Integrand size = 70, antiderivative size = 28 \[ \int \frac {-50+6 e^4-6 \log (4)+6 \log \left (x^2\right )+\left (-8+e^4-\log (4)+\log \left (x^2\right )\right ) \log \left (-\frac {5 x}{-8+e^4-\log (4)+\log \left (x^2\right )}\right )}{-24+3 e^4-3 \log (4)+3 \log \left (x^2\right )} \, dx=\frac {1}{3} x \left (5+\log \left (\frac {5 x}{8-e^4+\log (4)-\log \left (x^2\right )}\right )\right ) \]
Time = 0.46 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {-50+6 e^4-6 \log (4)+6 \log \left (x^2\right )+\left (-8+e^4-\log (4)+\log \left (x^2\right )\right ) \log \left (-\frac {5 x}{-8+e^4-\log (4)+\log \left (x^2\right )}\right )}{-24+3 e^4-3 \log (4)+3 \log \left (x^2\right )} \, dx=\frac {1}{3} x \left (5+\log \left (-\frac {5 x}{-8+e^4+\log \left (\frac {x^2}{4}\right )}\right )\right ) \]
Integrate[(-50 + 6*E^4 - 6*Log[4] + 6*Log[x^2] + (-8 + E^4 - Log[4] + Log[ x^2])*Log[(-5*x)/(-8 + E^4 - Log[4] + Log[x^2])])/(-24 + 3*E^4 - 3*Log[4] + 3*Log[x^2]),x]
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.80 (sec) , antiderivative size = 182, normalized size of antiderivative = 6.50, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {7292, 27, 7292, 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.
\(\displaystyle \int \frac {6 \log \left (x^2\right )+\left (\log \left (x^2\right )+e^4-8-\log (4)\right ) \log \left (-\frac {5 x}{\log \left (x^2\right )+e^4-8-\log (4)}\right )+6 e^4-50-6 \log (4)}{3 \log \left (x^2\right )+3 e^4-24-3 \log (4)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-6 \log \left (x^2\right )-\left (\log \left (x^2\right )+e^4-8-\log (4)\right ) \log \left (-\frac {5 x}{\log \left (x^2\right )+e^4-8-\log (4)}\right )+50 \left (1-\frac {3}{25} \left (e^4-\log (4)\right )\right )}{3 \left (8 \left (1-\frac {e^4}{8}\right )-\log \left (\frac {x^2}{4}\right )\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \int \frac {-6 \log \left (x^2\right )+\left (-\log \left (x^2\right )+\log (4)-e^4+8\right ) \log \left (\frac {5 x}{-\log \left (x^2\right )+\log (4)-e^4+8}\right )+2 \left (25-3 e^4+3 \log (4)\right )}{-\log \left (\frac {x^2}{4}\right )-e^4+8}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \frac {1}{3} \int \frac {-6 \log \left (x^2\right )+\left (-\log \left (x^2\right )+\log (4)-e^4+8\right ) \log \left (\frac {5 x}{-\log \left (x^2\right )+\log (4)-e^4+8}\right )+2 \left (25-3 e^4+3 \log (4)\right )}{8 \left (1-\frac {e^4}{8}\right )-\log \left (\frac {x^2}{4}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{3} \int \left (\frac {2 \left (25 \left (1-\frac {3}{25} \left (e^4-\log (4)\right )\right )-3 \log \left (x^2\right )\right )}{8 \left (1-\frac {e^4}{8}\right )-\log \left (\frac {x^2}{4}\right )}+\log \left (-\frac {5 x}{\log \left (\frac {x^2}{4}\right )-8 \left (1-\frac {e^4}{8}\right )}\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (\frac {2 e^{4-\frac {e^4}{2}} x \operatorname {ExpIntegralEi}\left (\frac {1}{2} \left (\log \left (\frac {x^2}{4}\right )+e^4-8\right )\right )}{\sqrt {x^2}}+\frac {6 e^{4-\frac {e^4}{2}} x \left (-\log \left (\frac {x^2}{4}\right )-e^4+8\right ) \operatorname {ExpIntegralEi}\left (\frac {1}{2} \left (\log \left (\frac {x^2}{4}\right )+e^4-8\right )\right )}{\sqrt {x^2}}-\frac {2 e^{4-\frac {e^4}{2}} x \left (-3 \log \left (x^2\right )-3 e^4+25+\log (64)\right ) \operatorname {ExpIntegralEi}\left (\frac {1}{2} \left (\log \left (\frac {x^2}{4}\right )+e^4-8\right )\right )}{\sqrt {x^2}}+x \log \left (\frac {5 x}{-\log \left (\frac {x^2}{4}\right )-e^4+8}\right )+5 x\right )\) |
