Integrand size = 60, antiderivative size = 27 \[ \int \frac {3 x^3+e^{3/x} (15+e (-48-16 x)+5 x)}{-60 x^3+15 x^4+e^{3/x} \left (-25 x^2+80 e x^2\right )} \, dx=\frac {1}{5} \log \left (-4+\frac {e^{3/x} (-5+16 e)}{3 x}+x\right ) \] Output:
1/5*ln(1/3*exp(3/x)*(16*exp(1)-5)/x-4+x)
Time = 0.38 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44 \[ \int \frac {3 x^3+e^{3/x} (15+e (-48-16 x)+5 x)}{-60 x^3+15 x^4+e^{3/x} \left (-25 x^2+80 e x^2\right )} \, dx=\frac {1}{5} \left (-\log (x)+\log \left (16 e^{1+\frac {3}{x}}-5 e^{3/x}-12 x+3 x^2\right )\right ) \] Input:
Integrate[(3*x^3 + E^(3/x)*(15 + E*(-48 - 16*x) + 5*x))/(-60*x^3 + 15*x^4 + E^(3/x)*(-25*x^2 + 80*E*x^2)),x]
Output:
(-Log[x] + Log[16*E^(1 + 3/x) - 5*E^(3/x) - 12*x + 3*x^2])/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.
\(\displaystyle \int \frac {3 x^3+e^{3/x} (e (-16 x-48)+5 x+15)}{15 x^4-60 x^3+e^{3/x} \left (80 e x^2-25 x^2\right )} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {-x-3}{5 x^2}+\frac {3 \left (2 x^2-x-12\right )}{5 x \left (3 x^2-12 x+16 \left (1-\frac {5}{16 e}\right ) e^{\frac {3}{x}+1}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3}{5} \int \frac {1}{-3 x^2+12 x-16 \left (1-\frac {5}{16 e}\right ) e^{1+\frac {3}{x}}}dx+\frac {36}{5} \int \frac {1}{x \left (-3 x^2+12 x-16 \left (1-\frac {5}{16 e}\right ) e^{1+\frac {3}{x}}\right )}dx+\frac {6}{5} \int \frac {x}{3 x^2-12 x+16 \left (1-\frac {5}{16 e}\right ) e^{1+\frac {3}{x}}}dx+\frac {3}{5 x}-\frac {\log (x)}{5}\) |
Input:
Int[(3*x^3 + E^(3/x)*(15 + E*(-48 - 16*x) + 5*x))/(-60*x^3 + 15*x^4 + E^(3 /x)*(-25*x^2 + 80*E*x^2)),x]
Output:
$Aborted
Time = 0.69 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11
method | result | size |
risch | \(-\frac {\ln \left (x \right )}{5}+\frac {\ln \left ({\mathrm e}^{\frac {3}{x}}+\frac {3 \left (x -4\right ) x}{16 \,{\mathrm e}-5}\right )}{5}\) | \(30\) |
parallelrisch | \(-\frac {\ln \left (x \right )}{5}+\frac {\ln \left (\frac {16 \,{\mathrm e}^{\frac {3}{x}} {\mathrm e}}{3}+x^{2}-\frac {5 \,{\mathrm e}^{\frac {3}{x}}}{3}-4 x \right )}{5}\) | \(34\) |
norman | \(-\frac {\ln \left (x \right )}{5}+\frac {\ln \left (16 \,{\mathrm e}^{\frac {3}{x}} {\mathrm e}+3 x^{2}-5 \,{\mathrm e}^{\frac {3}{x}}-12 x \right )}{5}\) | \(36\) |
Input:
int((((-16*x-48)*exp(1)+5*x+15)*exp(3/x)+3*x^3)/((80*x^2*exp(1)-25*x^2)*ex p(3/x)+15*x^4-60*x^3),x,method=_RETURNVERBOSE)
Output:
-1/5*ln(x)+1/5*ln(exp(3/x)+3*(x-4)*x/(16*exp(1)-5))
Time = 0.09 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {3 x^3+e^{3/x} (15+e (-48-16 x)+5 x)}{-60 x^3+15 x^4+e^{3/x} \left (-25 x^2+80 e x^2\right )} \, dx=\frac {1}{5} \, \log \left (3 \, x^{2} + {\left (16 \, e - 5\right )} e^{\frac {3}{x}} - 12 \, x\right ) - \frac {1}{5} \, \log \left (x\right ) \] Input:
integrate((((-16*x-48)*exp(1)+5*x+15)*exp(3/x)+3*x^3)/((80*x^2*exp(1)-25*x ^2)*exp(3/x)+15*x^4-60*x^3),x, algorithm="fricas")
Output:
1/5*log(3*x^2 + (16*e - 5)*e^(3/x) - 12*x) - 1/5*log(x)
Time = 0.22 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {3 x^3+e^{3/x} (15+e (-48-16 x)+5 x)}{-60 x^3+15 x^4+e^{3/x} \left (-25 x^2+80 e x^2\right )} \, dx=- \frac {\log {\left (x \right )}}{5} + \frac {\log {\left (\frac {3 x^{2} - 12 x}{-5 + 16 e} + e^{\frac {3}{x}} \right )}}{5} \] Input:
