Integrand size = 158, antiderivative size = 34 \[ \int \frac {e^{3/x} \left (-12+3 e^3+3 x\right ) \log (5)+\left (-4 x^2+e^3 x^2+x^3\right ) \log (5)+\left (-3 e^{3/x} \log (5)-x^2 \log (5)\right ) \log (x)+e^{\frac {4+x+\log (5) \log \left (4-e^3-x+\log (x)\right )}{\log (5)}} \left (4 x^2-e^3 x^2-x^3+\left (x-x^2\right ) \log (5)+x^2 \log (x)\right )}{\left (4 x^2-e^3 x^2-x^3\right ) \log (5)+x^2 \log (5) \log (x)} \, dx=e^{3/x}-x+e^{\frac {4+x}{\log (5)}} \left (4-e^3-x+\log (x)\right ) \] Output:
exp(ln(ln(x)+4-exp(3)-x)+(4+x)/ln(5))+exp(3/x)-x
Time = 0.52 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.32 \[ \int \frac {e^{3/x} \left (-12+3 e^3+3 x\right ) \log (5)+\left (-4 x^2+e^3 x^2+x^3\right ) \log (5)+\left (-3 e^{3/x} \log (5)-x^2 \log (5)\right ) \log (x)+e^{\frac {4+x+\log (5) \log \left (4-e^3-x+\log (x)\right )}{\log (5)}} \left (4 x^2-e^3 x^2-x^3+\left (x-x^2\right ) \log (5)+x^2 \log (x)\right )}{\left (4 x^2-e^3 x^2-x^3\right ) \log (5)+x^2 \log (5) \log (x)} \, dx=e^{3/x}-x-e^{\frac {4+x}{\log (5)}} \left (-4+125^{\frac {1}{\log (5)}}+x\right )+e^{\frac {4+x}{\log (5)}} \log (x) \] Input:
Integrate[(E^(3/x)*(-12 + 3*E^3 + 3*x)*Log[5] + (-4*x^2 + E^3*x^2 + x^3)*L og[5] + (-3*E^(3/x)*Log[5] - x^2*Log[5])*Log[x] + E^((4 + x + Log[5]*Log[4 - E^3 - x + Log[x]])/Log[5])*(4*x^2 - E^3*x^2 - x^3 + (x - x^2)*Log[5] + x^2*Log[x]))/((4*x^2 - E^3*x^2 - x^3)*Log[5] + x^2*Log[5]*Log[x]),x]
Output:
E^(3/x) - x - E^((4 + x)/Log[5])*(-4 + 125^Log[5]^(-1) + x) + E^((4 + x)/L og[5])*Log[x]
Leaf count is larger than twice the leaf count of optimal. \(95\) vs. \(2(34)=68\).
Time = 1.33 (sec) , antiderivative size = 95, normalized size of antiderivative = 2.79, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.013, Rules used = {7239, 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.
\(\displaystyle \int \frac {\left (x^2 (-\log (5))-3 e^{3/x} \log (5)\right ) \log (x)+e^{\frac {x+\log (5) \log \left (-x+\log (x)-e^3+4\right )+4}{\log (5)}} \left (-x^3-e^3 x^2+4 x^2+x^2 \log (x)+\left (x-x^2\right ) \log (5)\right )+\left (x^3+e^3 x^2-4 x^2\right ) \log (5)+e^{3/x} \left (3 x+3 e^3-12\right ) \log (5)}{x^2 \log (5) \log (x)+\left (-x^3-e^3 x^2+4 x^2\right ) \log (5)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \left (-\frac {3 e^{3/x}}{x^2}-\frac {e^{\frac {x}{\log (5)}+\frac {4}{\log (5)}} \left (x^2-x \left (4-125^{\frac {1}{\log (5)}}-\log (5)\right )-\log (5)\right )}{x \log (5)}+\frac {e^{\frac {x}{\log (5)}+\frac {4}{\log (5)}} \log (x)}{\log (5)}-1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -x+e^{3/x}+x \left (-e^{\frac {x}{\log (5)}+\frac {4}{\log (5)}}\right )+e^{\frac {x}{\log (5)}+\frac {4}{\log (5)}} \log (x)+\log (5) e^{\frac {x}{\log (5)}+\frac {4}{\log (5)}}+\left (4-125^{\frac {1}{\log (5)}}-\log (5)\right ) e^{\frac {x}{\log (5)}+\frac {4}{\log (5)}}\) |
