Integrand size = 79, antiderivative size = 24 \[ \int \frac {\left (15 e^5-15 x\right ) \log ^2(x)+\left (\left (30 e^5-30 x\right ) \log (x)+\left (-15 e^5+45 x\right ) \log ^2(x)\right ) \log (2 x)}{e^{15} x^2-3 e^{10} x^3+3 e^5 x^4-x^5} \, dx=-5+\frac {15 \log ^2(x) \log (2 x)}{\left (e^5-x\right )^2 x} \] Output:
15/x*ln(x)^2*ln(2*x)/(exp(5)-x)^2-5
Time = 5.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {\left (15 e^5-15 x\right ) \log ^2(x)+\left (\left (30 e^5-30 x\right ) \log (x)+\left (-15 e^5+45 x\right ) \log ^2(x)\right ) \log (2 x)}{e^{15} x^2-3 e^{10} x^3+3 e^5 x^4-x^5} \, dx=\frac {15 \log ^2(x) \log (2 x)}{\left (e^5-x\right )^2 x} \] Input:
Integrate[((15*E^5 - 15*x)*Log[x]^2 + ((30*E^5 - 30*x)*Log[x] + (-15*E^5 + 45*x)*Log[x]^2)*Log[2*x])/(E^15*x^2 - 3*E^10*x^3 + 3*E^5*x^4 - x^5),x]
Output:
(15*Log[x]^2*Log[2*x])/((E^5 - 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.
\(\displaystyle \int \frac {\left (15 e^5-15 x\right ) \log ^2(x)+\left (\left (45 x-15 e^5\right ) \log ^2(x)+\left (30 e^5-30 x\right ) \log (x)\right ) \log (2 x)}{-x^5+3 e^5 x^4-3 e^{10} x^3+e^{15} x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (15 e^5-15 x\right ) \log ^2(x)+\left (\left (45 x-15 e^5\right ) \log ^2(x)+\left (30 e^5-30 x\right ) \log (x)\right ) \log (2 x)}{x^2 \left (-x^3+3 e^5 x^2-3 e^{10} x+e^{15}\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {\left (15 e^5-15 x\right ) \log ^2(x)+\left (\left (45 x-15 e^5\right ) \log ^2(x)+\left (30 e^5-30 x\right ) \log (x)\right ) \log (2 x)}{\left (e^5-x\right )^3 x^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {15 \log ^2(x)}{\left (e^5-x\right )^2 x^2}-\frac {15 \log (x) \left (2 x-3 x \log (x)+e^5 \log (x)-2 e^5\right ) \log (2 x)}{\left (e^5-x\right )^3 x^2}\right )dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \int \left (\frac {15 \log ^2(x)}{\left (e^5-x\right )^2 x^2}-\frac {15 \log (x) \left (2 x-3 x \log (x)+e^5 \log (x)-2 e^5\right ) \log (2 x)}{\left (e^5-x\right )^3 x^2}\right )dx\) |
Input:
Int[((15*E^5 - 15*x)*Log[x]^2 + ((30*E^5 - 30*x)*Log[x] + (-15*E^5 + 45*x) *Log[x]^2)*Log[2*x])/(E^15*x^2 - 3*E^10*x^3 + 3*E^5*x^4 - x^5),x]
Output:
$Aborted
Time = 5.17 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21
method | result | size |
parallelrisch | \(\frac {15 \ln \left (2 x \right ) \ln \left (x \right )^{2}}{x \left ({\mathrm e}^{10}-2 x \,{\mathrm e}^{5}+x^{2}\right )}\) | \(29\) |
