Integrand size = 70, antiderivative size = 26 \[ \int \frac {25+e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )} \left (-25+25 x-30 x^3\right )}{6-12 e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )}+6 e^{\frac {2}{5} \left (-7+5 x-2 x^3\right )}} \, dx=\frac {25 x}{6 \left (1-e^{x+\frac {1}{5} \left (-7-2 x^3\right )}\right )} \] Output:
25/6*x/(1-exp(-2/5*x^3+x-7/5))
Time = 3.19 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50 \[ \int \frac {25+e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )} \left (-25+25 x-30 x^3\right )}{6-12 e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )}+6 e^{\frac {2}{5} \left (-7+5 x-2 x^3\right )}} \, dx=\frac {25 e^{\frac {7}{5}+\frac {2 x^3}{5}} x}{6 \left (-e^x+e^{\frac {7}{5}+\frac {2 x^3}{5}}\right )} \] Input:
Integrate[(25 + E^((-7 + 5*x - 2*x^3)/5)*(-25 + 25*x - 30*x^3))/(6 - 12*E^ ((-7 + 5*x - 2*x^3)/5) + 6*E^((2*(-7 + 5*x - 2*x^3))/5)),x]
Output:
(25*E^(7/5 + (2*x^3)/5)*x)/(6*(-E^x + E^(7/5 + (2*x^3)/5)))
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.
\(\displaystyle \int \frac {e^{\frac {1}{5} \left (-2 x^3+5 x-7\right )} \left (-30 x^3+25 x-25\right )+25}{-12 e^{\frac {1}{5} \left (-2 x^3+5 x-7\right )}+6 e^{\frac {2}{5} \left (-2 x^3+5 x-7\right )}+6} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{\frac {4 x^3}{5}+\frac {14}{5}} \left (e^{\frac {1}{5} \left (-2 x^3+5 x-7\right )} \left (-30 x^3+25 x-25\right )+25\right )}{6 \left (e^x-e^{\frac {2 x^3}{5}+\frac {7}{5}}\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{6} \int \frac {5 e^{\frac {4 x^3}{5}+\frac {14}{5}} \left (5-e^{\frac {1}{5} \left (-2 x^3+5 x-7\right )} \left (6 x^3-5 x+5\right )\right )}{\left (e^x-e^{\frac {2 x^3}{5}+\frac {7}{5}}\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {5}{6} \int \frac {e^{\frac {4 x^3}{5}+\frac {14}{5}} \left (5-e^{\frac {1}{5} \left (-2 x^3+5 x-7\right )} \left (6 x^3-5 x+5\right )\right )}{\left (e^x-e^{\frac {2 x^3}{5}+\frac {7}{5}}\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {5}{6} \int \left (-\frac {e^{\frac {4 x^3}{5}+\frac {14}{5}} x \left (6 x^2-5\right )}{\left (e^x-e^{\frac {2 x^3}{5}+\frac {7}{5}}\right )^2}-e^{\frac {2 x^3}{5}-x+\frac {7}{5}} \left (6 x^3-5 x+5\right )-\frac {e^{\frac {4 x^3}{5}-x+\frac {14}{5}} \left (6 x^3-5 x+5\right )}{e^x-e^{\frac {2 x^3}{5}+\frac {7}{5}}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {5}{6} \left (-5 \int \frac {e^{\frac {4 x^3}{5}-x+\frac {14}{5}}}{e^x-e^{\frac {2 x^3}{5}+\frac {7}{5}}}dx+5 \int \frac {e^{\frac {4 x^3}{5}+\frac {14}{5}} x}{\left (e^x-e^{\frac {2 x^3}{5}+\frac {7}{5}}\right )^2}dx+5 \int \frac {e^{\frac {4 x^3}{5}-x+\frac {14}{5}} x}{e^x-e^{\frac {2 x^3}{5}+\frac {7}{5}}}dx-6 \int \frac {e^{\frac {4 x^3}{5}+\frac {14}{5}} x^3}{\left (e^x-e^{\frac {2 x^3}{5}+\frac {7}{5}}\right )^2}dx-6 \int \frac {e^{\frac {4 x^3}{5}-x+\frac {14}{5}} x^3}{e^x-e^{\frac {2 x^3}{5}+\frac {7}{5}}}dx-\frac {5 e^{\frac {2 x^3}{5}-x+\frac {7}{5}} \left (5 x-6 x^3\right )}{5-6 x^2}\right )\) |
Input:
Int[(25 + E^((-7 + 5*x - 2*x^3)/5)*(-25 + 25*x - 30*x^3))/(6 - 12*E^((-7 + 5*x - 2*x^3)/5) + 6*E^((2*(-7 + 5*x - 2*x^3))/5)),x]
Output:
$Aborted
Time = 0.16 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65
method | result | size |
norman | \(-\frac {25 x}{6 \left ({\mathrm e}^{-\frac {2}{5} x^{3}+x -\frac {7}{5}}-1\right )}\) | \(17\) |
risch | \(-\frac {25 x}{6 \left ({\mathrm e}^{-\frac {2}{5} x^{3}+x -\frac {7}{5}}-1\right )}\) | \(17\) |
