = 3.61, size = 0, normalized size = 0. \begin{align*} \int{\frac{\arctan \left ( x \right ) \ln \left ({x}^{2}+1 \right ) }{{x}^{5}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(x)*ln(x^2+1)/x^5,x)

[Out]

int(arctan(x)*ln(x^2+1)/x^5,x)

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Maxima [A]  time = 1.68755, size = 120, normalized size = 1.18 \begin{align*} -\frac{12 \, x^{4} \arctan \left (x\right ) \log \left (x\right ) - 6 i \, x^{4}{\rm Li}_2\left (i \, x + 1\right ) + 6 i \, x^{4}{\rm Li}_2\left (-i \, x + 1\right ) + 10 \, x^{3} + 2 \,{\left (11 \, x^{4} + 3 \, x^{2}\right )} \arctan \left (x\right ) -{\left (3 \, \pi x^{4} + 6 \, x^{3} + 6 \,{\left (x^{4} - 1\right )} \arctan \left (x\right ) - 2 \, x\right )} \log \left (x^{2} + 1\right )}{24 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(x)*log(x^2+1)/x^5,x, algorithm="maxima")

[Out]

-1/24*(12*x^4*arctan(x)*log(x) - 6*I*x^4*dilog(I*x + 1) + 6*I*x^4*dilog(-I*x + 1) + 10*x^3 + 2*(11*x^4 + 3*x^2
)*arctan(x) - (3*pi*x^4 + 6*x^3 + 6*(x^4 - 1)*arctan(x) - 2*x)*log(x^2 + 1))/x^4

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\arctan \left (x\right ) \log \left (x^{2} + 1\right )}{x^{5}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(x)*log(x^2+1)/x^5,x, algorithm="fricas")

[Out]

integral(arctan(x)*log(x^2 + 1)/x^5, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\log{\left (x^{2} + 1 \right )} \operatorname{atan}{\left (x \right )}}{x^{5}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atan(x)*ln(x**2+1)/x**5,x)

[Out]

Integral(log(x**2 + 1)*atan(x)/x**5, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\arctan \left (x\right ) \log \left (x^{2} + 1\right )}{x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(x)*log(x^2+1)/x^5,x, algorithm="giac")

[Out]

integrate(arctan(x)*log(x^2 + 1)/x^5, x)