3.7.78 \(\int \frac {(1+x^2)^2 \tan ^{-1}(x)}{x^5} \, dx\) [678]

Optimal. Leaf size=63 \[ -\frac {1}{12 x^3}-\frac {3}{4 x}-\frac {3}{4} \tan ^{-1}(x)-\frac {\tan ^{-1}(x)}{4 x^4}-\frac {\tan ^{-1}(x)}{x^2}+\frac {1}{2} i \text {Li}_2(-i x)-\frac {1}{2} i \text {Li}_2(i x) \]

[Out]

-1/12/x^3-3/4/x-3/4*arctan(x)-1/4*arctan(x)/x^4-arctan(x)/x^2+1/2*I*polylog(2,-I*x)-1/2*I*polylog(2,I*x)

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Rubi [A]
time = 0.06, 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 = {5068, 4946, 331, 209, 4940, 2438} \begin {gather*} \frac {1}{2} i \text {PolyLog}(2,-i x)-\frac {1}{2} i \text {PolyLog}(2,i x)-\frac {\text {ArcTan}(x)}{4 x^4}-\frac {\text {ArcTan}(x)}{x^2}-\frac {3 \text {ArcTan}(x)}{4}-\frac {1}{12 x^3}-\frac {3}{4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((1 + x^2)^2*ArcTan[x])/x^5,x]

[Out]

-1/12*1/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 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 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 5068

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] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.01, size = 81, normalized size = 1.29 \begin {gather*} -\frac {\tan ^{-1}(x)}{4 x^4}-\frac {\tan ^{-1}(x)}{x^2}-\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 {1}{2} i \text {Li}_2(-i x)-\frac {1}{2} i \text {Li}_2(i x) \end {gather*}

Antiderivative was successfully verified.

[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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Maple [A]
time = 0.10, size = 79, normalized size = 1.25

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

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)-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)-1/12/x^3-3/4/x-3/4*arctan(x)

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Maxima [A]
time = 1.53, size = 71, normalized size = 1.13 \begin {gather*} -\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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x^{2} + 1\right )^{2} \operatorname {atan}{\left (x \right )}}{x^{5}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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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Mupad [B]
time = 0.50, size = 53, normalized size = 0.84 \begin {gather*} \frac {x^2-\frac {1}{3}}{4\,x^3}-\frac {\mathrm {atan}\left (x\right )}{x^2}-\frac {\mathrm {atan}\left (x\right )}{4\,x^4}-\frac {3\,\mathrm {atan}\left (x\right )}{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} \end {gather*}

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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