Integrand size = 39, antiderivative size = 26 \[ \int \frac {e^{\frac {1}{25} \left (125+x+\log \left (-14 x+2 x^2\right )\right )} \left (-7-5 x+x^2\right )}{-175 x+25 x^2} \, dx=e^{5+\frac {1}{25} (x+\log (2 (-2 x+(-5+x) x)))}+\log (\log (5)) \] Output:
ln(ln(5))+exp(5+1/25*x+1/25*ln(2*(-5+x)*x-4*x))
\[ \int \frac {e^{\frac {1}{25} \left (125+x+\log \left (-14 x+2 x^2\right )\right )} \left (-7-5 x+x^2\right )}{-175 x+25 x^2} \, dx=\int \frac {e^{\frac {1}{25} \left (125+x+\log \left (-14 x+2 x^2\right )\right )} \left (-7-5 x+x^2\right )}{-175 x+25 x^2} \, dx \] Input:
Integrate[(E^((125 + x + Log[-14*x + 2*x^2])/25)*(-7 - 5*x + x^2))/(-175*x + 25*x^2),x]
Output:
Integrate[(E^((125 + x + Log[-14*x + 2*x^2])/25)*(-7 - 5*x + x^2))/(-175*x + 25*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.
\(\displaystyle \int \frac {\left (x^2-5 x-7\right ) e^{\frac {1}{25} \left (\log \left (2 x^2-14 x\right )+x+125\right )}}{25 x^2-175 x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (x^2-5 x-7\right ) e^{\frac {1}{25} \left (\log \left (2 x^2-14 x\right )+x+125\right )}}{x (25 x-175)}dx\) |
\(\Big \downarrow \) 2704 |
\(\displaystyle \int \frac {e^{\frac {x+125}{25}} \left (x^2-5 x-7\right ) \sqrt [25]{2 x^2-14 x}}{x (25 x-175)}dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt [25]{x^2-7 x} \int -\frac {\sqrt [25]{2} e^{\frac {x+125}{25}} \left (-x^2+5 x+7\right )}{25 (x-7)^{24/25} x^{24/25}}dx}{\sqrt [25]{x-7} \sqrt [25]{x}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\sqrt [25]{2} \sqrt [25]{x^2-7 x} \int \frac {e^{\frac {x+125}{25}} \left (-x^2+5 x+7\right )}{(x-7)^{24/25} x^{24/25}}dx}{25 \sqrt [25]{x-7} \sqrt [25]{x}}\) |
\(\Big \downarrow \) 7284 |
\(\displaystyle -\frac {\sqrt [25]{2} \sqrt [25]{x^2-7 x} \int \frac {e^{\frac {x+125}{25}} \left (-x^2+5 x+7\right )}{(x-7)^{24/25}}d\sqrt [25]{x}}{\sqrt [25]{x-7} \sqrt [25]{x}}\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle -\frac {\sqrt [25]{2} \sqrt [25]{x^2-7 x} \int \frac {e^{\frac {x}{25}+5} \left (-x^2+5 x+7\right )}{(x-7)^{24/25}}d\sqrt [25]{x}}{\sqrt [25]{x-7} \sqrt [25]{x}}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {\sqrt [25]{2} \sqrt [25]{x^2-7 x} \int \left (-\frac {e^{\frac {x}{25}+5} x^2}{(x-7)^{24/25}}+\frac {5 e^{\frac {x}{25}+5} x}{(x-7)^{24/25}}+\frac {7 e^{\frac {x}{25}+5}}{(x-7)^{24/25}}\right )d\sqrt [25]{x}}{\sqrt [25]{x-7} \sqrt [25]{x}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\sqrt [25]{2} \sqrt [25]{x^2-7 x} \left (-\int \frac {e^{\frac {x}{25}+5} x^2}{(x-7)^{24/25}}d\sqrt [25]{x}+7 \int \frac {e^{\frac {x}{25}+5}}{(x-7)^{24/25}}d\sqrt [25]{x}+5 \int \frac {e^{\frac {x}{25}+5} x}{(x-7)^{24/25}}d\sqrt [25]{x}\right )}{\sqrt [25]{x-7} \sqrt [25]{x}}\) |
Input:
Int[(E^((125 + x + Log[-14*x + 2*x^2])/25)*(-7 - 5*x + x^2))/(-175*x + 25* x^2),x]
Output:
$Aborted
Time = 0.13 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73
method | result | size |
gosper | \({\mathrm e}^{\frac {\ln \left (2 x^{2}-14 x \right )}{25}+\frac {x}{25}+5}\) | \(19\) |
default | \({\mathrm e}^{\frac {\ln \left (2 x^{2}-14 x \right )}{25}+\frac {x}{25}+5}\) | \(19\) |
