Integrand size = 57, antiderivative size = 26 \begin {dmath*} \int \frac {e^{\frac {2}{5} (7-5 x+5 \log (\log (5)))} (-1-2 x)+e^{\frac {1}{5} (7-5 x+5 \log (\log (5)))} (2+2 x) \log (4)-\log ^2(4)}{x^2} \, dx=\frac {-2 x+\left (\log (4)-e^{\frac {7}{5}-x} \log (5)\right )^2}{x} \end {dmath*}
Time = 0.27 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.65 \begin {dmath*} \int \frac {e^{\frac {2}{5} (7-5 x+5 \log (\log (5)))} (-1-2 x)+e^{\frac {1}{5} (7-5 x+5 \log (\log (5)))} (2+2 x) \log (4)-\log ^2(4)}{x^2} \, dx=\frac {2 \log ^2(4)+e^{\frac {14}{5}-2 x} \log (5) \log (25)-e^{\frac {7}{5}-x} \log (16) \log (25)}{2 x} \end {dmath*}
Integrate[(E^((2*(7 - 5*x + 5*Log[Log[5]]))/5)*(-1 - 2*x) + E^((7 - 5*x + 5*Log[Log[5]])/5)*(2 + 2*x)*Log[4] - Log[4]^2)/x^2,x]
Time = 0.31 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.69, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {2010, 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.
\(\displaystyle \int \frac {(-2 x-1) e^{\frac {2}{5} (-5 x+7+5 \log (\log (5)))}+(2 x+2) \log (4) e^{\frac {1}{5} (-5 x+7+5 \log (\log (5)))}-\log ^2(4)}{x^2} \, dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (-\frac {e^{\frac {14}{5}-2 x} (2 x+1) \log ^2(5)}{x^2}-\frac {\log ^2(4)}{x^2}+\frac {2 e^{\frac {7}{5}-x} (x+1) \log (4) \log (5)}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^{\frac {14}{5}-2 x} \log ^2(5)}{x}+\frac {\log ^2(4)}{x}-\frac {2 e^{\frac {7}{5}-x} \log (4) \log (5)}{x}\) |
Int[(E^((2*(7 - 5*x + 5*Log[Log[5]]))/5)*(-1 - 2*x) + E^((7 - 5*x + 5*Log[ Log[5]])/5)*(2 + 2*x)*Log[4] - Log[4]^2)/x^2,x]
3.2.28.3.1 Defintions of rubi rules used
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Time = 0.60 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38
method | result | size |
norman | \(\frac {\ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}+4 \ln \left (2\right )^{2}-4 \ln \left (2\right ) {\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}-x}}{x}\) | \(36\) |
parallelrisch | \(\frac {\ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}+4 \ln \left (2\right )^{2}-4 \ln \left (2\right ) {\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}-x}}{x}\) | \(36\) |
risch | \(\frac {4 \ln \left (2\right )^{2}}{x}+\frac {\ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}}{x}-\frac {4 \ln \left (2\right ) \ln \left (5\right ) {\mathrm e}^{\frac {7}{5}-x}}{x}\) | \(40\) |
parts | \(\frac {4 \ln \left (2\right )^{2}}{x}+\frac {\ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}}{x}-\frac {4 \ln \left (2\right ) {\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}-x}}{x}\) | \(42\) |
derivativedivides | \(\frac {19 \ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}}{5 x}-\frac {28 \,{\mathrm e}^{\frac {14}{5}+2 \ln \left (\ln \left (5\right )\right )} \operatorname {Ei}_{1}\left (2 x \right )}{5}+\frac {4 \ln \left (2\right )^{2}}{x}-10 \left (5 \ln \left (\ln \left (5\right )\right )+7\right ) \left (\frac {\ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}}{25 x}-\frac {2 \,{\mathrm e}^{\frac {14}{5}+2 \ln \left (\ln \left (5\right )\right )} \operatorname {Ei}_{1}\left (2 x \right )}{25}\right )+50 \ln \left (\ln \left (5\right )\right ) \left (\frac {\ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}}{25 x}-\frac {2 \,{\mathrm e}^{\frac {14}{5}+2 \ln \left (\ln \left (5\right )\right )} \operatorname {Ei}_{1}\left (2 x \right )}{25}\right )-240 \ln \left (2\right ) \left (\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}-x}}{25 x}-\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}} \operatorname {Ei}_{1}\left (x \right )}{25}\right )+100 \ln \left (2\right ) \left (\frac {\left (5 \ln \left (\ln \left (5\right )\right )+7\right ) \left (\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}-x}}{25 x}-\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}} \operatorname {Ei}_{1}\left (x \right )}{25}\right )}{5}-\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}} \operatorname {Ei}_{1}\left (x \right )}{25}\right )-100 \ln \left (2\right ) \ln \left (\ln \left (5\right )\right ) \left (\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}-x}}{25 x}-\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}} \operatorname {Ei}_{1}\left (x \right )}{25}\right )\) | \(234\) |
default | \(\frac {19 \ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}}{5 x}-\frac {28 \,{\mathrm e}^{\frac {14}{5}+2 \ln \left (\ln \left (5\right )\right )} \operatorname {Ei}_{1}\left (2 x \right )}{5}+\frac {4 \ln \left (2\right )^{2}}{x}-10 \left (5 \ln \left (\ln \left (5\right )\right )+7\right ) \left (\frac {\ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}}{25 x}-\frac {2 \,{\mathrm e}^{\frac {14}{5}+2 \ln \left (\ln \left (5\right )\right )} \operatorname {Ei}_{1}\left (2 x \right )}{25}\right )+50 \ln \left (\ln \left (5\right )\right ) \left (\frac {\ln \left (5\right )^{2} {\mathrm e}^{\frac {14}{5}-2 x}}{25 x}-\frac {2 \,{\mathrm e}^{\frac {14}{5}+2 \ln \left (\ln \left (5\right )\right )} \operatorname {Ei}_{1}\left (2 x \right )}{25}\right )-240 \ln \left (2\right ) \left (\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}-x}}{25 x}-\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}} \operatorname {Ei}_{1}\left (x \right )}{25}\right )+100 \ln \left (2\right ) \left (\frac {\left (5 \ln \left (\ln \left (5\right )\right )+7\right ) \left (\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}-x}}{25 x}-\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}} \operatorname {Ei}_{1}\left (x \right )}{25}\right )}{5}-\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}} \operatorname {Ei}_{1}\left (x \right )}{25}\right )-100 \ln \left (2\right ) \ln \left (\ln \left (5\right )\right ) \left (\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}-x}}{25 x}-\frac {{\mathrm e}^{\ln \left (\ln \left (5\right )\right )+\frac {7}{5}} \operatorname {Ei}_{1}\left (x \right )}{25}\right )\) | \(234\) |
int(((-1-2*x)*exp(ln(ln(5))+7/5-x)^2+2*(2+2*x)*ln(2)*exp(ln(ln(5))+7/5-x)- 4*ln(2)^2)/x^2,x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46 \begin {dmath*} \int \frac {e^{\frac {2}{5} (7-5 x+5 \log (\log (5)))} (-1-2 x)+e^{\frac {1}{5} (7-5 x+5 \log (\log (5)))} (2+2 x) \log (4)-\log ^2(4)}{x^2} \, dx=-\frac {4 \, e^{\left (-x + \log \left (\log \left (5\right )\right ) + \frac {7}{5}\right )} \log \left (2\right ) - 4 \, \log \left (2\right )^{2} - e^{\left (-2 \, x + 2 \, \log \left (\log \left (5\right )\right ) + \frac {14}{5}\right )}}{x} \end {dmath*}
integrate(((-1-2*x)*exp(log(log(5))+7/5-x)^2+2*(2+2*x)*log(2)*exp(log(log( 5))+7/5-x)-4*log(2)^2)/x^2,x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (20) = 40\).
