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

3.7.74.1 Optimal result

Integrand size = 13, antiderivative size = 79 \[ \int \frac {x^3 \arctan (x)}{\left (1+x^2\right )^2} \, dx=-\frac {x}{4 \left (1+x^2\right )}-\frac {\arctan (x)}{4}+\frac {\arctan (x)}{2 \left (1+x^2\right )}-\frac {1}{2} i \arctan (x)^2-\arctan (x) \log \left (\frac {2}{1+i x}\right )-\frac {1}{2} i \operatorname {PolyLog}\left (2,1-\frac {2}{1+i x}\right ) \]

output

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

3.7.74.2 Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.81 \[ \int \frac {x^3 \arctan (x)}{\left (1+x^2\right )^2} \, dx=\frac {1}{2} i \arctan (x)^2+\frac {1}{4} \arctan (x) \cos (2 \arctan (x))-\arctan (x) \log \left (1+e^{2 i \arctan (x)}\right )+\frac {1}{2} i \operatorname {PolyLog}\left (2,-e^{2 i \arctan (x)}\right )-\frac {1}{8} \sin (2 \arctan (x)) \]

input

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

output

(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
 

3.7.74.3 Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {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.

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

\(\Big \downarrow \) 5499

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

\(\Big \downarrow \) 5455

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

\(\Big \downarrow \) 5379

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

\(\Big \downarrow \) 2849

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

\(\Big \downarrow \) 2752

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

\(\Big \downarrow \) 5465

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

\(\Big \downarrow \) 215

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

\(\Big \downarrow \) 216

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

input

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

output

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

3.7.74.3.1 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 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 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 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]
 

3.7.74.4 Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 136 vs. \(2 (65 ) = 130\).

Time = 0.35 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.73

method	result	size
default	\(\frac {\arctan \left (x \right )}{2 x^{2}+2}+\frac {\arctan \left (x \right ) \ln \left (x^{2}+1\right )}{2}-\frac {x}{4 \left (x^{2}+1\right )}-\frac {\arctan \left (x \right )}{4}+\frac {i \left (\ln \left (x -i\right ) \ln \left (x^{2}+1\right )-\frac {\ln \left (x -i\right )^{2}}{2}-\operatorname {dilog}\left (-\frac {i \left (x +i\right )}{2}\right )-\ln \left (x -i\right ) \ln \left (-\frac {i \left (x +i\right )}{2}\right )\right )}{4}-\frac {i \left (\ln \left (x +i\right ) \ln \left (x^{2}+1\right )-\frac {\ln \left (x +i\right )^{2}}{2}-\operatorname {dilog}\left (\frac {i \left (x -i\right )}{2}\right )-\ln \left (x +i\right ) \ln \left (\frac {i \left (x -i\right )}{2}\right )\right )}{4}\)	\(137\)
parts	\(\frac {\arctan \left (x \right )}{2 x^{2}+2}+\frac {\arctan \left (x \right ) \ln \left (x^{2}+1\right )}{2}-\frac {x}{4 \left (x^{2}+1\right )}-\frac {\arctan \left (x \right )}{4}+\frac {i \left (\ln \left (x -i\right ) \ln \left (x^{2}+1\right )-\frac {\ln \left (x -i\right )^{2}}{2}-\operatorname {dilog}\left (-\frac {i \left (x +i\right )}{2}\right )-\ln \left (x -i\right ) \ln \left (-\frac {i \left (x +i\right )}{2}\right )\right )}{4}-\frac {i \left (\ln \left (x +i\right ) \ln \left (x^{2}+1\right )-\frac {\ln \left (x +i\right )^{2}}{2}-\operatorname {dilog}\left (\frac {i \left (x -i\right )}{2}\right )-\ln \left (x +i\right ) \ln \left (\frac {i \left (x -i\right )}{2}\right )\right )}{4}\)	\(137\)
risch	\(\frac {i \ln \left (-i x +1\right )^{2}}{8}+\frac {i \operatorname {dilog}\left (\frac {1}{2}+\frac {i x}{2}\right )}{4}+\frac {i \ln \left (i x +1\right )}{16 i x -16}+\frac {i}{-8 i x +8}+\frac {i \ln \left (\frac {1}{2}+\frac {i x}{2}\right ) \ln \left (-i x +1\right )}{4}-\frac {i \operatorname {dilog}\left (\frac {1}{2}-\frac {i x}{2}\right )}{4}-\frac {i}{8 \left (i x +1\right )}-\frac {\arctan \left (x \right )}{8}-\frac {\ln \left (-i x +1\right ) x}{16 \left (-i x -1\right )}-\frac {i \ln \left (-i x +1\right )}{16 \left (-i x -1\right )}-\frac {i \ln \left (i x +1\right )}{8 \left (i x +1\right )}-\frac {i \ln \left (i x +1\right )^{2}}{8}+\frac {i \ln \left (-i x +1\right )}{-8 i x +8}-\frac {i \ln \left (\frac {1}{2}-\frac {i x}{2}\right ) \ln \left (i x +1\right )}{4}-\frac {\ln \left (i x +1\right ) x}{16 \left (i x -1\right )}\)	\(214\)

input

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

output

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

3.7.74.5 Fricas [F]

\[ \int \frac {x^3 \arctan (x)}{\left (1+x^2\right )^2} \, dx=\int { \frac {x^{3} \arctan \left (x\right )}{{\left (x^{2} + 1\right )}^{2}} \,d x } \]

input

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

output

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

3.7.74.6 Sympy [F(-2)]

Exception generated. \[ \int \frac {x^3 \arctan (x)}{\left (1+x^2\right )^2} \, dx=\text {Exception raised: RecursionError} \]

input

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

output

Exception raised: RecursionError >> maximum recursion depth exceeded while 
 calling a Python object
 

3.7.74.7 Maxima [F]

\[ \int \frac {x^3 \arctan (x)}{\left (1+x^2\right )^2} \, dx=\int { \frac {x^{3} \arctan \left (x\right )}{{\left (x^{2} + 1\right )}^{2}} \,d x } \]

input

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

output

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

3.7.74.8 Giac [F]

\[ \int \frac {x^3 \arctan (x)}{\left (1+x^2\right )^2} \, dx=\int { \frac {x^{3} \arctan \left (x\right )}{{\left (x^{2} + 1\right )}^{2}} \,d x } \]

input

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

output

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

3.7.74.9 Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 \arctan (x)}{\left (1+x^2\right )^2} \, dx=\int \frac {x^3\,\mathrm {atan}\left (x\right )}{{\left (x^2+1\right )}^2} \,d x \]

input

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

output

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

3.7.74.10 Reduce [F]

\[ \int \frac {x^3 \arctan (x)}{\left (1+x^2\right )^2} \, dx=\int \frac {\mathit {atan} \left (x \right ) x^{3}}{x^{4}+2 x^{2}+1}d x \]

input

int((atan(x)*x**3)/(x**4 + 2*x**2 + 1),x)
 

output

int((atan(x)*x**3)/(x**4 + 2*x**2 + 1),x)