Integrand size = 150, antiderivative size = 24 \[ \int \frac {e^{\frac {15}{-12+5 x}} \left (-351 x+180 x^2-100 x^3\right )+\left (6480-22680 x+27045 x^2-12600 x^3+2000 x^4+e^{\frac {15}{-12+5 x}} \left (-432+936 x-555 x^2+100 x^3\right )\right ) \log \left (\frac {-15+e^{\frac {15}{-12+5 x}}+20 x}{-3+4 x}\right )}{6480-22680 x+27045 x^2-12600 x^3+2000 x^4+e^{\frac {15}{-12+5 x}} \left (-432+936 x-555 x^2+100 x^3\right )} \, dx=x \log \left (5+\frac {e^{\frac {3}{-\frac {12}{5}+x}}}{-3+4 x}\right ) \]
Time = 1.32 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {e^{\frac {15}{-12+5 x}} \left (-351 x+180 x^2-100 x^3\right )+\left (6480-22680 x+27045 x^2-12600 x^3+2000 x^4+e^{\frac {15}{-12+5 x}} \left (-432+936 x-555 x^2+100 x^3\right )\right ) \log \left (\frac {-15+e^{\frac {15}{-12+5 x}}+20 x}{-3+4 x}\right )}{6480-22680 x+27045 x^2-12600 x^3+2000 x^4+e^{\frac {15}{-12+5 x}} \left (-432+936 x-555 x^2+100 x^3\right )} \, dx=x \log \left (\frac {-15+e^{\frac {15}{-12+5 x}}+20 x}{-3+4 x}\right ) \]
Integrate[(E^(15/(-12 + 5*x))*(-351*x + 180*x^2 - 100*x^3) + (6480 - 22680 *x + 27045*x^2 - 12600*x^3 + 2000*x^4 + E^(15/(-12 + 5*x))*(-432 + 936*x - 555*x^2 + 100*x^3))*Log[(-15 + E^(15/(-12 + 5*x)) + 20*x)/(-3 + 4*x)])/(6 480 - 22680*x + 27045*x^2 - 12600*x^3 + 2000*x^4 + E^(15/(-12 + 5*x))*(-43 2 + 936*x - 555*x^2 + 100*x^3)),x]
Time = 4.32 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.013, Rules used = {7239, 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.
\(\displaystyle \int \frac {e^{\frac {15}{5 x-12}} \left (-100 x^3+180 x^2-351 x\right )+\left (2000 x^4-12600 x^3+27045 x^2+e^{\frac {15}{5 x-12}} \left (100 x^3-555 x^2+936 x-432\right )-22680 x+6480\right ) \log \left (\frac {20 x+e^{\frac {15}{5 x-12}}-15}{4 x-3}\right )}{2000 x^4-12600 x^3+27045 x^2+e^{\frac {15}{5 x-12}} \left (100 x^3-555 x^2+936 x-432\right )-22680 x+6480} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \left (\log \left (\frac {20 x+e^{\frac {15}{5 x-12}}-15}{4 x-3}\right )-\frac {e^{\frac {15}{5 x-12}} x \left (100 x^2-180 x+351\right )}{(12-5 x)^2 (4 x-3) \left (20 x+e^{\frac {15}{5 x-12}}-15\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x \log \left (\frac {-20 x-e^{-\frac {15}{12-5 x}}+15}{3-4 x}\right )\) |
Int[(E^(15/(-12 + 5*x))*(-351*x + 180*x^2 - 100*x^3) + (6480 - 22680*x + 2 7045*x^2 - 12600*x^3 + 2000*x^4 + E^(15/(-12 + 5*x))*(-432 + 936*x - 555*x ^2 + 100*x^3))*Log[(-15 + E^(15/(-12 + 5*x)) + 20*x)/(-3 + 4*x)])/(6480 - 22680*x + 27045*x^2 - 12600*x^3 + 2000*x^4 + E^(15/(-12 + 5*x))*(-432 + 93 6*x - 555*x^2 + 100*x^3)),x]
3.5.92.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 3.97 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12
method | result | size |
parallelrisch | \(\ln \left (\frac {{\mathrm e}^{\frac {15}{5 x -12}}+20 x -15}{-3+4 x}\right ) x\) | \(27\) |
