Integrand size = 114, antiderivative size = 32 \[ \int \frac {4 e^5 x^2 \log ^2(x)+e^{\frac {3}{e^5 x^2 \log (x)}} \log \left (2+e^2\right ) \left (6+12 \log (x)-4 e^5 x^2 \log ^2(x)\right )+e^{\frac {6}{e^5 x^2 \log (x)}} \log ^2\left (2+e^2\right ) \left (-3-6 \log (x)+e^5 x^2 \log ^2(x)\right )}{2 e^5 x \log ^2\left (2+e^2\right ) \log ^2(x)} \, dx=\left (-\frac {1}{2} e^{\frac {3}{e^5 x^2 \log (x)}} x+\frac {x}{\log \left (2+e^2\right )}\right )^2 \]
Time = 0.13 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.25 \[ \int \frac {4 e^5 x^2 \log ^2(x)+e^{\frac {3}{e^5 x^2 \log (x)}} \log \left (2+e^2\right ) \left (6+12 \log (x)-4 e^5 x^2 \log ^2(x)\right )+e^{\frac {6}{e^5 x^2 \log (x)}} \log ^2\left (2+e^2\right ) \left (-3-6 \log (x)+e^5 x^2 \log ^2(x)\right )}{2 e^5 x \log ^2\left (2+e^2\right ) \log ^2(x)} \, dx=\frac {x^2 \left (-2+e^{\frac {3}{e^5 x^2 \log (x)}} \log \left (2+e^2\right )\right )^2}{4 \log ^2\left (2+e^2\right )} \]
Integrate[(4*E^5*x^2*Log[x]^2 + E^(3/(E^5*x^2*Log[x]))*Log[2 + E^2]*(6 + 1 2*Log[x] - 4*E^5*x^2*Log[x]^2) + E^(6/(E^5*x^2*Log[x]))*Log[2 + E^2]^2*(-3 - 6*Log[x] + E^5*x^2*Log[x]^2))/(2*E^5*x*Log[2 + E^2]^2*Log[x]^2),x]
Leaf count is larger than twice the leaf count of optimal. \(150\) vs. \(2(32)=64\).
Time = 1.02 (sec) , antiderivative size = 150, normalized size of antiderivative = 4.69, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {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.
\(\displaystyle \int \frac {4 e^5 x^2 \log ^2(x)+\log \left (2+e^2\right ) e^{\frac {3}{e^5 x^2 \log (x)}} \left (-4 e^5 x^2 \log ^2(x)+12 \log (x)+6\right )+\log ^2\left (2+e^2\right ) e^{\frac {6}{e^5 x^2 \log (x)}} \left (e^5 x^2 \log ^2(x)-6 \log (x)-3\right )}{2 e^5 x \log ^2\left (2+e^2\right ) \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {4 e^5 x^2 \log ^2(x)+2 e^{\frac {3}{e^5 x^2 \log (x)}} \log \left (2+e^2\right ) \left (-2 e^5 x^2 \log ^2(x)+6 \log (x)+3\right )-e^{\frac {6}{e^5 x^2 \log (x)}} \log ^2\left (2+e^2\right ) \left (-e^5 x^2 \log ^2(x)+6 \log (x)+3\right )}{x \log ^2(x)}dx}{2 e^5 \log ^2\left (2+e^2\right )}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {\int \left (4 e^5 x+\frac {e^{\frac {6}{e^5 x^2 \log (x)}} \log ^2\left (2+e^2\right ) \left (e^5 x^2 \log ^2(x)-6 \log (x)-3\right )}{\log ^2(x) x}-\frac {2 e^{\frac {3}{e^5 x^2 \log (x)}} \log \left (2+e^2\right ) \left (2 e^5 x^2 \log ^2(x)-6 \log (x)-3\right )}{\log ^2(x) x}\right )dx}{2 e^5 \log ^2\left (2+e^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 e^5 x^2+\frac {\log ^2\left (2+e^2\right ) e^{\frac {6}{e^5 x^2 \log (x)}} (2 \log (x)+1)}{2 x \left (\frac {1}{e^5 x^3 \log ^2(x)}+\frac {2}{e^5 x^3 \log (x)}\right ) \log ^2(x)}-\frac {2 \log \left (2+e^2\right ) e^{\frac {3}{e^5 x^2 \log (x)}} (2 \log (x)+1)}{x \left (\frac {1}{e^5 x^3 \log ^2(x)}+\frac {2}{e^5 x^3 \log (x)}\right ) \log ^2(x)}}{2 e^5 \log ^2\left (2+e^2\right )}\) |
