Integrand size = 102, antiderivative size = 18 \[ \int \frac {e^x \left (-160 x-32 x^2\right )-512 \log (x)+\left (e^{2 x} \left (50 x^2+20 x^3+2 x^4\right )+e^x \left (160 x-160 x^2-32 x^3\right ) \log (x)\right ) \log (\log (x))+e^{2 x} \left (60 x^3+22 x^4+2 x^5\right ) \log (x) \log ^2(\log (x))}{x^3 \log (x)} \, dx=\left (-\frac {16}{x}+e^x (5+x) \log (\log (x))\right )^2 \] Output:
(ln(ln(x))*(5+x)*exp(x)-16/x)^2
Time = 0.29 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {e^x \left (-160 x-32 x^2\right )-512 \log (x)+\left (e^{2 x} \left (50 x^2+20 x^3+2 x^4\right )+e^x \left (160 x-160 x^2-32 x^3\right ) \log (x)\right ) \log (\log (x))+e^{2 x} \left (60 x^3+22 x^4+2 x^5\right ) \log (x) \log ^2(\log (x))}{x^3 \log (x)} \, dx=\frac {\left (-16+e^x x (5+x) \log (\log (x))\right )^2}{x^2} \] Input:
Integrate[(E^x*(-160*x - 32*x^2) - 512*Log[x] + (E^(2*x)*(50*x^2 + 20*x^3 + 2*x^4) + E^x*(160*x - 160*x^2 - 32*x^3)*Log[x])*Log[Log[x]] + E^(2*x)*(6 0*x^3 + 22*x^4 + 2*x^5)*Log[x]*Log[Log[x]]^2)/(x^3*Log[x]),x]
Output:
(-16 + E^x*x*(5 + x)*Log[Log[x]])^2/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.
\(\displaystyle \int \frac {e^x \left (-32 x^2-160 x\right )+e^{2 x} \left (2 x^5+22 x^4+60 x^3\right ) \log (x) \log ^2(\log (x))+\left (e^x \left (-32 x^3-160 x^2+160 x\right ) \log (x)+e^{2 x} \left (2 x^4+20 x^3+50 x^2\right )\right ) \log (\log (x))-512 \log (x)}{x^3 \log (x)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {2 \left (16-e^x x (x+5) \log (\log (x))\right ) \left (-\log (x) \left (e^x (x+6) x^2 \log (\log (x))+16\right )-e^x x (x+5)\right )}{x^3 \log (x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int -\frac {\left (16-e^x x (x+5) \log (\log (x))\right ) \left (e^x x (x+5)+\log (x) \left (e^x (x+6) \log (\log (x)) x^2+16\right )\right )}{x^3 \log (x)}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 \int \frac {\left (16-e^x x (x+5) \log (\log (x))\right ) \left (e^x x (x+5)+\log (x) \left (e^x (x+6) \log (\log (x)) x^2+16\right )\right )}{x^3 \log (x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -2 \int \left (\frac {16 e^x \left (\log (x) \log (\log (x)) x^2+5 \log (x) \log (\log (x)) x+x-5 \log (x) \log (\log (x))+5\right )}{x^2 \log (x)}-\frac {e^{2 x} (x+5) \log (\log (x)) \left (\log (x) \log (\log (x)) x^2+6 \log (x) \log (\log (x)) x+x+5\right )}{x \log (x)}+\frac {256}{x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \left (-\int e^{2 x} x^2 \log ^2(\log (x))dx-30 \int e^{2 x} \log ^2(\log (x))dx-11 \int e^{2 x} x \log ^2(\log (x))dx-10 \int \frac {e^{2 x} \log (\log (x))}{\log (x)}dx-25 \int \frac {e^{2 x} \log (\log (x))}{x \log (x)}dx-\int \frac {e^{2 x} x \log (\log (x))}{\log (x)}dx-\frac {128}{x^2}+\frac {16 e^x \left (x^2 \log (x) \log (\log (x))+5 x \log (x) \log (\log (x))\right )}{x^2 \log (x)}\right )\) |
Input:
Int[(E^x*(-160*x - 32*x^2) - 512*Log[x] + (E^(2*x)*(50*x^2 + 20*x^3 + 2*x^ 4) + E^x*(160*x - 160*x^2 - 32*x^3)*Log[x])*Log[Log[x]] + E^(2*x)*(60*x^3 + 22*x^4 + 2*x^5)*Log[x]*Log[Log[x]]^2)/(x^3*Log[x]),x]
Output:
$Aborted
Leaf count of result is larger than twice the leaf count of optimal. \(37\) vs. \(2(17)=34\).
