Integrand size = 110, antiderivative size = 25 \[ \int \frac {e^{4+x} \left (-150-50 x^2\right )+e^4 \left (300+100 x^2\right )+\left (-200 e^4 x^2+e^{4+x} \left (150 x+100 x^2+50 x^3\right )\right ) \log (x)}{36 x+24 x^3+4 x^5+e^x \left (-36 x-24 x^3-4 x^5\right )+e^{2 x} \left (9 x+6 x^3+x^5\right )} \, dx=\frac {25 e^4 \log (x)}{\left (1-\frac {e^x}{2}\right ) \left (3+x^2\right )} \] Output:
5*exp(4)*ln(x)/(1-1/2*exp(x))/(1/5*x^2+3/5)
Time = 0.31 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {e^{4+x} \left (-150-50 x^2\right )+e^4 \left (300+100 x^2\right )+\left (-200 e^4 x^2+e^{4+x} \left (150 x+100 x^2+50 x^3\right )\right ) \log (x)}{36 x+24 x^3+4 x^5+e^x \left (-36 x-24 x^3-4 x^5\right )+e^{2 x} \left (9 x+6 x^3+x^5\right )} \, dx=-\frac {50 e^4 \log (x)}{\left (-2+e^x\right ) \left (3+x^2\right )} \] Input:
Integrate[(E^(4 + x)*(-150 - 50*x^2) + E^4*(300 + 100*x^2) + (-200*E^4*x^2 + E^(4 + x)*(150*x + 100*x^2 + 50*x^3))*Log[x])/(36*x + 24*x^3 + 4*x^5 + E^x*(-36*x - 24*x^3 - 4*x^5) + E^(2*x)*(9*x + 6*x^3 + x^5)),x]
Output:
(-50*E^4*Log[x])/((-2 + E^x)*(3 + x^2))
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.
\(\displaystyle \int \frac {e^{x+4} \left (-50 x^2-150\right )+e^4 \left (100 x^2+300\right )+\left (e^{x+4} \left (50 x^3+100 x^2+150 x\right )-200 e^4 x^2\right ) \log (x)}{4 x^5+24 x^3+e^x \left (-4 x^5-24 x^3-36 x\right )+e^{2 x} \left (x^5+6 x^3+9 x\right )+36 x} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {50 e^4 \left (x \left (e^x \left (x^2+2 x+3\right )-4 x\right ) \log (x)-\left (e^x-2\right ) \left (x^2+3\right )\right )}{\left (2-e^x\right )^2 x \left (x^2+3\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 50 e^4 \int \frac {\left (2-e^x\right ) \left (x^2+3\right )-x \left (4 x-e^x \left (x^2+2 x+3\right )\right ) \log (x)}{\left (2-e^x\right )^2 x \left (x^2+3\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 50 e^4 \int \left (\frac {2 \log (x)}{\left (-2+e^x\right )^2 \left (x^2+3\right )}+\frac {\log (x) x^3+2 \log (x) x^2-x^2+3 \log (x) x-3}{\left (-2+e^x\right ) x \left (x^2+3\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 50 e^4 \left (-2 \int \frac {\int \frac {x}{\left (-2+e^x\right ) \left (x^2+3\right )^2}dx}{x}dx+2 \log (x) \int \frac {x}{\left (-2+e^x\right ) \left (x^2+3\right )^2}dx-\frac {1}{6} \int \frac {1}{\left (-2+e^x\right ) \left (i \sqrt {3}-x\right )}dx-\frac {1}{3} \int \frac {1}{\left (-2+e^x\right ) x}dx+\frac {1}{6} \int \frac {1}{\left (-2+e^x\right ) \left (x+i \sqrt {3}\right )}dx-\frac {i \int \frac {\int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}-x\right )}dx}{x}dx}{\sqrt {3}}-\frac {i \int \frac {\int \frac {1}{\left (-2+e^x\right ) \left (i \sqrt {3}-x\right )}dx}{x}dx}{2 \sqrt {3}}-\frac {i \int \frac {\int \frac {1}{\left (-2+e^x\right )^2 \left (x+i \sqrt {3}\right )}dx}{x}dx}{\sqrt {3}}-\frac {i \int \frac {\int \frac {1}{\left (-2+e^x\right ) \left (x+i \sqrt {3}\right )}dx}{x}dx}{2 \sqrt {3}}+\frac {i \log (x) \int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}-x\right )}dx}{\sqrt {3}}+\frac {i \log (x) \int \frac {1}{\left (-2+e^x\right ) \left (i \sqrt {3}-x\right )}dx}{2 \sqrt {3}}+\frac {i \log (x) \int \frac {1}{\left (-2+e^x\right )^2 \left (x+i \sqrt {3}\right )}dx}{\sqrt {3}}+\frac {i \log (x) \int \frac {1}{\left (-2+e^x\right ) \left (x+i \sqrt {3}\right )}dx}{2 \sqrt {3}}\right )\) |
Input:
Int[(E^(4 + x)*(-150 - 50*x^2) + E^4*(300 + 100*x^2) + (-200*E^4*x^2 + E^( 4 + x)*(150*x + 100*x^2 + 50*x^3))*Log[x])/(36*x + 24*x^3 + 4*x^5 + E^x*(- 36*x - 24*x^3 - 4*x^5) + E^(2*x)*(9*x + 6*x^3 + x^5)),x]
Output:
$Aborted
Time = 1.20 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80
method | result | size |
risch | \(-\frac {50 \,{\mathrm e}^{4} \ln \left (x \right )}{\left (x^{2}+3\right ) \left ({\mathrm e}^{x}-2\right )}\) | \(20\) |
parallelrisch | \(-\frac {50 \,{\mathrm e}^{4} \ln \left (x \right )}{\left (x^{2}+3\right ) \left ({\mathrm e}^{x}-2\right )}\) | \(20\) |
Input:
int((((50*x^3+100*x^2+150*x)*exp(4)*exp(x)-200*x^2*exp(4))*ln(x)+(-50*x^2- 150)*exp(4)*exp(x)+(100*x^2+300)*exp(4))/((x^5+6*x^3+9*x)*exp(x)^2+(-4*x^5 -24*x^3-36*x)*exp(x)+4*x^5+24*x^3+36*x),x,method=_RETURNVERBOSE)
Output:
-50*exp(4)/(x^2+3)/(exp(x)-2)*ln(x)
Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {e^{4+x} \left (-150-50 x^2\right )+e^4 \left (300+100 x^2\right )+\left (-200 e^4 x^2+e^{4+x} \left (150 x+100 x^2+50 x^3\right )\right ) \log (x)}{36 x+24 x^3+4 x^5+e^x \left (-36 x-24 x^3-4 x^5\right )+e^{2 x} \left (9 x+6 x^3+x^5\right )} \, dx=\frac {50 \, e^{8} \log \left (x\right )}{2 \, {\left (x^{2} + 3\right )} e^{4} - {\left (x^{2} + 3\right )} e^{\left (x + 4\right )}} \] Input:
integrate((((50*x^3+100*x^2+150*x)*exp(4)*exp(x)-200*x^2*exp(4))*log(x)+(- 50*x^2-150)*exp(4)*exp(x)+(100*x^2+300)*exp(4))/((x^5+6*x^3+9*x)*exp(x)^2+ (-4*x^5-24*x^3-36*x)*exp(x)+4*x^5+24*x^3+36*x),x, algorithm="fricas")
Output:
50*e^8*log(x)/(2*(x^2 + 3)*e^4 - (x^2 + 3)*e^(x + 4))
Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{4+x} \left (-150-50 x^2\right )+e^4 \left (300+100 x^2\right )+\left (-200 e^4 x^2+e^{4+x} \left (150 x+100 x^2+50 x^3\right )\right ) \log (x)}{36 x+24 x^3+4 x^5+e^x \left (-36 x-24 x^3-4 x^5\right )+e^{2 x} \left (9 x+6 x^3+x^5\right )} \, dx=- \frac {50 e^{4} \log {\left (x \right )}}{- 2 x^{2} + \left (x^{2} + 3\right ) e^{x} - 6} \] Input:
integrate((((50*x**3+100*x**2+150*x)*exp(4)*exp(x)-200*x**2*exp(4))*ln(x)+ (-50*x**2-150)*exp(4)*exp(x)+(100*x**2+300)*exp(4))/((x**5+6*x**3+9*x)*exp (x)**2+(-4*x**5-24*x**3-36*x)*exp(x)+4*x**5+24*x**3+36*x),x)