Int[(4*E^5*x^2*Log[x]^2 + E^(3/(E^5*x^2*Log[x]))*Log[2 + E^2]*(6 + 12*Log[ x] - 4*E^5*x^2*Log[x]^2) + E^(6/(E^5*x^2*Log[x]))*Log[2 + E^2]^2*(-3 - 6*L og[x] + E^5*x^2*Log[x]^2))/(2*E^5*x*Log[2 + E^2]^2*Log[x]^2),x]
(2*E^5*x^2 - (2*E^(3/(E^5*x^2*Log[x]))*Log[2 + E^2]*(1 + 2*Log[x]))/(x*(1/ (E^5*x^3*Log[x]^2) + 2/(E^5*x^3*Log[x]))*Log[x]^2) + (E^(6/(E^5*x^2*Log[x] ))*Log[2 + E^2]^2*(1 + 2*Log[x]))/(2*x*(1/(E^5*x^3*Log[x]^2) + 2/(E^5*x^3* Log[x]))*Log[x]^2))/(2*E^5*Log[2 + E^2]^2)
3.29.93.3.1 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 = 3.66 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.69
method | result | size |
risch | \(\frac {x^{2}}{\ln \left ({\mathrm e}^{2}+2\right )^{2}}+\frac {x^{2} {\mathrm e}^{\frac {6 \,{\mathrm e}^{-5}}{x^{2} \ln \left (x \right )}}}{4}-\frac {x^{2} {\mathrm e}^{\frac {3 \,{\mathrm e}^{-5}}{x^{2} \ln \left (x \right )}}}{\ln \left ({\mathrm e}^{2}+2\right )}\) | \(54\) |
default | \(\frac {{\mathrm e}^{-5} \left (-2 \,{\mathrm e}^{5} \ln \left ({\mathrm e}^{2}+2\right ) x^{2} {\mathrm e}^{\frac {3 \,{\mathrm e}^{-5}}{x^{2} \ln \left (x \right )}}+\frac {{\mathrm e}^{5} \ln \left ({\mathrm e}^{2}+2\right )^{2} x^{2} {\mathrm e}^{\frac {6 \,{\mathrm e}^{-5}}{x^{2} \ln \left (x \right )}}}{2}+2 x^{2} {\mathrm e}^{5}\right )}{2 \ln \left ({\mathrm e}^{2}+2\right )^{2}}\) | \(72\) |
parallelrisch | \(\frac {{\mathrm e}^{-5} \left (-2 \,{\mathrm e}^{5} \ln \left ({\mathrm e}^{2}+2\right ) x^{2} {\mathrm e}^{\frac {3 \,{\mathrm e}^{-5}}{x^{2} \ln \left (x \right )}}+\frac {{\mathrm e}^{5} \ln \left ({\mathrm e}^{2}+2\right )^{2} x^{2} {\mathrm e}^{\frac {6 \,{\mathrm e}^{-5}}{x^{2} \ln \left (x \right )}}}{2}+2 x^{2} {\mathrm e}^{5}\right )}{2 \ln \left ({\mathrm e}^{2}+2\right )^{2}}\) | \(78\) |
int(1/2*((x^2*exp(5)*ln(x)^2-6*ln(x)-3)*ln(exp(2)+2)^2*exp(3/x^2/exp(5)/ln (x))^2+(-4*x^2*exp(5)*ln(x)^2+12*ln(x)+6)*ln(exp(2)+2)*exp(3/x^2/exp(5)/ln (x))+4*x^2*exp(5)*ln(x)^2)/x/exp(5)/ln(x)^2/ln(exp(2)+2)^2,x,method=_RETUR NVERBOSE)
1/ln(exp(2)+2)^2*x^2+1/4*x^2*exp(6/x^2*exp(-5)/ln(x))-1/ln(exp(2)+2)*x^2*e xp(3/x^2*exp(-5)/ln(x))
Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (29) = 58\).
