Integrand size = 106, antiderivative size = 26 \[ \int \frac {\left (160+160 x+40 x^2+e^x \left (-8 x^2-4 x^3\right )\right ) \log (2 x) \log \left (x^2\right )+\left (40+40 x+10 x^2+e^x \left (-2 x^2-x^3\right )+e^x \left (-4 x^2-3 x^3-x^4\right ) \log (2 x)\right ) \log ^2\left (x^2\right )}{4 x+4 x^2+x^3} \, dx=\left (10-\frac {e^x x^2}{2+x}\right ) \log (2 x) \log ^2\left (x^2\right ) \]
Time = 0.53 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {\left (160+160 x+40 x^2+e^x \left (-8 x^2-4 x^3\right )\right ) \log (2 x) \log \left (x^2\right )+\left (40+40 x+10 x^2+e^x \left (-2 x^2-x^3\right )+e^x \left (-4 x^2-3 x^3-x^4\right ) \log (2 x)\right ) \log ^2\left (x^2\right )}{4 x+4 x^2+x^3} \, dx=-\frac {\left (-20-10 x+e^x x^2\right ) \log (2 x) \log ^2\left (x^2\right )}{2+x} \]
Integrate[((160 + 160*x + 40*x^2 + E^x*(-8*x^2 - 4*x^3))*Log[2*x]*Log[x^2] + (40 + 40*x + 10*x^2 + E^x*(-2*x^2 - x^3) + E^x*(-4*x^2 - 3*x^3 - x^4)*L og[2*x])*Log[x^2]^2)/(4*x + 4*x^2 + x^3),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.
\(\displaystyle \int \frac {\left (40 x^2+e^x \left (-4 x^3-8 x^2\right )+160 x+160\right ) \log (2 x) \log \left (x^2\right )+\left (10 x^2+e^x \left (-x^3-2 x^2\right )+e^x \left (-x^4-3 x^3-4 x^2\right ) \log (2 x)+40 x+40\right ) \log ^2\left (x^2\right )}{x^3+4 x^2+4 x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (40 x^2+e^x \left (-4 x^3-8 x^2\right )+160 x+160\right ) \log (2 x) \log \left (x^2\right )+\left (10 x^2+e^x \left (-x^3-2 x^2\right )+e^x \left (-x^4-3 x^3-4 x^2\right ) \log (2 x)+40 x+40\right ) \log ^2\left (x^2\right )}{x \left (x^2+4 x+4\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {\left (40 x^2+e^x \left (-4 x^3-8 x^2\right )+160 x+160\right ) \log (2 x) \log \left (x^2\right )+\left (10 x^2+e^x \left (-x^3-2 x^2\right )+e^x \left (-x^4-3 x^3-4 x^2\right ) \log (2 x)+40 x+40\right ) \log ^2\left (x^2\right )}{x (x+2)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {10 x \log ^2\left (x^2\right )}{(x+2)^2}+\frac {40 \log ^2\left (x^2\right )}{x (x+2)^2}+\frac {40 \log ^2\left (x^2\right )}{(x+2)^2}+\frac {40 x \log (2 x) \log \left (x^2\right )}{(x+2)^2}+\frac {160 \log (2 x) \log \left (x^2\right )}{x (x+2)^2}+\frac {160 \log (2 x) \log \left (x^2\right )}{(x+2)^2}-\frac {e^x x \left (x^2 \log (2 x) \log \left (x^2\right )+3 x \log (2 x) \log \left (x^2\right )+x \log \left (x^2\right )+4 \log (2 x) \log \left (x^2\right )+2 \log \left (x^2\right )+4 x \log (2 x)+8 \log (2 x)\right ) \log \left (x^2\right )}{(x+2)^2}\right )dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {\log \left (x^2\right ) \left (-\left ((x+2) \left (e^x x^2-10 x-20\right ) \log \left (x^2\right )\right )-\log (2 x) \left (4 (x+2) \left (e^x x^2-10 x-20\right )+e^x \left (x^2+3 x+4\right ) x^2 \log \left (x^2\right )\right )\right )}{x (x+2)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {10 \log \left (x^2\right ) \left (\log \left (x^2\right )+4 \log (2 x)\right )}{x}-\frac {e^x x \log \left (x^2\right ) \left (x^2 \log (2 x) \log \left (x^2\right )+3 x \log (2 x) \log \left (x^2\right )+x \log \left (x^2\right )+4 \log (2 x) \log \left (x^2\right )+2 \log \left (x^2\right )+4 x \log (2 x)+8 \log (2 x)\right )}{(x+2)^2}\right )dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \int \left (\frac {10 \log \left (x^2\right ) \left (\log \left (x^2\right )+4 \log (2 x)\right )}{x}-\frac {e^x x \log \left (x^2\right ) \left (x^2 \log (2 x) \log \left (x^2\right )+3 x \log (2 x) \log \left (x^2\right )+x \log \left (x^2\right )+4 \log (2 x) \log \left (x^2\right )+2 \log \left (x^2\right )+4 x \log (2 x)+8 \log (2 x)\right )}{(x+2)^2}\right )dx\) |
Int[((160 + 160*x + 40*x^2 + E^x*(-8*x^2 - 4*x^3))*Log[2*x]*Log[x^2] + (40 + 40*x + 10*x^2 + E^x*(-2*x^2 - x^3) + E^x*(-4*x^2 - 3*x^3 - x^4)*Log[2*x ])*Log[x^2]^2)/(4*x + 4*x^2 + x^3),x]
3.13.37.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Leaf count of result is larger than twice the leaf count of optimal. \(84\) vs. \(2(25)=50\).
Time = 1.72 (sec) , antiderivative size = 85, normalized size of antiderivative = 3.27
method | result | size |
parallelrisch | \(-\frac {-268800 \ln \left (2 x \right )^{2} x \ln \left (x^{2}\right )-22400 x \ln \left (x^{2}\right )^{3}+13440 \ln \left (x^{2}\right )^{2} {\mathrm e}^{x} \ln \left (2 x \right ) x^{2}+179200 x \ln \left (2 x \right )^{3}-44800 \ln \left (x^{2}\right )^{3}+358400 \ln \left (2 x \right )^{3}-537600 \ln \left (x^{2}\right ) \ln \left (2 x \right )^{2}}{13440 \left (2+x \right )}\) | \(85\) |
risch | \(\text {Expression too large to display}\) | \(806\) |
int((((-x^4-3*x^3-4*x^2)*exp(x)*ln(2*x)+(-x^3-2*x^2)*exp(x)+10*x^2+40*x+40 )*ln(x^2)^2+((-4*x^3-8*x^2)*exp(x)+40*x^2+160*x+160)*ln(2*x)*ln(x^2))/(x^3 +4*x^2+4*x),x,method=_RETURNVERBOSE)
-1/13440*(-268800*ln(2*x)^2*x*ln(x^2)-22400*x*ln(x^2)^3+13440*ln(x^2)^2*ex p(x)*ln(2*x)*x^2+179200*x*ln(2*x)^3-44800*ln(x^2)^3+358400*ln(2*x)^3-53760 0*ln(x^2)*ln(2*x)^2)/(2+x)
Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (25) = 50\).