Int[(-50 + 6*E^4 - 6*Log[4] + 6*Log[x^2] + (-8 + E^4 - Log[4] + Log[x^2])* Log[(-5*x)/(-8 + E^4 - Log[4] + Log[x^2])])/(-24 + 3*E^4 - 3*Log[4] + 3*Lo g[x^2]),x]
(5*x + (2*E^(4 - E^4/2)*x*ExpIntegralEi[(-8 + E^4 + Log[x^2/4])/2])/Sqrt[x ^2] + (6*E^(4 - E^4/2)*x*ExpIntegralEi[(-8 + E^4 + Log[x^2/4])/2]*(8 - E^4 - Log[x^2/4]))/Sqrt[x^2] - (2*E^(4 - E^4/2)*x*ExpIntegralEi[(-8 + E^4 + L og[x^2/4])/2]*(25 - 3*E^4 + Log[64] - 3*Log[x^2]))/Sqrt[x^2] + x*Log[(5*x) /(8 - E^4 - Log[x^2/4])])/3
3.18.60.3.1 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 = 0.70 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93
method | result | size |
norman | \(\frac {5 x}{3}+\frac {x \ln \left (-\frac {5 x}{\ln \left (x^{2}\right )-2 \ln \left (2\right )+{\mathrm e}^{4}-8}\right )}{3}\) | \(26\) |
parallelrisch | \(\frac {5 x}{3}+\frac {x \ln \left (-\frac {5 x}{\ln \left (x^{2}\right )-2 \ln \left (2\right )+{\mathrm e}^{4}-8}\right )}{3}\) | \(26\) |
int(((ln(x^2)-2*ln(2)+exp(4)-8)*ln(-5*x/(ln(x^2)-2*ln(2)+exp(4)-8))+6*ln(x ^2)-12*ln(2)+6*exp(4)-50)/(3*ln(x^2)-6*ln(2)+3*exp(4)-24),x,method=_RETURN VERBOSE)
Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {-50+6 e^4-6 \log (4)+6 \log \left (x^2\right )+\left (-8+e^4-\log (4)+\log \left (x^2\right )\right ) \log \left (-\frac {5 x}{-8+e^4-\log (4)+\log \left (x^2\right )}\right )}{-24+3 e^4-3 \log (4)+3 \log \left (x^2\right )} \, dx=\frac {1}{3} \, x \log \left (-\frac {5 \, x}{e^{4} - 2 \, \log \left (2\right ) + \log \left (x^{2}\right ) - 8}\right ) + \frac {5}{3} \, x \]
integrate(((log(x^2)-2*log(2)+exp(4)-8)*log(-5*x/(log(x^2)-2*log(2)+exp(4) -8))+6*log(x^2)-12*log(2)+6*exp(4)-50)/(3*log(x^2)-6*log(2)+3*exp(4)-24),x , algorithm=\
Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {-50+6 e^4-6 \log (4)+6 \log \left (x^2\right )+\left (-8+e^4-\log (4)+\log \left (x^2\right )\right ) \log \left (-\frac {5 x}{-8+e^4-\log (4)+\log \left (x^2\right )}\right )}{-24+3 e^4-3 \log (4)+3 \log \left (x^2\right )} \, dx=\frac {x \log {\left (- \frac {5 x}{\log {\left (x^{2} \right )} - 8 - 2 \log {\left (2 \right )} + e^{4}} \right )}}{3} + \frac {5 x}{3} \]
integrate(((ln(x**2)-2*ln(2)+exp(4)-8)*ln(-5*x/(ln(x**2)-2*ln(2)+exp(4)-8) )+6*ln(x**2)-12*ln(2)+6*exp(4)-50)/(3*ln(x**2)-6*ln(2)+3*exp(4)-24),x)
Time = 0.30 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {-50+6 e^4-6 \log (4)+6 \log \left (x^2\right )+\left (-8+e^4-\log (4)+\log \left (x^2\right )\right ) \log \left (-\frac {5 x}{-8+e^4-\log (4)+\log \left (x^2\right )}\right )}{-24+3 e^4-3 \log (4)+3 \log \left (x^2\right )} \, dx=\frac {1}{3} \, x {\left (\log \left (5\right ) + 5\right )} + \frac {1}{3} \, x \log \left (x\right ) - \frac {1}{3} \, x \log \left (-e^{4} + 2 \, \log \left (2\right ) - 2 \, \log \left (x\right ) + 8\right ) \]
integrate(((log(x^2)-2*log(2)+exp(4)-8)*log(-5*x/(log(x^2)-2*log(2)+exp(4) -8))+6*log(x^2)-12*log(2)+6*exp(4)-50)/(3*log(x^2)-6*log(2)+3*exp(4)-24),x , algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 123 vs. \(2 (23) = 46\).
