Integrand size = 104, antiderivative size = 23 \[ \int \frac {100 x \log (x)+100 x \log ^2(x)+\left (2 x^4+2 e^x x^4\right ) \log ^4(x)}{625+\left (250 x^2-50 e^x x^2-50 x^3\right ) \log ^2(x)+\left (25 x^4+e^{2 x} x^4-10 x^5+x^6+e^x \left (-10 x^4+2 x^5\right )\right ) \log ^4(x)} \, dx=\frac {2}{5-e^x-x+\frac {25}{x^2 \log ^2(x)}} \]
Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {100 x \log (x)+100 x \log ^2(x)+\left (2 x^4+2 e^x x^4\right ) \log ^4(x)}{625+\left (250 x^2-50 e^x x^2-50 x^3\right ) \log ^2(x)+\left (25 x^4+e^{2 x} x^4-10 x^5+x^6+e^x \left (-10 x^4+2 x^5\right )\right ) \log ^4(x)} \, dx=-\frac {2 x^2 \log ^2(x)}{-25+x^2 \left (-5+e^x+x\right ) \log ^2(x)} \]
Integrate[(100*x*Log[x] + 100*x*Log[x]^2 + (2*x^4 + 2*E^x*x^4)*Log[x]^4)/( 625 + (250*x^2 - 50*E^x*x^2 - 50*x^3)*Log[x]^2 + (25*x^4 + E^(2*x)*x^4 - 1 0*x^5 + x^6 + E^x*(-10*x^4 + 2*x^5))*Log[x]^4),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.
\(\displaystyle \int \frac {\left (2 e^x x^4+2 x^4\right ) \log ^4(x)+100 x \log ^2(x)+100 x \log (x)}{\left (-50 x^3-50 e^x x^2+250 x^2\right ) \log ^2(x)+\left (x^6-10 x^5+e^{2 x} x^4+25 x^4+e^x \left (2 x^5-10 x^4\right )\right ) \log ^4(x)+625} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {2 x \log (x) \left (\left (e^x+1\right ) x^3 \log ^3(x)+50 \log (x)+50\right )}{\left (25-x^2 \left (x+e^x-5\right ) \log ^2(x)\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int \frac {x \log (x) \left (\left (1+e^x\right ) x^3 \log ^3(x)+50 \log (x)+50\right )}{\left (\left (-x-e^x+5\right ) x^2 \log ^2(x)+25\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 2 \int \left (\frac {x^2 \log ^2(x)}{\log ^2(x) x^3+e^x \log ^2(x) x^2-5 \log ^2(x) x^2-25}-\frac {x \log (x) \left (\log ^3(x) x^4-6 \log ^3(x) x^3-25 \log (x) x-50 \log (x)-50\right )}{\left (\log ^2(x) x^3+e^x \log ^2(x) x^2-5 \log ^2(x) x^2-25\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (50 \int \frac {x \log (x)}{\left (25-x^2 \left (x+e^x-5\right ) \log ^2(x)\right )^2}dx+50 \int \frac {x \log ^2(x)}{\left (25-x^2 \left (x+e^x-5\right ) \log ^2(x)\right )^2}dx+25 \int \frac {x^2 \log ^2(x)}{\left (25-x^2 \left (x+e^x-5\right ) \log ^2(x)\right )^2}dx+\int \frac {x^2 \log ^2(x)}{x^2 \left (x+e^x-5\right ) \log ^2(x)-25}dx-\int \frac {x^5 \log ^4(x)}{\left (25-x^2 \left (x+e^x-5\right ) \log ^2(x)\right )^2}dx+6 \int \frac {x^4 \log ^4(x)}{\left (25-x^2 \left (x+e^x-5\right ) \log ^2(x)\right )^2}dx\right )\) |
Int[(100*x*Log[x] + 100*x*Log[x]^2 + (2*x^4 + 2*E^x*x^4)*Log[x]^4)/(625 + (250*x^2 - 50*E^x*x^2 - 50*x^3)*Log[x]^2 + (25*x^4 + E^(2*x)*x^4 - 10*x^5 + x^6 + E^x*(-10*x^4 + 2*x^5))*Log[x]^4),x]
3.2.7.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 1.33 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.78
