Integrand size = 53, antiderivative size = 27 \[ \int \frac {5 \log (x)+\left (10+\left (-15+x^2\right ) \log (x)\right ) \log (2 x)-5 \log (x) \log (2 x) \log \left (x \log ^2(x) \log (2 x)\right )}{x^2 \log (x) \log (2 x)} \, dx=x-\log (3)+\frac {5 \left (4-x+\log \left (x \log ^2(x) \log (2 x)\right )\right )}{x} \] Output:
x+5*(ln(x*ln(x)^2*ln(2*x))-x+4)/x-ln(3)
Time = 0.11 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {5 \log (x)+\left (10+\left (-15+x^2\right ) \log (x)\right ) \log (2 x)-5 \log (x) \log (2 x) \log \left (x \log ^2(x) \log (2 x)\right )}{x^2 \log (x) \log (2 x)} \, dx=\frac {20}{x}+x+\frac {5 \log \left (x \log ^2(x) \log (2 x)\right )}{x} \] Input:
Integrate[(5*Log[x] + (10 + (-15 + x^2)*Log[x])*Log[2*x] - 5*Log[x]*Log[2* x]*Log[x*Log[x]^2*Log[2*x]])/(x^2*Log[x]*Log[2*x]),x]
Output:
20/x + x + (5*Log[x*Log[x]^2*Log[2*x]])/x
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 1.18 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.85, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {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.
\(\displaystyle \int \frac {\left (\left (x^2-15\right ) \log (x)+10\right ) \log (2 x)-5 \log (2 x) \log \left (x \log ^2(x) \log (2 x)\right ) \log (x)+5 \log (x)}{x^2 \log (x) \log (2 x)} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x^2 \log (x) \log (2 x)+5 \log (x)-15 \log (x) \log (2 x)+10 \log (2 x)}{x^2 \log (x) \log (2 x)}-\frac {5 \log \left (x \log ^2(x) \log (2 x)\right )}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 10 \operatorname {ExpIntegralEi}(-\log (x))+5 \log (x) \operatorname {ExpIntegralEi}(-\log (x))-5 (\log (x)+2) \operatorname {ExpIntegralEi}(-\log (x))+x+\frac {20}{x}+\frac {5 \log \left (x \log ^2(x) \log (2 x)\right )}{x}\) |
Input:
Int[(5*Log[x] + (10 + (-15 + x^2)*Log[x])*Log[2*x] - 5*Log[x]*Log[2*x]*Log [x*Log[x]^2*Log[2*x]])/(x^2*Log[x]*Log[2*x]),x]
Output:
20/x + x + 10*ExpIntegralEi[-Log[x]] + 5*ExpIntegralEi[-Log[x]]*Log[x] - 5 *ExpIntegralEi[-Log[x]]*(2 + Log[x]) + (5*Log[x*Log[x]^2*Log[2*x]])/x
Time = 1.71 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85
method | result | size |
parallelrisch | \(\frac {20+x^{2}+5 \ln \left (x \ln \left (x \right )^{2} \ln \left (2 x \right )\right )}{x}\) | \(23\) |
risch | \(\frac {5 \ln \left (2 i \ln \left (x \right )+2 i \ln \left (2\right )\right )}{x}-\frac {5 i \pi \,\operatorname {csgn}\left (i x \left (2 i \ln \left (x \right )+2 i \ln \left (2\right )\right ) \ln \left (x \right )^{2}\right ) \operatorname {csgn}\left (x \ln \left (x \right )^{2} \left (2 i \ln \left (x \right )+2 i \ln \left (2\right )\right )\right )+5 i \pi \operatorname {csgn}\left (i x \left (2 i \ln \left (x \right )+2 i \ln \left (2\right )\right ) \ln \left (x \right )^{2}\right )^{3}-5 i \pi \operatorname {csgn}\left (i x \left (2 i \ln \left (x \right )+2 i \ln \left (2\right )\right ) \ln \left (x \right )^{2}\right )^{2} \operatorname {csgn}\left (i x \right )+5 i \pi \,\operatorname {csgn}\left (2 \ln \left (x \right )+2 \ln \left (2\right )\right ) \operatorname {csgn}\left (i \ln \left (x \right )^{2} \left (2 i \ln \left (x \right )+2 i \ln \left (2\right )\right )\right )^{2}+5 i \pi \operatorname {csgn}\left (i \ln \left (x \right )\right )^{2} \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )+5 i \pi \operatorname {csgn}\left (x \ln \left (x \right )^{2} \left (2 i \ln \left (x \right )+2 i \ln \left (2\right )\right )\right )^{3}-5 i \pi \,\operatorname {csgn}\left (i \ln \left (x \right )^{2} \left (2 i \ln \left (x \right )+2 i \ln \left (2\right )\right )\right ) \operatorname {csgn}\left (i x \left (2 i \ln \left (x \right )+2 i \ln \left (2\right )\right ) \ln \left (x \right )^{2}\right )^{2}-10 i \pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{2}+5 i \pi \,\operatorname {csgn}\left (i \ln \left (x \right )^{2} \left (2 i \ln \left (x \right )+2 i \ln \left (2\right )\right )\right ) \operatorname {csgn}\left (i x \left (2 i \ln \left (x \right )+2 i \ln \left (2\right )\right ) \ln \left (x \right )^{2}\right ) \operatorname {csgn}\left (i x \right )-5 i \pi \operatorname {csgn}\left (x \ln \left (x \right )^{2} \left (2 i \ln \left (x \right )+2 i \ln \left (2\right )\right )\right )^{2}+5 i \pi \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{3}-5 i \pi \,\operatorname {csgn}\left (i x \left (2 i \ln \left (x \right )+2 i \ln \left (2\right )\right ) \ln \left (x \right )^{2}\right ) \operatorname {csgn}\left (x \ln \left (x \right )^{2} \left (2 i \ln \left (x \right )+2 i \ln \left (2\right )\right )\right )^{2}+5 i \pi \operatorname {csgn}\left (i \ln \left (x \right )^{2} \left (2 i \ln \left (x \right )+2 i \ln \left (2\right )\right )\right )^{3}+5 i \pi -5 i \pi \,\operatorname {csgn}\left (i \ln \left (x \right )^{2}\right ) \operatorname {csgn}\left (i \ln \left (x \right )^{2} \left (2 i \ln \left (x \right )+2 i \ln \left (2\right )\right )\right )^{2}-5 i \pi \,\operatorname {csgn}\left (i \ln \left (x \right )^{2}\right ) \operatorname {csgn}\left (2 \ln \left (x \right )+2 \ln \left (2\right )\right ) \operatorname {csgn}\left (i \ln \left (x \right )^{2} \left (2 i \ln \left (x \right )+2 i \ln \left (2\right )\right )\right )-2 x^{2}+10 \ln \left (2\right )-10 \ln \left (x \right )-20 \ln \left (\ln \left (x \right )\right )-40}{2 x}\) | \(520\) |
Input:
int((-5*ln(x)*ln(2*x)*ln(x*ln(x)^2*ln(2*x))+((x^2-15)*ln(x)+10)*ln(2*x)+5* ln(x))/x^2/ln(x)/ln(2*x),x,method=_RETURNVERBOSE)
Output:
1/x*(20+x^2+5*ln(x*ln(x)^2*ln(2*x)))
Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {5 \log (x)+\left (10+\left (-15+x^2\right ) \log (x)\right ) \log (2 x)-5 \log (x) \log (2 x) \log \left (x \log ^2(x) \log (2 x)\right )}{x^2 \log (x) \log (2 x)} \, dx=\frac {x^{2} + 5 \, \log \left (x \log \left (2\right ) \log \left (x\right )^{2} + x \log \left (x\right )^{3}\right ) + 20}{x} \] Input:
integrate((-5*log(x)*log(2*x)*log(x*log(x)^2*log(2*x))+((x^2-15)*log(x)+10 )*log(2*x)+5*log(x))/x^2/log(x)/log(2*x),x, algorithm="fricas")
Output:
(x^2 + 5*log(x*log(2)*log(x)^2 + x*log(x)^3) + 20)/x