norman | \(\frac {-12 \ln \left (\frac {{\mathrm e}^{\frac {15}{5 x -12}}+20 x -15}{-3+4 x}\right ) x +5 \ln \left (\frac {{\mathrm e}^{\frac {15}{5 x -12}}+20 x -15}{-3+4 x}\right ) x^{2}}{5 x -12}\) | \(66\) |
risch | \(x \ln \left (\frac {{\mathrm e}^{\frac {15}{5 x -12}}}{20}+x -\frac {3}{4}\right )-x \ln \left (x -\frac {3}{4}\right )-\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{x -\frac {3}{4}}\right ) \operatorname {csgn}\left (i \left (\frac {{\mathrm e}^{\frac {15}{5 x -12}}}{20}+x -\frac {3}{4}\right )\right ) \operatorname {csgn}\left (\frac {i \left (\frac {{\mathrm e}^{\frac {15}{5 x -12}}}{20}+x -\frac {3}{4}\right )}{x -\frac {3}{4}}\right )}{2}+\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{x -\frac {3}{4}}\right ) {\operatorname {csgn}\left (\frac {i \left (\frac {{\mathrm e}^{\frac {15}{5 x -12}}}{20}+x -\frac {3}{4}\right )}{x -\frac {3}{4}}\right )}^{2}}{2}+\frac {i \pi x \,\operatorname {csgn}\left (i \left (\frac {{\mathrm e}^{\frac {15}{5 x -12}}}{20}+x -\frac {3}{4}\right )\right ) {\operatorname {csgn}\left (\frac {i \left (\frac {{\mathrm e}^{\frac {15}{5 x -12}}}{20}+x -\frac {3}{4}\right )}{x -\frac {3}{4}}\right )}^{2}}{2}-\frac {i \pi x {\operatorname {csgn}\left (\frac {i \left (\frac {{\mathrm e}^{\frac {15}{5 x -12}}}{20}+x -\frac {3}{4}\right )}{x -\frac {3}{4}}\right )}^{3}}{2}+x \ln \left (5\right )\) | \(209\) |
int((((100*x^3-555*x^2+936*x-432)*exp(15/(5*x-12))+2000*x^4-12600*x^3+2704 5*x^2-22680*x+6480)*ln((exp(15/(5*x-12))+20*x-15)/(-3+4*x))+(-100*x^3+180* x^2-351*x)*exp(15/(5*x-12)))/((100*x^3-555*x^2+936*x-432)*exp(15/(5*x-12)) +2000*x^4-12600*x^3+27045*x^2-22680*x+6480),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{\frac {15}{-12+5 x}} \left (-351 x+180 x^2-100 x^3\right )+\left (6480-22680 x+27045 x^2-12600 x^3+2000 x^4+e^{\frac {15}{-12+5 x}} \left (-432+936 x-555 x^2+100 x^3\right )\right ) \log \left (\frac {-15+e^{\frac {15}{-12+5 x}}+20 x}{-3+4 x}\right )}{6480-22680 x+27045 x^2-12600 x^3+2000 x^4+e^{\frac {15}{-12+5 x}} \left (-432+936 x-555 x^2+100 x^3\right )} \, dx=x \log \left (\frac {20 \, x + e^{\left (\frac {15}{5 \, x - 12}\right )} - 15}{4 \, x - 3}\right ) \]
integrate((((100*x^3-555*x^2+936*x-432)*exp(15/(5*x-12))+2000*x^4-12600*x^ 3+27045*x^2-22680*x+6480)*log((exp(15/(5*x-12))+20*x-15)/(-3+4*x))+(-100*x ^3+180*x^2-351*x)*exp(15/(5*x-12)))/((100*x^3-555*x^2+936*x-432)*exp(15/(5 *x-12))+2000*x^4-12600*x^3+27045*x^2-22680*x+6480),x, algorithm=\
Time = 0.34 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {e^{\frac {15}{-12+5 x}} \left (-351 x+180 x^2-100 x^3\right )+\left (6480-22680 x+27045 x^2-12600 x^3+2000 x^4+e^{\frac {15}{-12+5 x}} \left (-432+936 x-555 x^2+100 x^3\right )\right ) \log \left (\frac {-15+e^{\frac {15}{-12+5 x}}+20 x}{-3+4 x}\right )}{6480-22680 x+27045 x^2-12600 x^3+2000 x^4+e^{\frac {15}{-12+5 x}} \left (-432+936 x-555 x^2+100 x^3\right )} \, dx=x \log {\left (\frac {20 x + e^{\frac {15}{5 x - 12}} - 15}{4 x - 3} \right )} \]