parallelrisch | \(-\frac {25 x}{6 \left ({\mathrm e}^{-\frac {2}{5} x^{3}+x -\frac {7}{5}}-1\right )}\) | \(17\) |
Input:
int(((-30*x^3+25*x-25)*exp(-2/5*x^3+x-7/5)+25)/(6*exp(-2/5*x^3+x-7/5)^2-12 *exp(-2/5*x^3+x-7/5)+6),x,method=_RETURNVERBOSE)
Output:
-25/6*x/(exp(-2/5*x^3+x-7/5)-1)
Time = 0.09 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.62 \[ \int \frac {25+e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )} \left (-25+25 x-30 x^3\right )}{6-12 e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )}+6 e^{\frac {2}{5} \left (-7+5 x-2 x^3\right )}} \, dx=-\frac {25 \, x}{6 \, {\left (e^{\left (-\frac {2}{5} \, x^{3} + x - \frac {7}{5}\right )} - 1\right )}} \] Input:
integrate(((-30*x^3+25*x-25)*exp(-2/5*x^3+x-7/5)+25)/(6*exp(-2/5*x^3+x-7/5 )^2-12*exp(-2/5*x^3+x-7/5)+6),x, algorithm="fricas")
Output:
-25/6*x/(e^(-2/5*x^3 + x - 7/5) - 1)
Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {25+e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )} \left (-25+25 x-30 x^3\right )}{6-12 e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )}+6 e^{\frac {2}{5} \left (-7+5 x-2 x^3\right )}} \, dx=- \frac {25 x}{6 e^{- \frac {2 x^{3}}{5} + x - \frac {7}{5}} - 6} \] Input:
integrate(((-30*x**3+25*x-25)*exp(-2/5*x**3+x-7/5)+25)/(6*exp(-2/5*x**3+x- 7/5)**2-12*exp(-2/5*x**3+x-7/5)+6),x)
Output:
-25*x/(6*exp(-2*x**3/5 + x - 7/5) - 6)
Exception generated. \[ \int \frac {25+e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )} \left (-25+25 x-30 x^3\right )}{6-12 e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )}+6 e^{\frac {2}{5} \left (-7+5 x-2 x^3\right )}} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(((-30*x^3+25*x-25)*exp(-2/5*x^3+x-7/5)+25)/(6*exp(-2/5*x^3+x-7/5 )^2-12*exp(-2/5*x^3+x-7/5)+6),x, algorithm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un defined.
Time = 0.15 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.62 \[ \int \frac {25+e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )} \left (-25+25 x-30 x^3\right )}{6-12 e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )}+6 e^{\frac {2}{5} \left (-7+5 x-2 x^3\right )}} \, dx=-\frac {25 \, x}{6 \, {\left (e^{\left (-\frac {2}{5} \, x^{3} + x - \frac {7}{5}\right )} - 1\right )}} \] Input:
integrate(((-30*x^3+25*x-25)*exp(-2/5*x^3+x-7/5)+25)/(6*exp(-2/5*x^3+x-7/5 )^2-12*exp(-2/5*x^3+x-7/5)+6),x, algorithm="giac")
Output:
-25/6*x/(e^(-2/5*x^3 + x - 7/5) - 1)
Time = 2.98 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.69 \[ \int \frac {25+e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )} \left (-25+25 x-30 x^3\right )}{6-12 e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )}+6 e^{\frac {2}{5} \left (-7+5 x-2 x^3\right )}} \, dx=-\frac {25\,x}{6\,\left ({\mathrm {e}}^{-\frac {2\,x^3}{5}+x-\frac {7}{5}}-1\right )} \] Input:
int(-(exp(x - (2*x^3)/5 - 7/5)*(30*x^3 - 25*x + 25) - 25)/(6*exp(2*x - (4* x^3)/5 - 14/5) - 12*exp(x - (2*x^3)/5 - 7/5) + 6),x)
Output:
-(25*x)/(6*(exp(x - (2*x^3)/5 - 7/5) - 1))
Time = 0.16 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27 \[ \int \frac {25+e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )} \left (-25+25 x-30 x^3\right )}{6-12 e^{\frac {1}{5} \left (-7+5 x-2 x^3\right )}+6 e^{\frac {2}{5} \left (-7+5 x-2 x^3\right )}} \, dx=\frac {25 e^{\frac {2 x^{3}}{5}+\frac {2}{5}} e x}{6 e^{\frac {2 x^{3}}{5}+\frac {2}{5}} e -6 e^{x}} \] Input:
int(((-30*x^3+25*x-25)*exp(-2/5*x^3+x-7/5)+25)/(6*exp(-2/5*x^3+x-7/5)^2-12 *exp(-2/5*x^3+x-7/5)+6),x)
Output:
(25*e**((2*x**3 + 2)/5)*e*x)/(6*(e**((2*x**3 + 2)/5)*e - e**x))