method | result | size |
parallelrisch | \(-\frac {2 \ln \left (x \right )^{2} x^{2}}{x^{2} {\mathrm e}^{x} \ln \left (x \right )^{2}+x^{3} \ln \left (x \right )^{2}-5 x^{2} \ln \left (x \right )^{2}-25}\) | \(41\) |
risch | \(-\frac {2}{{\mathrm e}^{x}+x -5}-\frac {50}{\left ({\mathrm e}^{x}+x -5\right ) \left (x^{2} {\mathrm e}^{x} \ln \left (x \right )^{2}+x^{3} \ln \left (x \right )^{2}-5 x^{2} \ln \left (x \right )^{2}-25\right )}\) | \(51\) |
int(((2*exp(x)*x^4+2*x^4)*ln(x)^4+100*x*ln(x)^2+100*x*ln(x))/((exp(x)^2*x^ 4+(2*x^5-10*x^4)*exp(x)+x^6-10*x^5+25*x^4)*ln(x)^4+(-50*exp(x)*x^2-50*x^3+ 250*x^2)*ln(x)^2+625),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43 \[ \int \frac {100 x \log (x)+100 x \log ^2(x)+\left (2 x^4+2 e^x x^4\right ) \log ^4(x)}{625+\left (250 x^2-50 e^x x^2-50 x^3\right ) \log ^2(x)+\left (25 x^4+e^{2 x} x^4-10 x^5+x^6+e^x \left (-10 x^4+2 x^5\right )\right ) \log ^4(x)} \, dx=-\frac {2 \, x^{2} \log \left (x\right )^{2}}{{\left (x^{3} + x^{2} e^{x} - 5 \, x^{2}\right )} \log \left (x\right )^{2} - 25} \]
integrate(((2*exp(x)*x^4+2*x^4)*log(x)^4+100*x*log(x)^2+100*x*log(x))/((ex p(x)^2*x^4+(2*x^5-10*x^4)*exp(x)+x^6-10*x^5+25*x^4)*log(x)^4+(-50*exp(x)*x ^2-50*x^3+250*x^2)*log(x)^2+625),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (17) = 34\).
Time = 0.14 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.83 \[ \int \frac {100 x \log (x)+100 x \log ^2(x)+\left (2 x^4+2 e^x x^4\right ) \log ^4(x)}{625+\left (250 x^2-50 e^x x^2-50 x^3\right ) \log ^2(x)+\left (25 x^4+e^{2 x} x^4-10 x^5+x^6+e^x \left (-10 x^4+2 x^5\right )\right ) \log ^4(x)} \, dx=- \frac {2 x^{2} \log {\left (x \right )}^{2}}{x^{3} \log {\left (x \right )}^{2} + x^{2} e^{x} \log {\left (x \right )}^{2} - 5 x^{2} \log {\left (x \right )}^{2} - 25} \]
integrate(((2*exp(x)*x**4+2*x**4)*ln(x)**4+100*x*ln(x)**2+100*x*ln(x))/((e xp(x)**2*x**4+(2*x**5-10*x**4)*exp(x)+x**6-10*x**5+25*x**4)*ln(x)**4+(-50* exp(x)*x**2-50*x**3+250*x**2)*ln(x)**2+625),x)
Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (18) = 36\).
Time = 0.24 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \[ \int \frac {100 x \log (x)+100 x \log ^2(x)+\left (2 x^4+2 e^x x^4\right ) \log ^4(x)}{625+\left (250 x^2-50 e^x x^2-50 x^3\right ) \log ^2(x)+\left (25 x^4+e^{2 x} x^4-10 x^5+x^6+e^x \left (-10 x^4+2 x^5\right )\right ) \log ^4(x)} \, dx=-\frac {2 \, x^{2} \log \left (x\right )^{2}}{x^{2} e^{x} \log \left (x\right )^{2} + {\left (x^{3} - 5 \, x^{2}\right )} \log \left (x\right )^{2} - 25} \]
integrate(((2*exp(x)*x^4+2*x^4)*log(x)^4+100*x*log(x)^2+100*x*log(x))/((ex p(x)^2*x^4+(2*x^5-10*x^4)*exp(x)+x^6-10*x^5+25*x^4)*log(x)^4+(-50*exp(x)*x ^2-50*x^3+250*x^2)*log(x)^2+625),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 843 vs. \(2 (18) = 36\).
