Integrand size = 82, antiderivative size = 26 \[ \int \frac {\left (-4+4 e^{4-x}\right ) \log ^2(x)+e^{\frac {5 x^2}{4 \log (x)}} (-5 x+10 x \log (x))}{4 e^{\frac {5 x^2}{4 \log (x)}} \log ^2(x)+\left (-28-4 e^{4-x}-4 x\right ) \log ^2(x)} \, dx=\log \left (7+e^{4-x}-e^{\frac {5 x^2}{4 \log (x)}}+x\right ) \]
Time = 0.78 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.54 \[ \int \frac {\left (-4+4 e^{4-x}\right ) \log ^2(x)+e^{\frac {5 x^2}{4 \log (x)}} (-5 x+10 x \log (x))}{4 e^{\frac {5 x^2}{4 \log (x)}} \log ^2(x)+\left (-28-4 e^{4-x}-4 x\right ) \log ^2(x)} \, dx=\frac {1}{4} \left (-4 x+4 \log \left (-e^4+e^{x+\frac {5 x^2}{4 \log (x)}}-e^x (7+x)\right )\right ) \]
Integrate[((-4 + 4*E^(4 - x))*Log[x]^2 + E^((5*x^2)/(4*Log[x]))*(-5*x + 10 *x*Log[x]))/(4*E^((5*x^2)/(4*Log[x]))*Log[x]^2 + (-28 - 4*E^(4 - x) - 4*x) *Log[x]^2),x]
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.
| \(\displaystyle \int \frac {e^{\frac {5 x^2}{4 \log (x)}} (10 x \log (x)-5 x)+\left (4 e^{4-x}-4\right ) \log ^2(x)}{4 e^{\frac {5 x^2}{4 \log (x)}} \log ^2(x)+\left (-4 x-4 e^{4-x}-28\right ) \log ^2(x)} \, dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {-5 e^x x^2+10 e^x x^2 \log (x)-35 e^x x-5 e^4 x-4 e^x \log ^2(x)+4 e^4 \log ^2(x)+70 e^x x \log (x)+10 e^4 x \log (x)}{4 \left (e^{\frac {5 x^2}{4 \log (x)}+x}-e^x x-7 e^x-e^4\right ) \log ^2(x)}+\frac {5 x (2 \log (x)-1)}{4 \log ^2(x)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {5}{4} e^4 \int \frac {x}{\left (e^x x+7 e^x-e^{\frac {5 x^2}{4 \log (x)}+x}+e^4\right ) \log ^2(x)}dx+\frac {35}{4} \int \frac {e^x x}{\left (e^x x+7 e^x-e^{\frac {5 x^2}{4 \log (x)}+x}+e^4\right ) \log ^2(x)}dx+\frac {5}{4} \int \frac {e^x x^2}{\left (e^x x+7 e^x-e^{\frac {5 x^2}{4 \log (x)}+x}+e^4\right ) \log ^2(x)}dx-\int \frac {e^x}{-e^x x-7 e^x+e^{\frac {5 x^2}{4 \log (x)}+x}-e^4}dx-e^4 \int \frac {1}{e^x x+7 e^x-e^{\frac {5 x^2}{4 \log (x)}+x}+e^4}dx-\frac {5}{2} e^4 \int \frac {x}{\left (e^x x+7 e^x-e^{\frac {5 x^2}{4 \log (x)}+x}+e^4\right ) \log (x)}dx-\frac {35}{2} \int \frac {e^x x}{\left (e^x x+7 e^x-e^{\frac {5 x^2}{4 \log (x)}+x}+e^4\right ) \log (x)}dx-\frac {5}{2} \int \frac {e^x x^2}{\left (e^x x+7 e^x-e^{\frac {5 x^2}{4 \log (x)}+x}+e^4\right ) \log (x)}dx+\frac {5}{2} \operatorname {ExpIntegralEi}(2 \log (x))-\frac {5}{2} (1-2 \log (x)) \operatorname {ExpIntegralEi}(2 \log (x))-5 \log (x) \operatorname {ExpIntegralEi}(2 \log (x))+\frac {5 x^2}{2}+\frac {5 x^2 (1-2 \log (x))}{4 \log (x)}\) |
Int[((-4 + 4*E^(4 - x))*Log[x]^2 + E^((5*x^2)/(4*Log[x]))*(-5*x + 10*x*Log [x]))/(4*E^((5*x^2)/(4*Log[x]))*Log[x]^2 + (-28 - 4*E^(4 - x) - 4*x)*Log[x ]^2),x]
3.23.21.3.1 Defintions of rubi rules used
Time = 0.50 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88
| method | result | size |
