\(\int \frac {\arctan (x)^2}{x^5} \, dx\) [648]

Optimal result

Integrand size = 8, antiderivative size = 61 \[ \int \frac {\arctan (x)^2}{x^5} \, dx=-\frac {1}{12 x^2}-\frac {\arctan (x)}{6 x^3}+\frac {\arctan (x)}{2 x}+\frac {\arctan (x)^2}{4}-\frac {\arctan (x)^2}{4 x^4}-\frac {2 \log (x)}{3}+\frac {1}{3} \log \left (1+x^2\right ) \] Output:

-1/12/x^2-1/6*arctan(x)/x^3+1/2*arctan(x)/x+1/4*arctan(x)^2-1/4*arctan(x)^ 
2/x^4-2/3*ln(x)+1/3*ln(x^2+1)
 

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.92 \[ \int \frac {\arctan (x)^2}{x^5} \, dx=-\frac {1}{12 x^2}+\frac {\left (-1+3 x^2\right ) \arctan (x)}{6 x^3}+\frac {\left (-1+x^4\right ) \arctan (x)^2}{4 x^4}-\frac {2 \log (x)}{3}+\frac {1}{3} \log \left (1+x^2\right ) \] Input:

Integrate[ArcTan[x]^2/x^5,x]
 

Output:

-1/12*1/x^2 + ((-1 + 3*x^2)*ArcTan[x])/(6*x^3) + ((-1 + x^4)*ArcTan[x]^2)/ 
(4*x^4) - (2*Log[x])/3 + Log[1 + x^2]/3
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.30, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.625, Rules used = {5361, 5453, 5361, 243, 54, 2009, 5453, 5361, 243, 47, 14, 16, 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.

\(\displaystyle \int \frac {\arctan (x)^2}{x^5} \, dx\)

\(\Big \downarrow \) 5361

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

\(\Big \downarrow \) 5453

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

\(\Big \downarrow \) 5361

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

\(\Big \downarrow \) 243

\(\displaystyle \frac {1}{2} \left (-\int \frac {\arctan (x)}{x^2 \left (x^2+1\right )}dx+\frac {1}{6} \int \frac {1}{x^4 \left (x^2+1\right )}dx^2-\frac {\arctan (x)}{3 x^3}\right )-\frac {\arctan (x)^2}{4 x^4}\)

\(\Big \downarrow \) 54

\(\displaystyle \frac {1}{2} \left (-\int \frac {\arctan (x)}{x^2 \left (x^2+1\right )}dx+\frac {1}{6} \int \left (-\frac {1}{x^2}+\frac {1}{x^4}+\frac {1}{x^2+1}\right )dx^2-\frac {\arctan (x)}{3 x^3}\right )-\frac {\arctan (x)^2}{4 x^4}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (-\int \frac {\arctan (x)}{x^2 \left (x^2+1\right )}dx-\frac {\arctan (x)}{3 x^3}+\frac {1}{6} \left (-\frac {1}{x^2}-\log \left (x^2\right )+\log \left (x^2+1\right )\right )\right )-\frac {\arctan (x)^2}{4 x^4}\)

\(\Big \downarrow \) 5453

\(\displaystyle \frac {1}{2} \left (-\int \frac {\arctan (x)}{x^2}dx+\int \frac {\arctan (x)}{x^2+1}dx-\frac {\arctan (x)}{3 x^3}+\frac {1}{6} \left (-\frac {1}{x^2}-\log \left (x^2\right )+\log \left (x^2+1\right )\right )\right )-\frac {\arctan (x)^2}{4 x^4}\)

\(\Big \downarrow \) 5361

\(\displaystyle \frac {1}{2} \left (\int \frac {\arctan (x)}{x^2+1}dx-\int \frac {1}{x \left (x^2+1\right )}dx-\frac {\arctan (x)}{3 x^3}+\frac {\arctan (x)}{x}+\frac {1}{6} \left (-\frac {1}{x^2}-\log \left (x^2\right )+\log \left (x^2+1\right )\right )\right )-\frac {\arctan (x)^2}{4 x^4}\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {1}{2} \left (\int \frac {\arctan (x)}{x^2+1}dx-\frac {1}{2} \int \frac {1}{x^2 \left (x^2+1\right )}dx^2-\frac {\arctan (x)}{3 x^3}+\frac {\arctan (x)}{x}+\frac {1}{6} \left (-\frac {1}{x^2}-\log \left (x^2\right )+\log \left (x^2+1\right )\right )\right )-\frac {\arctan (x)^2}{4 x^4}\)

