Integrand size = 13, antiderivative size = 63 \[ \int \frac {\left (1+x^2\right )^2 \arctan (x)}{x^5} \, dx=-\frac {1}{12 x^3}-\frac {3}{4 x}-\frac {3 \arctan (x)}{4}-\frac {\arctan (x)}{4 x^4}-\frac {\arctan (x)}{x^2}+\frac {1}{2} i \operatorname {PolyLog}(2,-i x)-\frac {1}{2} i \operatorname {PolyLog}(2,i x) \]
-1/12/x^3-3/4/x-3/4*arctan(x)-1/4*arctan(x)/x^4-arctan(x)/x^2+1/2*I*polylo g(2,-I*x)-1/2*I*polylog(2,I*x)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.00 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.29 \[ \int \frac {\left (1+x^2\right )^2 \arctan (x)}{x^5} \, dx=-\frac {\arctan (x)}{4 x^4}-\frac {\arctan (x)}{x^2}-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-x^2\right )}{12 x^3}-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-x^2\right )}{x}+\frac {1}{2} i \operatorname {PolyLog}(2,-i x)-\frac {1}{2} i \operatorname {PolyLog}(2,i x) \]
-1/4*ArcTan[x]/x^4 - ArcTan[x]/x^2 - Hypergeometric2F1[-3/2, 1, -1/2, -x^2 ]/(12*x^3) - Hypergeometric2F1[-1/2, 1, 1/2, -x^2]/x + (I/2)*PolyLog[2, (- I)*x] - (I/2)*PolyLog[2, I*x]
Time = 0.25 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {5483, 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.
\(\displaystyle \int \frac {\left (x^2+1\right )^2 \arctan (x)}{x^5} \, dx\) |
\(\Big \downarrow \) 5483 |
\(\displaystyle \int \left (\frac {\arctan (x)}{x^5}+\frac {2 \arctan (x)}{x^3}+\frac {\arctan (x)}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\arctan (x)}{4 x^4}-\frac {\arctan (x)}{x^2}-\frac {3 \arctan (x)}{4}+\frac {1}{2} i \operatorname {PolyLog}(2,-i x)-\frac {1}{2} i \operatorname {PolyLog}(2,i x)-\frac {1}{12 x^3}-\frac {3}{4 x}\) |
-1/12*1/x^3 - 3/(4*x) - (3*ArcTan[x])/4 - ArcTan[x]/(4*x^4) - ArcTan[x]/x^ 2 + (I/2)*PolyLog[2, (-I)*x] - (I/2)*PolyLog[2, I*x]
3.7.78.3.1 Defintions of rubi rules used
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_)*((d_) + (e_. )*(x_)^2)^(q_), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*ArcTan[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2* d] && IGtQ[p, 0] && IGtQ[q, 1] && (EqQ[p, 1] || IntegerQ[m])
Time = 0.36 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.25
method | result | size |
default | \(\arctan \left (x \right ) \ln \left (x \right )-\frac {\arctan \left (x \right )}{x^{2}}-\frac {\arctan \left (x \right )}{4 x^{4}}+\frac {i \ln \left (x \right ) \ln \left (i x +1\right )}{2}-\frac {i \ln \left (x \right ) \ln \left (-i x +1\right )}{2}+\frac {i \operatorname {dilog}\left (i x +1\right )}{2}-\frac {i \operatorname {dilog}\left (-i x +1\right )}{2}-\frac {1}{12 x^{3}}-\frac {3}{4 x}-\frac {3 \arctan \left (x \right )}{4}\) | \(79\) |
parts | \(\arctan \left (x \right ) \ln \left (x \right )-\frac {\arctan \left (x \right )}{x^{2}}-\frac {\arctan \left (x \right )}{4 x^{4}}+\frac {i \ln \left (x \right ) \ln \left (i x +1\right )}{2}-\frac {i \ln \left (x \right ) \ln \left (-i x +1\right )}{2}+\frac {i \operatorname {dilog}\left (i x +1\right )}{2}-\frac {i \operatorname {dilog}\left (-i x +1\right )}{2}-\frac {1}{12 x^{3}}-\frac {3}{4 x}-\frac {3 \arctan \left (x \right )}{4}\) | \(79\) |
