Integrand size = 72, antiderivative size = 25 \[ \int \frac {-25+e^{2 x}-10 x+(5+2 x) \log (3)+\left (-10 x^2+2 x^2 \log (3)\right ) \log (x)+\left (-10 x+2 e^{2 x} x+2 x \log (3)\right ) \log (x) \log (\log (x))}{(-5 x+x \log (3)) \log (x)} \, dx=x^2+\left (5+2 x+\frac {e^{2 x}}{-5+\log (3)}\right ) \log (\log (x)) \] Output:
x^2+ln(ln(x))*(2*x+5+exp(2*x)/(ln(3)-5))
Time = 0.30 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.56 \[ \int \frac {-25+e^{2 x}-10 x+(5+2 x) \log (3)+\left (-10 x^2+2 x^2 \log (3)\right ) \log (x)+\left (-10 x+2 e^{2 x} x+2 x \log (3)\right ) \log (x) \log (\log (x))}{(-5 x+x \log (3)) \log (x)} \, dx=\frac {x^2 \left (-5+\frac {\log (9)}{2}\right )+\left (-25+e^{2 x}+x (-10+\log (9))+\log (243)\right ) \log (\log (x))}{-5+\log (3)} \] Input:
Integrate[(-25 + E^(2*x) - 10*x + (5 + 2*x)*Log[3] + (-10*x^2 + 2*x^2*Log[ 3])*Log[x] + (-10*x + 2*E^(2*x)*x + 2*x*Log[3])*Log[x]*Log[Log[x]])/((-5*x + x*Log[3])*Log[x]),x]
Output:
(x^2*(-5 + Log[9]/2) + (-25 + E^(2*x) + x*(-10 + Log[9]) + Log[243])*Log[L og[x]])/(-5 + Log[3])
Leaf count is larger than twice the leaf count of optimal. \(53\) vs. \(2(25)=50\).
Time = 0.89 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.12, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {6, 27, 25, 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.
\(\displaystyle \int \frac {\left (2 x^2 \log (3)-10 x^2\right ) \log (x)-10 x+e^{2 x}+\left (2 e^{2 x} x-10 x+2 x \log (3)\right ) \log (x) \log (\log (x))+(2 x+5) \log (3)-25}{(x \log (3)-5 x) \log (x)} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {\left (2 x^2 \log (3)-10 x^2\right ) \log (x)-10 x+e^{2 x}+\left (2 e^{2 x} x-10 x+2 x \log (3)\right ) \log (x) \log (\log (x))+(2 x+5) \log (3)-25}{x (\log (3)-5) \log (x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int -\frac {10 x-e^{2 x}+2 \left (5 x^2-x^2 \log (3)\right ) \log (x)+2 \left (-e^{2 x} x-\log (3) x+5 x\right ) \log (x) \log (\log (x))-(2 x+5) \log (3)+25}{x \log (x)}dx}{5-\log (3)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {10 x-e^{2 x}+2 \left (5 x^2-x^2 \log (3)\right ) \log (x)+2 \left (-e^{2 x} x-\log (3) x+5 x\right ) \log (x) \log (\log (x))-(2 x+5) \log (3)+25}{x \log (x)}dx}{5-\log (3)}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {\int \left (-\frac {e^{2 x} (2 x \log (x) \log (\log (x))+1)}{x \log (x)}-\frac {(-5+\log (3)) \left (2 \log (x) x^2+2 \log (x) \log (\log (x)) x+2 x+5\right )}{x \log (x)}\right )dx}{5-\log (3)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {x^2 (5-\log (3))+2 x (5-\log (3)) \log (\log (x))-e^{2 x} \log (\log (x))+5 (5-\log (3)) \log (\log (x))}{5-\log (3)}\) |
Input:
Int[(-25 + E^(2*x) - 10*x + (5 + 2*x)*Log[3] + (-10*x^2 + 2*x^2*Log[3])*Lo g[x] + (-10*x + 2*E^(2*x)*x + 2*x*Log[3])*Log[x]*Log[Log[x]])/((-5*x + x*L og[3])*Log[x]),x]
Output:
(x^2*(5 - Log[3]) - E^(2*x)*Log[Log[x]] + 5*(5 - Log[3])*Log[Log[x]] + 2*x *(5 - Log[3])*Log[Log[x]])/(5 - Log[3])
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 1.14 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32
method | result | size |
risch | \(\frac {\left (2 x \ln \left (3\right )-10 x +{\mathrm e}^{2 x}\right ) \ln \left (\ln \left (x \right )\right )}{\ln \left (3\right )-5}+x^{2}+5 \ln \left (\ln \left (x \right )\right )\) | \(33\) |