Time = 0.27 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.88 \[ \int \frac {\left (160+160 x+40 x^2+e^x \left (-8 x^2-4 x^3\right )\right ) \log (2 x) \log \left (x^2\right )+\left (40+40 x+10 x^2+e^x \left (-2 x^2-x^3\right )+e^x \left (-4 x^2-3 x^3-x^4\right ) \log (2 x)\right ) \log ^2\left (x^2\right )}{4 x+4 x^2+x^3} \, dx=-\frac {4 \, {\left ({\left (x^{2} e^{x} - 10 \, x - 20\right )} \log \left (2 \, x\right )^{3} - 2 \, {\left (x^{2} e^{x} \log \left (2\right ) - 10 \, {\left (x + 2\right )} \log \left (2\right )\right )} \log \left (2 \, x\right )^{2} + {\left (x^{2} e^{x} \log \left (2\right )^{2} - 10 \, {\left (x + 2\right )} \log \left (2\right )^{2}\right )} \log \left (2 \, x\right )\right )}}{x + 2} \]
integrate((((-x^4-3*x^3-4*x^2)*exp(x)*log(2*x)+(-x^3-2*x^2)*exp(x)+10*x^2+ 40*x+40)*log(x^2)^2+((-4*x^3-8*x^2)*exp(x)+40*x^2+160*x+160)*log(2*x)*log( x^2))/(x^3+4*x^2+4*x),x, algorithm=\
-4*((x^2*e^x - 10*x - 20)*log(2*x)^3 - 2*(x^2*e^x*log(2) - 10*(x + 2)*log( 2))*log(2*x)^2 + (x^2*e^x*log(2)^2 - 10*(x + 2)*log(2)^2)*log(2*x))/(x + 2 )
Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (22) = 44\).
Time = 0.21 (sec) , antiderivative size = 78, normalized size of antiderivative = 3.00 \[ \int \frac {\left (160+160 x+40 x^2+e^x \left (-8 x^2-4 x^3\right )\right ) \log (2 x) \log \left (x^2\right )+\left (40+40 x+10 x^2+e^x \left (-2 x^2-x^3\right )+e^x \left (-4 x^2-3 x^3-x^4\right ) \log (2 x)\right ) \log ^2\left (x^2\right )}{4 x+4 x^2+x^3} \, dx=40 \log {\left (2 \right )}^{2} \log {\left (x \right )} + 40 \log {\left (2 x \right )}^{3} - 80 \log {\left (2 \right )} \log {\left (2 x \right )}^{2} + \frac {\left (- 4 x^{2} \log {\left (2 x \right )}^{3} + 8 x^{2} \log {\left (2 \right )} \log {\left (2 x \right )}^{2} - 4 x^{2} \log {\left (2 \right )}^{2} \log {\left (2 x \right )}\right ) e^{x}}{x + 2} \]
integrate((((-x**4-3*x**3-4*x**2)*exp(x)*ln(2*x)+(-x**3-2*x**2)*exp(x)+10* x**2+40*x+40)*ln(x**2)**2+((-4*x**3-8*x**2)*exp(x)+40*x**2+160*x+160)*ln(2 *x)*ln(x**2))/(x**3+4*x**2+4*x),x)
40*log(2)**2*log(x) + 40*log(2*x)**3 - 80*log(2)*log(2*x)**2 + (-4*x**2*lo g(2*x)**3 + 8*x**2*log(2)*log(2*x)**2 - 4*x**2*log(2)**2*log(2*x))*exp(x)/ (x + 2)
Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (25) = 50\).
Time = 0.31 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.12 \[ \int \frac {\left (160+160 x+40 x^2+e^x \left (-8 x^2-4 x^3\right )\right ) \log (2 x) \log \left (x^2\right )+\left (40+40 x+10 x^2+e^x \left (-2 x^2-x^3\right )+e^x \left (-4 x^2-3 x^3-x^4\right ) \log (2 x)\right ) \log ^2\left (x^2\right )}{4 x+4 x^2+x^3} \, dx=\frac {4 \, {\left (10 \, {\left (x + 2\right )} \log \left (x\right )^{3} + 10 \, {\left (x \log \left (2\right ) + 2 \, \log \left (2\right )\right )} \log \left (x\right )^{2} - {\left (x^{2} \log \left (2\right ) \log \left (x\right )^{2} + x^{2} \log \left (x\right )^{3}\right )} e^{x}\right )}}{x + 2} \]
integrate((((-x^4-3*x^3-4*x^2)*exp(x)*log(2*x)+(-x^3-2*x^2)*exp(x)+10*x^2+ 40*x+40)*log(x^2)^2+((-4*x^3-8*x^2)*exp(x)+40*x^2+160*x+160)*log(2*x)*log( x^2))/(x^3+4*x^2+4*x),x, algorithm=\
4*(10*(x + 2)*log(x)^3 + 10*(x*log(2) + 2*log(2))*log(x)^2 - (x^2*log(2)*l og(x)^2 + x^2*log(x)^3)*e^x)/(x + 2)
Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (25) = 50\).
