\(\int \frac {x^3 \arctan (x)^2}{(1+x^2)^3} \, dx\) [682]

Optimal result

Integrand size = 15, antiderivative size = 79 \[ \int \frac {x^3 \arctan (x)^2}{\left (1+x^2\right )^3} \, dx=-\frac {1}{32 \left (1+x^2\right )^2}+\frac {5}{32 \left (1+x^2\right )}+\frac {x^3 \arctan (x)}{8 \left (1+x^2\right )^2}+\frac {3 x \arctan (x)}{16 \left (1+x^2\right )}-\frac {3 \arctan (x)^2}{32}+\frac {x^4 \arctan (x)^2}{4 \left (1+x^2\right )^2} \] Output:

-1/32/(x^2+1)^2+5/32/(x^2+1)+1/8*x^3*arctan(x)/(x^2+1)^2+3/16*x*arctan(x)/ 
(x^2+1)-3/32*arctan(x)^2+1/4*x^4*arctan(x)^2/(x^2+1)^2
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.59 \[ \int \frac {x^3 \arctan (x)^2}{\left (1+x^2\right )^3} \, dx=\frac {4+5 x^2+2 x \left (3+5 x^2\right ) \arctan (x)+\left (-3-6 x^2+5 x^4\right ) \arctan (x)^2}{32 \left (1+x^2\right )^2} \] Input:

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

Output:

(4 + 5*x^2 + 2*x*(3 + 5*x^2)*ArcTan[x] + (-3 - 6*x^2 + 5*x^4)*ArcTan[x]^2) 
/(32*(1 + x^2)^2)
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.16, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {5479, 5473, 5469, 5419}

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.

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

\(\Big \downarrow \) 5479

\(\displaystyle \frac {x^4 \arctan (x)^2}{4 \left (x^2+1\right )^2}-\frac {1}{2} \int \frac {x^4 \arctan (x)}{\left (x^2+1\right )^3}dx\)

\(\Big \downarrow \) 5473

\(\displaystyle \frac {1}{2} \left (-\frac {3}{4} \int \frac {x^2 \arctan (x)}{\left (x^2+1\right )^2}dx+\frac {x^3 \arctan (x)}{4 \left (x^2+1\right )^2}-\frac {x^4}{16 \left (x^2+1\right )^2}\right )+\frac {x^4 \arctan (x)^2}{4 \left (x^2+1\right )^2}\)

\(\Big \downarrow \) 5469

\(\displaystyle \frac {1}{2} \left (-\frac {3}{4} \left (\frac {1}{2} \int \frac {\arctan (x)}{x^2+1}dx-\frac {x \arctan (x)}{2 \left (x^2+1\right )}-\frac {1}{4 \left (x^2+1\right )}\right )+\frac {x^3 \arctan (x)}{4 \left (x^2+1\right )^2}-\frac {x^4}{16 \left (x^2+1\right )^2}\right )+\frac {x^4 \arctan (x)^2}{4 \left (x^2+1\right )^2}\)

\(\Big \downarrow \) 5419

\(\displaystyle \frac {x^4 \arctan (x)^2}{4 \left (x^2+1\right )^2}+\frac {1}{2} \left (-\frac {3}{4} \left (-\frac {x \arctan (x)}{2 \left (x^2+1\right )}+\frac {\arctan (x)^2}{4}-\frac {1}{4 \left (x^2+1\right )}\right )+\frac {x^3 \arctan (x)}{4 \left (x^2+1\right )^2}-\frac {x^4}{16 \left (x^2+1\right )^2}\right )\)

Input:

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

Output:

(x^4*ArcTan[x]^2)/(4*(1 + x^2)^2) + (-1/16*x^4/(1 + x^2)^2 + (x^3*ArcTan[x 
])/(4*(1 + x^2)^2) - (3*(-1/4*1/(1 + x^2) - (x*ArcTan[x])/(2*(1 + x^2)) + 
ArcTan[x]^2/4))/4)/2
 