parallelrisch | \(\frac {2 \ln \left (3\right ) \ln \left (\ln \left (x \right )\right ) x +x^{2} \ln \left (3\right )+\ln \left (\ln \left (x \right )\right ) {\mathrm e}^{2 x}+5 \ln \left (\ln \left (x \right )\right ) \ln \left (3\right )-10 x \ln \left (\ln \left (x \right )\right )-5 x^{2}-25 \ln \left (\ln \left (x \right )\right )}{\ln \left (3\right )-5}\) | \(54\) |
Input:
int(((2*x*exp(2*x)+2*x*ln(3)-10*x)*ln(x)*ln(ln(x))+(2*x^2*ln(3)-10*x^2)*ln (x)+exp(2*x)+(5+2*x)*ln(3)-10*x-25)/(x*ln(3)-5*x)/ln(x),x,method=_RETURNVE RBOSE)
Output:
(2*x*ln(3)-10*x+exp(2*x))/(ln(3)-5)*ln(ln(x))+x^2+5*ln(ln(x))
Time = 0.09 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.60 \[ \int \frac {-25+e^{2 x}-10 x+(5+2 x) \log (3)+\left (-10 x^2+2 x^2 \log (3)\right ) \log (x)+\left (-10 x+2 e^{2 x} x+2 x \log (3)\right ) \log (x) \log (\log (x))}{(-5 x+x \log (3)) \log (x)} \, dx=\frac {x^{2} \log \left (3\right ) - 5 \, x^{2} + {\left ({\left (2 \, x + 5\right )} \log \left (3\right ) - 10 \, x + e^{\left (2 \, x\right )} - 25\right )} \log \left (\log \left (x\right )\right )}{\log \left (3\right ) - 5} \] Input:
integrate(((2*x*exp(2*x)+2*x*log(3)-10*x)*log(x)*log(log(x))+(2*x^2*log(3) -10*x^2)*log(x)+exp(2*x)+(5+2*x)*log(3)-10*x-25)/(x*log(3)-5*x)/log(x),x, algorithm="fricas")
Output:
(x^2*log(3) - 5*x^2 + ((2*x + 5)*log(3) - 10*x + e^(2*x) - 25)*log(log(x)) )/(log(3) - 5)
Time = 1.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {-25+e^{2 x}-10 x+(5+2 x) \log (3)+\left (-10 x^2+2 x^2 \log (3)\right ) \log (x)+\left (-10 x+2 e^{2 x} x+2 x \log (3)\right ) \log (x) \log (\log (x))}{(-5 x+x \log (3)) \log (x)} \, dx=x^{2} + 2 x \log {\left (\log {\left (x \right )} \right )} + \frac {e^{2 x} \log {\left (\log {\left (x \right )} \right )}}{-5 + \log {\left (3 \right )}} + 5 \log {\left (\log {\left (x \right )} \right )} \] Input:
integrate(((2*x*exp(2*x)+2*x*ln(3)-10*x)*ln(x)*ln(ln(x))+(2*x**2*ln(3)-10* x**2)*ln(x)+exp(2*x)+(5+2*x)*ln(3)-10*x-25)/(x*ln(3)-5*x)/ln(x),x)
Output:
x**2 + 2*x*log(log(x)) + exp(2*x)*log(log(x))/(-5 + log(3)) + 5*log(log(x) )
Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (24) = 48\).
Time = 0.17 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.28 \[ \int \frac {-25+e^{2 x}-10 x+(5+2 x) \log (3)+\left (-10 x^2+2 x^2 \log (3)\right ) \log (x)+\left (-10 x+2 e^{2 x} x+2 x \log (3)\right ) \log (x) \log (\log (x))}{(-5 x+x \log (3)) \log (x)} \, dx=\frac {5 \, \log \left (3\right ) \log \left (\log \left (x\right )\right )}{\log \left (3\right ) - 5} + \frac {x^{2} {\left (\log \left (3\right ) - 5\right )} + {\left (2 \, x {\left (\log \left (3\right ) - 5\right )} + e^{\left (2 \, x\right )}\right )} \log \left (\log \left (x\right )\right )}{\log \left (3\right ) - 5} - \frac {25 \, \log \left (\log \left (x\right )\right )}{\log \left (3\right ) - 5} \] Input:
integrate(((2*x*exp(2*x)+2*x*log(3)-10*x)*log(x)*log(log(x))+(2*x^2*log(3) -10*x^2)*log(x)+exp(2*x)+(5+2*x)*log(3)-10*x-25)/(x*log(3)-5*x)/log(x),x, algorithm="maxima")
Output:
5*log(3)*log(log(x))/(log(3) - 5) + (x^2*(log(3) - 5) + (2*x*(log(3) - 5) + e^(2*x))*log(log(x)))/(log(3) - 5) - 25*log(log(x))/(log(3) - 5)
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (24) = 48\).
