Integrand size = 62, antiderivative size = 27 \[ \int \frac {\left (-15-5 e x^2\right ) \log \left (\frac {4}{3 x \log (2)}\right )+\log (x) \left (-15-5 e x^2+10 e x^2 \log \left (\frac {4}{3 x \log (2)}\right )\right )}{e x \log ^2(x)} \, dx=\frac {5 \left (\frac {3}{e}+x^2\right ) \log \left (\frac {4}{3 x \log (2)}\right )}{\log (x)} \]
Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {\left (-15-5 e x^2\right ) \log \left (\frac {4}{3 x \log (2)}\right )+\log (x) \left (-15-5 e x^2+10 e x^2 \log \left (\frac {4}{3 x \log (2)}\right )\right )}{e x \log ^2(x)} \, dx=\frac {15+\frac {5 \left (3+e x^2\right ) \log \left (\frac {4}{x \log (8)}\right )}{\log (x)}}{e} \]
Integrate[((-15 - 5*E*x^2)*Log[4/(3*x*Log[2])] + Log[x]*(-15 - 5*E*x^2 + 1 0*E*x^2*Log[4/(3*x*Log[2])]))/(E*x*Log[x]^2),x]
Time = 0.96 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.56, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {27, 27, 7293, 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.
\(\displaystyle \int \frac {\left (-5 e x^2-15\right ) \log \left (\frac {4}{3 x \log (2)}\right )+\log (x) \left (-5 e x^2+10 e x^2 \log \left (\frac {4}{3 x \log (2)}\right )-15\right )}{e x \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {5 \left (\left (e x^2+3\right ) \log \left (\frac {4}{3 x \log (2)}\right )+\log (x) \left (-2 e \log \left (\frac {4}{3 x \log (2)}\right ) x^2+e x^2+3\right )\right )}{x \log ^2(x)}dx}{e}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {5 \int \frac {\left (e x^2+3\right ) \log \left (\frac {4}{3 x \log (2)}\right )+\log (x) \left (-2 e \log \left (\frac {4}{3 x \log (2)}\right ) x^2+e x^2+3\right )}{x \log ^2(x)}dx}{e}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {5 \int \left (\frac {e x^2+3}{x \log (x)}-\frac {\left (2 e \log (x) x^2-e x^2-3\right ) \log \left (\frac {4}{x \log (8)}\right )}{x \log ^2(x)}\right )dx}{e}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {5 \left (-\frac {e x^2 \log \left (\frac {4}{x \log (8)}\right )}{\log (x)}-\frac {3 \log \left (\frac {4}{x \log (8)}\right )}{\log (x)}\right )}{e}\) |
Int[((-15 - 5*E*x^2)*Log[4/(3*x*Log[2])] + Log[x]*(-15 - 5*E*x^2 + 10*E*x^ 2*Log[4/(3*x*Log[2])]))/(E*x*Log[x]^2),x]
3.13.48.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 4.66 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.48
method | result | size |
parallelrisch | \(\frac {{\mathrm e}^{-1} \left (5 x^{2} {\mathrm e} \ln \left (\frac {4}{3 x \ln \left (2\right )}\right )+15 \ln \left (\frac {4}{3 x \ln \left (2\right )}\right )\right )}{\ln \left (x \right )}\) | \(40\) |
risch | \(-5 x^{2}+\frac {5 \,{\mathrm e}^{-1} \left (4 \ln \left (2\right ) {\mathrm e} x^{2}-2 x^{2} {\mathrm e} \ln \left (3\right )-2 x^{2} {\mathrm e} \ln \left (\ln \left (2\right )\right )+12 \ln \left (2\right )-6 \ln \left (3\right )-6 \ln \left (\ln \left (2\right )\right )\right )}{2 \ln \left (x \right )}\) | \(57\) |