Defintions of rubi rules used

rule 5419

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

rule 5469

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*(x_)^2*((d_) + (e_.)*(x_)^2)^(q_), x 
_Symbol] :> Simp[(-b)*((d + e*x^2)^(q + 1)/(4*c^3*d*(q + 1)^2)), x] + (Simp 
[x*(d + e*x^2)^(q + 1)*((a + b*ArcTan[c*x])/(2*c^2*d*(q + 1))), x] - Simp[1 
/(2*c^2*d*(q + 1))   Int[(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x]), x], x]) / 
; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && LtQ[q, -1] && NeQ[q, -5/2]
 

rule 5473

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

rule 5479

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_ 
.)*(x_)^2)^(q_.), x_Symbol] :> Simp[(f*x)^(m + 1)*(d + e*x^2)^(q + 1)*((a + 
 b*ArcTan[c*x])^p/(d*f*(m + 1))), x] - Simp[b*c*(p/(f*(m + 1)))   Int[(f*x) 
^(m + 1)*(d + e*x^2)^q*(a + b*ArcTan[c*x])^(p - 1), x], x] /; FreeQ[{a, b, 
c, d, e, f, m, q}, x] && EqQ[e, c^2*d] && EqQ[m + 2*q + 3, 0] && GtQ[p, 0] 
&& NeQ[m, -1]
 

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.99

method	result	size
default	\(\frac {\arctan \left (x \right )^{2}}{4 \left (x^{2}+1\right )^{2}}-\frac {\arctan \left (x \right )^{2}}{2 \left (x^{2}+1\right )}+\frac {5 x^{3} \arctan \left (x \right )}{16 \left (x^{2}+1\right )^{2}}+\frac {3 x \arctan \left (x \right )}{16 \left (x^{2}+1\right )^{2}}+\frac {5 \arctan \left (x \right )^{2}}{32}+\frac {5}{32 \left (x^{2}+1\right )}-\frac {1}{32 \left (x^{2}+1\right )^{2}}\)	\(78\)
parts	\(\frac {\arctan \left (x \right )^{2}}{4 \left (x^{2}+1\right )^{2}}-\frac {\arctan \left (x \right )^{2}}{2 \left (x^{2}+1\right )}+\frac {5 x^{3} \arctan \left (x \right )}{16 \left (x^{2}+1\right )^{2}}+\frac {3 x \arctan \left (x \right )}{16 \left (x^{2}+1\right )^{2}}+\frac {5 \arctan \left (x \right )^{2}}{32}+\frac {5}{32 \left (x^{2}+1\right )}-\frac {1}{32 \left (x^{2}+1\right )^{2}}\)	\(78\)
risch	\(-\frac {\left (5 x^{4}-6 x^{2}-3\right ) \ln \left (i x +1\right )^{2}}{128 \left (x^{2}+1\right )^{2}}+\frac {\left (-6 x^{2} \ln \left (-i x +1\right )-3 \ln \left (-i x +1\right )+5 x^{4} \ln \left (-i x +1\right )-10 i x^{3}-6 i x \right ) \ln \left (i x +1\right )}{64 \left (x +i\right )^{2} \left (x -i\right )^{2}}-\frac {5 x^{4} \ln \left (-i x +1\right )^{2}-6 x^{2} \ln \left (-i x +1\right )^{2}-3 \ln \left (-i x +1\right )^{2}-20 i x^{3} \ln \left (-i x +1\right )-12 i x \ln \left (-i x +1\right )-20 x^{2}-16}{128 \left (x +i\right )^{2} \left (x -i\right )^{2}}\)	\(181\)
orering	\(\frac {\left (25 x^{6}-9 x^{4}-3 x^{2}+12\right ) \arctan \left (x \right )^{2}}{16 \left (x^{2}+1\right )^{2} x^{2}}+\frac {\left (x^{2}+1\right )^{2} \left (25 x^{4}+4 x^{2}-12\right ) \left (\frac {3 x^{2} \arctan \left (x \right )^{2}}{\left (x^{2}+1\right )^{3}}+\frac {2 x^{3} \arctan \left (x \right )}{\left (x^{2}+1\right )^{4}}-\frac {6 x^{4} \arctan \left (x \right )^{2}}{\left (x^{2}+1\right )^{4}}\right )}{32 x^{4}}+\frac {\left (5 x^{2}+4\right ) \left (x^{2}+1\right )^{3} \left (\frac {6 x \arctan \left (x \right )^{2}}{\left (x^{2}+1\right )^{3}}+\frac {12 x^{2} \arctan \left (x \right )}{\left (x^{2}+1\right )^{4}}-\frac {42 x^{3} \arctan \left (x \right )^{2}}{\left (x^{2}+1\right )^{4}}+\frac {2 x^{3}}{\left (x^{2}+1\right )^{5}}-\frac {28 x^{4} \arctan \left (x \right )}{\left (x^{2}+1\right )^{5}}+\frac {48 x^{5} \arctan \left (x \right )^{2}}{\left (x^{2}+1\right )^{5}}\right )}{64 x^{3}}\)	\(212\)

