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

Optimal. Leaf size=63 \[ \frac{1}{2} i \text{PolyLog}(2,-i x)-\frac{1}{2} i \text{PolyLog}(2,i x)-\frac{1}{12 x^3}-\frac{\tan ^{-1}(x)}{x^2}-\frac{\tan ^{-1}(x)}{4 x^4}-\frac{3}{4 x}-\frac{3}{4} \tan ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.0836894, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {4948, 4852, 325, 203, 4848, 2391} \[ \frac{1}{2} i \text{PolyLog}(2,-i x)-\frac{1}{2} i \text{PolyLog}(2,i x)-\frac{1}{12 x^3}-\frac{\tan ^{-1}(x)}{x^2}-\frac{\tan ^{-1}(x)}{4 x^4}-\frac{3}{4 x}-\frac{3}{4} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[((1 + x^2)^2*ArcTan[x])/x^5,x]

[Out]

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

Rule 4948

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

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

Mathematica [C]  time = 0.0070159, size = 81, normalized size = 1.29 \[ \frac{1}{2} i \text{PolyLog}(2,-i x)-\frac{1}{2} i \text{PolyLog}(2,i x)-\frac{\, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};-x^2\right )}{12 x^3}-\frac{\, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};-x^2\right )}{x}-\frac{\tan ^{-1}(x)}{x^2}-\frac{\tan ^{-1}(x)}{4 x^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-ArcTan[x]/(4*x^4) - ArcTan[x]/x^2 - Hypergeometric2F1[-3/2, 1, -1/2, -x^2]/(12*x^3) - Hypergeometric2F1[-1/2,
1, 1/2, -x^2]/x + (I/2)*PolyLog[2, (-I)*x] - (I/2)*PolyLog[2, I*x]

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Maple [A]  time = 0.019, size = 79, normalized size = 1.3 \begin{align*} \arctan \left ( x \right ) \ln \left ( x \right ) -{\frac{\arctan \left ( x \right ) }{4\,{x}^{4}}}-{\frac{\arctan \left ( x \right ) }{{x}^{2}}}+{\frac{i}{2}}\ln \left ( x \right ) \ln \left ( 1+ix \right ) -{\frac{i}{2}}\ln \left ( x \right ) \ln \left ( 1-ix \right ) +{\frac{i}{2}}{\it dilog} \left ( 1+ix \right ) -{\frac{i}{2}}{\it dilog} \left ( 1-ix \right ) -{\frac{3\,\arctan \left ( x \right ) }{4}}-{\frac{1}{12\,{x}^{3}}}-{\frac{3}{4\,x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(x)*ln(x)-1/4*arctan(x)/x^4-arctan(x)/x^2+1/2*I*ln(x)*ln(1+I*x)-1/2*I*ln(x)*ln(1-I*x)+1/2*I*dilog(1+I*x)
-1/2*I*dilog(1-I*x)-3/4*arctan(x)-1/12/x^3-3/4/x

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Maxima [A]  time = 1.57451, size = 96, normalized size = 1.52 \begin{align*} -\frac{3 \, \pi x^{4} \log \left (x^{2} + 1\right ) - 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 ) + 9 \, x^{3} + 3 \,{\left (3 \, x^{4} + 4 \, x^{2} + 1\right )} \arctan \left (x\right ) + x}{12 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*pi*x^4*log(x^2 + 1) - 12*x^4*arctan(x)*log(x) + 6*I*x^4*dilog(I*x + 1) - 6*I*x^4*dilog(-I*x + 1) + 9*
x^3 + 3*(3*x^4 + 4*x^2 + 1)*arctan(x) + x)/x^4

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (x^{4} + 2 \, x^{2} + 1\right )} \arctan \left (x\right )}{x^{5}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((x^4 + 2*x^2 + 1)*arctan(x)/x^5, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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