risch | \(\frac {15 \ln \left (x \right )^{3}}{x \left ({\mathrm e}^{10}-2 x \,{\mathrm e}^{5}+x^{2}\right )}+\frac {15 \ln \left (2\right ) \ln \left (x \right )^{2}}{x \left ({\mathrm e}^{10}-2 x \,{\mathrm e}^{5}+x^{2}\right )}\) | \(48\) |
Input:
int((((-15*exp(5)+45*x)*ln(x)^2+(30*exp(5)-30*x)*ln(x))*ln(2*x)+(15*exp(5) -15*x)*ln(x)^2)/(x^2*exp(5)^3-3*x^3*exp(5)^2+3*x^4*exp(5)-x^5),x,method=_R ETURNVERBOSE)
Output:
15/x*ln(2*x)*ln(x)^2/(exp(5)^2-2*x*exp(5)+x^2)
Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {\left (15 e^5-15 x\right ) \log ^2(x)+\left (\left (30 e^5-30 x\right ) \log (x)+\left (-15 e^5+45 x\right ) \log ^2(x)\right ) \log (2 x)}{e^{15} x^2-3 e^{10} x^3+3 e^5 x^4-x^5} \, dx=\frac {15 \, {\left (\log \left (2\right ) \log \left (x\right )^{2} + \log \left (x\right )^{3}\right )}}{x^{3} - 2 \, x^{2} e^{5} + x e^{10}} \] Input:
integrate((((-15*exp(5)+45*x)*log(x)^2+(30*exp(5)-30*x)*log(x))*log(2*x)+( 15*exp(5)-15*x)*log(x)^2)/(x^2*exp(5)^3-3*x^3*exp(5)^2+3*x^4*exp(5)-x^5),x , algorithm="fricas")
Output:
15*(log(2)*log(x)^2 + log(x)^3)/(x^3 - 2*x^2*e^5 + x*e^10)
Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (20) = 40\).
Time = 0.17 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.04 \[ \int \frac {\left (15 e^5-15 x\right ) \log ^2(x)+\left (\left (30 e^5-30 x\right ) \log (x)+\left (-15 e^5+45 x\right ) \log ^2(x)\right ) \log (2 x)}{e^{15} x^2-3 e^{10} x^3+3 e^5 x^4-x^5} \, dx=\frac {15 \log {\left (x \right )}^{3}}{x^{3} - 2 x^{2} e^{5} + x e^{10}} + \frac {15 \log {\left (2 \right )} \log {\left (x \right )}^{2}}{x^{3} - 2 x^{2} e^{5} + x e^{10}} \] Input:
integrate((((-15*exp(5)+45*x)*ln(x)**2+(30*exp(5)-30*x)*ln(x))*ln(2*x)+(15 *exp(5)-15*x)*ln(x)**2)/(x**2*exp(5)**3-3*x**3*exp(5)**2+3*x**4*exp(5)-x** 5),x)
Output:
15*log(x)**3/(x**3 - 2*x**2*exp(5) + x*exp(10)) + 15*log(2)*log(x)**2/(x** 3 - 2*x**2*exp(5) + x*exp(10))
Time = 0.20 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {\left (15 e^5-15 x\right ) \log ^2(x)+\left (\left (30 e^5-30 x\right ) \log (x)+\left (-15 e^5+45 x\right ) \log ^2(x)\right ) \log (2 x)}{e^{15} x^2-3 e^{10} x^3+3 e^5 x^4-x^5} \, dx=\frac {15 \, {\left (\log \left (2\right ) \log \left (x\right )^{2} + \log \left (x\right )^{3}\right )}}{x^{3} - 2 \, x^{2} e^{5} + x e^{10}} \] Input:
integrate((((-15*exp(5)+45*x)*log(x)^2+(30*exp(5)-30*x)*log(x))*log(2*x)+( 15*exp(5)-15*x)*log(x)^2)/(x^2*exp(5)^3-3*x^3*exp(5)^2+3*x^4*exp(5)-x^5),x , algorithm="maxima")