Input:
Int[(E^(3/x)*(-12 + 3*E^3 + 3*x)*Log[5] + (-4*x^2 + E^3*x^2 + x^3)*Log[5] + (-3*E^(3/x)*Log[5] - x^2*Log[5])*Log[x] + E^((4 + x + Log[5]*Log[4 - E^3 - x + Log[x]])/Log[5])*(4*x^2 - E^3*x^2 - x^3 + (x - x^2)*Log[5] + x^2*Lo g[x]))/((4*x^2 - E^3*x^2 - x^3)*Log[5] + x^2*Log[5]*Log[x]),x]
Output:
E^(3/x) - x - E^(4/Log[5] + x/Log[5])*x + E^(4/Log[5] + x/Log[5])*(4 - 125 ^Log[5]^(-1) - Log[5]) + E^(4/Log[5] + x/Log[5])*Log[5] + E^(4/Log[5] + x/ Log[5])*Log[x]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 32.84 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94
method | result | size |
risch | \({\mathrm e}^{\frac {3}{x}}-x +\left (\ln \left (x \right )+4-{\mathrm e}^{3}-x \right ) {\mathrm e}^{\frac {4+x}{\ln \left (5\right )}}\) | \(32\) |
parallelrisch | \(-\frac {x \ln \left (5\right )-\ln \left (5\right ) {\mathrm e}^{\frac {3}{x}}-\ln \left (5\right ) {\mathrm e}^{\frac {\ln \left (5\right ) \ln \left (\ln \left (x \right )+4-{\mathrm e}^{3}-x \right )+4+x}{\ln \left (5\right )}}}{\ln \left (5\right )}\) | \(50\) |
Input:
int(((x^2*ln(x)+(-x^2+x)*ln(5)-x^2*exp(3)-x^3+4*x^2)*exp((ln(5)*ln(ln(x)+4 -exp(3)-x)+4+x)/ln(5))+(-3*ln(5)*exp(3/x)-x^2*ln(5))*ln(x)+(3*exp(3)+3*x-1 2)*ln(5)*exp(3/x)+(x^2*exp(3)+x^3-4*x^2)*ln(5))/(x^2*ln(5)*ln(x)+(-x^2*exp (3)-x^3+4*x^2)*ln(5)),x,method=_RETURNVERBOSE)
Output:
exp(3/x)-x+(ln(x)+4-exp(3)-x)*exp((4+x)/ln(5))
Time = 0.10 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {e^{3/x} \left (-12+3 e^3+3 x\right ) \log (5)+\left (-4 x^2+e^3 x^2+x^3\right ) \log (5)+\left (-3 e^{3/x} \log (5)-x^2 \log (5)\right ) \log (x)+e^{\frac {4+x+\log (5) \log \left (4-e^3-x+\log (x)\right )}{\log (5)}} \left (4 x^2-e^3 x^2-x^3+\left (x-x^2\right ) \log (5)+x^2 \log (x)\right )}{\left (4 x^2-e^3 x^2-x^3\right ) \log (5)+x^2 \log (5) \log (x)} \, dx=-x + e^{\left (\frac {\log \left (5\right ) \log \left (-x - e^{3} + \log \left (x\right ) + 4\right ) + x + 4}{\log \left (5\right )}\right )} + e^{\frac {3}{x}} \] Input:
integrate(((x^2*log(x)+(-x^2+x)*log(5)-x^2*exp(3)-x^3+4*x^2)*exp((log(5)*l og(log(x)+4-exp(3)-x)+4+x)/log(5))+(-3*log(5)*exp(3/x)-x^2*log(5))*log(x)+ (3*exp(3)+3*x-12)*log(5)*exp(3/x)+(x^2*exp(3)+x^3-4*x^2)*log(5))/(x^2*log( 5)*log(x)+(-x^2*exp(3)-x^3+4*x^2)*log(5)),x, algorithm="fricas")
Output:
-x + e^((log(5)*log(-x - e^3 + log(x) + 4) + x + 4)/log(5)) + e^(3/x)
Time = 2.36 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {e^{3/x} \left (-12+3 e^3+3 x\right ) \log (5)+\left (-4 x^2+e^3 x^2+x^3\right ) \log (5)+\left (-3 e^{3/x} \log (5)-x^2 \log (5)\right ) \log (x)+e^{\frac {4+x+\log (5) \log \left (4-e^3-x+\log (x)\right )}{\log (5)}} \left (4 x^2-e^3 x^2-x^3+\left (x-x^2\right ) \log (5)+x^2 \log (x)\right )}{\left (4 x^2-e^3 x^2-x^3\right ) \log (5)+x^2 \log (5) \log (x)} \, dx=- x + e^{\frac {3}{x}} + e^{\frac {x + \log {\left (5 \right )} \log {\left (- x + \log {\left (x \right )} - e^{3} + 4 \right )} + 4}{\log {\left (5 \right )}}} \] Input:
integrate(((x**2*ln(x)+(-x**2+x)*ln(5)-x**2*exp(3)-x**3+4*x**2)*exp((ln(5) *ln(ln(x)+4-exp(3)-x)+4+x)/ln(5))+(-3*ln(5)*exp(3/x)-x**2*ln(5))*ln(x)+(3* exp(3)+3*x-12)*ln(5)*exp(3/x)+(x**2*exp(3)+x**3-4*x**2)*ln(5))/(x**2*ln(5) *ln(x)+(-x**2*exp(3)-x**3+4*x**2)*ln(5)),x)
Output:
-x + exp(3/x) + exp((x + log(5)*log(-x + log(x) - exp(3) + 4) + 4)/log(5))
Time = 0.15 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.71 \[ \int \frac {e^{3/x} \left (-12+3 e^3+3 x\right ) \log (5)+\left (-4 x^2+e^3 x^2+x^3\right ) \log (5)+\left (-3 e^{3/x} \log (5)-x^2 \log (5)\right ) \log (x)+e^{\frac {4+x+\log (5) \log \left (4-e^3-x+\log (x)\right )}{\log (5)}} \left (4 x^2-e^3 x^2-x^3+\left (x-x^2\right ) \log (5)+x^2 \log (x)\right )}{\left (4 x^2-e^3 x^2-x^3\right ) \log (5)+x^2 \log (5) \log (x)} \, dx=-{\left (x e^{\frac {4}{\log \left (5\right )}} - e^{\frac {4}{\log \left (5\right )}} \log \left (x\right ) - 4 \, e^{\frac {4}{\log \left (5\right )}} + e^{\left (\frac {4}{\log \left (5\right )} + 3\right )}\right )} e^{\frac {x}{\log \left (5\right )}} - x + e^{\frac {3}{x}} \] Input:
integrate(((x^2*log(x)+(-x^2+x)*log(5)-x^2*exp(3)-x^3+4*x^2)*exp((log(5)*l og(log(x)+4-exp(3)-x)+4+x)/log(5))+(-3*log(5)*exp(3/x)-x^2*log(5))*log(x)+ (3*exp(3)+3*x-12)*log(5)*exp(3/x)+(x^2*exp(3)+x^3-4*x^2)*log(5))/(x^2*log( 5)*log(x)+(-x^2*exp(3)-x^3+4*x^2)*log(5)),x, algorithm="maxima")
Output:
-(x*e^(4/log(5)) - e^(4/log(5))*log(x) - 4*e^(4/log(5)) + e^(4/log(5) + 3) )*e^(x/log(5)) - x + e^(3/x)
Leaf count of result is larger than twice the leaf count of optimal. 77 vs. \(2 (32) = 64\).
Time = 0.16 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.26 \[ \int \frac {e^{3/x} \left (-12+3 e^3+3 x\right ) \log (5)+\left (-4 x^2+e^3 x^2+x^3\right ) \log (5)+\left (-3 e^{3/x} \log (5)-x^2 \log (5)\right ) \log (x)+e^{\frac {4+x+\log (5) \log \left (4-e^3-x+\log (x)\right )}{\log (5)}} \left (4 x^2-e^3 x^2-x^3+\left (x-x^2\right ) \log (5)+x^2 \log (x)\right )}{\left (4 x^2-e^3 x^2-x^3\right ) \log (5)+x^2 \log (5) \log (x)} \, dx=-x e^{\left (\frac {x}{\log \left (5\right )} + \frac {4}{\log \left (5\right )}\right )} + e^{\left (\frac {x}{\log \left (5\right )} + \frac {4}{\log \left (5\right )}\right )} \log \left (x\right ) - x - e^{\left (\frac {x}{\log \left (5\right )} + \frac {4}{\log \left (5\right )} + 3\right )} + 4 \, e^{\left (\frac {x}{\log \left (5\right )} + \frac {4}{\log \left (5\right )}\right )} + e^{\frac {3}{x}} \] Input:
integrate(((x^2*log(x)+(-x^2+x)*log(5)-x^2*exp(3)-x^3+4*x^2)*exp((log(5)*l og(log(x)+4-exp(3)-x)+4+x)/log(5))+(-3*log(5)*exp(3/x)-x^2*log(5))*log(x)+ (3*exp(3)+3*x-12)*log(5)*exp(3/x)+(x^2*exp(3)+x^3-4*x^2)*log(5))/(x^2*log( 5)*log(x)+(-x^2*exp(3)-x^3+4*x^2)*log(5)),x, algorithm="giac")
Output:
-x*e^(x/log(5) + 4/log(5)) + e^(x/log(5) + 4/log(5))*log(x) - x - e^(x/log (5) + 4/log(5) + 3) + 4*e^(x/log(5) + 4/log(5)) + e^(3/x)
Time = 4.23 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.26 \[ \int \frac {e^{3/x} \left (-12+3 e^3+3 x\right ) \log (5)+\left (-4 x^2+e^3 x^2+x^3\right ) \log (5)+\left (-3 e^{3/x} \log (5)-x^2 \log (5)\right ) \log (x)+e^{\frac {4+x+\log (5) \log \left (4-e^3-x+\log (x)\right )}{\log (5)}} \left (4 x^2-e^3 x^2-x^3+\left (x-x^2\right ) \log (5)+x^2 \log (x)\right )}{\left (4 x^2-e^3 x^2-x^3\right ) \log (5)+x^2 \log (5) \log (x)} \, dx=4\,{\mathrm {e}}^{\frac {x}{\ln \left (5\right )}+\frac {4}{\ln \left (5\right )}}-x-{\mathrm {e}}^{\frac {x}{\ln \left (5\right )}+\frac {4}{\ln \left (5\right )}+3}+{\mathrm {e}}^{3/x}-x\,{\mathrm {e}}^{\frac {x}{\ln \left (5\right )}+\frac {4}{\ln \left (5\right )}}+{\mathrm {e}}^{\frac {x}{\ln \left (5\right )}+\frac {4}{\ln \left (5\right )}}\,\ln \left (x\right ) \] Input:
int(-(exp((x + log(log(x) - exp(3) - x + 4)*log(5) + 4)/log(5))*(x^2*log(x ) - x^2*exp(3) + 4*x^2 - x^3 + log(5)*(x - x^2)) - log(x)*(3*exp(3/x)*log( 5) + x^2*log(5)) + log(5)*(x^2*exp(3) - 4*x^2 + x^3) + exp(3/x)*log(5)*(3* x + 3*exp(3) - 12))/(log(5)*(x^2*exp(3) - 4*x^2 + x^3) - x^2*log(5)*log(x) ),x)
Output:
4*exp(x/log(5) + 4/log(5)) - x - exp(x/log(5) + 4/log(5) + 3) + exp(3/x) - x*exp(x/log(5) + 4/log(5)) + exp(x/log(5) + 4/log(5))*log(x)
Time = 0.17 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.88 \[ \int \frac {e^{3/x} \left (-12+3 e^3+3 x\right ) \log (5)+\left (-4 x^2+e^3 x^2+x^3\right ) \log (5)+\left (-3 e^{3/x} \log (5)-x^2 \log (5)\right ) \log (x)+e^{\frac {4+x+\log (5) \log \left (4-e^3-x+\log (x)\right )}{\log (5)}} \left (4 x^2-e^3 x^2-x^3+\left (x-x^2\right ) \log (5)+x^2 \log (x)\right )}{\left (4 x^2-e^3 x^2-x^3\right ) \log (5)+x^2 \log (5) \log (x)} \, dx=e^{\frac {x +4}{\mathrm {log}\left (5\right )}} \mathrm {log}\left (x \right )-e^{\frac {x +4}{\mathrm {log}\left (5\right )}} e^{3}-e^{\frac {x +4}{\mathrm {log}\left (5\right )}} x +4 e^{\frac {x +4}{\mathrm {log}\left (5\right )}}+e^{\frac {3}{x}}-x \] Input:
int(((x^2*log(x)+(-x^2+x)*log(5)-x^2*exp(3)-x^3+4*x^2)*exp((log(5)*log(log (x)+4-exp(3)-x)+4+x)/log(5))+(-3*log(5)*exp(3/x)-x^2*log(5))*log(x)+(3*exp (3)+3*x-12)*log(5)*exp(3/x)+(x^2*exp(3)+x^3-4*x^2)*log(5))/(x^2*log(5)*log (x)+(-x^2*exp(3)-x^3+4*x^2)*log(5)),x)
Output:
e**((x + 4)/log(5))*log(x) - e**((x + 4)/log(5))*e**3 - e**((x + 4)/log(5) )*x + 4*e**((x + 4)/log(5)) + e**(3/x) - x