Time = 234.92 (sec) , antiderivative size = 843, normalized size of antiderivative = 36.65 \[ \int \frac {100 x \log (x)+100 x \log ^2(x)+\left (2 x^4+2 e^x x^4\right ) \log ^4(x)}{625+\left (250 x^2-50 e^x x^2-50 x^3\right ) \log ^2(x)+\left (25 x^4+e^{2 x} x^4-10 x^5+x^6+e^x \left (-10 x^4+2 x^5\right )\right ) \log ^4(x)} \, dx=\text {Too large to display} \]
integrate(((2*exp(x)*x^4+2*x^4)*log(x)^4+100*x*log(x)^2+100*x*log(x))/((ex p(x)^2*x^4+(2*x^5-10*x^4)*exp(x)+x^6-10*x^5+25*x^4)*log(x)^4+(-50*exp(x)*x ^2-50*x^3+250*x^2)*log(x)^2+625),x, algorithm=\
-2*(4*x^8*log(x)^2 + 16*x^7*e^x*log(x)^2 - 80*x^7*log(x)^2 + 24*x^6*e^(2*x )*log(x)^2 - 240*x^6*e^x*log(x)^2 + 600*x^6*log(x)^2 + 16*x^5*e^(3*x)*log( x)^2 - 265*x^5*e^(2*x)*log(x)^2 + 1050*x^5*e^x*log(x)^2 + 50*x^5*e^x*log(x ) - 2225*x^5*log(x)^2 + 4*x^4*e^(4*x)*log(x)^2 - 105*x^4*e^(3*x)*log(x)^2 + 475*x^4*e^(2*x)*log(x)^2 - 1275*x^4*e^x*log(x)^2 + 100*x^4*e^(2*x)*log(x ) - 500*x^4*e^x*log(x) + 5125*x^4*log(x)^2 - 100*x^3*e^(3*x)*log(x)^2 + 60 0*x^3*e^(2*x)*log(x)^2 + 1500*x^3*e^x*log(x)^2 + 50*x^3*e^(3*x)*log(x) - 5 00*x^3*e^(2*x)*log(x) + 1250*x^3*e^x*log(x) - 10000*x^3*log(x)^2 - 100*x^2 *e^(3*x)*log(x)^2 + 1500*x^2*e^(2*x)*log(x)^2 - 7500*x^2*e^x*log(x)^2 + 12 500*x^2*log(x)^2 - 625*x^2*e^(2*x) - 1875*x^2*e^x - 1250*x*e^(2*x) + 6250* x*e^x)/(4*x^9*log(x)^2 + 20*x^8*e^x*log(x)^2 - 100*x^8*log(x)^2 + 40*x^7*e ^(2*x)*log(x)^2 - 400*x^7*e^x*log(x)^2 + 1000*x^7*log(x)^2 + 40*x^6*e^(3*x )*log(x)^2 - 625*x^6*e^(2*x)*log(x)^2 + 2850*x^6*e^x*log(x)^2 - 5225*x^6*l og(x)^2 + 20*x^5*e^(4*x)*log(x)^2 - 450*x^5*e^(3*x)*log(x)^2 + 2850*x^5*e^ (2*x)*log(x)^2 - 8750*x^5*e^x*log(x)^2 + 16250*x^5*log(x)^2 + 4*x^4*e^(5*x )*log(x)^2 - 125*x^4*e^(4*x)*log(x)^2 + 900*x^4*e^(3*x)*log(x)^2 - 3050*x^ 4*e^(2*x)*log(x)^2 + 13000*x^4*e^x*log(x)^2 - 100*x^6 - 400*x^5*e^x - 3562 5*x^4*log(x)^2 - 100*x^3*e^(4*x)*log(x)^2 + 1000*x^3*e^(3*x)*log(x)^2 - 25 000*x^3*e^x*log(x)^2 + 2000*x^5 - 600*x^4*e^(2*x) + 6000*x^4*e^x + 62500*x ^3*log(x)^2 - 100*x^2*e^(4*x)*log(x)^2 + 2000*x^2*e^(3*x)*log(x)^2 - 15...
Timed out. \[ \int \frac {100 x \log (x)+100 x \log ^2(x)+\left (2 x^4+2 e^x x^4\right ) \log ^4(x)}{625+\left (250 x^2-50 e^x x^2-50 x^3\right ) \log ^2(x)+\left (25 x^4+e^{2 x} x^4-10 x^5+x^6+e^x \left (-10 x^4+2 x^5\right )\right ) \log ^4(x)} \, dx=\int \frac {\left (2\,x^4\,{\mathrm {e}}^x+2\,x^4\right )\,{\ln \left (x\right )}^4+100\,x\,{\ln \left (x\right )}^2+100\,x\,\ln \left (x\right )}{\left (x^4\,{\mathrm {e}}^{2\,x}-{\mathrm {e}}^x\,\left (10\,x^4-2\,x^5\right )+25\,x^4-10\,x^5+x^6\right )\,{\ln \left (x\right )}^4+\left (250\,x^2-50\,x^2\,{\mathrm {e}}^x-50\,x^3\right )\,{\ln \left (x\right )}^2+625} \,d x \]
int((100*x*log(x)^2 + log(x)^4*(2*x^4*exp(x) + 2*x^4) + 100*x*log(x))/(log (x)^4*(x^4*exp(2*x) - exp(x)*(10*x^4 - 2*x^5) + 25*x^4 - 10*x^5 + x^6) - l og(x)^2*(50*x^2*exp(x) - 250*x^2 + 50*x^3) + 625),x)