Time = 0.30 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.88 \[ \int \frac {4 e^5 x^2 \log ^2(x)+e^{\frac {3}{e^5 x^2 \log (x)}} \log \left (2+e^2\right ) \left (6+12 \log (x)-4 e^5 x^2 \log ^2(x)\right )+e^{\frac {6}{e^5 x^2 \log (x)}} \log ^2\left (2+e^2\right ) \left (-3-6 \log (x)+e^5 x^2 \log ^2(x)\right )}{2 e^5 x \log ^2\left (2+e^2\right ) \log ^2(x)} \, dx=\frac {x^{2} e^{\left (\frac {6 \, e^{\left (-5\right )}}{x^{2} \log \left (x\right )}\right )} \log \left (e^{2} + 2\right )^{2} - 4 \, x^{2} e^{\left (\frac {3 \, e^{\left (-5\right )}}{x^{2} \log \left (x\right )}\right )} \log \left (e^{2} + 2\right ) + 4 \, x^{2}}{4 \, \log \left (e^{2} + 2\right )^{2}} \]
integrate(1/2*((x^2*exp(5)*log(x)^2-6*log(x)-3)*log(exp(2)+2)^2*exp(3/x^2/ exp(5)/log(x))^2+(-4*x^2*exp(5)*log(x)^2+12*log(x)+6)*log(exp(2)+2)*exp(3/ x^2/exp(5)/log(x))+4*x^2*exp(5)*log(x)^2)/x/exp(5)/log(x)^2/log(exp(2)+2)^ 2,x, algorithm=\
1/4*(x^2*e^(6*e^(-5)/(x^2*log(x)))*log(e^2 + 2)^2 - 4*x^2*e^(3*e^(-5)/(x^2 *log(x)))*log(e^2 + 2) + 4*x^2)/log(e^2 + 2)^2
Exception generated. \[ \int \frac {4 e^5 x^2 \log ^2(x)+e^{\frac {3}{e^5 x^2 \log (x)}} \log \left (2+e^2\right ) \left (6+12 \log (x)-4 e^5 x^2 \log ^2(x)\right )+e^{\frac {6}{e^5 x^2 \log (x)}} \log ^2\left (2+e^2\right ) \left (-3-6 \log (x)+e^5 x^2 \log ^2(x)\right )}{2 e^5 x \log ^2\left (2+e^2\right ) \log ^2(x)} \, dx=\text {Exception raised: TypeError} \]
integrate(1/2*((x**2*exp(5)*ln(x)**2-6*ln(x)-3)*ln(exp(2)+2)**2*exp(3/x**2 /exp(5)/ln(x))**2+(-4*x**2*exp(5)*ln(x)**2+12*ln(x)+6)*ln(exp(2)+2)*exp(3/ x**2/exp(5)/ln(x))+4*x**2*exp(5)*ln(x)**2)/x/exp(5)/ln(x)**2/ln(exp(2)+2)* *2,x)
Exception generated. \[ \int \frac {4 e^5 x^2 \log ^2(x)+e^{\frac {3}{e^5 x^2 \log (x)}} \log \left (2+e^2\right ) \left (6+12 \log (x)-4 e^5 x^2 \log ^2(x)\right )+e^{\frac {6}{e^5 x^2 \log (x)}} \log ^2\left (2+e^2\right ) \left (-3-6 \log (x)+e^5 x^2 \log ^2(x)\right )}{2 e^5 x \log ^2\left (2+e^2\right ) \log ^2(x)} \, dx=\text {Exception raised: RuntimeError} \]
integrate(1/2*((x^2*exp(5)*log(x)^2-6*log(x)-3)*log(exp(2)+2)^2*exp(3/x^2/ exp(5)/log(x))^2+(-4*x^2*exp(5)*log(x)^2+12*log(x)+6)*log(exp(2)+2)*exp(3/ x^2/exp(5)/log(x))+4*x^2*exp(5)*log(x)^2)/x/exp(5)/log(x)^2/log(exp(2)+2)^ 2,x, algorithm=\
Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is 0which is not of the expected type LIST
\[ \int \frac {4 e^5 x^2 \log ^2(x)+e^{\frac {3}{e^5 x^2 \log (x)}} \log \left (2+e^2\right ) \left (6+12 \log (x)-4 e^5 x^2 \log ^2(x)\right )+e^{\frac {6}{e^5 x^2 \log (x)}} \log ^2\left (2+e^2\right ) \left (-3-6 \log (x)+e^5 x^2 \log ^2(x)\right )}{2 e^5 x \log ^2\left (2+e^2\right ) \log ^2(x)} \, dx=\int { \frac {{\left (4 \, x^{2} e^{5} \log \left (x\right )^{2} + {\left (x^{2} e^{5} \log \left (x\right )^{2} - 6 \, \log \left (x\right ) - 3\right )} e^{\left (\frac {6 \, e^{\left (-5\right )}}{x^{2} \log \left (x\right )}\right )} \log \left (e^{2} + 2\right )^{2} - 2 \, {\left (2 \, x^{2} e^{5} \log \left (x\right )^{2} - 6 \, \log \left (x\right ) - 3\right )} e^{\left (\frac {3 \, e^{\left (-5\right )}}{x^{2} \log \left (x\right )}\right )} \log \left (e^{2} + 2\right )\right )} e^{\left (-5\right )}}{2 \, x \log \left (x\right )^{2} \log \left (e^{2} + 2\right )^{2}} \,d x } \]
integrate(1/2*((x^2*exp(5)*log(x)^2-6*log(x)-3)*log(exp(2)+2)^2*exp(3/x^2/ exp(5)/log(x))^2+(-4*x^2*exp(5)*log(x)^2+12*log(x)+6)*log(exp(2)+2)*exp(3/ x^2/exp(5)/log(x))+4*x^2*exp(5)*log(x)^2)/x/exp(5)/log(x)^2/log(exp(2)+2)^ 2,x, algorithm=\
integrate(1/2*(4*x^2*e^5*log(x)^2 + (x^2*e^5*log(x)^2 - 6*log(x) - 3)*e^(6 *e^(-5)/(x^2*log(x)))*log(e^2 + 2)^2 - 2*(2*x^2*e^5*log(x)^2 - 6*log(x) - 3)*e^(3*e^(-5)/(x^2*log(x)))*log(e^2 + 2))*e^(-5)/(x*log(x)^2*log(e^2 + 2) ^2), x)
Time = 13.96 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {4 e^5 x^2 \log ^2(x)+e^{\frac {3}{e^5 x^2 \log (x)}} \log \left (2+e^2\right ) \left (6+12 \log (x)-4 e^5 x^2 \log ^2(x)\right )+e^{\frac {6}{e^5 x^2 \log (x)}} \log ^2\left (2+e^2\right ) \left (-3-6 \log (x)+e^5 x^2 \log ^2(x)\right )}{2 e^5 x \log ^2\left (2+e^2\right ) \log ^2(x)} \, dx=\frac {x^2\,{\left (\ln \left ({\mathrm {e}}^2+2\right )\,{\mathrm {e}}^{\frac {3\,{\mathrm {e}}^{-5}}{x^2\,\ln \left (x\right )}}-2\right )}^2}{4\,{\ln \left ({\mathrm {e}}^2+2\right )}^2} \]