default | \({\mathrm e}^{-1} \left (-\frac {5 \ln \left (3\right ) {\mathrm e} x^{2}}{\ln \left (x \right )}-\frac {15 \ln \left (3\right )}{\ln \left (x \right )}+\frac {10 \ln \left (2\right ) {\mathrm e} x^{2}}{\ln \left (x \right )}+\frac {30 \ln \left (2\right )}{\ln \left (x \right )}-5 \ln \left (\ln \left (2\right )\right ) \left (-{\mathrm e} \left (-\frac {x^{2}}{\ln \left (x \right )}-2 \,\operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )\right )+\frac {3}{\ln \left (x \right )}\right )-5 \,{\mathrm e} \left (\left (\ln \left (x \right )+\ln \left (\frac {1}{x}\right )\right ) \left (-\frac {x^{2}}{\ln \left (x \right )}-2 \,\operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )\right )+\operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )\right )+\frac {15 \ln \left (\frac {1}{x}\right )}{\ln \left (x \right )}+10 \ln \left (\ln \left (2\right )\right ) {\mathrm e} \,\operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )+5 \,{\mathrm e} \,\operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )-10 \,{\mathrm e} \left (\left (\ln \left (x \right )+\ln \left (\frac {1}{x}\right )\right ) \operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )+\frac {x^{2}}{2}\right )\right )\) | \(177\) |
parts | \(10 \ln \left (\ln \left (2\right )\right ) \operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )+5 \,\operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )-15 \,{\mathrm e}^{-1} \ln \left (\ln \left (x \right )\right )-10 \left (\ln \left (x \right )+\ln \left (\frac {1}{x}\right )\right ) \operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )-5 x^{2}-20 \ln \left (2\right ) \operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )+10 \ln \left (3\right ) \operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )-5 \,{\mathrm e}^{-1} \left (\ln \left (\ln \left (2\right )\right ) \left (-{\mathrm e} \left (-\frac {x^{2}}{\ln \left (x \right )}-2 \,\operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )\right )+\frac {3}{\ln \left (x \right )}\right )+{\mathrm e} \left (\left (\ln \left (x \right )+\ln \left (\frac {1}{x}\right )\right ) \left (-\frac {x^{2}}{\ln \left (x \right )}-2 \,\operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )\right )+\operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )\right )-\frac {3 \ln \left (\frac {1}{x}\right )}{\ln \left (x \right )}-3 \ln \left (\ln \left (x \right )\right )+2 \ln \left (2\right ) \left ({\mathrm e} \left (-\frac {x^{2}}{\ln \left (x \right )}-2 \,\operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )\right )-\frac {3}{\ln \left (x \right )}\right )-\ln \left (3\right ) \left ({\mathrm e} \left (-\frac {x^{2}}{\ln \left (x \right )}-2 \,\operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )\right )-\frac {3}{\ln \left (x \right )}\right )\right )\) | \(225\) |
int(((10*x^2*exp(1)*ln(4/3/x/ln(2))-5*x^2*exp(1)-15)*ln(x)+(-5*x^2*exp(1)- 15)*ln(4/3/x/ln(2)))/x/exp(1)/ln(x)^2,x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (24) = 48\).
Time = 0.25 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.04 \[ \int \frac {\left (-15-5 e x^2\right ) \log \left (\frac {4}{3 x \log (2)}\right )+\log (x) \left (-15-5 e x^2+10 e x^2 \log \left (\frac {4}{3 x \log (2)}\right )\right )}{e x \log ^2(x)} \, dx=\frac {5 \, {\left (x^{2} e \log \left (\frac {4}{3 \, x \log \left (2\right )}\right ) + 3 \, \log \left (\frac {4}{3 \, \log \left (2\right )}\right )\right )}}{e \log \left (\frac {4}{3 \, \log \left (2\right )}\right ) - e \log \left (\frac {4}{3 \, x \log \left (2\right )}\right )} \]
integrate(((10*x^2*exp(1)*log(4/3/x/log(2))-5*x^2*exp(1)-15)*log(x)+(-5*x^ 2*exp(1)-15)*log(4/3/x/log(2)))/x/exp(1)/log(x)^2,x, algorithm=\
5*(x^2*e*log(4/3/(x*log(2))) + 3*log(4/3/log(2)))/(e*log(4/3/log(2)) - e*l og(4/3/(x*log(2))))
Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (22) = 44\).
