Integrand size = 76, antiderivative size = 21 \[ \int \frac {e^5 (3-12 x)+x-4 x^2+\left (-x+2 x^2\right ) \log \left (x-2 x^2\right )}{-x^3+2 x^4+e^{10} \left (-9 x+18 x^2\right )+e^5 \left (-6 x^2+12 x^3\right )} \, dx=\frac {\log (-x (-1+2 x))}{-3 e^5-x} \]
Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {e^5 (3-12 x)+x-4 x^2+\left (-x+2 x^2\right ) \log \left (x-2 x^2\right )}{-x^3+2 x^4+e^{10} \left (-9 x+18 x^2\right )+e^5 \left (-6 x^2+12 x^3\right )} \, dx=-\frac {\log ((1-2 x) x)}{3 e^5+x} \]
Integrate[(E^5*(3 - 12*x) + x - 4*x^2 + (-x + 2*x^2)*Log[x - 2*x^2])/(-x^3 + 2*x^4 + E^10*(-9*x + 18*x^2) + E^5*(-6*x^2 + 12*x^3)),x]
Leaf count is larger than twice the leaf count of optimal. \(319\) vs. \(2(21)=42\).
Time = 1.29 (sec) , antiderivative size = 319, normalized size of antiderivative = 15.19, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {2026, 2463, 2009}
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.
\(\displaystyle \int \frac {-4 x^2+\left (2 x^2-x\right ) \log \left (x-2 x^2\right )+x+e^5 (3-12 x)}{2 x^4-x^3+e^{10} \left (18 x^2-9 x\right )+e^5 \left (12 x^3-6 x^2\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {-4 x^2+\left (2 x^2-x\right ) \log \left (x-2 x^2\right )+x+e^5 (3-12 x)}{x \left (2 x^3-\left (1-12 e^5\right ) x^2-6 e^5 \left (1-3 e^5\right ) x-9 e^{10}\right )}dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (-\frac {2 \left (-4 x^2+\left (2 x^2-x\right ) \log \left (x-2 x^2\right )+x+e^5 (3-12 x)\right )}{\left (1+6 e^5\right )^2 x \left (x+3 e^5\right )}+\frac {4 \left (-4 x^2+\left (2 x^2-x\right ) \log \left (x-2 x^2\right )+x+e^5 (3-12 x)\right )}{\left (1+6 e^5\right )^2 x (2 x-1)}-\frac {-4 x^2+\left (2 x^2-x\right ) \log \left (x-2 x^2\right )+x+e^5 (3-12 x)}{\left (1+6 e^5\right ) x \left (x+3 e^5\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 (1-2 x) \log (1-2 x)}{\left (1+6 e^5\right )^2}-\frac {(1-2 x) \log (1-2 x)}{\left (1+6 e^5\right ) \left (x+3 e^5\right )}-\frac {2 \log (1-2 x)}{1+6 e^5}-\frac {4 x \log (x)}{\left (1+6 e^5\right )^2}+\frac {x \log (x)}{3 e^5 \left (x+3 e^5\right )}-\frac {\log (x)}{3 e^5 \left (1+6 e^5\right )}-\frac {12 e^5 \log (x)}{\left (1+6 e^5\right )^2}+\frac {4 x (\log (1-2 x)+\log (x)-\log ((1-2 x) x))}{\left (1+6 e^5\right )^2}+\frac {\log (1-2 x)+\log (x)-\log ((1-2 x) x)}{x+3 e^5}-\frac {2 (1-2 x) \log ((1-2 x) x)}{\left (1+6 e^5\right )^2}-\frac {2 \left (1+\log \left (1+6 e^5\right )\right ) \log \left (x+3 e^5\right )}{1+6 e^5}+\frac {2 \log \left (1+6 e^5\right ) \log \left (x+3 e^5\right )}{1+6 e^5}+\frac {\left (12+\frac {1}{e^5}\right ) \log \left (x+3 e^5\right )}{3 \left (1+6 e^5\right )}-\frac {\log \left (x+3 e^5\right )}{3 e^5}\) |
Int[(E^5*(3 - 12*x) + x - 4*x^2 + (-x + 2*x^2)*Log[x - 2*x^2])/(-x^3 + 2*x ^4 + E^10*(-9*x + 18*x^2) + E^5*(-6*x^2 + 12*x^3)),x]
(-2*Log[1 - 2*x])/(1 + 6*E^5) + (2*(1 - 2*x)*Log[1 - 2*x])/(1 + 6*E^5)^2 - ((1 - 2*x)*Log[1 - 2*x])/((1 + 6*E^5)*(3*E^5 + x)) - (12*E^5*Log[x])/(1 + 6*E^5)^2 - Log[x]/(3*E^5*(1 + 6*E^5)) - (4*x*Log[x])/(1 + 6*E^5)^2 + (x*L og[x])/(3*E^5*(3*E^5 + x)) + (4*x*(Log[1 - 2*x] + Log[x] - Log[(1 - 2*x)*x ]))/(1 + 6*E^5)^2 + (Log[1 - 2*x] + Log[x] - Log[(1 - 2*x)*x])/(3*E^5 + x) - (2*(1 - 2*x)*Log[(1 - 2*x)*x])/(1 + 6*E^5)^2 - Log[3*E^5 + x]/(3*E^5) + ((12 + E^(-5))*Log[3*E^5 + x])/(3*(1 + 6*E^5)) + (2*Log[1 + 6*E^5]*Log[3* E^5 + x])/(1 + 6*E^5) - (2*(1 + Log[1 + 6*E^5])*Log[3*E^5 + x])/(1 + 6*E^5 )