integrate((((100*x**3-555*x**2+936*x-432)*exp(15/(5*x-12))+2000*x**4-12600 *x**3+27045*x**2-22680*x+6480)*ln((exp(15/(5*x-12))+20*x-15)/(-3+4*x))+(-1 00*x**3+180*x**2-351*x)*exp(15/(5*x-12)))/((100*x**3-555*x**2+936*x-432)*e xp(15/(5*x-12))+2000*x**4-12600*x**3+27045*x**2-22680*x+6480),x)
Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {e^{\frac {15}{-12+5 x}} \left (-351 x+180 x^2-100 x^3\right )+\left (6480-22680 x+27045 x^2-12600 x^3+2000 x^4+e^{\frac {15}{-12+5 x}} \left (-432+936 x-555 x^2+100 x^3\right )\right ) \log \left (\frac {-15+e^{\frac {15}{-12+5 x}}+20 x}{-3+4 x}\right )}{6480-22680 x+27045 x^2-12600 x^3+2000 x^4+e^{\frac {15}{-12+5 x}} \left (-432+936 x-555 x^2+100 x^3\right )} \, dx=x \log \left (20 \, x + e^{\left (\frac {15}{5 \, x - 12}\right )} - 15\right ) - x \log \left (4 \, x - 3\right ) \]
integrate((((100*x^3-555*x^2+936*x-432)*exp(15/(5*x-12))+2000*x^4-12600*x^ 3+27045*x^2-22680*x+6480)*log((exp(15/(5*x-12))+20*x-15)/(-3+4*x))+(-100*x ^3+180*x^2-351*x)*exp(15/(5*x-12)))/((100*x^3-555*x^2+936*x-432)*exp(15/(5 *x-12))+2000*x^4-12600*x^3+27045*x^2-22680*x+6480),x, algorithm=\
Time = 0.39 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {e^{\frac {15}{-12+5 x}} \left (-351 x+180 x^2-100 x^3\right )+\left (6480-22680 x+27045 x^2-12600 x^3+2000 x^4+e^{\frac {15}{-12+5 x}} \left (-432+936 x-555 x^2+100 x^3\right )\right ) \log \left (\frac {-15+e^{\frac {15}{-12+5 x}}+20 x}{-3+4 x}\right )}{6480-22680 x+27045 x^2-12600 x^3+2000 x^4+e^{\frac {15}{-12+5 x}} \left (-432+936 x-555 x^2+100 x^3\right )} \, dx=x \log \left (\frac {20 \, x + e^{\left (\frac {25 \, x}{4 \, {\left (5 \, x - 12\right )}} - \frac {5}{4}\right )} - 15}{4 \, x - 3}\right ) \]
integrate((((100*x^3-555*x^2+936*x-432)*exp(15/(5*x-12))+2000*x^4-12600*x^ 3+27045*x^2-22680*x+6480)*log((exp(15/(5*x-12))+20*x-15)/(-3+4*x))+(-100*x ^3+180*x^2-351*x)*exp(15/(5*x-12)))/((100*x^3-555*x^2+936*x-432)*exp(15/(5 *x-12))+2000*x^4-12600*x^3+27045*x^2-22680*x+6480),x, algorithm=\
Time = 0.44 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{\frac {15}{-12+5 x}} \left (-351 x+180 x^2-100 x^3\right )+\left (6480-22680 x+27045 x^2-12600 x^3+2000 x^4+e^{\frac {15}{-12+5 x}} \left (-432+936 x-555 x^2+100 x^3\right )\right ) \log \left (\frac {-15+e^{\frac {15}{-12+5 x}}+20 x}{-3+4 x}\right )}{6480-22680 x+27045 x^2-12600 x^3+2000 x^4+e^{\frac {15}{-12+5 x}} \left (-432+936 x-555 x^2+100 x^3\right )} \, dx=x\,\ln \left (\frac {20\,x+{\mathrm {e}}^{\frac {15}{5\,x-12}}-15}{4\,x-3}\right ) \]
int(-(exp(15/(5*x - 12))*(351*x - 180*x^2 + 100*x^3) - log((20*x + exp(15/ (5*x - 12)) - 15)/(4*x - 3))*(exp(15/(5*x - 12))*(936*x - 555*x^2 + 100*x^ 3 - 432) - 22680*x + 27045*x^2 - 12600*x^3 + 2000*x^4 + 6480))/(exp(15/(5* x - 12))*(936*x - 555*x^2 + 100*x^3 - 432) - 22680*x + 27045*x^2 - 12600*x ^3 + 2000*x^4 + 6480),x)