3.7.74 \(\int \frac {x^3 \tan ^{-1}(x)}{(1+x^2)^2} \, dx\) [674]

Optimal. Leaf size=79 \[ -\frac {x}{4 \left (1+x^2\right )}-\frac {1}{4} \tan ^{-1}(x)+\frac {\tan ^{-1}(x)}{2 \left (1+x^2\right )}-\frac {1}{2} i \tan ^{-1}(x)^2-\tan ^{-1}(x) \log \left (\frac {2}{1+i x}\right )-\frac {1}{2} i \text {Li}_2\left (1-\frac {2}{1+i x}\right ) \]

[Out]

-1/4*x/(x^2+1)-1/4*arctan(x)+1/2*arctan(x)/(x^2+1)-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 = 0.08, antiderivative size = 79, 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 = {5084, 5040, 4964, 2449, 2352, 5050, 205, 209} \begin {gather*} -\frac {1}{2} i \text {PolyLog}\left (2,1-\frac {2}{1+i x}\right )+\frac {\text {ArcTan}(x)}{2 \left (x^2+1\right )}-\frac {1}{2} i \text {ArcTan}(x)^2-\frac {\text {ArcTan}(x)}{4}-\text {ArcTan}(x) \log \left (\frac {2}{1+i x}\right )-\frac {x}{4 \left (x^2+1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3*ArcTan[x])/(1 + x^2)^2,x]

[Out]

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

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

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 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 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 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]

Rule 5050

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

Rule 5084

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[1/e, Int[
x^(m - 2)*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x])^p, x], x] - Dist[d/e, Int[x^(m - 2)*(d + e*x^2)^q*(a + b*Arc
Tan[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IntegersQ[p, 2*q] && LtQ[q, -1] && IGtQ[m
, 1] && NeQ[p, -1]

Rubi steps

\begin {align*} \int \frac {x^3 \tan ^{-1}(x)}{\left (1+x^2\right )^2} \, dx &=-\int \frac {x \tan ^{-1}(x)}{\left (1+x^2\right )^2} \, dx+\int \frac {x \tan ^{-1}(x)}{1+x^2} \, dx\\ &=\frac {\tan ^{-1}(x)}{2 \left (1+x^2\right )}-\frac {1}{2} i \tan ^{-1}(x)^2-\frac {1}{2} \int \frac {1}{\left (1+x^2\right )^2} \, dx-\int \frac {\tan ^{-1}(x)}{i-x} \, dx\\ &=-\frac {x}{4 \left (1+x^2\right )}+\frac {\tan ^{-1}(x)}{2 \left (1+x^2\right )}-\frac {1}{2} i \tan ^{-1}(x)^2-\tan ^{-1}(x) \log \left (\frac {2}{1+i x}\right )-\frac {1}{4} \int \frac {1}{1+x^2} \, dx+\int \frac {\log \left (\frac {2}{1+i x}\right )}{1+x^2} \, dx\\ &=-\frac {x}{4 \left (1+x^2\right )}-\frac {1}{4} \tan ^{-1}(x)+\frac {\tan ^{-1}(x)}{2 \left (1+x^2\right )}-\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}{4 \left (1+x^2\right )}-\frac {1}{4} \tan ^{-1}(x)+\frac {\tan ^{-1}(x)}{2 \left (1+x^2\right )}-\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 = 0.04, size = 64, normalized size = 0.81 \begin {gather*} \frac {1}{2} i \tan ^{-1}(x)^2+\frac {1}{4} \tan ^{-1}(x) \cos \left (2 \tan ^{-1}(x)\right )-\tan ^{-1}(x) \log \left (1+e^{2 i \tan ^{-1}(x)}\right )+\frac {1}{2} i \text {Li}_2\left (-e^{2 i \tan ^{-1}(x)}\right )-\frac {1}{8} \sin \left (2 \tan ^{-1}(x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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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. 138 vs. \(2 (65 ) = 130\).
time = 0.09, size = 139, normalized size = 1.76

method	result	size
default	\(\frac {\arctan \left (x \right )}{2 x^{2}+2}+\frac {\arctan \left (x \right ) \ln \left (x^{2}+1\right )}{2}-\frac {x}{4 \left (x^{2}+1\right )}-\frac {\arctan \left (x \right )}{4}+\frac {i \ln \left (x -i\right ) \ln \left (x^{2}+1\right )}{4}-\frac {i \dilog \left (-\frac {i \left (x +i\right )}{2}\right )}{4}-\frac {i \ln \left (x -i\right ) \ln \left (-\frac {i \left (x +i\right )}{2}\right )}{4}-\frac {i \ln \left (x -i\right )^{2}}{8}-\frac {i \ln \left (x +i\right ) \ln \left (x^{2}+1\right )}{4}+\frac {i \dilog \left (\frac {i \left (x -i\right )}{2}\right )}{4}+\frac {i \ln \left (x +i\right ) \ln \left (\frac {i \left (x -i\right )}{2}\right )}{4}+\frac {i \ln \left (x +i\right )^{2}}{8}\)	\(139\)
risch	\(\frac {i}{-8 i x +8}+\frac {i \ln \left (-i x +1\right )^{2}}{8}-\frac {i \ln \left (\frac {1}{2}-\frac {i x}{2}\right ) \ln \left (i x +1\right )}{4}+\frac {i \ln \left (\frac {1}{2}+\frac {i x}{2}\right ) \ln \left (-i x +1\right )}{4}-\frac {i \ln \left (i x +1\right )}{8 \left (i x +1\right )}-\frac {\arctan \left (x \right )}{8}-\frac {\ln \left (-i x +1\right ) x}{16 \left (-i x -1\right )}+\frac {i \ln \left (-i x +1\right )}{-8 i x +8}-\frac {i \dilog \left (\frac {1}{2}-\frac {i x}{2}\right )}{4}-\frac {i}{8 \left (i x +1\right )}+\frac {i \ln \left (i x +1\right )}{16 i x -16}-\frac {i \ln \left (i x +1\right )^{2}}{8}+\frac {i \dilog \left (\frac {1}{2}+\frac {i x}{2}\right )}{4}-\frac {i \ln \left (-i x +1\right )}{16 \left (-i x -1\right )}-\frac {\ln \left (i x +1\right ) x}{16 \left (i x -1\right )}\)	\(214\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*arctan(x)/(x^2+1)+1/2*arctan(x)*ln(x^2+1)-1/4*x/(x^2+1)-1/4*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)^2,x, algorithm="maxima")

[Out]

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

[Out]

integral(x^3*arctan(x)/(x^4 + 2*x^2 + 1), x)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RecursionError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RecursionError >> maximum recursion depth exceeded

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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)^2,x, algorithm="giac")

[Out]

integrate(x^3*arctan(x)/(x^2 + 1)^2, 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 )}{{\left (x^2+1\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*atan(x))/(x^2 + 1)^2,x)

[Out]

int((x^3*atan(x))/(x^2 + 1)^2, x)

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