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

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

[Out]

1/2*x+1/2*x^2*arccot(x)-1/2*I*arccot(x)^2-1/2*arctan(x)+arccot(x)*ln(2/(1+I*x))-1/2*I*polylog(2,1-2/(1+I*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 = {4917, 4853, 321, 203, 4921, 4855, 2402, 2315} \[ -\frac {1}{2} i \text {PolyLog}\left (2,1-\frac {2}{1+i x}\right )+\frac {1}{2} x^2 \cot ^{-1}(x)+\frac {x}{2}-\frac {1}{2} \tan ^{-1}(x)-\frac {1}{2} i \cot ^{-1}(x)^2+\log \left (\frac {2}{1+i x}\right ) \cot ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(x^3*ArcCot[x])/(1 + x^2),x]

[Out]

x/2 + (x^2*ArcCot[x])/2 - (I/2)*ArcCot[x]^2 - ArcTan[x]/2 + ArcCot[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 4853

Int[((a_.) + ArcCot[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcCo
t[c*x])^p)/(d*(m + 1)), x] + Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcCot[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 4855

Int[((a_.) + ArcCot[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcCot[c*x])^p*Lo
g[2/(1 + (e*x)/d)])/e, x] - Dist[(b*c*p)/e, Int[((a + b*ArcCot[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 4917

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

Rule 4921

Int[(((a_.) + ArcCot[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(I*(a + b*ArcCot[
c*x])^(p + 1))/(b*e*(p + 1)), x] - Dist[1/(c*d), Int[(a + b*ArcCot[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 \cot ^{-1}(x)}{1+x^2} \, dx &=\int x \cot ^{-1}(x) \, dx-\int \frac {x \cot ^{-1}(x)}{1+x^2} \, dx\\ &=\frac {1}{2} x^2 \cot ^{-1}(x)-\frac {1}{2} i \cot ^{-1}(x)^2+\frac {1}{2} \int \frac {x^2}{1+x^2} \, dx+\int \frac {\cot ^{-1}(x)}{i-x} \, dx\\ &=\frac {x}{2}+\frac {1}{2} x^2 \cot ^{-1}(x)-\frac {1}{2} i \cot ^{-1}(x)^2+\cot ^{-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} x^2 \cot ^{-1}(x)-\frac {1}{2} i \cot ^{-1}(x)^2-\frac {1}{2} \tan ^{-1}(x)+\cot ^{-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} x^2 \cot ^{-1}(x)-\frac {1}{2} i \cot ^{-1}(x)^2-\frac {1}{2} \tan ^{-1}(x)+\cot ^{-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 [B]  time = 0.07, size = 241, normalized size = 3.60 \[ -\frac {1}{4} i \text {Li}_2\left (-\frac {1}{2} i (i-x)\right )+\frac {1}{4} i \text {Li}_2\left (-\frac {1}{2} i (x+i)\right )+\frac {1}{2} x^2 \cot ^{-1}(x)+\frac {x}{2}+\frac {1}{8} i \log ^2(-x+i)-\frac {1}{8} i \log ^2(x+i)-\frac {1}{4} i \log (-x+i) \log \left (-\frac {-x+i}{x}\right )-\frac {1}{4} i \log (-x+i) \log \left (-\frac {1}{2} i (x+i)\right )+\frac {1}{4} i \log \left (-\frac {1}{2} i (-x+i)\right ) \log (x+i)-\frac {1}{4} i \log \left (-\frac {-x+i}{x}\right ) \log (x+i)+\frac {1}{4} i \log (-x+i) \log \left (\frac {x+i}{x}\right )+\frac {1}{4} i \log (x+i) \log \left (\frac {x+i}{x}\right )-\frac {1}{2} \tan ^{-1}(x) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

x/2 + (x^2*ArcCot[x])/2 - ArcTan[x]/2 + (I/8)*Log[I - x]^2 - (I/4)*Log[I - x]*Log[-((I - x)/x)] - (I/4)*Log[I
- x]*Log[(-1/2*I)*(I + x)] + (I/4)*Log[(-1/2*I)*(I - x)]*Log[I + x] - (I/4)*Log[-((I - x)/x)]*Log[I + x] - (I/
8)*Log[I + x]^2 + (I/4)*Log[I - x]*Log[(I + x)/x] + (I/4)*Log[I + x]*Log[(I + x)/x] - (I/4)*PolyLog[2, (-1/2*I
)*(I - x)] + (I/4)*PolyLog[2, (-1/2*I)*(I + x)]

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fricas [F]  time = 0.39, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{3} \operatorname {arccot}\relax (x)}{x^{2} + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^3*arccot(x)/(x^2 + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3*arccot(x)/(x^2 + 1), x)

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maple [B]  time = 0.18, size = 128, normalized size = 1.91 \[ \frac {x^{2} \mathrm {arccot}\relax (x )}{2}-\frac {\mathrm {arccot}\relax (x ) \ln \left (x^{2}+1\right )}{2}+\frac {x}{2}-\frac {\arctan \relax (x )}{2}+\frac {i \ln \left (x -i\right ) \ln \left (x^{2}+1\right )}{4}-\frac {i \ln \left (x -i\right )^{2}}{8}-\frac {i \dilog \left (-\frac {i \left (x +i\right )}{2}\right )}{4}-\frac {i \ln \left (x -i\right ) \ln \left (-\frac {i \left (x +i\right )}{2}\right )}{4}-\frac {i \ln \left (x +i\right ) \ln \left (x^{2}+1\right )}{4}+\frac {i \ln \left (x +i\right )^{2}}{8}+\frac {i \dilog \left (\frac {i \left (x -i\right )}{2}\right )}{4}+\frac {i \ln \left (x +i\right ) \ln \left (\frac {i \left (x -i\right )}{2}\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arccot(x)/(x^2+1),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3*arccot(x)/(x^2 + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*acot(x))/(x^2 + 1),x)

[Out]

int((x^3*acot(x))/(x^2 + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*acot(x)/(x**2 + 1), x)

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