\(\Big \downarrow \) 47

\(\displaystyle \frac {1}{2} \left (\int \frac {\arctan (x)}{x^2+1}dx+\frac {1}{2} \left (\int \frac {1}{x^2+1}dx^2-\int \frac {1}{x^2}dx^2\right )-\frac {\arctan (x)}{3 x^3}+\frac {\arctan (x)}{x}+\frac {1}{6} \left (-\frac {1}{x^2}-\log \left (x^2\right )+\log \left (x^2+1\right )\right )\right )-\frac {\arctan (x)^2}{4 x^4}\)

\(\Big \downarrow \) 14

\(\displaystyle \frac {1}{2} \left (\int \frac {\arctan (x)}{x^2+1}dx+\frac {1}{2} \left (\int \frac {1}{x^2+1}dx^2-\log \left (x^2\right )\right )-\frac {\arctan (x)}{3 x^3}+\frac {\arctan (x)}{x}+\frac {1}{6} \left (-\frac {1}{x^2}-\log \left (x^2\right )+\log \left (x^2+1\right )\right )\right )-\frac {\arctan (x)^2}{4 x^4}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {1}{2} \left (\int \frac {\arctan (x)}{x^2+1}dx-\frac {\arctan (x)}{3 x^3}+\frac {\arctan (x)}{x}+\frac {1}{2} \left (\log \left (x^2+1\right )-\log \left (x^2\right )\right )+\frac {1}{6} \left (-\frac {1}{x^2}-\log \left (x^2\right )+\log \left (x^2+1\right )\right )\right )-\frac {\arctan (x)^2}{4 x^4}\)

\(\Big \downarrow \) 5419

\(\displaystyle \frac {1}{2} \left (-\frac {\arctan (x)}{3 x^3}+\frac {\arctan (x)^2}{2}+\frac {\arctan (x)}{x}+\frac {1}{2} \left (\log \left (x^2+1\right )-\log \left (x^2\right )\right )+\frac {1}{6} \left (-\frac {1}{x^2}-\log \left (x^2\right )+\log \left (x^2+1\right )\right )\right )-\frac {\arctan (x)^2}{4 x^4}\)

Input:

Int[ArcTan[x]^2/x^5,x]
 

Output:

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

Defintions of rubi rules used

rule 14

Int[(a_.)/(x_), x_Symbol] :> Simp[a*Log[x], x] /; FreeQ[a, x]
 

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 47

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[b/(b*c 
 - a*d)   Int[1/(a + b*x), x], x] - Simp[d/(b*c - a*d)   Int[1/(c + d*x), x 
], x] /; FreeQ[{a, b, c, d}, x]
 

rule 54

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E 
xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && 
 ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

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 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 5453

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e 
_.)*(x_)^2), x_Symbol] :> Simp[1/d   Int[(f*x)^m*(a + b*ArcTan[c*x])^p, x], 
 x] - Simp[e/(d*f^2)   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] && LtQ[m, -1]
 

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.79

method	result	size
default	\(-\frac {1}{12 x^{2}}-\frac {\arctan \left (x \right )}{6 x^{3}}+\frac {\arctan \left (x \right )}{2 x}+\frac {\arctan \left (x \right )^{2}}{4}-\frac {\arctan \left (x \right )^{2}}{4 x^{4}}-\frac {2 \ln \left (x \right )}{3}+\frac {\ln \left (x^{2}+1\right )}{3}\)	\(48\)
parts	\(-\frac {1}{12 x^{2}}-\frac {\arctan \left (x \right )}{6 x^{3}}+\frac {\arctan \left (x \right )}{2 x}+\frac {\arctan \left (x \right )^{2}}{4}-\frac {\arctan \left (x \right )^{2}}{4 x^{4}}-\frac {2 \ln \left (x \right )}{3}+\frac {\ln \left (x^{2}+1\right )}{3}\)	\(48\)
parallelrisch	\(-\frac {-3 x^{4} \arctan \left (x \right )^{2}+8 x^{4} \ln \left (x \right )-4 \ln \left (x^{2}+1\right ) x^{4}-6 x^{3} \arctan \left (x \right )+x^{2}+2 x \arctan \left (x \right )+3 \arctan \left (x \right )^{2}}{12 x^{4}}\)	\(55\)
risch	\(-\frac {\left (x^{4}-1\right ) \ln \left (i x +1\right )^{2}}{16 x^{4}}+\frac {\left (3 x^{4} \ln \left (-i x +1\right )-6 i x^{3}+2 i x -3 \ln \left (-i x +1\right )\right ) \ln \left (i x +1\right )}{24 x^{4}}-\frac {3 x^{4} \ln \left (-i x +1\right )^{2}+32 x^{4} \ln \left (x \right )-16 \ln \left (x^{2}+1\right ) x^{4}-12 i x^{3} \ln \left (-i x +1\right )+4 i x \ln \left (-i x +1\right )+4 x^{2}-3 \ln \left (-i x +1\right )^{2}}{48 x^{4}}\)	\(143\)

