\(\int \frac {x^5 \arctan (x)}{(1+x^2)^2} \, dx\) [675]

Optimal result

Integrand size = 13, antiderivative size = 89 \[ \int \frac {x^5 \arctan (x)}{\left (1+x^2\right )^2} \, dx=-\frac {x}{2}+\frac {x}{4 \left (1+x^2\right )}+\frac {3 \arctan (x)}{4}+\frac {1}{2} x^2 \arctan (x)-\frac {\arctan (x)}{2 \left (1+x^2\right )}+i \arctan (x)^2+2 \arctan (x) \log \left (\frac {2}{1+i x}\right )+i \operatorname {PolyLog}\left (2,1-\frac {2}{1+i x}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.846, Rules used = {5084, 5036, 4946, 327, 209, 5040, 4964, 2449, 2352, 5050, 205} \[ \int \frac {x^5 \arctan (x)}{\left (1+x^2\right )^2} \, dx=\frac {1}{2} x^2 \arctan (x)-\frac {\arctan (x)}{2 \left (x^2+1\right )}+i \arctan (x)^2+\frac {3 \arctan (x)}{4}+2 \arctan (x) \log \left (\frac {2}{1+i x}\right )+i \operatorname {PolyLog}\left (2,1-\frac {2}{i x+1}\right )+\frac {x}{4 \left (x^2+1\right )}-\frac {x}{2} \]

[In]

Int[(x^5*ArcTan[x])/(1 + x^2)^2,x]

[Out]

-1/2*x + 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*A
rcTan[x]*Log[2/(1 + I*x)] + I*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 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]

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*} \text {integral}& = -\int \frac {x^3 \arctan (x)}{\left (1+x^2\right )^2} \, dx+\int \frac {x^3 \arctan (x)}{1+x^2} \, dx \\ & = \int x \arctan (x) \, dx+\int \frac {x \arctan (x)}{\left (1+x^2\right )^2} \, dx-2 \int \frac {x \arctan (x)}{1+x^2} \, dx \\ & = \frac {1}{2} x^2 \arctan (x)-\frac {\arctan (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 \arctan (x)^2-\int \frac {\arctan (x)}{i-x} \, dx\right ) \\ & = -\frac {x}{2}+\frac {x}{4 \left (1+x^2\right )}+\frac {1}{2} x^2 \arctan (x)-\frac {\arctan (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 \arctan (x)^2-\arctan (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 \arctan (x)}{4}+\frac {1}{2} x^2 \arctan (x)-\frac {\arctan (x)}{2 \left (1+x^2\right )}-2 \left (-\frac {1}{2} i \arctan (x)^2-\arctan (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 )\right ) \\ & = -\frac {x}{2}+\frac {x}{4 \left (1+x^2\right )}+\frac {3 \arctan (x)}{4}+\frac {1}{2} x^2 \arctan (x)-\frac {\arctan (x)}{2 \left (1+x^2\right )}-2 \left (-\frac {1}{2} i \arctan (x)^2-\arctan (x) \log \left (\frac {2}{1+i x}\right )-\frac {1}{2} i \operatorname {PolyLog}\left (2,1-\frac {2}{1+i x}\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.79 \[ \int \frac {x^5 \arctan (x)}{\left (1+x^2\right )^2} \, dx=\frac {1}{8} \left (-4 x+4 \left (1+x^2\right ) \arctan (x)-8 i \arctan (x)^2-2 \arctan (x) \cos (2 \arctan (x))+16 \arctan (x) \log \left (1+e^{2 i \arctan (x)}\right )-8 i \operatorname {PolyLog}\left (2,-e^{2 i \arctan (x)}\right )+\sin (2 \arctan (x))\right ) \]