Input:

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

Output:

1/4*arctan(x)^2/(x^2+1)^2-1/2*arctan(x)^2/(x^2+1)+5/16*x^3*arctan(x)/(x^2+ 
1)^2+3/16*x*arctan(x)/(x^2+1)^2+5/32*arctan(x)^2+5/32/(x^2+1)-1/32/(x^2+1) 
^2
 

Fricas [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.65 \[ \int \frac {x^3 \arctan (x)^2}{\left (1+x^2\right )^3} \, dx=\frac {{\left (5 \, x^{4} - 6 \, x^{2} - 3\right )} \arctan \left (x\right )^{2} + 5 \, x^{2} + 2 \, {\left (5 \, x^{3} + 3 \, x\right )} \arctan \left (x\right ) + 4}{32 \, {\left (x^{4} + 2 \, x^{2} + 1\right )}} \] Input:

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

Output:

1/32*((5*x^4 - 6*x^2 - 3)*arctan(x)^2 + 5*x^2 + 2*(5*x^3 + 3*x)*arctan(x) 
+ 4)/(x^4 + 2*x^2 + 1)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.19 \[ \int \frac {x^3 \arctan (x)^2}{\left (1+x^2\right )^3} \, dx=\frac {1}{16} \, {\left (\frac {5 \, x^{3} + 3 \, x}{x^{4} + 2 \, x^{2} + 1} + 5 \, \arctan \left (x\right )\right )} \arctan \left (x\right ) - \frac {{\left (2 \, x^{2} + 1\right )} \arctan \left (x\right )^{2}}{4 \, {\left (x^{4} + 2 \, x^{2} + 1\right )}} - \frac {5 \, {\left (x^{4} + 2 \, x^{2} + 1\right )} \arctan \left (x\right )^{2} - 5 \, x^{2} - 4}{32 \, {\left (x^{4} + 2 \, x^{2} + 1\right )}} \] Input:

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

Output:

1/16*((5*x^3 + 3*x)/(x^4 + 2*x^2 + 1) + 5*arctan(x))*arctan(x) - 1/4*(2*x^ 
2 + 1)*arctan(x)^2/(x^4 + 2*x^2 + 1) - 1/32*(5*(x^4 + 2*x^2 + 1)*arctan(x) 
^2 - 5*x^2 - 4)/(x^4 + 2*x^2 + 1)
 

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.71 \[ \int \frac {x^3 \arctan (x)^2}{\left (1+x^2\right )^3} \, dx=-\frac {-5\,x^4\,{\mathrm {atan}\left (x\right )}^2+4\,x^4-10\,x^3\,\mathrm {atan}\left (x\right )+6\,x^2\,{\mathrm {atan}\left (x\right )}^2+3\,x^2-6\,x\,\mathrm {atan}\left (x\right )+3\,{\mathrm {atan}\left (x\right )}^2}{32\,{\left (x^2+1\right )}^2} \] Input:

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

Output:

-(3*atan(x)^2 - 10*x^3*atan(x) + 6*x^2*atan(x)^2 - 5*x^4*atan(x)^2 - 6*x*a 
tan(x) + 3*x^2 + 4*x^4)/(32*(x^2 + 1)^2)
 

Reduce [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.73 \[ \int \frac {x^3 \arctan (x)^2}{\left (1+x^2\right )^3} \, dx=\frac {10 \mathit {atan} \left (x \right )^{2} x^{4}-12 \mathit {atan} \left (x \right )^{2} x^{2}-6 \mathit {atan} \left (x \right )^{2}+20 \mathit {atan} \left (x \right ) x^{3}+12 \mathit {atan} \left (x \right ) x -5 x^{4}+3}{64 x^{4}+128 x^{2}+64} \] Input:

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

Output:

(10*atan(x)**2*x**4 - 12*atan(x)**2*x**2 - 6*atan(x)**2 + 20*atan(x)*x**3 
+ 12*atan(x)*x - 5*x**4 + 3)/(64*(x**4 + 2*x**2 + 1))