Time = 1.11 (sec) , antiderivative size = 38, normalized size of antiderivative = 2.11
method | result | size |
risch | \(\left (x^{2}+10 x +25\right ) {\mathrm e}^{2 x} \ln \left (\ln \left (x \right )\right )^{2}-\frac {32 \left (5+x \right ) {\mathrm e}^{x} \ln \left (\ln \left (x \right )\right )}{x}+\frac {256}{x^{2}}\) | \(38\) |
parallelrisch | \(-\frac {-{\mathrm e}^{2 x} \ln \left (\ln \left (x \right )\right )^{2} x^{4}-10 \,{\mathrm e}^{2 x} \ln \left (\ln \left (x \right )\right )^{2} x^{3}-25 \,{\mathrm e}^{2 x} \ln \left (\ln \left (x \right )\right )^{2} x^{2}+32 \,{\mathrm e}^{x} \ln \left (\ln \left (x \right )\right ) x^{2}+160 \,{\mathrm e}^{x} \ln \left (\ln \left (x \right )\right ) x -256}{x^{2}}\) | \(68\) |
Input:
int(((2*x^5+22*x^4+60*x^3)*exp(x)^2*ln(x)*ln(ln(x))^2+((-32*x^3-160*x^2+16 0*x)*exp(x)*ln(x)+(2*x^4+20*x^3+50*x^2)*exp(x)^2)*ln(ln(x))-512*ln(x)+(-32 *x^2-160*x)*exp(x))/x^3/ln(x),x,method=_RETURNVERBOSE)
Output:
(x^2+10*x+25)*exp(x)^2*ln(ln(x))^2-32/x*(5+x)*exp(x)*ln(ln(x))+256/x^2
Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (17) = 34\).
Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.44 \[ \int \frac {e^x \left (-160 x-32 x^2\right )-512 \log (x)+\left (e^{2 x} \left (50 x^2+20 x^3+2 x^4\right )+e^x \left (160 x-160 x^2-32 x^3\right ) \log (x)\right ) \log (\log (x))+e^{2 x} \left (60 x^3+22 x^4+2 x^5\right ) \log (x) \log ^2(\log (x))}{x^3 \log (x)} \, dx=\frac {{\left (x^{4} + 10 \, x^{3} + 25 \, x^{2}\right )} e^{\left (2 \, x\right )} \log \left (\log \left (x\right )\right )^{2} - 32 \, {\left (x^{2} + 5 \, x\right )} e^{x} \log \left (\log \left (x\right )\right ) + 256}{x^{2}} \] Input:
integrate(((2*x^5+22*x^4+60*x^3)*exp(x)^2*log(x)*log(log(x))^2+((-32*x^3-1 60*x^2+160*x)*exp(x)*log(x)+(2*x^4+20*x^3+50*x^2)*exp(x)^2)*log(log(x))-51 2*log(x)+(-32*x^2-160*x)*exp(x))/x^3/log(x),x, algorithm="fricas")
Output:
((x^4 + 10*x^3 + 25*x^2)*e^(2*x)*log(log(x))^2 - 32*(x^2 + 5*x)*e^x*log(lo g(x)) + 256)/x^2
Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (15) = 30\).
