3.7.77 \(\int \frac {(1+x^2) \arctan (x)}{x^5} \, dx\) [677]

3.7.77.1 Optimal result

Integrand size = 11, antiderivative size = 31 \[ \int \frac {\left (1+x^2\right ) \arctan (x)}{x^5} \, dx=-\frac {1}{12 x^3}-\frac {1}{4 x}-\frac {\left (1+x^2\right )^2 \arctan (x)}{4 x^4} \]

output

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

3.7.77.2 Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.01 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.90 \[ \int \frac {\left (1+x^2\right ) \arctan (x)}{x^5} \, dx=-\frac {\arctan (x)}{4 x^4}-\frac {\arctan (x)}{2 x^2}-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-x^2\right )}{12 x^3}-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-x^2\right )}{2 x} \]

input

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

output

-1/4*ArcTan[x]/x^4 - ArcTan[x]/(2*x^2) - Hypergeometric2F1[-3/2, 1, -1/2, 
-x^2]/(12*x^3) - Hypergeometric2F1[-1/2, 1, 1/2, -x^2]/(2*x)
 

3.7.77.3 Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {5479, 244, 2009}

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.

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

\(\Big \downarrow \) 5479

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

\(\Big \downarrow \) 244

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

\(\Big \downarrow \) 2009

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

input

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

output

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

3.7.77.3.1 Defintions of rubi rules used

rule 244

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand 
Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p 
, 0]
 

rule 2009

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

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]
 

3.7.77.4 Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97

method	result	size
default	\(-\frac {\arctan \left (x \right )}{2 x^{2}}-\frac {\arctan \left (x \right )}{4 x^{4}}-\frac {\arctan \left (x \right )}{4}-\frac {1}{12 x^{3}}-\frac {1}{4 x}\)	\(30\)
parts	\(-\frac {\arctan \left (x \right )}{2 x^{2}}-\frac {\arctan \left (x \right )}{4 x^{4}}-\frac {\arctan \left (x \right )}{4}-\frac {1}{12 x^{3}}-\frac {1}{4 x}\)	\(30\)
parallelrisch	\(-\frac {3 \arctan \left (x \right ) x^{4}+3 x^{3}+6 x^{2} \arctan \left (x \right )+x +3 \arctan \left (x \right )}{12 x^{4}}\)	\(31\)
meijerg	\(-\frac {1}{12 x^{3}}-\frac {1}{4 x}-\frac {2 \left (-\frac {3 x^{4}}{8}+\frac {3}{8}\right ) \arctan \left (\sqrt {x^{2}}\right )}{3 x^{3} \sqrt {x^{2}}}-\frac {\left (x^{2}+1\right ) \arctan \left (x \right )}{2 x^{2}}\)	\(47\)
risch	\(\frac {i \left (2 x^{2}+1\right ) \ln \left (i x +1\right )}{8 x^{4}}+\frac {i \left (3 \ln \left (x -i\right ) x^{4}-3 \ln \left (x +i\right ) x^{4}+6 i x^{3}-6 x^{2} \ln \left (-i x +1\right )+2 i x -3 \ln \left (-i x +1\right )\right )}{24 x^{4}}\)	\(80\)

input

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

output

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

3.7.77.5 Fricas [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {\left (1+x^2\right ) \arctan (x)}{x^5} \, dx=-\frac {3 \, x^{3} + 3 \, {\left (x^{4} + 2 \, x^{2} + 1\right )} \arctan \left (x\right ) + x}{12 \, x^{4}} \]

input

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

output

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

3.7.77.6 Sympy [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {\left (1+x^2\right ) \arctan (x)}{x^5} \, dx=- \frac {\operatorname {atan}{\left (x \right )}}{4} - \frac {1}{4 x} - \frac {\operatorname {atan}{\left (x \right )}}{2 x^{2}} - \frac {1}{12 x^{3}} - \frac {\operatorname {atan}{\left (x \right )}}{4 x^{4}} \]

input

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

output

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

3.7.77.7 Maxima [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {\left (1+x^2\right ) \arctan (x)}{x^5} \, dx=-\frac {3 \, x^{2} + 1}{12 \, x^{3}} - \frac {{\left (2 \, x^{2} + 1\right )} \arctan \left (x\right )}{4 \, x^{4}} - \frac {1}{4} \, \arctan \left (x\right ) \]

input

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

output

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

3.7.77.8 Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {\left (1+x^2\right ) \arctan (x)}{x^5} \, dx=-\frac {3 \, x^{2} + 1}{12 \, x^{3}} - \frac {{\left (2 \, x^{2} + 1\right )} \arctan \left (x\right )}{4 \, x^{4}} - \frac {1}{4} \, \arctan \left (x\right ) \]

input

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

output

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

3.7.77.9 Mupad [B] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {\left (1+x^2\right ) \arctan (x)}{x^5} \, dx=-\frac {\mathrm {atan}\left (x\right )}{4}-\frac {\frac {x}{12}+\frac {\mathrm {atan}\left (x\right )}{4}+\frac {x^2\,\mathrm {atan}\left (x\right )}{2}+\frac {x^3}{4}}{x^4} \]

input

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

output

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

3.7.77.10 Reduce [B] (verification not implemented)

Time = 0.00 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {\left (1+x^2\right ) \arctan (x)}{x^5} \, dx=\frac {-3 \mathit {atan} \left (x \right ) x^{4}-6 \mathit {atan} \left (x \right ) x^{2}-3 \mathit {atan} \left (x \right )-3 x^{3}-x}{12 x^{4}} \]

input

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

output

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