Integrand size = 52, antiderivative size = 24 \[ \int \frac {1}{5} e^{\frac {1}{5} (5-7 x)} x^{e^{\frac {1}{5} (5-7 x)} \left (x+x^2\right )} \left (5+5 x+\left (5+3 x-7 x^2\right ) \log (x)\right ) \, dx=5+x^{e^{\frac {1}{5} (5-2 x)-x} x (1+x)} \] Output:
5+exp((1+x)/exp(7/5*x-1)*x*ln(x))
Time = 1.15 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {1}{5} e^{\frac {1}{5} (5-7 x)} x^{e^{\frac {1}{5} (5-7 x)} \left (x+x^2\right )} \left (5+5 x+\left (5+3 x-7 x^2\right ) \log (x)\right ) \, dx=x^{e^{1-\frac {7 x}{5}} x (1+x)} \] Input:
Integrate[(E^((5 - 7*x)/5)*x^(E^((5 - 7*x)/5)*(x + x^2))*(5 + 5*x + (5 + 3 *x - 7*x^2)*Log[x]))/5,x]
Output:
x^(E^(1 - (7*x)/5)*x*(1 + 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.
\(\displaystyle \int \frac {1}{5} e^{\frac {1}{5} (5-7 x)} x^{e^{\frac {1}{5} (5-7 x)} \left (x^2+x\right )} \left (\left (-7 x^2+3 x+5\right ) \log (x)+5 x+5\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int e^{\frac {1}{5} (5-7 x)} x^{e^{\frac {1}{5} (5-7 x)} \left (x^2+x\right )} \left (5 x+\left (-7 x^2+3 x+5\right ) \log (x)+5\right )dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \frac {1}{5} \int e^{1-\frac {7 x}{5}} x^{e^{1-\frac {7 x}{5}} x (x+1)} \left (5 x+\left (-7 x^2+3 x+5\right ) \log (x)+5\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{5} \int \left (5 e^{1-\frac {7 x}{5}} x^{e^{1-\frac {7 x}{5}} x (x+1)}-e^{1-\frac {7 x}{5}} \left (7 x^2-3 x-5\right ) \log (x) x^{e^{1-\frac {7 x}{5}} x (x+1)}+5 e^{1-\frac {7 x}{5}} x^{e^{1-\frac {7 x}{5}} x (x+1)+1}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (5 \int e^{1-\frac {7 x}{5}} x^{e^{1-\frac {7 x}{5}} x (x+1)}dx+5 \int e^{1-\frac {7 x}{5}} x^{e^{1-\frac {7 x}{5}} x (x+1)+1}dx-5 \int \frac {\int e^{1-\frac {7 x}{5}} x^{e^{1-\frac {7 x}{5}} x (x+1)}dx}{x}dx-3 \int \frac {\int e^{1-\frac {7 x}{5}} x^{e^{1-\frac {7 x}{5}} x (x+1)+1}dx}{x}dx+7 \int \frac {\int e^{1-\frac {7 x}{5}} x^{e^{1-\frac {7 x}{5}} x (x+1)+2}dx}{x}dx+5 \log (x) \int e^{1-\frac {7 x}{5}} x^{e^{1-\frac {7 x}{5}} x (x+1)}dx+3 \log (x) \int e^{1-\frac {7 x}{5}} x^{e^{1-\frac {7 x}{5}} x (x+1)+1}dx-7 \log (x) \int e^{1-\frac {7 x}{5}} x^{e^{1-\frac {7 x}{5}} x (x+1)+2}dx\right )\) |
Input:
Int[(E^((5 - 7*x)/5)*x^(E^((5 - 7*x)/5)*(x + x^2))*(5 + 5*x + (5 + 3*x - 7 *x^2)*Log[x]))/5,x]
Output:
$Aborted
Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.58
method | result | size |
risch | \(x^{\left (1+x \right ) x \,{\mathrm e}^{-\frac {7 x}{5}+1}}\) | \(14\) |
parallelrisch | \({\mathrm e}^{\left (1+x \right ) x \ln \left (x \right ) {\mathrm e}^{-\frac {7 x}{5}+1}}\) | \(17\) |
Input:
int(1/5*((-7*x^2+3*x+5)*ln(x)+5*x+5)*exp((x^2+x)*ln(x)/exp(7/5*x-1))/exp(7 /5*x-1),x,method=_RETURNVERBOSE)
Output:
x^((1+x)*x*exp(-7/5*x+1))
Time = 0.09 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.58 \[ \int \frac {1}{5} e^{\frac {1}{5} (5-7 x)} x^{e^{\frac {1}{5} (5-7 x)} \left (x+x^2\right )} \left (5+5 x+\left (5+3 x-7 x^2\right ) \log (x)\right ) \, dx=x^{{\left (x^{2} + x\right )} e^{\left (-\frac {7}{5} \, x + 1\right )}} \] Input:
integrate(1/5*((-7*x^2+3*x+5)*log(x)+5*x+5)*exp((x^2+x)*log(x)/exp(7/5*x-1 ))/exp(7/5*x-1),x, algorithm="fricas")
Output:
x^((x^2 + x)*e^(-7/5*x + 1))
Time = 0.22 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {1}{5} e^{\frac {1}{5} (5-7 x)} x^{e^{\frac {1}{5} (5-7 x)} \left (x+x^2\right )} \left (5+5 x+\left (5+3 x-7 x^2\right ) \log (x)\right ) \, dx=e^{\left (x^{2} + x\right ) e^{1 - \frac {7 x}{5}} \log {\left (x \right )}} \] Input:
integrate(1/5*((-7*x**2+3*x+5)*ln(x)+5*x+5)*exp((x**2+x)*ln(x)/exp(7/5*x-1 ))/exp(7/5*x-1),x)
Output:
exp((x**2 + x)*exp(1 - 7*x/5)*log(x))
Time = 0.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {1}{5} e^{\frac {1}{5} (5-7 x)} x^{e^{\frac {1}{5} (5-7 x)} \left (x+x^2\right )} \left (5+5 x+\left (5+3 x-7 x^2\right ) \log (x)\right ) \, dx=e^{\left (x^{2} e^{\left (-\frac {7}{5} \, x + 1\right )} \log \left (x\right ) + x e^{\left (-\frac {7}{5} \, x + 1\right )} \log \left (x\right )\right )} \] Input:
integrate(1/5*((-7*x^2+3*x+5)*log(x)+5*x+5)*exp((x^2+x)*log(x)/exp(7/5*x-1 ))/exp(7/5*x-1),x, algorithm="maxima")
Output:
e^(x^2*e^(-7/5*x + 1)*log(x) + x*e^(-7/5*x + 1)*log(x))
Time = 0.22 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {1}{5} e^{\frac {1}{5} (5-7 x)} x^{e^{\frac {1}{5} (5-7 x)} \left (x+x^2\right )} \left (5+5 x+\left (5+3 x-7 x^2\right ) \log (x)\right ) \, dx=x^{x^{2} e^{\left (-\frac {7}{5} \, x + 1\right )} + x e^{\left (-\frac {7}{5} \, x + 1\right )}} \] Input:
integrate(1/5*((-7*x^2+3*x+5)*log(x)+5*x+5)*exp((x^2+x)*log(x)/exp(7/5*x-1 ))/exp(7/5*x-1),x, algorithm="giac")
Output:
x^(x^2*e^(-7/5*x + 1) + x*e^(-7/5*x + 1))
Time = 3.13 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.58 \[ \int \frac {1}{5} e^{\frac {1}{5} (5-7 x)} x^{e^{\frac {1}{5} (5-7 x)} \left (x+x^2\right )} \left (5+5 x+\left (5+3 x-7 x^2\right ) \log (x)\right ) \, dx=x^{{\mathrm {e}}^{1-\frac {7\,x}{5}}\,\left (x^2+x\right )} \] Input:
int((exp(1 - (7*x)/5)*exp(exp(1 - (7*x)/5)*log(x)*(x + x^2))*(5*x + log(x) *(3*x - 7*x^2 + 5) + 5))/5,x)
Output:
x^(exp(1 - (7*x)/5)*(x + x^2))
\[ \int \frac {1}{5} e^{\frac {1}{5} (5-7 x)} x^{e^{\frac {1}{5} (5-7 x)} \left (x+x^2\right )} \left (5+5 x+\left (5+3 x-7 x^2\right ) \log (x)\right ) \, dx=\frac {e \left (5 \left (\int \frac {e^{\frac {2 e^{\frac {7 x}{5}} x +\mathrm {log}\left (x \right ) e \,x^{2}+\mathrm {log}\left (x \right ) e x}{e^{\frac {7 x}{5}}}}}{e^{\frac {17 x}{5}}}d x \right )-7 \left (\int \frac {e^{\frac {2 e^{\frac {7 x}{5}} x +\mathrm {log}\left (x \right ) e \,x^{2}+\mathrm {log}\left (x \right ) e x}{e^{\frac {7 x}{5}}}} \mathrm {log}\left (x \right ) x^{2}}{e^{\frac {17 x}{5}}}d x \right )+3 \left (\int \frac {e^{\frac {2 e^{\frac {7 x}{5}} x +\mathrm {log}\left (x \right ) e \,x^{2}+\mathrm {log}\left (x \right ) e x}{e^{\frac {7 x}{5}}}} \mathrm {log}\left (x \right ) x}{e^{\frac {17 x}{5}}}d x \right )+5 \left (\int \frac {e^{\frac {2 e^{\frac {7 x}{5}} x +\mathrm {log}\left (x \right ) e \,x^{2}+\mathrm {log}\left (x \right ) e x}{e^{\frac {7 x}{5}}}} \mathrm {log}\left (x \right )}{e^{\frac {17 x}{5}}}d x \right )+5 \left (\int \frac {e^{\frac {2 e^{\frac {7 x}{5}} x +\mathrm {log}\left (x \right ) e \,x^{2}+\mathrm {log}\left (x \right ) e x}{e^{\frac {7 x}{5}}}} x}{e^{\frac {17 x}{5}}}d x \right )\right )}{5} \] Input:
int(1/5*((-7*x^2+3*x+5)*log(x)+5*x+5)*exp((x^2+x)*log(x)/exp(7/5*x-1))/exp (7/5*x-1),x)
Output:
(e*(5*int(e**((2*e**((7*x)/5)*x + log(x)*e*x**2 + log(x)*e*x)/e**((7*x)/5) )/e**((17*x)/5),x) - 7*int((e**((2*e**((7*x)/5)*x + log(x)*e*x**2 + log(x) *e*x)/e**((7*x)/5))*log(x)*x**2)/e**((17*x)/5),x) + 3*int((e**((2*e**((7*x )/5)*x + log(x)*e*x**2 + log(x)*e*x)/e**((7*x)/5))*log(x)*x)/e**((17*x)/5) ,x) + 5*int((e**((2*e**((7*x)/5)*x + log(x)*e*x**2 + log(x)*e*x)/e**((7*x) /5))*log(x))/e**((17*x)/5),x) + 5*int((e**((2*e**((7*x)/5)*x + log(x)*e*x* *2 + log(x)*e*x)/e**((7*x)/5))*x)/e**((17*x)/5),x)))/5