Time = 0.16 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {5 \log (x)+\left (10+\left (-15+x^2\right ) \log (x)\right ) \log (2 x)-5 \log (x) \log (2 x) \log \left (x \log ^2(x) \log (2 x)\right )}{x^2 \log (x) \log (2 x)} \, dx=x + \frac {5 \log {\left (x \left (\log {\left (x \right )} + \log {\left (2 \right )}\right ) \log {\left (x \right )}^{2} \right )}}{x} + \frac {20}{x} \] Input:
integrate((-5*ln(x)*ln(2*x)*ln(x*ln(x)**2*ln(2*x))+((x**2-15)*ln(x)+10)*ln (2*x)+5*ln(x))/x**2/ln(x)/ln(2*x),x)
Output:
x + 5*log(x*(log(x) + log(2))*log(x)**2)/x + 20/x
\[ \int \frac {5 \log (x)+\left (10+\left (-15+x^2\right ) \log (x)\right ) \log (2 x)-5 \log (x) \log (2 x) \log \left (x \log ^2(x) \log (2 x)\right )}{x^2 \log (x) \log (2 x)} \, dx=\int { -\frac {5 \, \log \left (x \log \left (2 \, x\right ) \log \left (x\right )^{2}\right ) \log \left (2 \, x\right ) \log \left (x\right ) - {\left ({\left (x^{2} - 15\right )} \log \left (x\right ) + 10\right )} \log \left (2 \, x\right ) - 5 \, \log \left (x\right )}{x^{2} \log \left (2 \, x\right ) \log \left (x\right )} \,d x } \] Input:
integrate((-5*log(x)*log(2*x)*log(x*log(x)^2*log(2*x))+((x^2-15)*log(x)+10 )*log(2*x)+5*log(x))/x^2/log(x)/log(2*x),x, algorithm="maxima")
Output:
x + 5*(log(x) + log(log(2) + log(x)) + 2*log(log(x)) + 1)/x + 15/x + 10*Ei (-log(2*x)) + 10*Ei(-log(x)) - 5*integrate(1/(x^2*log(2) + x^2*log(x)), x) - 10*integrate(1/(x^2*log(x)), x)
Time = 0.15 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {5 \log (x)+\left (10+\left (-15+x^2\right ) \log (x)\right ) \log (2 x)-5 \log (x) \log (2 x) \log \left (x \log ^2(x) \log (2 x)\right )}{x^2 \log (x) \log (2 x)} \, dx=x + \frac {5 \, \log \left (\log \left (2\right ) \log \left (x\right )^{2} + \log \left (x\right )^{3}\right )}{x} + \frac {5 \, \log \left (x\right )}{x} + \frac {20}{x} \] Input:
integrate((-5*log(x)*log(2*x)*log(x*log(x)^2*log(2*x))+((x^2-15)*log(x)+10 )*log(2*x)+5*log(x))/x^2/log(x)/log(2*x),x, algorithm="giac")
Output:
x + 5*log(log(2)*log(x)^2 + log(x)^3)/x + 5*log(x)/x + 20/x
Time = 3.58 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {5 \log (x)+\left (10+\left (-15+x^2\right ) \log (x)\right ) \log (2 x)-5 \log (x) \log (2 x) \log \left (x \log ^2(x) \log (2 x)\right )}{x^2 \log (x) \log (2 x)} \, dx=x+\frac {5\,\ln \left (x\,\ln \left (2\,x\right )\,{\ln \left (x\right )}^2\right )+20}{x} \] Input:
int((5*log(x) + log(2*x)*(log(x)*(x^2 - 15) + 10) - 5*log(2*x)*log(x*log(2 *x)*log(x)^2)*log(x))/(x^2*log(2*x)*log(x)),x)
Output:
x + (5*log(x*log(2*x)*log(x)^2) + 20)/x
Time = 0.18 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {5 \log (x)+\left (10+\left (-15+x^2\right ) \log (x)\right ) \log (2 x)-5 \log (x) \log (2 x) \log \left (x \log ^2(x) \log (2 x)\right )}{x^2 \log (x) \log (2 x)} \, dx=\frac {5 \,\mathrm {log}\left (\mathrm {log}\left (2 x \right ) \mathrm {log}\left (x \right )^{2} x \right )+x^{2}+20}{x} \] Input:
int((-5*log(x)*log(2*x)*log(x*log(x)^2*log(2*x))+((x^2-15)*log(x)+10)*log( 2*x)+5*log(x))/x^2/log(x)/log(2*x),x)
Output:
(5*log(log(2*x)*log(x)**2*x) + x**2 + 20)/x