Output:
-50*exp(4)*log(x)/(-2*x**2 + (x**2 + 3)*exp(x) - 6)
Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{4+x} \left (-150-50 x^2\right )+e^4 \left (300+100 x^2\right )+\left (-200 e^4 x^2+e^{4+x} \left (150 x+100 x^2+50 x^3\right )\right ) \log (x)}{36 x+24 x^3+4 x^5+e^x \left (-36 x-24 x^3-4 x^5\right )+e^{2 x} \left (9 x+6 x^3+x^5\right )} \, dx=\frac {50 \, e^{4} \log \left (x\right )}{2 \, x^{2} - {\left (x^{2} + 3\right )} e^{x} + 6} \] Input:
integrate((((50*x^3+100*x^2+150*x)*exp(4)*exp(x)-200*x^2*exp(4))*log(x)+(- 50*x^2-150)*exp(4)*exp(x)+(100*x^2+300)*exp(4))/((x^5+6*x^3+9*x)*exp(x)^2+ (-4*x^5-24*x^3-36*x)*exp(x)+4*x^5+24*x^3+36*x),x, algorithm="maxima")
Output:
50*e^4*log(x)/(2*x^2 - (x^2 + 3)*e^x + 6)
Time = 0.13 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^{4+x} \left (-150-50 x^2\right )+e^4 \left (300+100 x^2\right )+\left (-200 e^4 x^2+e^{4+x} \left (150 x+100 x^2+50 x^3\right )\right ) \log (x)}{36 x+24 x^3+4 x^5+e^x \left (-36 x-24 x^3-4 x^5\right )+e^{2 x} \left (9 x+6 x^3+x^5\right )} \, dx=-\frac {50 \, e^{4} \log \left (x\right )}{x^{2} e^{x} - 2 \, x^{2} + 3 \, e^{x} - 6} \] Input:
integrate((((50*x^3+100*x^2+150*x)*exp(4)*exp(x)-200*x^2*exp(4))*log(x)+(- 50*x^2-150)*exp(4)*exp(x)+(100*x^2+300)*exp(4))/((x^5+6*x^3+9*x)*exp(x)^2+ (-4*x^5-24*x^3-36*x)*exp(x)+4*x^5+24*x^3+36*x),x, algorithm="giac")
Output:
-50*e^4*log(x)/(x^2*e^x - 2*x^2 + 3*e^x - 6)
Time = 3.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {e^{4+x} \left (-150-50 x^2\right )+e^4 \left (300+100 x^2\right )+\left (-200 e^4 x^2+e^{4+x} \left (150 x+100 x^2+50 x^3\right )\right ) \log (x)}{36 x+24 x^3+4 x^5+e^x \left (-36 x-24 x^3-4 x^5\right )+e^{2 x} \left (9 x+6 x^3+x^5\right )} \, dx=-\frac {50\,{\mathrm {e}}^4\,\ln \left (x\right )}{\left (x^2+3\right )\,\left ({\mathrm {e}}^x-2\right )} \] Input:
int(-(log(x)*(200*x^2*exp(4) - exp(4)*exp(x)*(150*x + 100*x^2 + 50*x^3)) - exp(4)*(100*x^2 + 300) + exp(4)*exp(x)*(50*x^2 + 150))/(36*x + 24*x^3 + 4 *x^5 + exp(2*x)*(9*x + 6*x^3 + x^5) - exp(x)*(36*x + 24*x^3 + 4*x^5)),x)
Output:
-(50*exp(4)*log(x))/((x^2 + 3)*(exp(x) - 2))
Time = 0.17 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {e^{4+x} \left (-150-50 x^2\right )+e^4 \left (300+100 x^2\right )+\left (-200 e^4 x^2+e^{4+x} \left (150 x+100 x^2+50 x^3\right )\right ) \log (x)}{36 x+24 x^3+4 x^5+e^x \left (-36 x-24 x^3-4 x^5\right )+e^{2 x} \left (9 x+6 x^3+x^5\right )} \, dx=-\frac {50 \,\mathrm {log}\left (x \right ) e^{4}}{e^{x} x^{2}+3 e^{x}-2 x^{2}-6} \] Input:
int((((50*x^3+100*x^2+150*x)*exp(4)*exp(x)-200*x^2*exp(4))*log(x)+(-50*x^2 -150)*exp(4)*exp(x)+(100*x^2+300)*exp(4))/((x^5+6*x^3+9*x)*exp(x)^2+(-4*x^ 5-24*x^3-36*x)*exp(x)+4*x^5+24*x^3+36*x),x)
Output:
( - 50*log(x)*e**4)/(e**x*x**2 + 3*e**x - 2*x**2 - 6)