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

3.7.75.1 Optimal result

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

output

-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))
 

3.7.75.2 Mathematica [A] (verified)

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

input

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

output

(-4*x + 4*(1 + x^2)*ArcTan[x] - (8*I)*ArcTan[x]^2 - 2*ArcTan[x]*Cos[2*ArcT 
an[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
 

3.7.75.3 Rubi [A] (verified)

Time = 1.08 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.11, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.308, Rules used = {5499, 5451, 5361, 262, 216, 5455, 5379, 2849, 2752, 5499, 5455, 5379, 2849, 2752, 5465, 215, 216}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x^5 \arctan (x)}{\left (x^2+1\right )^2} \, dx\)

\(\Big \downarrow \) 5499

\(\displaystyle \int \frac {x^3 \arctan (x)}{x^2+1}dx-\int \frac {x^3 \arctan (x)}{\left (x^2+1\right )^2}dx\)

\(\Big \downarrow \) 5451

\(\displaystyle -\int \frac {x \arctan (x)}{x^2+1}dx-\int \frac {x^3 \arctan (x)}{\left (x^2+1\right )^2}dx+\int x \arctan (x)dx\)

\(\Big \downarrow \) 5361

\(\displaystyle -\int \frac {x \arctan (x)}{x^2+1}dx-\int \frac {x^3 \arctan (x)}{\left (x^2+1\right )^2}dx-\frac {1}{2} \int \frac {x^2}{x^2+1}dx+\frac {1}{2} x^2 \arctan (x)\)

\(\Big \downarrow \) 262

\(\displaystyle -\int \frac {x \arctan (x)}{x^2+1}dx-\int \frac {x^3 \arctan (x)}{\left (x^2+1\right )^2}dx+\frac {1}{2} \left (\int \frac {1}{x^2+1}dx-x\right )+\frac {1}{2} x^2 \arctan (x)\)

\(\Big \downarrow \) 216

\(\displaystyle -\int \frac {x \arctan (x)}{x^2+1}dx-\int \frac {x^3 \arctan (x)}{\left (x^2+1\right )^2}dx+\frac {1}{2} x^2 \arctan (x)+\frac {1}{2} (\arctan (x)-x)\)

\(\Big \downarrow \) 5455

\(\displaystyle -\int \frac {x^3 \arctan (x)}{\left (x^2+1\right )^2}dx+\int \frac {\arctan (x)}{i-x}dx+\frac {1}{2} x^2 \arctan (x)+\frac {1}{2} i \arctan (x)^2+\frac {1}{2} (\arctan (x)-x)\)

\(\Big \downarrow \) 5379

\(\displaystyle -\int \frac {x^3 \arctan (x)}{\left (x^2+1\right )^2}dx-\int \frac {\log \left (\frac {2}{i x+1}\right )}{x^2+1}dx+\frac {1}{2} x^2 \arctan (x)+\frac {1}{2} i \arctan (x)^2+\frac {1}{2} (\arctan (x)-x)+\arctan (x) \log \left (\frac {2}{1+i x}\right )\)

\(\Big \downarrow \) 2849

\(\displaystyle -\int \frac {x^3 \arctan (x)}{\left (x^2+1\right )^2}dx+i \int \frac {\log \left (\frac {2}{i x+1}\right )}{1-\frac {2}{i x+1}}d\frac {1}{i x+1}+\frac {1}{2} x^2 \arctan (x)+\frac {1}{2} i \arctan (x)^2+\frac {1}{2} (\arctan (x)-x)+\arctan (x) \log \left (\frac {2}{1+i x}\right )\)

\(\Big \downarrow \) 2752

\(\displaystyle -\int \frac {x^3 \arctan (x)}{\left (x^2+1\right )^2}dx+\frac {1}{2} x^2 \arctan (x)+\frac {1}{2} i \arctan (x)^2+\frac {1}{2} (\arctan (x)-x)+\arctan (x) \log \left (\frac {2}{1+i x}\right )+\frac {1}{2} i \operatorname {PolyLog}\left (2,1-\frac {2}{i x+1}\right )\)