Input:

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

Output:

-1/12/x^2-1/6/x^3*arctan(x)+1/2/x*arctan(x)+1/4*arctan(x)^2-1/4*arctan(x)^ 
2/x^4-2/3*ln(x)+1/3*ln(x^2+1)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.87 \[ \int \frac {\arctan (x)^2}{x^5} \, dx=\frac {4 \, x^{4} \log \left (x^{2} + 1\right ) - 8 \, x^{4} \log \left (x\right ) + 3 \, {\left (x^{4} - 1\right )} \arctan \left (x\right )^{2} - x^{2} + 2 \, {\left (3 \, x^{3} - x\right )} \arctan \left (x\right )}{12 \, x^{4}} \] Input:

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

Output:

1/12*(4*x^4*log(x^2 + 1) - 8*x^4*log(x) + 3*(x^4 - 1)*arctan(x)^2 - x^2 + 
2*(3*x^3 - x)*arctan(x))/x^4
 

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.87 \[ \int \frac {\arctan (x)^2}{x^5} \, dx=- \frac {2 \log {\left (x \right )}}{3} + \frac {\log {\left (x^{2} + 1 \right )}}{3} + \frac {\operatorname {atan}^{2}{\left (x \right )}}{4} + \frac {\operatorname {atan}{\left (x \right )}}{2 x} - \frac {1}{12 x^{2}} - \frac {\operatorname {atan}{\left (x \right )}}{6 x^{3}} - \frac {\operatorname {atan}^{2}{\left (x \right )}}{4 x^{4}} \] Input:

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

Output:

-2*log(x)/3 + log(x**2 + 1)/3 + atan(x)**2/4 + atan(x)/(2*x) - 1/(12*x**2) 
 - atan(x)/(6*x**3) - atan(x)**2/(4*x**4)
                                                                                    
                                                                                    
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.05 \[ \int \frac {\arctan (x)^2}{x^5} \, dx=\frac {1}{6} \, {\left (\frac {3 \, x^{2} - 1}{x^{3}} + 3 \, \arctan \left (x\right )\right )} \arctan \left (x\right ) - \frac {3 \, x^{2} \arctan \left (x\right )^{2} - 4 \, x^{2} \log \left (x^{2} + 1\right ) + 8 \, x^{2} \log \left (x\right ) + 1}{12 \, x^{2}} - \frac {\arctan \left (x\right )^{2}}{4 \, x^{4}} \] Input:

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

Output:

1/6*((3*x^2 - 1)/x^3 + 3*arctan(x))*arctan(x) - 1/12*(3*x^2*arctan(x)^2 - 
4*x^2*log(x^2 + 1) + 8*x^2*log(x) + 1)/x^2 - 1/4*arctan(x)^2/x^4
 

Giac [F]

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

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

Output:

integrate(arctan(x)^2/x^5, x)
 

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.72 \[ \int \frac {\arctan (x)^2}{x^5} \, dx=\frac {\ln \left (x^2+1\right )}{3}-\frac {2\,\ln \left (x\right )}{3}-{\mathrm {atan}\left (x\right )}^2\,\left (\frac {1}{4\,x^4}-\frac {1}{4}\right )-\frac {1}{12\,x^2}+\frac {\mathrm {atan}\left (x\right )\,\left (\frac {x^2}{2}-\frac {1}{6}\right )}{x^3} \] Input:

int(atan(x)^2/x^5,x)
 

Output:

log(x^2 + 1)/3 - (2*log(x))/3 - atan(x)^2*(1/(4*x^4) - 1/4) - 1/(12*x^2) + 
 (atan(x)*(x^2/2 - 1/6))/x^3
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.92 \[ \int \frac {\arctan (x)^2}{x^5} \, dx=\frac {3 \mathit {atan} \left (x \right )^{2} x^{4}-3 \mathit {atan} \left (x \right )^{2}+6 \mathit {atan} \left (x \right ) x^{3}-2 \mathit {atan} \left (x \right ) x +4 \,\mathrm {log}\left (x^{2}+1\right ) x^{4}-8 \,\mathrm {log}\left (x \right ) x^{4}-x^{2}}{12 x^{4}} \] Input:

int(atan(x)^2/x^5,x)
 

Output:

(3*atan(x)**2*x**4 - 3*atan(x)**2 + 6*atan(x)*x**3 - 2*atan(x)*x + 4*log(x 
**2 + 1)*x**4 - 8*log(x)*x**4 - x**2)/(12*x**4)