Time = 0.32 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.31 \[ \int \frac {\left (160+160 x+40 x^2+e^x \left (-8 x^2-4 x^3\right )\right ) \log (2 x) \log \left (x^2\right )+\left (40+40 x+10 x^2+e^x \left (-2 x^2-x^3\right )+e^x \left (-4 x^2-3 x^3-x^4\right ) \log (2 x)\right ) \log ^2\left (x^2\right )}{4 x+4 x^2+x^3} \, dx=-\frac {4 \, {\left (x^{2} e^{x} \log \left (2\right ) \log \left (x\right )^{2} + x^{2} e^{x} \log \left (x\right )^{3} - 10 \, x \log \left (2\right ) \log \left (x\right )^{2} - 10 \, x \log \left (x\right )^{3} - 20 \, \log \left (2\right ) \log \left (x\right )^{2} - 20 \, \log \left (x\right )^{3}\right )}}{x + 2} \]
integrate((((-x^4-3*x^3-4*x^2)*exp(x)*log(2*x)+(-x^3-2*x^2)*exp(x)+10*x^2+ 40*x+40)*log(x^2)^2+((-4*x^3-8*x^2)*exp(x)+40*x^2+160*x+160)*log(2*x)*log( x^2))/(x^3+4*x^2+4*x),x, algorithm=\
-4*(x^2*e^x*log(2)*log(x)^2 + x^2*e^x*log(x)^3 - 10*x*log(2)*log(x)^2 - 10 *x*log(x)^3 - 20*log(2)*log(x)^2 - 20*log(x)^3)/(x + 2)
Time = 8.45 (sec) , antiderivative size = 87, normalized size of antiderivative = 3.35 \[ \int \frac {\left (160+160 x+40 x^2+e^x \left (-8 x^2-4 x^3\right )\right ) \log (2 x) \log \left (x^2\right )+\left (40+40 x+10 x^2+e^x \left (-2 x^2-x^3\right )+e^x \left (-4 x^2-3 x^3-x^4\right ) \log (2 x)\right ) \log ^2\left (x^2\right )}{4 x+4 x^2+x^3} \, dx=40\,\ln \left (x^2\right )\,{\ln \left (2\right )}^2-80\,{\ln \left (2\right )}^2\,\ln \left (x\right )-40\,\ln \left (2\right )\,{\ln \left (x\right )}^2+10\,{\ln \left (x^2\right )}^2\,\ln \left (x\right )+40\,\ln \left (x^2\right )\,\ln \left (2\right )\,\ln \left (x\right )-\frac {x^2\,{\ln \left (x^2\right )}^2\,{\mathrm {e}}^x\,\ln \left (2\right )}{x+2}-\frac {x^2\,{\ln \left (x^2\right )}^2\,{\mathrm {e}}^x\,\ln \left (x\right )}{x+2} \]
int((log(x^2)^2*(40*x - exp(x)*(2*x^2 + x^3) + 10*x^2 - log(2*x)*exp(x)*(4 *x^2 + 3*x^3 + x^4) + 40) + log(2*x)*log(x^2)*(160*x - exp(x)*(8*x^2 + 4*x ^3) + 40*x^2 + 160))/(4*x + 4*x^2 + x^3),x)