\(\int \frac {\arctan (x) \log (1+x^2)}{x^5} \, dx\) [1284]

Optimal result

Integrand size = 12, antiderivative size = 102 \[ \int \frac {\arctan (x) \log \left (1+x^2\right )}{x^5} \, dx=-\frac {5}{12 x}-\frac {11 \arctan (x)}{12}-\frac {\arctan (x)}{4 x^2}-\frac {\log \left (1+x^2\right )}{12 x^3}+\frac {\log \left (1+x^2\right )}{4 x}+\frac {1}{4} \arctan (x) \log \left (1+x^2\right )-\frac {\arctan (x) \log \left (1+x^2\right )}{4 x^4}-\frac {1}{4} i \operatorname {PolyLog}(2,-i x)+\frac {1}{4} i \operatorname {PolyLog}(2,i x) \]

[Out]

-5/12/x-11/12*arctan(x)-1/4*arctan(x)/x^2-1/12*ln(x^2+1)/x^3+1/4*ln(x^2+1)/x+1/4*arctan(x)*ln(x^2+1)-1/4*arcta
n(x)*ln(x^2+1)/x^4-1/4*I*polylog(2,-I*x)+1/4*I*polylog(2,I*x)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {4946, 331, 209, 5141, 464, 5100, 4940, 2438} \[ \int \frac {\arctan (x) \log \left (1+x^2\right )}{x^5} \, dx=-\frac {\arctan (x)}{4 x^2}+\frac {1}{4} \arctan (x) \log \left (x^2+1\right )-\frac {\arctan (x) \log \left (x^2+1\right )}{4 x^4}-\frac {11 \arctan (x)}{12}-\frac {1}{4} i \operatorname {PolyLog}(2,-i x)+\frac {1}{4} i \operatorname {PolyLog}(2,i x)+\frac {\log \left (x^2+1\right )}{4 x}-\frac {\log \left (x^2+1\right )}{12 x^3}-\frac {5}{12 x} \]

[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 209

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

Rule 331

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 464

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 2438

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 4940

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 4946

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

Rule 5100

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 5141

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {\log \left (1+x^2\right )}{12 x^3}+\frac {\log \left (1+x^2\right )}{4 x}+\frac {1}{4} \arctan (x) \log \left (1+x^2\right )-\frac {\arctan (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 ) \arctan (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} \arctan (x) \log \left (1+x^2\right )-\frac {\arctan (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 ) \arctan (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} \arctan (x) \log \left (1+x^2\right )-\frac {\arctan (x) \log \left (1+x^2\right )}{4 x^4}-\frac {1}{2} \int \left (-\frac {\arctan (x)}{x^3}+\frac {\arctan (x)}{x}\right ) \, dx-\frac {2}{3} \int \frac {1}{1+x^2} \, dx \\ & = -\frac {1}{6 x}-\frac {2 \arctan (x)}{3}-\frac {\log \left (1+x^2\right )}{12 x^3}+\frac {\log \left (1+x^2\right )}{4 x}+\frac {1}{4} \arctan (x) \log \left (1+x^2\right )-\frac {\arctan (x) \log \left (1+x^2\right )}{4 x^4}+\frac {1}{2} \int \frac {\arctan (x)}{x^3} \, dx-\frac {1}{2} \int \frac {\arctan (x)}{x} \, dx \\ & = -\frac {1}{6 x}-\frac {2 \arctan (x)}{3}-\frac {\arctan (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} \arctan (x) \log \left (1+x^2\right )-\frac {\arctan (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 \arctan (x)}{3}-\frac {\arctan (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} \arctan (x) \log \left (1+x^2\right )-\frac {\arctan (x) \log \left (1+x^2\right )}{4 x^4}-\frac {1}{4} i \operatorname {PolyLog}(2,-i x)+\frac {1}{4} i \operatorname {PolyLog}(2,i x)-\frac {1}{4} \int \frac {1}{1+x^2} \, dx \\ & = -\frac {5}{12 x}-\frac {11 \arctan (x)}{12}-\frac {\arctan (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} \arctan (x) \log \left (1+x^2\right )-\frac {\arctan (x) \log \left (1+x^2\right )}{4 x^4}-\frac {1}{4} i \operatorname {PolyLog}(2,-i x)+\frac {1}{4} i \operatorname {PolyLog}(2,i x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.96 \[ \int \frac {\arctan (x) \log \left (1+x^2\right )}{x^5} \, dx=-\frac {1}{6 x}-\frac {2 \arctan (x)}{3}+\frac {1}{2} \left (\frac {1}{2} \left (-\frac {1}{x}-\arctan (x)\right )-\frac {\arctan (x)}{2 x^2}\right )+\frac {\left (-x+3 x^3-3 \arctan (x)+3 x^4 \arctan (x)\right ) \log \left (1+x^2\right )}{12 x^4}-\frac {1}{4} i (\operatorname {PolyLog}(2,-i x)-\operatorname {PolyLog}(2,i x)) \]

[In]

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

[Out]

-1/6*1/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])

Maple [F]

\[\int \frac {\arctan \left (x \right ) \ln \left (x^{2}+1\right )}{x^{5}}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {\arctan (x) \log \left (1+x^2\right )}{x^5} \, dx=\int { \frac {\arctan \left (x\right ) \log \left (x^{2} + 1\right )}{x^{5}} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\arctan (x) \log \left (1+x^2\right )}{x^5} \, dx=\int \frac {\log {\left (x^{2} + 1 \right )} \operatorname {atan}{\left (x \right )}}{x^{5}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.87 \[ \int \frac {\arctan (x) \log \left (1+x^2\right )}{x^5} \, dx=-\frac {12 \, x^{4} \arctan \left (x\right ) \log \left (x\right ) - 6 i \, x^{4} {\rm Li}_2\left (i \, x + 1\right ) + 6 i \, x^{4} {\rm Li}_2\left (-i \, x + 1\right ) + 10 \, x^{3} + 2 \, {\left (11 \, x^{4} + 3 \, x^{2}\right )} \arctan \left (x\right ) - {\left (3 \, \pi x^{4} + 6 \, x^{3} + 6 \, {\left (x^{4} - 1\right )} \arctan \left (x\right ) - 2 \, x\right )} \log \left (x^{2} + 1\right )}{24 \, x^{4}} \]

[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

Giac [F]

\[ \int \frac {\arctan (x) \log \left (1+x^2\right )}{x^5} \, dx=\int { \frac {\arctan \left (x\right ) \log \left (x^{2} + 1\right )}{x^{5}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\arctan (x) \log \left (1+x^2\right )}{x^5} \, dx=\int \frac {\ln \left (x^2+1\right )\,\mathrm {atan}\left (x\right )}{x^5} \,d x \]

[In]

int((log(x^2 + 1)*atan(x))/x^5,x)

[Out]

int((log(x^2 + 1)*atan(x))/x^5, x)