meijerg | \(-\frac {1}{12 x^{3}}-\frac {3}{4 x}-\frac {2 \left (-\frac {3 x^{4}}{8}+\frac {3}{8}\right ) \arctan \left (\sqrt {x^{2}}\right )}{3 x^{3} \sqrt {x^{2}}}-\frac {i x \,\operatorname {Li}_{2}\left (i \sqrt {x^{2}}\right )}{2 \sqrt {x^{2}}}+\frac {i x \,\operatorname {Li}_{2}\left (-i \sqrt {x^{2}}\right )}{2 \sqrt {x^{2}}}-\frac {\left (x^{2}+1\right ) \arctan \left (x \right )}{x^{2}}\) | \(85\) |
risch | \(-\frac {3}{4 x}-\frac {1}{12 x^{3}}+\frac {3 i \ln \left (-i x \right )}{8}-\frac {3 \arctan \left (x \right )}{4}-\frac {i \ln \left (-i x +1\right )}{8 x^{4}}-\frac {i \operatorname {dilog}\left (-i x +1\right )}{2}-\frac {i \ln \left (-i x +1\right )}{2 x^{2}}-\frac {3 i \ln \left (i x \right )}{8}+\frac {i \ln \left (i x +1\right )}{8 x^{4}}+\frac {i \operatorname {dilog}\left (i x +1\right )}{2}+\frac {i \ln \left (i x +1\right )}{2 x^{2}}\) | \(104\) |
arctan(x)*ln(x)-arctan(x)/x^2-1/4*arctan(x)/x^4+1/2*I*ln(x)*ln(1+I*x)-1/2* I*ln(x)*ln(1-I*x)+1/2*I*dilog(1+I*x)-1/2*I*dilog(1-I*x)-1/12/x^3-3/4/x-3/4 *arctan(x)
\[ \int \frac {\left (1+x^2\right )^2 \arctan (x)}{x^5} \, dx=\int { \frac {{\left (x^{2} + 1\right )}^{2} \arctan \left (x\right )}{x^{5}} \,d x } \]
\[ \int \frac {\left (1+x^2\right )^2 \arctan (x)}{x^5} \, dx=\int \frac {\left (x^{2} + 1\right )^{2} \operatorname {atan}{\left (x \right )}}{x^{5}}\, dx \]
Time = 0.32 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.13 \[ \int \frac {\left (1+x^2\right )^2 \arctan (x)}{x^5} \, dx=-\frac {3 \, \pi x^{4} \log \left (x^{2} + 1\right ) - 12 \, x^{4} \arctan \left (x\right ) \log \left (x\right ) + 6 i \, x^{4} {\rm Li}_2\left (i \, x + 1\right ) - 6 i \, x^{4} {\rm Li}_2\left (-i \, x + 1\right ) + 9 \, x^{3} + 3 \, {\left (3 \, x^{4} + 4 \, x^{2} + 1\right )} \arctan \left (x\right ) + x}{12 \, x^{4}} \]
-1/12*(3*pi*x^4*log(x^2 + 1) - 12*x^4*arctan(x)*log(x) + 6*I*x^4*dilog(I*x + 1) - 6*I*x^4*dilog(-I*x + 1) + 9*x^3 + 3*(3*x^4 + 4*x^2 + 1)*arctan(x) + x)/x^4
\[ \int \frac {\left (1+x^2\right )^2 \arctan (x)}{x^5} \, dx=\int { \frac {{\left (x^{2} + 1\right )}^{2} \arctan \left (x\right )}{x^{5}} \,d x } \]
Time = 0.51 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.84 \[ \int \frac {\left (1+x^2\right )^2 \arctan (x)}{x^5} \, dx=\frac {x^2-\frac {1}{3}}{4\,x^3}-\frac {\mathrm {atan}\left (x\right )}{x^2}-\frac {\mathrm {atan}\left (x\right )}{4\,x^4}-\frac {3\,\mathrm {atan}\left (x\right )}{4}-\frac {1}{x}-\frac {{\mathrm {Li}}_{\mathrm {2}}\left (1-x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\frac {\mathrm {polylog}\left (2,-x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2} \]
(polylog(2, -x*1i)*1i)/2 - (3*atan(x))/4 - atan(x)/x^2 - atan(x)/(4*x^4) - (dilog(1 - x*1i)*1i)/2 + (x^2 - 1/3)/(4*x^3) - 1/x
\[ \int \frac {\left (1+x^2\right )^2 \arctan (x)}{x^5} \, dx=\frac {-9 \mathit {atan} \left (x \right ) x^{4}-12 \mathit {atan} \left (x \right ) x^{2}-3 \mathit {atan} \left (x \right )+12 \left (\int \frac {\mathit {atan} \left (x \right )}{x}d x \right ) x^{4}-9 x^{3}-x}{12 x^{4}} \]