3.7.72 \(\int \frac {x^3 \tan ^{-1}(x)}{1+x^2} \, dx\) [672]

Optimal. Leaf size=67 \[ -\frac {x}{2}+\frac {1}{2} \tan ^{-1}(x)+\frac {1}{2} x^2 \tan ^{-1}(x)+\frac {1}{2} i \tan ^{-1}(x)^2+\tan ^{-1}(x) \log \left (\frac {2}{1+i x}\right )+\frac {1}{2} i \text {Li}_2\left (1-\frac {2}{1+i x}\right ) \]

[Out]

-1/2*x+1/2*arctan(x)+1/2*x^2*arctan(x)+1/2*I*arctan(x)^2+arctan(x)*ln(2/(1+I*x))+1/2*I*polylog(2,1-2/(1+I*x))

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Rubi [A]
time = 0.07, 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 = {5036, 4946, 327, 209, 5040, 4964, 2449, 2352} \begin {gather*} \frac {1}{2} i \text {PolyLog}\left (2,1-\frac {2}{1+i x}\right )+\frac {1}{2} x^2 \text {ArcTan}(x)+\frac {1}{2} i \text {ArcTan}(x)^2+\frac {\text {ArcTan}(x)}{2}+\text {ArcTan}(x) \log \left (\frac {2}{1+i x}\right )-\frac {x}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3*ArcTan[x])/(1 + x^2),x]

[Out]

-1/2*x + 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 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 327

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 2352

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

Rule 2449

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 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 4964

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTan[c*x])^p)*(
Log[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 5036

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 5040

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-I)*((a + b*ArcT
an[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 \text {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 = 0.02, size = 57, normalized size = 0.85 \begin {gather*} \frac {1}{2} \left (-x+i \tan ^{-1}(x)^2+\tan ^{-1}(x) \left (1+x^2+2 \log \left (-\frac {2 i}{-i+x}\right )\right )+i \text {Li}_2\left (\frac {i+x}{-i+x}\right )\right ) \end {gather*}

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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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 127 vs. \(2 (53 ) = 106\).
time = 0.08, size = 128, normalized size = 1.91

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

[Out]

integral(x^3*arctan(x)/(x^2 + 1), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3} \operatorname {atan}{\left (x \right )}}{x^{2} + 1}\, dx \end {gather*}

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

[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 \begin {gather*} \int \frac {x^3\,\mathrm {atan}\left (x\right )}{x^2+1} \,d x \end {gather*}

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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