3.675 \(\int \frac {x^5 \tan ^{-1}(x)}{(1+x^2)^2} \, dx\)

Optimal. Leaf size=89 \[ i \operatorname {PolyLog}\left (2,1-\frac {2}{1+i x}\right )+\frac {x}{4 \left (x^2+1\right )}+\frac {1}{2} x^2 \tan ^{-1}(x)-\frac {\tan ^{-1}(x)}{2 \left (x^2+1\right )}-\frac {x}{2}+i \tan ^{-1}(x)^2+\frac {3}{4} \tan ^{-1}(x)+2 \log \left (\frac {2}{1+i x}\right ) \tan ^{-1}(x) \]

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Rubi [A]  time = 0.23, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 11, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.846, Rules used = {4964, 4916, 4852, 321, 203, 4920, 4854, 2402, 2315, 4930, 199} \[ i \text {PolyLog}\left (2,1-\frac {2}{1+i x}\right )+\frac {x}{4 \left (x^2+1\right )}+\frac {1}{2} x^2 \tan ^{-1}(x)-\frac {\tan ^{-1}(x)}{2 \left (x^2+1\right )}-\frac {x}{2}+i \tan ^{-1}(x)^2+\frac {3}{4} \tan ^{-1}(x)+2 \log \left (\frac {2}{1+i x}\right ) \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(x^5*ArcTan[x])/(1 + x^2)^2,x]

[Out]

-x/2 + x/(4*(1 + x^2)) + (3*ArcTan[x])/4 + (x^2*ArcTan[x])/2 - ArcTan[x]/(2*(1 + x^2)) + I*ArcTan[x]^2 + 2*Arc
Tan[x]*Log[2/(1 + I*x)] + I*PolyLog[2, 1 - 2/(1 + I*x)]

Rule 199

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] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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]

Rule 4930

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 4964

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^5 \tan ^{-1}(x)}{\left (1+x^2\right )^2} \, dx &=-\int \frac {x^3 \tan ^{-1}(x)}{\left (1+x^2\right )^2} \, dx+\int \frac {x^3 \tan ^{-1}(x)}{1+x^2} \, dx\\ &=\int x \tan ^{-1}(x) \, dx+\int \frac {x \tan ^{-1}(x)}{\left (1+x^2\right )^2} \, dx-2 \int \frac {x \tan ^{-1}(x)}{1+x^2} \, dx\\ &=\frac {1}{2} x^2 \tan ^{-1}(x)-\frac {\tan ^{-1}(x)}{2 \left (1+x^2\right )}+\frac {1}{2} \int \frac {1}{\left (1+x^2\right )^2} \, dx-\frac {1}{2} \int \frac {x^2}{1+x^2} \, dx-2 \left (-\frac {1}{2} i \tan ^{-1}(x)^2-\int \frac {\tan ^{-1}(x)}{i-x} \, dx\right )\\ &=-\frac {x}{2}+\frac {x}{4 \left (1+x^2\right )}+\frac {1}{2} x^2 \tan ^{-1}(x)-\frac {\tan ^{-1}(x)}{2 \left (1+x^2\right )}+\frac {1}{4} \int \frac {1}{1+x^2} \, dx+\frac {1}{2} \int \frac {1}{1+x^2} \, dx-2 \left (-\frac {1}{2} i \tan ^{-1}(x)^2-\tan ^{-1}(x) \log \left (\frac {2}{1+i x}\right )+\int \frac {\log \left (\frac {2}{1+i x}\right )}{1+x^2} \, dx\right )\\ &=-\frac {x}{2}+\frac {x}{4 \left (1+x^2\right )}+\frac {3}{4} \tan ^{-1}(x)+\frac {1}{2} x^2 \tan ^{-1}(x)-\frac {\tan ^{-1}(x)}{2 \left (1+x^2\right )}-2 \left (-\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 )\right )\\ &=-\frac {x}{2}+\frac {x}{4 \left (1+x^2\right )}+\frac {3}{4} \tan ^{-1}(x)+\frac {1}{2} x^2 \tan ^{-1}(x)-\frac {\tan ^{-1}(x)}{2 \left (1+x^2\right )}-2 \left (-\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 )\right )\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 70, normalized size = 0.79 \[ \frac {1}{8} \left (-8 i \operatorname {PolyLog}\left (2,-e^{2 i \tan ^{-1}(x)}\right )+4 \left (x^2+1\right ) \tan ^{-1}(x)-4 x-8 i \tan ^{-1}(x)^2+16 \tan ^{-1}(x) \log \left (1+e^{2 i \tan ^{-1}(x)}\right )+\sin \left (2 \tan ^{-1}(x)\right )-2 \tan ^{-1}(x) \cos \left (2 \tan ^{-1}(x)\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^5*ArcTan[x])/(1 + x^2)^2,x]

[Out]