Time = 0.21 (sec) , antiderivative size = 63, normalized size of antiderivative = 3.50 \[ \int \frac {e^x \left (-160 x-32 x^2\right )-512 \log (x)+\left (e^{2 x} \left (50 x^2+20 x^3+2 x^4\right )+e^x \left (160 x-160 x^2-32 x^3\right ) \log (x)\right ) \log (\log (x))+e^{2 x} \left (60 x^3+22 x^4+2 x^5\right ) \log (x) \log ^2(\log (x))}{x^3 \log (x)} \, dx=\frac {\left (- 32 x \log {\left (\log {\left (x \right )} \right )} - 160 \log {\left (\log {\left (x \right )} \right )}\right ) e^{x} + \left (x^{3} \log {\left (\log {\left (x \right )} \right )}^{2} + 10 x^{2} \log {\left (\log {\left (x \right )} \right )}^{2} + 25 x \log {\left (\log {\left (x \right )} \right )}^{2}\right ) e^{2 x}}{x} + \frac {256}{x^{2}} \] Input:
integrate(((2*x**5+22*x**4+60*x**3)*exp(x)**2*ln(x)*ln(ln(x))**2+((-32*x** 3-160*x**2+160*x)*exp(x)*ln(x)+(2*x**4+20*x**3+50*x**2)*exp(x)**2)*ln(ln(x ))-512*ln(x)+(-32*x**2-160*x)*exp(x))/x**3/ln(x),x)
Output:
((-32*x*log(log(x)) - 160*log(log(x)))*exp(x) + (x**3*log(log(x))**2 + 10* x**2*log(log(x))**2 + 25*x*log(log(x))**2)*exp(2*x))/x + 256/x**2
Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (17) = 34\).
Time = 0.08 (sec) , antiderivative size = 43, normalized size of antiderivative = 2.39 \[ \int \frac {e^x \left (-160 x-32 x^2\right )-512 \log (x)+\left (e^{2 x} \left (50 x^2+20 x^3+2 x^4\right )+e^x \left (160 x-160 x^2-32 x^3\right ) \log (x)\right ) \log (\log (x))+e^{2 x} \left (60 x^3+22 x^4+2 x^5\right ) \log (x) \log ^2(\log (x))}{x^3 \log (x)} \, dx=\frac {{\left (x^{3} + 10 \, x^{2} + 25 \, x\right )} e^{\left (2 \, x\right )} \log \left (\log \left (x\right )\right )^{2} - 32 \, {\left (x + 5\right )} e^{x} \log \left (\log \left (x\right )\right )}{x} + \frac {256}{x^{2}} \] Input:
integrate(((2*x^5+22*x^4+60*x^3)*exp(x)^2*log(x)*log(log(x))^2+((-32*x^3-1 60*x^2+160*x)*exp(x)*log(x)+(2*x^4+20*x^3+50*x^2)*exp(x)^2)*log(log(x))-51 2*log(x)+(-32*x^2-160*x)*exp(x))/x^3/log(x),x, algorithm="maxima")
Output:
((x^3 + 10*x^2 + 25*x)*e^(2*x)*log(log(x))^2 - 32*(x + 5)*e^x*log(log(x))) /x + 256/x^2
Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (17) = 34\).
Time = 0.16 (sec) , antiderivative size = 65, normalized size of antiderivative = 3.61 \[ \int \frac {e^x \left (-160 x-32 x^2\right )-512 \log (x)+\left (e^{2 x} \left (50 x^2+20 x^3+2 x^4\right )+e^x \left (160 x-160 x^2-32 x^3\right ) \log (x)\right ) \log (\log (x))+e^{2 x} \left (60 x^3+22 x^4+2 x^5\right ) \log (x) \log ^2(\log (x))}{x^3 \log (x)} \, dx=\frac {x^{4} e^{\left (2 \, x\right )} \log \left (\log \left (x\right )\right )^{2} + 10 \, x^{3} e^{\left (2 \, x\right )} \log \left (\log \left (x\right )\right )^{2} + 25 \, x^{2} e^{\left (2 \, x\right )} \log \left (\log \left (x\right )\right )^{2} - 32 \, x^{2} e^{x} \log \left (\log \left (x\right )\right ) - 160 \, x e^{x} \log \left (\log \left (x\right )\right ) + 256}{x^{2}} \] Input:
integrate(((2*x^5+22*x^4+60*x^3)*exp(x)^2*log(x)*log(log(x))^2+((-32*x^3-1 60*x^2+160*x)*exp(x)*log(x)+(2*x^4+20*x^3+50*x^2)*exp(x)^2)*log(log(x))-51 2*log(x)+(-32*x^2-160*x)*exp(x))/x^3/log(x),x, algorithm="giac")
Output:
(x^4*e^(2*x)*log(log(x))^2 + 10*x^3*e^(2*x)*log(log(x))^2 + 25*x^2*e^(2*x) *log(log(x))^2 - 32*x^2*e^x*log(log(x)) - 160*x*e^x*log(log(x)) + 256)/x^2
Timed out. \[ \int \frac {e^x \left (-160 x-32 x^2\right )-512 \log (x)+\left (e^{2 x} \left (50 x^2+20 x^3+2 x^4\right )+e^x \left (160 x-160 x^2-32 x^3\right ) \log (x)\right ) \log (\log (x))+e^{2 x} \left (60 x^3+22 x^4+2 x^5\right ) \log (x) \log ^2(\log (x))}{x^3 \log (x)} \, dx=-\int \frac {-{\mathrm {e}}^{2\,x}\,\ln \left (x\right )\,\left (2\,x^5+22\,x^4+60\,x^3\right )\,{\ln \left (\ln \left (x\right )\right )}^2+\left ({\mathrm {e}}^x\,\ln \left (x\right )\,\left (32\,x^3+160\,x^2-160\,x\right )-{\mathrm {e}}^{2\,x}\,\left (2\,x^4+20\,x^3+50\,x^2\right )\right )\,\ln \left (\ln \left (x\right )\right )+512\,\ln \left (x\right )+{\mathrm {e}}^x\,\left (32\,x^2+160\,x\right )}{x^3\,\ln \left (x\right )} \,d x \] Input:
int(-(512*log(x) - log(log(x))*(exp(2*x)*(50*x^2 + 20*x^3 + 2*x^4) - exp(x )*log(x)*(160*x^2 - 160*x + 32*x^3)) + exp(x)*(160*x + 32*x^2) - log(log(x ))^2*exp(2*x)*log(x)*(60*x^3 + 22*x^4 + 2*x^5))/(x^3*log(x)),x)
Output:
-int((512*log(x) - log(log(x))*(exp(2*x)*(50*x^2 + 20*x^3 + 2*x^4) - exp(x )*log(x)*(160*x^2 - 160*x + 32*x^3)) + exp(x)*(160*x + 32*x^2) - log(log(x ))^2*exp(2*x)*log(x)*(60*x^3 + 22*x^4 + 2*x^5))/(x^3*log(x)), x)
Time = 0.20 (sec) , antiderivative size = 70, normalized size of antiderivative = 3.89 \[ \int \frac {e^x \left (-160 x-32 x^2\right )-512 \log (x)+\left (e^{2 x} \left (50 x^2+20 x^3+2 x^4\right )+e^x \left (160 x-160 x^2-32 x^3\right ) \log (x)\right ) \log (\log (x))+e^{2 x} \left (60 x^3+22 x^4+2 x^5\right ) \log (x) \log ^2(\log (x))}{x^3 \log (x)} \, dx=\frac {e^{2 x} \mathrm {log}\left (\mathrm {log}\left (x \right )\right )^{2} x^{4}+10 e^{2 x} \mathrm {log}\left (\mathrm {log}\left (x \right )\right )^{2} x^{3}+25 e^{2 x} \mathrm {log}\left (\mathrm {log}\left (x \right )\right )^{2} x^{2}-32 e^{x} \mathrm {log}\left (\mathrm {log}\left (x \right )\right ) x^{2}-160 e^{x} \mathrm {log}\left (\mathrm {log}\left (x \right )\right ) x +256}{x^{2}} \] Input:
int(((2*x^5+22*x^4+60*x^3)*exp(x)^2*log(x)*log(log(x))^2+((-32*x^3-160*x^2 +160*x)*exp(x)*log(x)+(2*x^4+20*x^3+50*x^2)*exp(x)^2)*log(log(x))-512*log( x)+(-32*x^2-160*x)*exp(x))/x^3/log(x),x)
Output:
(e**(2*x)*log(log(x))**2*x**4 + 10*e**(2*x)*log(log(x))**2*x**3 + 25*e**(2 *x)*log(log(x))**2*x**2 - 32*e**x*log(log(x))*x**2 - 160*e**x*log(log(x))* x + 256)/x**2