Output:
15*(log(2)*log(x)^2 + log(x)^3)/(x^3 - 2*x^2*e^5 + x*e^10)
Time = 0.12 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {\left (15 e^5-15 x\right ) \log ^2(x)+\left (\left (30 e^5-30 x\right ) \log (x)+\left (-15 e^5+45 x\right ) \log ^2(x)\right ) \log (2 x)}{e^{15} x^2-3 e^{10} x^3+3 e^5 x^4-x^5} \, dx=\frac {15 \, {\left (\log \left (2\right ) \log \left (x\right )^{2} + \log \left (x\right )^{3}\right )}}{x^{3} - 2 \, x^{2} e^{5} + x e^{10}} \] Input:
integrate((((-15*exp(5)+45*x)*log(x)^2+(30*exp(5)-30*x)*log(x))*log(2*x)+( 15*exp(5)-15*x)*log(x)^2)/(x^2*exp(5)^3-3*x^3*exp(5)^2+3*x^4*exp(5)-x^5),x , algorithm="giac")
Output:
15*(log(2)*log(x)^2 + log(x)^3)/(x^3 - 2*x^2*e^5 + x*e^10)
Time = 3.21 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {\left (15 e^5-15 x\right ) \log ^2(x)+\left (\left (30 e^5-30 x\right ) \log (x)+\left (-15 e^5+45 x\right ) \log ^2(x)\right ) \log (2 x)}{e^{15} x^2-3 e^{10} x^3+3 e^5 x^4-x^5} \, dx=\frac {15\,{\ln \left (x\right )}^2\,\left (\ln \left (2\right )+\ln \left (x\right )\right )}{x\,{\left (x-{\mathrm {e}}^5\right )}^2} \] Input:
int(-(log(x)^2*(15*x - 15*exp(5)) + log(2*x)*(log(x)*(30*x - 30*exp(5)) - log(x)^2*(45*x - 15*exp(5))))/(3*x^4*exp(5) - 3*x^3*exp(10) + x^2*exp(15) - x^5),x)
Output:
(15*log(x)^2*(log(2) + log(x)))/(x*(x - exp(5))^2)
Time = 0.21 (sec) , antiderivative size = 93, normalized size of antiderivative = 3.88 \[ \int \frac {\left (15 e^5-15 x\right ) \log ^2(x)+\left (\left (30 e^5-30 x\right ) \log (x)+\left (-15 e^5+45 x\right ) \log ^2(x)\right ) \log (2 x)}{e^{15} x^2-3 e^{10} x^3+3 e^5 x^4-x^5} \, dx=\frac {15 \,\mathrm {log}\left (2 x \right ) \mathrm {log}\left (x \right )^{2} e^{15}-\frac {20 \,\mathrm {log}\left (2 x \right ) e^{10} x}{3}+\frac {40 \,\mathrm {log}\left (2 x \right ) e^{5} x^{2}}{3}-\frac {20 \,\mathrm {log}\left (2 x \right ) x^{3}}{3}+\frac {20 \,\mathrm {log}\left (x \right ) e^{10} x}{3}-\frac {40 \,\mathrm {log}\left (x \right ) e^{5} x^{2}}{3}+\frac {20 \,\mathrm {log}\left (x \right ) x^{3}}{3}}{e^{15} x \left (e^{10}-2 e^{5} x +x^{2}\right )} \] Input:
int((((-15*exp(5)+45*x)*log(x)^2+(30*exp(5)-30*x)*log(x))*log(2*x)+(15*exp (5)-15*x)*log(x)^2)/(x^2*exp(5)^3-3*x^3*exp(5)^2+3*x^4*exp(5)-x^5),x)
Output:
(5*(9*log(2*x)*log(x)**2*e**15 - 4*log(2*x)*e**10*x + 8*log(2*x)*e**5*x**2 - 4*log(2*x)*x**3 + 4*log(x)*e**10*x - 8*log(x)*e**5*x**2 + 4*log(x)*x**3 ))/(3*e**15*x*(e**10 - 2*e**5*x + x**2))