Time = 0.40 (sec) , antiderivative size = 123, normalized size of antiderivative = 4.39 \[ \int \frac {-50+6 e^4-6 \log (4)+6 \log \left (x^2\right )+\left (-8+e^4-\log (4)+\log \left (x^2\right )\right ) \log \left (-\frac {5 x}{-8+e^4-\log (4)+\log \left (x^2\right )}\right )}{-24+3 e^4-3 \log (4)+3 \log \left (x^2\right )} \, dx=\frac {1}{3} \, x \log \left (5\right ) - \frac {1}{6} \, x \log \left (4 \, \pi ^{2} \left \lfloor -\frac {1}{2} \, \mathrm {sgn}\left (x\right ) + 1 \right \rfloor ^{2} + 4 \, \pi ^{2} \left \lfloor -\frac {1}{2} \, \mathrm {sgn}\left (x\right ) + 1 \right \rfloor \mathrm {sgn}\left (x\right ) - 4 \, \pi ^{2} \left \lfloor -\frac {1}{2} \, \mathrm {sgn}\left (x\right ) + 1 \right \rfloor - 2 \, \pi ^{2} \mathrm {sgn}\left (x\right ) + 2 \, \pi ^{2} - 4 \, e^{4} \log \left (2\right ) + 4 \, \log \left (2\right )^{2} + 2 \, e^{4} \log \left (x^{2}\right ) - 4 \, \log \left (2\right ) \log \left (x^{2}\right ) + \log \left (x^{2}\right )^{2} + e^{8} - 16 \, e^{4} + 32 \, \log \left (2\right ) - 16 \, \log \left (x^{2}\right ) + 64\right ) + \frac {1}{3} \, x \log \left ({\left | x \right |}\right ) + \frac {5}{3} \, x \]
integrate(((log(x^2)-2*log(2)+exp(4)-8)*log(-5*x/(log(x^2)-2*log(2)+exp(4) -8))+6*log(x^2)-12*log(2)+6*exp(4)-50)/(3*log(x^2)-6*log(2)+3*exp(4)-24),x , algorithm=\
1/3*x*log(5) - 1/6*x*log(4*pi^2*floor(-1/2*sgn(x) + 1)^2 + 4*pi^2*floor(-1 /2*sgn(x) + 1)*sgn(x) - 4*pi^2*floor(-1/2*sgn(x) + 1) - 2*pi^2*sgn(x) + 2* pi^2 - 4*e^4*log(2) + 4*log(2)^2 + 2*e^4*log(x^2) - 4*log(2)*log(x^2) + lo g(x^2)^2 + e^8 - 16*e^4 + 32*log(2) - 16*log(x^2) + 64) + 1/3*x*log(abs(x) ) + 5/3*x
Timed out. \[ \int \frac {-50+6 e^4-6 \log (4)+6 \log \left (x^2\right )+\left (-8+e^4-\log (4)+\log \left (x^2\right )\right ) \log \left (-\frac {5 x}{-8+e^4-\log (4)+\log \left (x^2\right )}\right )}{-24+3 e^4-3 \log (4)+3 \log \left (x^2\right )} \, dx=\int \frac {6\,\ln \left (x^2\right )+6\,{\mathrm {e}}^4-12\,\ln \left (2\right )+\ln \left (-\frac {5\,x}{\ln \left (x^2\right )+{\mathrm {e}}^4-2\,\ln \left (2\right )-8}\right )\,\left (\ln \left (x^2\right )+{\mathrm {e}}^4-2\,\ln \left (2\right )-8\right )-50}{3\,\ln \left (x^2\right )+3\,{\mathrm {e}}^4-6\,\ln \left (2\right )-24} \,d x \]
int((6*log(x^2) + 6*exp(4) - 12*log(2) + log(-(5*x)/(log(x^2) + exp(4) - 2 *log(2) - 8))*(log(x^2) + exp(4) - 2*log(2) - 8) - 50)/(3*log(x^2) + 3*exp (4) - 6*log(2) - 24),x)