integrate((((-16*x-48)*exp(1)+5*x+15)*exp(3/x)+3*x**3)/((80*x**2*exp(1)-25 *x**2)*exp(3/x)+15*x**4-60*x**3),x)
Output:
-log(x)/5 + log((3*x**2 - 12*x)/(-5 + 16*E) + exp(3/x))/5
Time = 0.06 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44 \[ \int \frac {3 x^3+e^{3/x} (15+e (-48-16 x)+5 x)}{-60 x^3+15 x^4+e^{3/x} \left (-25 x^2+80 e x^2\right )} \, dx=-\frac {1}{5} \, \log \left (x\right ) + \frac {1}{5} \, \log \left (\frac {3 \, x^{2} + {\left (16 \, e - 5\right )} e^{\frac {3}{x}} - 12 \, x}{16 \, e - 5}\right ) \] Input:
integrate((((-16*x-48)*exp(1)+5*x+15)*exp(3/x)+3*x^3)/((80*x^2*exp(1)-25*x ^2)*exp(3/x)+15*x^4-60*x^3),x, algorithm="maxima")
Output:
-1/5*log(x) + 1/5*log((3*x^2 + (16*e - 5)*e^(3/x) - 12*x)/(16*e - 5))
Time = 0.12 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.59 \[ \int \frac {3 x^3+e^{3/x} (15+e (-48-16 x)+5 x)}{-60 x^3+15 x^4+e^{3/x} \left (-25 x^2+80 e x^2\right )} \, dx=-\frac {1}{5} \, \log \left (\frac {3}{x}\right ) + \frac {1}{5} \, \log \left (-\frac {108}{x} - \frac {45 \, e^{\frac {3}{x}}}{x^{2}} + \frac {144 \, e^{\left (\frac {3}{x} + 1\right )}}{x^{2}} + 27\right ) \] Input:
integrate((((-16*x-48)*exp(1)+5*x+15)*exp(3/x)+3*x^3)/((80*x^2*exp(1)-25*x ^2)*exp(3/x)+15*x^4-60*x^3),x, algorithm="giac")
Output:
-1/5*log(3/x) + 1/5*log(-108/x - 45*e^(3/x)/x^2 + 144*e^(3/x + 1)/x^2 + 27 )
Timed out. \[ \int \frac {3 x^3+e^{3/x} (15+e (-48-16 x)+5 x)}{-60 x^3+15 x^4+e^{3/x} \left (-25 x^2+80 e x^2\right )} \, dx=\int \frac {{\mathrm {e}}^{3/x}\,\left (5\,x-\mathrm {e}\,\left (16\,x+48\right )+15\right )+3\,x^3}{{\mathrm {e}}^{3/x}\,\left (80\,x^2\,\mathrm {e}-25\,x^2\right )-60\,x^3+15\,x^4} \,d x \] Input:
int((exp(3/x)*(5*x - exp(1)*(16*x + 48) + 15) + 3*x^3)/(exp(3/x)*(80*x^2*e xp(1) - 25*x^2) - 60*x^3 + 15*x^4),x)
Output:
int((exp(3/x)*(5*x - exp(1)*(16*x + 48) + 15) + 3*x^3)/(exp(3/x)*(80*x^2*e xp(1) - 25*x^2) - 60*x^3 + 15*x^4), x)
\[ \int \frac {3 x^3+e^{3/x} (15+e (-48-16 x)+5 x)}{-60 x^3+15 x^4+e^{3/x} \left (-25 x^2+80 e x^2\right )} \, dx=-\frac {48 \left (\int \frac {e^{\frac {3}{x}}}{16 e^{\frac {3}{x}} e \,x^{2}-5 e^{\frac {3}{x}} x^{2}+3 x^{4}-12 x^{3}}d x \right ) e}{5}+3 \left (\int \frac {e^{\frac {3}{x}}}{16 e^{\frac {3}{x}} e \,x^{2}-5 e^{\frac {3}{x}} x^{2}+3 x^{4}-12 x^{3}}d x \right )-\frac {16 \left (\int \frac {e^{\frac {3}{x}}}{16 e^{\frac {3}{x}} e x -5 e^{\frac {3}{x}} x +3 x^{3}-12 x^{2}}d x \right ) e}{5}+\int \frac {e^{\frac {3}{x}}}{16 e^{\frac {3}{x}} e x -5 e^{\frac {3}{x}} x +3 x^{3}-12 x^{2}}d x +\frac {3 \left (\int \frac {x}{16 e^{\frac {3}{x}} e -5 e^{\frac {3}{x}}+3 x^{2}-12 x}d x \right )}{5} \] Input:
int((((-16*x-48)*exp(1)+5*x+15)*exp(3/x)+3*x^3)/((80*x^2*exp(1)-25*x^2)*ex p(3/x)+15*x^4-60*x^3),x)
Output:
( - 48*int(e**(3/x)/(16*e**(3/x)*e*x**2 - 5*e**(3/x)*x**2 + 3*x**4 - 12*x* *3),x)*e + 15*int(e**(3/x)/(16*e**(3/x)*e*x**2 - 5*e**(3/x)*x**2 + 3*x**4 - 12*x**3),x) - 16*int(e**(3/x)/(16*e**(3/x)*e*x - 5*e**(3/x)*x + 3*x**3 - 12*x**2),x)*e + 5*int(e**(3/x)/(16*e**(3/x)*e*x - 5*e**(3/x)*x + 3*x**3 - 12*x**2),x) + 3*int(x/(16*e**(3/x)*e - 5*e**(3/x) + 3*x**2 - 12*x),x))/5