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

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

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Rubi [A]  time = 0.08, antiderivative size = 63, normalized size of antiderivative = 1.00, 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 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 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 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 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 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 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])

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*}

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Mathematica [C]  time = 0.01, size = 81, normalized size = 1.29 \[ \frac {1}{2} i \operatorname {PolyLog}(2,-i x)-\frac {1}{2} i \operatorname {PolyLog}(2,i x)-\frac {\, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-x^2\right )}{x}-\frac {\, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};-x^2\right )}{12 x^3}-\frac {\tan ^{-1}(x)}{4 x^4}-\frac {\tan ^{-1}(x)}{x^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/4*ArcTan[x]/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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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (1+x^2\right )^2 \tan ^{-1}(x)}{x^5} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

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fricas [F]  time = 0.95, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (x^{4} + 2 \, x^{2} + 1\right )} \arctan \relax (x)}{x^{5}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (x^{2} + 1\right )}^{2} \arctan \relax (x)}{x^{5}}\,{d x} \]

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)

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maple [A]  time = 0.36, size = 79, normalized size = 1.25

method	result	size
default	\(\arctan \relax (x ) \ln \relax (x )-\frac {\arctan \relax (x )}{x^{2}}-\frac {\arctan \relax (x )}{4 x^{4}}+\frac {i \ln \relax (x ) \ln \left (i x +1\right )}{2}-\frac {i \ln \relax (x ) \ln \left (-i x +1\right )}{2}+\frac {i \dilog \left (i x +1\right )}{2}-\frac {i \dilog \left (-i x +1\right )}{2}-\frac {3 \arctan \relax (x )}{4}-\frac {1}{12 x^{3}}-\frac {3}{4 x}\)	\(79\)
meijerg	\(-\frac {1}{12 x^{3}}-\frac {3}{4 x}-\frac {2 \left (-\frac {3 x^{4}}{8}+\frac {3}{8}\right ) \arctan \left (\sqrt {x^{2}}\right )}{3 x^{3} \sqrt {x^{2}}}-\frac {i x \polylog \left (2, i \sqrt {x^{2}}\right )}{2 \sqrt {x^{2}}}+\frac {i x \polylog \left (2, -i \sqrt {x^{2}}\right )}{2 \sqrt {x^{2}}}-\frac {\left (x^{2}+1\right ) \arctan \relax (x )}{x^{2}}\)	\(85\)
risch	\(\frac {3 i \ln \left (-i x \right )}{8}-\frac {3}{4 x}-\frac {3 \arctan \relax (x )}{4}-\frac {i \ln \left (-i x +1\right )}{2 x^{2}}-\frac {i \dilog \left (-i x +1\right )}{2}-\frac {1}{12 x^{3}}-\frac {i \ln \left (-i x +1\right )}{8 x^{4}}-\frac {3 i \ln \left (i x \right )}{8}+\frac {i \ln \left (i x +1\right )}{2 x^{2}}+\frac {i \dilog \left (i x +1\right )}{2}+\frac {i \ln \left (i x +1\right )}{8 x^{4}}\)	\(104\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(x)*ln(x)-arctan(x)/x^2-1/4*arctan(x)/x^4+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 = 1.12, size = 71, normalized size = 1.13 \[ -\frac {3 \, \pi x^{4} \log \left (x^{2} + 1\right ) - 12 \, x^{4} \arctan \relax (x) \log \relax (x) + 6 i \, x^{4} {\rm Li}_2\left (i \, x + 1\right ) - 6 i \, x^{4} {\rm Li}_2\left (-i \, x + 1\right ) + 9 \, x^{3} + 3 \, {\left (3 \, x^{4} + 4 \, x^{2} + 1\right )} \arctan \relax (x) + x}{12 \, x^{4}} \]

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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mupad [B]  time = 0.50, size = 53, normalized size = 0.84 \[ \frac {x^2-\frac {1}{3}}{4\,x^3}-\frac {\mathrm {atan}\relax (x)}{x^2}-\frac {\mathrm {atan}\relax (x)}{4\,x^4}-\frac {3\,\mathrm {atan}\relax (x)}{4}-\frac {1}{x}-\frac {{\mathrm {Li}}_{\mathrm {2}}\left (1-x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\frac {\mathrm {polylog}\left (2,-x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((atan(x)*(x^2 + 1)^2)/x^5,x)

[Out]

(polylog(2, -x*1i)*1i)/2 - (3*atan(x))/4 - atan(x)/x^2 - atan(x)/(4*x^4) - (dilog(1 - x*1i)*1i)/2 + (x^2 - 1/3
)/(4*x^3) - 1/x

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x^{2} + 1\right )^{2} \operatorname {atan}{\relax (x )}}{x^{5}}\, dx \]

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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