| parallelrisch | \(\ln \left (7+{\mathrm e}^{-x +4}-{\mathrm e}^{\frac {5 x^{2}}{4 \ln \left (x \right )}}+x \right )\) | \(23\) |
| risch | \(\ln \left ({\mathrm e}^{\frac {5 x^{2}}{4 \ln \left (x \right )}}-{\mathrm e}^{-x +4}-x -7\right )\) | \(25\) |
int(((10*x*ln(x)-5*x)*exp(5/4*x^2/ln(x))+(4*exp(-x+4)-4)*ln(x)^2)/(4*ln(x) ^2*exp(5/4*x^2/ln(x))+(-4*exp(-x+4)-4*x-28)*ln(x)^2),x,method=_RETURNVERBO SE)
Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {\left (-4+4 e^{4-x}\right ) \log ^2(x)+e^{\frac {5 x^2}{4 \log (x)}} (-5 x+10 x \log (x))}{4 e^{\frac {5 x^2}{4 \log (x)}} \log ^2(x)+\left (-28-4 e^{4-x}-4 x\right ) \log ^2(x)} \, dx=\log \left (-x - e^{\left (-x + 4\right )} + e^{\left (\frac {5 \, x^{2}}{4 \, \log \left (x\right )}\right )} - 7\right ) \]
integrate(((10*x*log(x)-5*x)*exp(5/4*x^2/log(x))+(4*exp(-x+4)-4)*log(x)^2) /(4*log(x)^2*exp(5/4*x^2/log(x))+(-4*exp(-x+4)-4*x-28)*log(x)^2),x, algori thm=\
Time = 0.29 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {\left (-4+4 e^{4-x}\right ) \log ^2(x)+e^{\frac {5 x^2}{4 \log (x)}} (-5 x+10 x \log (x))}{4 e^{\frac {5 x^2}{4 \log (x)}} \log ^2(x)+\left (-28-4 e^{4-x}-4 x\right ) \log ^2(x)} \, dx=\log {\left (x - e^{\frac {5 x^{2}}{4 \log {\left (x \right )}}} + e^{4 - x} + 7 \right )} \]
integrate(((10*x*ln(x)-5*x)*exp(5/4*x**2/ln(x))+(4*exp(-x+4)-4)*ln(x)**2)/ (4*ln(x)**2*exp(5/4*x**2/ln(x))+(-4*exp(-x+4)-4*x-28)*ln(x)**2),x)
Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {\left (-4+4 e^{4-x}\right ) \log ^2(x)+e^{\frac {5 x^2}{4 \log (x)}} (-5 x+10 x \log (x))}{4 e^{\frac {5 x^2}{4 \log (x)}} \log ^2(x)+\left (-28-4 e^{4-x}-4 x\right ) \log ^2(x)} \, dx=\log \left (-{\left ({\left (x + 7\right )} e^{x} + e^{4} - e^{\left (x + \frac {5 \, x^{2}}{4 \, \log \left (x\right )}\right )}\right )} e^{\left (-x\right )}\right ) \]
integrate(((10*x*log(x)-5*x)*exp(5/4*x^2/log(x))+(4*exp(-x+4)-4)*log(x)^2) /(4*log(x)^2*exp(5/4*x^2/log(x))+(-4*exp(-x+4)-4*x-28)*log(x)^2),x, algori thm=\
Exception generated. \[ \int \frac {\left (-4+4 e^{4-x}\right ) \log ^2(x)+e^{\frac {5 x^2}{4 \log (x)}} (-5 x+10 x \log (x))}{4 e^{\frac {5 x^2}{4 \log (x)}} \log ^2(x)+\left (-28-4 e^{4-x}-4 x\right ) \log ^2(x)} \, dx=\text {Exception raised: TypeError} \]
integrate(((10*x*log(x)-5*x)*exp(5/4*x^2/log(x))+(4*exp(-x+4)-4)*log(x)^2) /(4*log(x)^2*exp(5/4*x^2/log(x))+(-4*exp(-x+4)-4*x-28)*log(x)^2),x, algori thm=\
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{62500,[0,17]%%%} / %%%{250000,[0,17]%%%} Error: Bad Argume nt Value
Time = 12.97 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {\left (-4+4 e^{4-x}\right ) \log ^2(x)+e^{\frac {5 x^2}{4 \log (x)}} (-5 x+10 x \log (x))}{4 e^{\frac {5 x^2}{4 \log (x)}} \log ^2(x)+\left (-28-4 e^{4-x}-4 x\right ) \log ^2(x)} \, dx=\ln \left (x+{\mathrm {e}}^{4-x}-{\mathrm {e}}^{\frac {5\,x^2}{4\,\ln \left (x\right )}}+7\right ) \]