\(\Big \downarrow \) 5499

\(\displaystyle \int \frac {x \arctan (x)}{\left (x^2+1\right )^2}dx-\int \frac {x \arctan (x)}{x^2+1}dx+\frac {1}{2} x^2 \arctan (x)+\frac {1}{2} i \arctan (x)^2+\frac {1}{2} (\arctan (x)-x)+\arctan (x) \log \left (\frac {2}{1+i x}\right )+\frac {1}{2} i \operatorname {PolyLog}\left (2,1-\frac {2}{i x+1}\right )\)

\(\Big \downarrow \) 5455

\(\displaystyle \int \frac {x \arctan (x)}{\left (x^2+1\right )^2}dx+\int \frac {\arctan (x)}{i-x}dx+\frac {1}{2} x^2 \arctan (x)+i \arctan (x)^2+\frac {1}{2} (\arctan (x)-x)+\arctan (x) \log \left (\frac {2}{1+i x}\right )+\frac {1}{2} i \operatorname {PolyLog}\left (2,1-\frac {2}{i x+1}\right )\)

\(\Big \downarrow \) 5379

\(\displaystyle \int \frac {x \arctan (x)}{\left (x^2+1\right )^2}dx-\int \frac {\log \left (\frac {2}{i x+1}\right )}{x^2+1}dx+\frac {1}{2} x^2 \arctan (x)+i \arctan (x)^2+\frac {1}{2} (\arctan (x)-x)+2 \arctan (x) \log \left (\frac {2}{1+i x}\right )+\frac {1}{2} i \operatorname {PolyLog}\left (2,1-\frac {2}{i x+1}\right )\)

\(\Big \downarrow \) 2849

\(\displaystyle \int \frac {x \arctan (x)}{\left (x^2+1\right )^2}dx+i \int \frac {\log \left (\frac {2}{i x+1}\right )}{1-\frac {2}{i x+1}}d\frac {1}{i x+1}+\frac {1}{2} x^2 \arctan (x)+i \arctan (x)^2+\frac {1}{2} (\arctan (x)-x)+2 \arctan (x) \log \left (\frac {2}{1+i x}\right )+\frac {1}{2} i \operatorname {PolyLog}\left (2,1-\frac {2}{i x+1}\right )\)

\(\Big \downarrow \) 2752

\(\displaystyle \int \frac {x \arctan (x)}{\left (x^2+1\right )^2}dx+\frac {1}{2} x^2 \arctan (x)+i \arctan (x)^2+\frac {1}{2} (\arctan (x)-x)+2 \arctan (x) \log \left (\frac {2}{1+i x}\right )+i \operatorname {PolyLog}\left (2,1-\frac {2}{i x+1}\right )\)

\(\Big \downarrow \) 5465

\(\displaystyle \frac {1}{2} \int \frac {1}{\left (x^2+1\right )^2}dx+\frac {1}{2} x^2 \arctan (x)-\frac {\arctan (x)}{2 \left (x^2+1\right )}+i \arctan (x)^2+\frac {1}{2} (\arctan (x)-x)+2 \arctan (x) \log \left (\frac {2}{1+i x}\right )+i \operatorname {PolyLog}\left (2,1-\frac {2}{i x+1}\right )\)

\(\Big \downarrow \) 215

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x^2+1}dx+\frac {x}{2 \left (x^2+1\right )}\right )+\frac {1}{2} x^2 \arctan (x)-\frac {\arctan (x)}{2 \left (x^2+1\right )}+i \arctan (x)^2+\frac {1}{2} (\arctan (x)-x)+2 \arctan (x) \log \left (\frac {2}{1+i x}\right )+i \operatorname {PolyLog}\left (2,1-\frac {2}{i x+1}\right )\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{2} x^2 \arctan (x)+\frac {1}{2} \left (\frac {\arctan (x)}{2}+\frac {x}{2 \left (x^2+1\right )}\right )-\frac {\arctan (x)}{2 \left (x^2+1\right )}+i \arctan (x)^2+\frac {1}{2} (\arctan (x)-x)+2 \arctan (x) \log \left (\frac {2}{1+i x}\right )+i \operatorname {PolyLog}\left (2,1-\frac {2}{i x+1}\right )\)

input

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

output

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

3.7.75.3.1 Defintions of rubi rules used

rule 215

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) 
/(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1))   Int[(a + b*x^2)^(p + 1 
), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 
*p])
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c 
, 2, m, p, x]
 

rule 2752

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

rule 2849

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Simp 
[-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 5361