Time = 0.08 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \begin {dmath*} \int \frac {e^{\frac {2}{5} (7-5 x+5 \log (\log (5)))} (-1-2 x)+e^{\frac {1}{5} (7-5 x+5 \log (\log (5)))} (2+2 x) \log (4)-\log ^2(4)}{x^2} \, dx=\frac {4 \log {\left (2 \right )}^{2}}{x} + \frac {- 4 x e^{\frac {7}{5} - x} \log {\left (2 \right )} \log {\left (5 \right )} + x e^{\frac {14}{5} - 2 x} \log {\left (5 \right )}^{2}}{x^{2}} \end {dmath*}
integrate(((-1-2*x)*exp(ln(ln(5))+7/5-x)**2+2*(2+2*x)*ln(2)*exp(ln(ln(5))+ 7/5-x)-4*ln(2)**2)/x**2,x)
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.23 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.23 \begin {dmath*} \int \frac {e^{\frac {2}{5} (7-5 x+5 \log (\log (5)))} (-1-2 x)+e^{\frac {1}{5} (7-5 x+5 \log (\log (5)))} (2+2 x) \log (4)-\log ^2(4)}{x^2} \, dx=-2 \, {\rm Ei}\left (-2 \, x\right ) e^{\frac {14}{5}} \log \left (5\right )^{2} + 2 \, e^{\frac {14}{5}} \Gamma \left (-1, 2 \, x\right ) \log \left (5\right )^{2} + 4 \, {\rm Ei}\left (-x\right ) e^{\frac {7}{5}} \log \left (5\right ) \log \left (2\right ) - 4 \, e^{\frac {7}{5}} \Gamma \left (-1, x\right ) \log \left (5\right ) \log \left (2\right ) + \frac {4 \, \log \left (2\right )^{2}}{x} \end {dmath*}
integrate(((-1-2*x)*exp(log(log(5))+7/5-x)^2+2*(2+2*x)*log(2)*exp(log(log( 5))+7/5-x)-4*log(2)^2)/x^2,x, algorithm=\
-2*Ei(-2*x)*e^(14/5)*log(5)^2 + 2*e^(14/5)*gamma(-1, 2*x)*log(5)^2 + 4*Ei( -x)*e^(7/5)*log(5)*log(2) - 4*e^(7/5)*gamma(-1, x)*log(5)*log(2) + 4*log(2 )^2/x
Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46 \begin {dmath*} \int \frac {e^{\frac {2}{5} (7-5 x+5 \log (\log (5)))} (-1-2 x)+e^{\frac {1}{5} (7-5 x+5 \log (\log (5)))} (2+2 x) \log (4)-\log ^2(4)}{x^2} \, dx=-\frac {4 \, e^{\left (-x + \log \left (\log \left (5\right )\right ) + \frac {7}{5}\right )} \log \left (2\right ) - 4 \, \log \left (2\right )^{2} - e^{\left (-2 \, x + 2 \, \log \left (\log \left (5\right )\right ) + \frac {14}{5}\right )}}{x} \end {dmath*}
integrate(((-1-2*x)*exp(log(log(5))+7/5-x)^2+2*(2+2*x)*log(2)*exp(log(log( 5))+7/5-x)-4*log(2)^2)/x^2,x, algorithm=\
Time = 0.15 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \begin {dmath*} \int \frac {e^{\frac {2}{5} (7-5 x+5 \log (\log (5)))} (-1-2 x)+e^{\frac {1}{5} (7-5 x+5 \log (\log (5)))} (2+2 x) \log (4)-\log ^2(4)}{x^2} \, dx=\frac {{\mathrm {e}}^{-2\,x}\,{\left ({\mathrm {e}}^{7/5}\,\ln \left (5\right )-2\,{\mathrm {e}}^x\,\ln \left (2\right )\right )}^2}{x} \end {dmath*}