(-4*x + 4*(1 + x^2)*ArcTan[x] - (8*I)*ArcTan[x]^2 - 2*ArcTan[x]*Cos[2*ArcTan[x]] + 16*ArcTan[x]*Log[1 + E^((2*
I)*ArcTan[x])] - (8*I)*PolyLog[2, -E^((2*I)*ArcTan[x])] + Sin[2*ArcTan[x]])/8

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^5 \tan ^{-1}(x)}{\left (1+x^2\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[(x^5*ArcTan[x])/(1 + x^2)^2,x]

[Out]

Could not integrate

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fricas [F]  time = 1.01, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{5} \arctan \relax (x)}{x^{4} + 2 \, x^{2} + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*arctan(x)/(x^2+1)^2,x, algorithm="fricas")

[Out]

integral(x^5*arctan(x)/(x^4 + 2*x^2 + 1), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{5} \arctan \relax (x)}{{\left (x^{2} + 1\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*arctan(x)/(x^2+1)^2,x, algorithm="giac")

[Out]

integrate(x^5*arctan(x)/(x^2 + 1)^2, x)

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maple [A]  time = 0.34, size = 149, normalized size = 1.67

method	result	size
default	\(\frac {x^{2} \arctan \relax (x )}{2}-\arctan \relax (x ) \ln \left (x^{2}+1\right )-\frac {\arctan \relax (x )}{2 \left (x^{2}+1\right )}-\frac {x}{2}+\frac {x}{4 x^{2}+4}+\frac {3 \arctan \relax (x )}{4}-\frac {i \ln \left (x -i\right ) \ln \left (x^{2}+1\right )}{2}+\frac {i \dilog \left (-\frac {i \left (x +i\right )}{2}\right )}{2}+\frac {i \ln \left (x -i\right ) \ln \left (-\frac {i \left (x +i\right )}{2}\right )}{2}+\frac {i \ln \left (x -i\right )^{2}}{4}+\frac {i \ln \left (x +i\right ) \ln \left (x^{2}+1\right )}{2}-\frac {i \dilog \left (\frac {i \left (x -i\right )}{2}\right )}{2}-\frac {i \ln \left (x +i\right ) \ln \left (\frac {i \left (x -i\right )}{2}\right )}{2}-\frac {i \ln \left (x +i\right )^{2}}{4}\)	\(149\)
risch	\(-\frac {x}{2}+\frac {5 \arctan \relax (x )}{8}-\frac {i \dilog \left (\frac {1}{2}+\frac {i x}{2}\right )}{2}-\frac {i \ln \left (i x +1\right ) x^{2}}{4}+\frac {i}{8 i x +8}-\frac {i \ln \left (-i x +1\right )}{8 \left (-i x +1\right )}+\frac {i \dilog \left (\frac {1}{2}-\frac {i x}{2}\right )}{2}+\frac {i \ln \left (-i x +1\right ) x^{2}}{4}-\frac {i \ln \left (-i x +1\right )^{2}}{4}+\frac {i \ln \left (i x +1\right )^{2}}{4}+\frac {\ln \left (i x +1\right ) x}{16 i x -16}+\frac {\ln \left (-i x +1\right ) x}{-16 i x -16}-\frac {i \ln \left (\frac {1}{2}+\frac {i x}{2}\right ) \ln \left (-i x +1\right )}{2}+\frac {i \ln \left (-i x +1\right )}{-16 i x -16}-\frac {i \ln \left (i x +1\right )}{16 \left (i x -1\right )}+\frac {i \ln \left (\frac {1}{2}-\frac {i x}{2}\right ) \ln \left (i x +1\right )}{2}-\frac {i}{8 \left (-i x +1\right )}+\frac {i \ln \left (i x +1\right )}{8 i x +8}\)	\(243\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*arctan(x)/(x^2+1)^2,x,method=_RETURNVERBOSE)

[Out]

1/2*x^2*arctan(x)-arctan(x)*ln(x^2+1)-1/2*arctan(x)/(x^2+1)-1/2*x+1/4*x/(x^2+1)+3/4*arctan(x)-1/2*I*ln(x-I)*ln
(x^2+1)+1/2*I*dilog(-1/2*I*(x+I))+1/2*I*ln(x-I)*ln(-1/2*I*(x+I))+1/4*I*ln(x-I)^2+1/2*I*ln(x+I)*ln(x^2+1)-1/2*I
*dilog(1/2*I*(x-I))-1/2*I*ln(x+I)*ln(1/2*I*(x-I))-1/4*I*ln(x+I)^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{5} \arctan \relax (x)}{{\left (x^{2} + 1\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*arctan(x)/(x^2+1)^2,x, algorithm="maxima")

[Out]

integrate(x^5*arctan(x)/(x^2 + 1)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^5\,\mathrm {atan}\relax (x)}{{\left (x^2+1\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5*atan(x))/(x^2 + 1)^2,x)

[Out]

int((x^5*atan(x))/(x^2 + 1)^2, x)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RecursionError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*atan(x)/(x**2+1)**2,x)

[Out]

Exception raised: RecursionError

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