norman | \({\mathrm e}^{\frac {\ln \left (2 x^{2}-14 x \right )}{25}+\frac {x}{25}+5}\) | \(19\) |
risch | \(\left (2 x^{2}-14 x \right )^{\frac {1}{25}} {\mathrm e}^{5+\frac {x}{25}}\) | \(19\) |
parallelrisch | \({\mathrm e}^{\frac {\ln \left (2 x^{2}-14 x \right )}{25}+\frac {x}{25}+5}\) | \(19\) |
orering | \(\frac {25 x \left (-7+x \right ) {\mathrm e}^{\frac {\ln \left (2 x^{2}-14 x \right )}{25}+\frac {x}{25}+5}}{25 x^{2}-175 x}\) | \(36\) |
Input:
int((x^2-5*x-7)*exp(1/25*ln(2*x^2-14*x)+1/25*x+5)/(25*x^2-175*x),x,method= _RETURNVERBOSE)
Output:
exp(1/25*ln(2*x^2-14*x)+1/25*x+5)
Time = 0.09 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.69 \[ \int \frac {e^{\frac {1}{25} \left (125+x+\log \left (-14 x+2 x^2\right )\right )} \left (-7-5 x+x^2\right )}{-175 x+25 x^2} \, dx=e^{\left (\frac {1}{25} \, x + \frac {1}{25} \, \log \left (2 \, x^{2} - 14 \, x\right ) + 5\right )} \] Input:
integrate((x^2-5*x-7)*exp(1/25*log(2*x^2-14*x)+1/25*x+5)/(25*x^2-175*x),x, algorithm="fricas")
Output:
e^(1/25*x + 1/25*log(2*x^2 - 14*x) + 5)
Timed out. \[ \int \frac {e^{\frac {1}{25} \left (125+x+\log \left (-14 x+2 x^2\right )\right )} \left (-7-5 x+x^2\right )}{-175 x+25 x^2} \, dx=\text {Timed out} \] Input:
integrate((x**2-5*x-7)*exp(1/25*ln(2*x**2-14*x)+1/25*x+5)/(25*x**2-175*x), x)
Output:
Timed out
\[ \int \frac {e^{\frac {1}{25} \left (125+x+\log \left (-14 x+2 x^2\right )\right )} \left (-7-5 x+x^2\right )}{-175 x+25 x^2} \, dx=\int { \frac {{\left (x^{2} - 5 \, x - 7\right )} e^{\left (\frac {1}{25} \, x + \frac {1}{25} \, \log \left (2 \, x^{2} - 14 \, x\right ) + 5\right )}}{25 \, {\left (x^{2} - 7 \, x\right )}} \,d x } \] Input:
integrate((x^2-5*x-7)*exp(1/25*log(2*x^2-14*x)+1/25*x+5)/(25*x^2-175*x),x, algorithm="maxima")
Output:
1/25*integrate((x^2 - 5*x - 7)*e^(1/25*x + 1/25*log(2*x^2 - 14*x) + 5)/(x^ 2 - 7*x), x)
Time = 0.11 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.69 \[ \int \frac {e^{\frac {1}{25} \left (125+x+\log \left (-14 x+2 x^2\right )\right )} \left (-7-5 x+x^2\right )}{-175 x+25 x^2} \, dx=e^{\left (\frac {1}{25} \, x + \frac {1}{25} \, \log \left (2 \, x^{2} - 14 \, x\right ) + 5\right )} \] Input:
integrate((x^2-5*x-7)*exp(1/25*log(2*x^2-14*x)+1/25*x+5)/(25*x^2-175*x),x, algorithm="giac")
Output:
e^(1/25*x + 1/25*log(2*x^2 - 14*x) + 5)
Time = 2.62 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int \frac {e^{\frac {1}{25} \left (125+x+\log \left (-14 x+2 x^2\right )\right )} \left (-7-5 x+x^2\right )}{-175 x+25 x^2} \, dx=2^{1/25}\,{\mathrm {e}}^{x/25}\,{\mathrm {e}}^5\,{\left (x^2-7\,x\right )}^{1/25} \] Input:
int((exp(x/25 + log(2*x^2 - 14*x)/25 + 5)*(5*x - x^2 + 7))/(175*x - 25*x^2 ),x)
Output:
2^(1/25)*exp(x/25)*exp(5)*(x^2 - 7*x)^(1/25)
Time = 0.19 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {e^{\frac {1}{25} \left (125+x+\log \left (-14 x+2 x^2\right )\right )} \left (-7-5 x+x^2\right )}{-175 x+25 x^2} \, dx=x^{\frac {1}{25}} e^{\frac {x}{25}} \left (x -7\right )^{\frac {1}{25}} 2^{\frac {1}{25}} e^{5} \] Input:
int((x^2-5*x-7)*exp(1/25*log(2*x^2-14*x)+1/25*x+5)/(25*x^2-175*x),x)
Output:
(x**(21/25)*e**(x/25)*(x - 7)**(1/25)*2**(1/25)*e**5)/x**(4/5)