Time = 0.07 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.41 \[ \int \frac {\left (-15-5 e x^2\right ) \log \left (\frac {4}{3 x \log (2)}\right )+\log (x) \left (-15-5 e x^2+10 e x^2 \log \left (\frac {4}{3 x \log (2)}\right )\right )}{e x \log ^2(x)} \, dx=- 5 x^{2} + \frac {- 5 e x^{2} \log {\left (3 \right )} - 5 e x^{2} \log {\left (\log {\left (2 \right )} \right )} + 10 e x^{2} \log {\left (2 \right )} - 15 \log {\left (3 \right )} - 15 \log {\left (\log {\left (2 \right )} \right )} + 30 \log {\left (2 \right )}}{e \log {\left (x \right )}} \]
integrate(((10*x**2*exp(1)*ln(4/3/x/ln(2))-5*x**2*exp(1)-15)*ln(x)+(-5*x** 2*exp(1)-15)*ln(4/3/x/ln(2)))/x/exp(1)/ln(x)**2,x)
-5*x**2 + (-5*E*x**2*log(3) - 5*E*x**2*log(log(2)) + 10*E*x**2*log(2) - 15 *log(3) - 15*log(log(2)) + 30*log(2))*exp(-1)/log(x)
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.22 (sec) , antiderivative size = 108, normalized size of antiderivative = 4.00 \[ \int \frac {\left (-15-5 e x^2\right ) \log \left (\frac {4}{3 x \log (2)}\right )+\log (x) \left (-15-5 e x^2+10 e x^2 \log \left (\frac {4}{3 x \log (2)}\right )\right )}{e x \log ^2(x)} \, dx=5 \, {\left (2 \, {\rm Ei}\left (2 \, \log \left (x\right )\right ) e \log \left (\frac {4}{3 \, x \log \left (2\right )}\right ) - 2 \, e \Gamma \left (-1, -2 \, \log \left (x\right )\right ) \log \left (\frac {4}{3 \, x \log \left (2\right )}\right ) - {\left (x^{2} - 2 \, {\rm Ei}\left (2 \, \log \left (x\right )\right ) \log \left (x\right )\right )} e - {\left (2 \, \Gamma \left (-1, -2 \, \log \left (x\right )\right ) \log \left (x\right ) - {\rm Ei}\left (2 \, \log \left (x\right )\right )\right )} e - {\rm Ei}\left (2 \, \log \left (x\right )\right ) e + \frac {3 \, \log \left (\frac {4}{3 \, x \log \left (2\right )}\right )}{\log \left (x\right )}\right )} e^{\left (-1\right )} \]
integrate(((10*x^2*exp(1)*log(4/3/x/log(2))-5*x^2*exp(1)-15)*log(x)+(-5*x^ 2*exp(1)-15)*log(4/3/x/log(2)))/x/exp(1)/log(x)^2,x, algorithm=\
5*(2*Ei(2*log(x))*e*log(4/3/(x*log(2))) - 2*e*gamma(-1, -2*log(x))*log(4/3 /(x*log(2))) - (x^2 - 2*Ei(2*log(x))*log(x))*e - (2*gamma(-1, -2*log(x))*l og(x) - Ei(2*log(x)))*e - Ei(2*log(x))*e + 3*log(4/3/(x*log(2)))/log(x))*e ^(-1)
Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (24) = 48\).
Time = 0.25 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.07 \[ \int \frac {\left (-15-5 e x^2\right ) \log \left (\frac {4}{3 x \log (2)}\right )+\log (x) \left (-15-5 e x^2+10 e x^2 \log \left (\frac {4}{3 x \log (2)}\right )\right )}{e x \log ^2(x)} \, dx=-\frac {5 \, {\left (x^{2} e \log \left (3\right ) - 2 \, x^{2} e \log \left (2\right ) + x^{2} e \log \left (x\right ) + x^{2} e \log \left (\log \left (2\right )\right ) + 3 \, \log \left (3\right ) - 6 \, \log \left (2\right ) + 3 \, \log \left (\log \left (2\right )\right )\right )} e^{\left (-1\right )}}{\log \left (x\right )} \]
integrate(((10*x^2*exp(1)*log(4/3/x/log(2))-5*x^2*exp(1)-15)*log(x)+(-5*x^ 2*exp(1)-15)*log(4/3/x/log(2)))/x/exp(1)/log(x)^2,x, algorithm=\
-5*(x^2*e*log(3) - 2*x^2*e*log(2) + x^2*e*log(x) + x^2*e*log(log(2)) + 3*l og(3) - 6*log(2) + 3*log(log(2)))*e^(-1)/log(x)
Time = 9.12 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.11 \[ \int \frac {\left (-15-5 e x^2\right ) \log \left (\frac {4}{3 x \log (2)}\right )+\log (x) \left (-15-5 e x^2+10 e x^2 \log \left (\frac {4}{3 x \log (2)}\right )\right )}{e x \log ^2(x)} \, dx=\frac {x^2\,\left (5\,\ln \left (\frac {1}{x}\right )+10\,\ln \left (2\right )-5\,\ln \left (3\right )-5\,\ln \left (\ln \left (2\right )\right )\right )}{\ln \left (x\right )}+\frac {15\,{\mathrm {e}}^{-1}\,\left (\ln \left (\frac {1}{x}\right )+2\,\ln \left (2\right )-\ln \left (3\right )-\ln \left (\ln \left (2\right )\right )+\ln \left (x\right )\right )}{\ln \left (x\right )} \]
int(-(exp(-1)*(log(x)*(5*x^2*exp(1) - 10*x^2*exp(1)*log(4/(3*x*log(2))) + 15) + log(4/(3*x*log(2)))*(5*x^2*exp(1) + 15)))/(x*log(x)^2),x)