[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

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.65

method	result	size
default	\(\frac {x^{2} \arctan \left (x \right )}{2}-\arctan \left (x \right ) \ln \left (x^{2}+1\right )-\frac {\arctan \left (x \right )}{2 \left (x^{2}+1\right )}-\frac {x}{2}+\frac {x}{4 x^{2}+4}+\frac {3 \arctan \left (x \right )}{4}-\frac {i \left (\ln \left (x -i\right ) \ln \left (x^{2}+1\right )-\frac {\ln \left (x -i\right )^{2}}{2}-\operatorname {dilog}\left (-\frac {i \left (x +i\right )}{2}\right )-\ln \left (x -i\right ) \ln \left (-\frac {i \left (x +i\right )}{2}\right )\right )}{2}+\frac {i \left (\ln \left (x +i\right ) \ln \left (x^{2}+1\right )-\frac {\ln \left (x +i\right )^{2}}{2}-\operatorname {dilog}\left (\frac {i \left (x -i\right )}{2}\right )-\ln \left (x +i\right ) \ln \left (\frac {i \left (x -i\right )}{2}\right )\right )}{2}\)	\(147\)
parts	\(\frac {x^{2} \arctan \left (x \right )}{2}-\arctan \left (x \right ) \ln \left (x^{2}+1\right )-\frac {\arctan \left (x \right )}{2 \left (x^{2}+1\right )}-\frac {x}{2}+\frac {x}{4 x^{2}+4}+\frac {3 \arctan \left (x \right )}{4}-\frac {i \left (\ln \left (x -i\right ) \ln \left (x^{2}+1\right )-\frac {\ln \left (x -i\right )^{2}}{2}-\operatorname {dilog}\left (-\frac {i \left (x +i\right )}{2}\right )-\ln \left (x -i\right ) \ln \left (-\frac {i \left (x +i\right )}{2}\right )\right )}{2}+\frac {i \left (\ln \left (x +i\right ) \ln \left (x^{2}+1\right )-\frac {\ln \left (x +i\right )^{2}}{2}-\operatorname {dilog}\left (\frac {i \left (x -i\right )}{2}\right )-\ln \left (x +i\right ) \ln \left (\frac {i \left (x -i\right )}{2}\right )\right )}{2}\)	\(147\)
risch	\(-\frac {x}{2}-\frac {i \ln \left (i x +1\right )}{4}-\frac {i \ln \left (-i x +1\right )^{2}}{4}+\frac {i \ln \left (\frac {1}{2}-\frac {i x}{2}\right ) \ln \left (i x +1\right )}{2}+\frac {\arctan \left (x \right )}{8}+\frac {\ln \left (-i x +1\right ) x}{-16 i x -16}-\frac {i}{8 \left (-i x +1\right )}+\frac {i \operatorname {dilog}\left (\frac {1}{2}-\frac {i x}{2}\right )}{2}-\frac {i \ln \left (\frac {1}{2}+\frac {i x}{2}\right ) \ln \left (-i x +1\right )}{2}+\frac {i \ln \left (i x +1\right )^{2}}{4}+\frac {i x^{2} \ln \left (-i x +1\right )}{4}-\frac {i \ln \left (-i x +1\right )}{8 \left (-i x +1\right )}+\frac {\ln \left (i x +1\right ) x}{16 i x -16}-\frac {i \operatorname {dilog}\left (\frac {1}{2}+\frac {i x}{2}\right )}{2}+\frac {i \ln \left (-i x +1\right )}{4}+\frac {i \ln \left (-i x +1\right )}{-16 i x -16}+\frac {i}{8 i x +8}-\frac {i \ln \left (i x +1\right )}{16 \left (i x -1\right )}+\frac {i \ln \left (i x +1\right )}{8 i x +8}-\frac {i x^{2} \ln \left (i x +1\right )}{4}\)	\(263\)

[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)*l
n(x^2+1)-1/2*ln(x-I)^2-dilog(-1/2*I*(x+I))-ln(x-I)*ln(-1/2*I*(x+I)))+1/2*I*(ln(x+I)*ln(x^2+1)-1/2*ln(x+I)^2-di
log(1/2*I*(x-I))-ln(x+I)*ln(1/2*I*(x-I)))

Fricas [F]

\[ \int \frac {x^5 \arctan (x)}{\left (1+x^2\right )^2} \, dx=\int { \frac {x^{5} \arctan \left (x\right )}{{\left (x^{2} + 1\right )}^{2}} \,d x } \]

[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)

Sympy [F(-2)]

Exception generated. \[ \int \frac {x^5 \arctan (x)}{\left (1+x^2\right )^2} \, dx=\text {Exception raised: RecursionError} \]

[In]

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

[Out]

Exception raised: RecursionError >> maximum recursion depth exceeded in comparison

Maxima [F]

\[ \int \frac {x^5 \arctan (x)}{\left (1+x^2\right )^2} \, dx=\int { \frac {x^{5} \arctan \left (x\right )}{{\left (x^{2} + 1\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {x^5 \arctan (x)}{\left (1+x^2\right )^2} \, dx=\int { \frac {x^{5} \arctan \left (x\right )}{{\left (x^{2} + 1\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^5 \arctan (x)}{\left (1+x^2\right )^2} \, dx=\int \frac {x^5\,\mathrm {atan}\left (x\right )}{{\left (x^2+1\right )}^2} \,d x \]

[In]

int((x^5*atan(x))/(x^2 + 1)^2,x)

[Out]

int((x^5*atan(x))/(x^2 + 1)^2, x)