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] - Simp[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 5379

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] + Simp[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 5451

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e 
_.)*(x_)^2), x_Symbol] :> Simp[f^2/e   Int[(f*x)^(m - 2)*(a + b*ArcTan[c*x] 
)^p, x], x] - Simp[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 5455

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] - Si 
mp[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 5465

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] - Simp[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 5499

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_)*((d_) + (e_.)*(x_)^2 
)^(q_), x_Symbol] :> Simp[1/e   Int[x^(m - 2)*(d + e*x^2)^(q + 1)*(a + b*Ar 
cTan[c*x])^p, x], x] - Simp[d/e   Int[x^(m - 2)*(d + e*x^2)^q*(a + b*ArcTan 
[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]
 

3.7.75.4 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	\(-\frac {x}{2}-\frac {i \ln \left (i x +1\right )}{4}-\frac {i \ln \left (-i x +1\right )^{2}}{4}+\frac {i \ln \left (\frac {1}{2}-\frac {i x}{2}\right ) \ln \left (i x +1\right )}{2}+\frac {\arctan \left (x \right )}{8}+\frac {\ln \left (-i x +1\right ) x}{-16 i x -16}-\frac {i}{8 \left (-i x +1\right )}+\frac {i \operatorname {dilog}\left (\frac {1}{2}-\frac {i x}{2}\right )}{2}-\frac {i \ln \left (\frac {1}{2}+\frac {i x}{2}\right ) \ln \left (-i x +1\right )}{2}+\frac {i \ln \left (i x +1\right )^{2}}{4}+\frac {i x^{2} \ln \left (-i x +1\right )}{4}-\frac {i \ln \left (-i x +1\right )}{8 \left (-i x +1\right )}+\frac {\ln \left (i x +1\right ) x}{16 i x -16}-\frac {i \operatorname {dilog}\left (\frac {1}{2}+\frac {i x}{2}\right )}{2}+\frac {i \ln \left (-i x +1\right )}{4}+\frac {i \ln \left (-i x +1\right )}{-16 i x -16}+\frac {i}{8 i x +8}-\frac {i \ln \left (i x +1\right )}{16 \left (i x -1\right )}+\frac {i \ln \left (i x +1\right )}{8 i x +8}-\frac {i x^{2} \ln \left (i x +1\right )}{4}\)	\(263\)

input

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

output

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*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-dilo 
g(1/2*I*(x-I))-ln(x+I)*ln(1/2*I*(x-I)))
 

3.7.75.5 Fricas [F]

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

input

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

output

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

3.7.75.6 Sympy [F(-2)]

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

input

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

output

Exception raised: RecursionError >> maximum recursion depth exceeded in co 
mparison
 

3.7.75.7 Maxima [F]

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

input

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

output

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

3.7.75.8 Giac [F]

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

input

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

output

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

3.7.75.9 Mupad [F(-1)]

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

input

int((x^5*atan(x))/(x^2 + 1)^2,x)
 

output

int((x^5*atan(x))/(x^2 + 1)^2, x)
 

3.7.75.10 Reduce [F]

\[ \int \frac {x^5 \arctan (x)}{\left (1+x^2\right )^2} \, dx=\int \frac {\mathit {atan} \left (x \right ) x^{5}}{x^{4}+2 x^{2}+1}d x \]

input

int((atan(x)*x**5)/(x**4 + 2*x**2 + 1),x)
 

output

int((atan(x)*x**5)/(x**4 + 2*x**2 + 1),x)