3.672 \(\int \frac {x^3 \tan ^{-1}(x)}{1+x^2} \, dx\)

Optimal. Leaf size=67 \[ \frac {1}{2} i \operatorname {PolyLog}\left (2,1-\frac {2}{1+i x}\right )+\frac {1}{2} x^2 \tan ^{-1}(x)-\frac {x}{2}+\frac {1}{2} i \tan ^{-1}(x)^2+\frac {1}{2} \tan ^{-1}(x)+\log \left (\frac {2}{1+i x}\right ) \tan ^{-1}(x) \]

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Rubi [A]  time = 0.09, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {4916, 4852, 321, 203, 4920, 4854, 2402, 2315} \[ \frac {1}{2} i \text {PolyLog}\left (2,1-\frac {2}{1+i x}\right )+\frac {1}{2} x^2 \tan ^{-1}(x)-\frac {x}{2}+\frac {1}{2} i \tan ^{-1}(x)^2+\frac {1}{2} \tan ^{-1}(x)+\log \left (\frac {2}{1+i x}\right ) \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(x^3*ArcTan[x])/(1 + x^2),x]

[Out]

-x/2 + ArcTan[x]/2 + (x^2*ArcTan[x])/2 + (I/2)*ArcTan[x]^2 + ArcTan[x]*Log[2/(1 + I*x)] + (I/2)*PolyLog[2, 1 -
2/(1 + 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 321

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

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

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 4854

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])^p*Lo
g[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcTan[c*x])^(p - 1)*Log[2/(1 + (e*x)/d)])/(1 + c^2*x^
2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e^2, 0]

Rule 4916

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[f^2/
e, Int[(f*x)^(m - 2)*(a + b*ArcTan[c*x])^p, x], x] - Dist[(d*f^2)/e, Int[((f*x)^(m - 2)*(a + b*ArcTan[c*x])^p)
/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && GtQ[m, 1]

Rule 4920

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[(I*(a + b*ArcTan
[c*x])^(p + 1))/(b*e*(p + 1)), x] - Dist[1/(c*d), Int[(a + b*ArcTan[c*x])^p/(I - c*x), x], x] /; FreeQ[{a, b,
c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 57, normalized size = 0.85 \[ \frac {1}{2} \left (i \operatorname {PolyLog}\left (2,\frac {x+i}{x-i}\right )+\left (x^2+2 \log \left (-\frac {2 i}{x-i}\right )+1\right ) \tan ^{-1}(x)-x+i \tan ^{-1}(x)^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*ArcTan[x])/(1 + x^2),x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^3 \tan ^{-1}(x)}{1+x^2} \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[(x^3*ArcTan[x])/(1 + x^2),x]

[Out]

Could not integrate

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fricas [F]  time = 1.02, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{3} \arctan \relax (x)}{x^{2} + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctan(x)/(x^2+1),x, algorithm="fricas")

[Out]

integral(x^3*arctan(x)/(x^2 + 1), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \arctan \relax (x)}{x^{2} + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctan(x)/(x^2+1),x, algorithm="giac")

[Out]

integrate(x^3*arctan(x)/(x^2 + 1), x)

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maple [B]  time = 0.33, size = 113, normalized size = 1.69

method	result	size
risch	\(\frac {i \ln \left (-i x +1\right ) x^{2}}{4}+\frac {\arctan \relax (x )}{2}-\frac {x}{2}-\frac {i \ln \left (\frac {1}{2}+\frac {i x}{2}\right ) \ln \left (-i x +1\right )}{4}+\frac {i \dilog \left (\frac {1}{2}-\frac {i x}{2}\right )}{4}-\frac {i \ln \left (-i x +1\right )^{2}}{8}-\frac {i \ln \left (i x +1\right ) x^{2}}{4}+\frac {i \ln \left (\frac {1}{2}-\frac {i x}{2}\right ) \ln \left (i x +1\right )}{4}-\frac {i \dilog \left (\frac {1}{2}+\frac {i x}{2}\right )}{4}+\frac {i \ln \left (i x +1\right )^{2}}{8}\)	\(113\)
default	\(\frac {x^{2} \arctan \relax (x )}{2}-\frac {\arctan \relax (x ) \ln \left (x^{2}+1\right )}{2}-\frac {x}{2}+\frac {\arctan \relax (x )}{2}-\frac {i \ln \left (x -i\right ) \ln \left (x^{2}+1\right )}{4}+\frac {i \dilog \left (-\frac {i \left (x +i\right )}{2}\right )}{4}+\frac {i \ln \left (x -i\right ) \ln \left (-\frac {i \left (x +i\right )}{2}\right )}{4}+\frac {i \ln \left (x -i\right )^{2}}{8}+\frac {i \ln \left (x +i\right ) \ln \left (x^{2}+1\right )}{4}-\frac {i \dilog \left (\frac {i \left (x -i\right )}{2}\right )}{4}-\frac {i \ln \left (x +i\right ) \ln \left (\frac {i \left (x -i\right )}{2}\right )}{4}-\frac {i \ln \left (x +i\right )^{2}}{8}\)	\(128\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arctan(x)/(x^2+1),x,method=_RETURNVERBOSE)

[Out]

1/4*I*ln(1-I*x)*x^2+1/2*arctan(x)-1/2*x-1/4*I*ln(1/2+1/2*I*x)*ln(1-I*x)+1/4*I*dilog(1/2-1/2*I*x)-1/8*I*ln(1-I*
x)^2-1/4*I*ln(1+I*x)*x^2+1/4*I*ln(1/2-1/2*I*x)*ln(1+I*x)-1/4*I*dilog(1/2+1/2*I*x)+1/8*I*ln(1+I*x)^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \arctan \relax (x)}{x^{2} + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctan(x)/(x^2+1),x, algorithm="maxima")

[Out]

integrate(x^3*arctan(x)/(x^2 + 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3\,\mathrm {atan}\relax (x)}{x^2+1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*atan(x))/(x^2 + 1),x)

[Out]

int((x^3*atan(x))/(x^2 + 1), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \operatorname {atan}{\relax (x )}}{x^{2} + 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*atan(x)/(x**2+1),x)

[Out]

Integral(x**3*atan(x)/(x**2 + 1), x)

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