Time = 0.13 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.12 \[ \int \frac {-25+e^{2 x}-10 x+(5+2 x) \log (3)+\left (-10 x^2+2 x^2 \log (3)\right ) \log (x)+\left (-10 x+2 e^{2 x} x+2 x \log (3)\right ) \log (x) \log (\log (x))}{(-5 x+x \log (3)) \log (x)} \, dx=\frac {x^{2} \log \left (3\right ) + 2 \, x \log \left (3\right ) \log \left (\log \left (x\right )\right ) - 5 \, x^{2} - 10 \, x \log \left (\log \left (x\right )\right ) + e^{\left (2 \, x\right )} \log \left (\log \left (x\right )\right ) + 5 \, \log \left (3\right ) \log \left (\log \left (x\right )\right ) - 25 \, \log \left (\log \left (x\right )\right )}{\log \left (3\right ) - 5} \] Input:
integrate(((2*x*exp(2*x)+2*x*log(3)-10*x)*log(x)*log(log(x))+(2*x^2*log(3) -10*x^2)*log(x)+exp(2*x)+(5+2*x)*log(3)-10*x-25)/(x*log(3)-5*x)/log(x),x, algorithm="giac")
Output:
(x^2*log(3) + 2*x*log(3)*log(log(x)) - 5*x^2 - 10*x*log(log(x)) + e^(2*x)* log(log(x)) + 5*log(3)*log(log(x)) - 25*log(log(x)))/(log(3) - 5)
Timed out. \[ \int \frac {-25+e^{2 x}-10 x+(5+2 x) \log (3)+\left (-10 x^2+2 x^2 \log (3)\right ) \log (x)+\left (-10 x+2 e^{2 x} x+2 x \log (3)\right ) \log (x) \log (\log (x))}{(-5 x+x \log (3)) \log (x)} \, dx=\int -\frac {{\mathrm {e}}^{2\,x}-10\,x+\ln \left (3\right )\,\left (2\,x+5\right )+\ln \left (x\right )\,\left (2\,x^2\,\ln \left (3\right )-10\,x^2\right )+\ln \left (\ln \left (x\right )\right )\,\ln \left (x\right )\,\left (2\,x\,{\mathrm {e}}^{2\,x}-10\,x+2\,x\,\ln \left (3\right )\right )-25}{\ln \left (x\right )\,\left (5\,x-x\,\ln \left (3\right )\right )} \,d x \] Input:
int(-(exp(2*x) - 10*x + log(3)*(2*x + 5) + log(x)*(2*x^2*log(3) - 10*x^2) + log(log(x))*log(x)*(2*x*exp(2*x) - 10*x + 2*x*log(3)) - 25)/(log(x)*(5*x - x*log(3))),x)
Output:
int(-(exp(2*x) - 10*x + log(3)*(2*x + 5) + log(x)*(2*x^2*log(3) - 10*x^2) + log(log(x))*log(x)*(2*x*exp(2*x) - 10*x + 2*x*log(3)) - 25)/(log(x)*(5*x - x*log(3))), x)
Time = 0.15 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.16 \[ \int \frac {-25+e^{2 x}-10 x+(5+2 x) \log (3)+\left (-10 x^2+2 x^2 \log (3)\right ) \log (x)+\left (-10 x+2 e^{2 x} x+2 x \log (3)\right ) \log (x) \log (\log (x))}{(-5 x+x \log (3)) \log (x)} \, dx=\frac {e^{2 x} \mathrm {log}\left (\mathrm {log}\left (x \right )\right )+2 \,\mathrm {log}\left (\mathrm {log}\left (x \right )\right ) \mathrm {log}\left (3\right ) x +5 \,\mathrm {log}\left (\mathrm {log}\left (x \right )\right ) \mathrm {log}\left (3\right )-10 \,\mathrm {log}\left (\mathrm {log}\left (x \right )\right ) x -25 \,\mathrm {log}\left (\mathrm {log}\left (x \right )\right )+\mathrm {log}\left (3\right ) x^{2}-5 x^{2}}{\mathrm {log}\left (3\right )-5} \] Input:
int(((2*x*exp(2*x)+2*x*log(3)-10*x)*log(x)*log(log(x))+(2*x^2*log(3)-10*x^ 2)*log(x)+exp(2*x)+(5+2*x)*log(3)-10*x-25)/(x*log(3)-5*x)/log(x),x)
Output:
(e**(2*x)*log(log(x)) + 2*log(log(x))*log(3)*x + 5*log(log(x))*log(3) - 10 *log(log(x))*x - 25*log(log(x)) + log(3)*x**2 - 5*x**2)/(log(3) - 5)