3.22.2.3.1 Defintions of rubi rules used
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_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u, Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt Q[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90
method | result | size |
norman | \(-\frac {\ln \left (-2 x^{2}+x \right )}{3 \,{\mathrm e}^{5}+x}\) | \(19\) |
risch | \(-\frac {\ln \left (-2 x^{2}+x \right )}{3 \,{\mathrm e}^{5}+x}\) | \(19\) |
parallelrisch | \(-\frac {\ln \left (-2 x^{2}+x \right )}{3 \,{\mathrm e}^{5}+x}\) | \(19\) |
default | \(\frac {\left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (9 \,{\mathrm e}^{10}+6 \textit {\_Z} \,{\mathrm e}^{5}+\textit {\_Z}^{2}\right )}{\sum }\left (-\frac {\underline {\hspace {1.25 ex}}\alpha }{-\left ({\mathrm e}^{5}\right )^{2}+{\mathrm e}^{10}}-\frac {3 \,{\mathrm e}^{5}}{-\left ({\mathrm e}^{5}\right )^{2}+{\mathrm e}^{10}}\right ) \left (\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (-2 x^{2}+x \right )-\left (\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right )-\ln \left (-\frac {x -\underline {\hspace {1.25 ex}}\alpha }{\underline {\hspace {1.25 ex}}\alpha }\right )\right ) \ln \left (\frac {x}{\underline {\hspace {1.25 ex}}\alpha }\right )+\operatorname {dilog}\left (-\frac {x -\underline {\hspace {1.25 ex}}\alpha }{\underline {\hspace {1.25 ex}}\alpha }\right )-\left (\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right )-\ln \left (\frac {2 x -2 \underline {\hspace {1.25 ex}}\alpha }{-2 \underline {\hspace {1.25 ex}}\alpha +1}\right )\right ) \ln \left (\frac {1-2 x}{-2 \underline {\hspace {1.25 ex}}\alpha +1}\right )+\operatorname {dilog}\left (\frac {2 x -2 \underline {\hspace {1.25 ex}}\alpha }{-2 \underline {\hspace {1.25 ex}}\alpha +1}\right )\right )\right )}{18}-\frac {2 \ln \left (-1+2 x \right )}{1+6 \,{\mathrm e}^{5}}-\frac {\left (-12 \,{\mathrm e}^{5}-1\right ) \ln \left (3 \,{\mathrm e}^{5}+x \right )}{3 \,{\mathrm e}^{5} \left (1+6 \,{\mathrm e}^{5}\right )}-\frac {\ln \left (x \right )}{3 \,{\mathrm e}^{5}}\) | \(213\) |
parts | \(\frac {\left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (9 \,{\mathrm e}^{10}+6 \textit {\_Z} \,{\mathrm e}^{5}+\textit {\_Z}^{2}\right )}{\sum }\left (-\frac {\underline {\hspace {1.25 ex}}\alpha }{-\left ({\mathrm e}^{5}\right )^{2}+{\mathrm e}^{10}}-\frac {3 \,{\mathrm e}^{5}}{-\left ({\mathrm e}^{5}\right )^{2}+{\mathrm e}^{10}}\right ) \left (\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (-2 x^{2}+x \right )-\left (\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right )-\ln \left (-\frac {x -\underline {\hspace {1.25 ex}}\alpha }{\underline {\hspace {1.25 ex}}\alpha }\right )\right ) \ln \left (\frac {x}{\underline {\hspace {1.25 ex}}\alpha }\right )+\operatorname {dilog}\left (-\frac {x -\underline {\hspace {1.25 ex}}\alpha }{\underline {\hspace {1.25 ex}}\alpha }\right )-\left (\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right )-\ln \left (\frac {2 x -2 \underline {\hspace {1.25 ex}}\alpha }{-2 \underline {\hspace {1.25 ex}}\alpha +1}\right )\right ) \ln \left (\frac {1-2 x}{-2 \underline {\hspace {1.25 ex}}\alpha +1}\right )+\operatorname {dilog}\left (\frac {2 x -2 \underline {\hspace {1.25 ex}}\alpha }{-2 \underline {\hspace {1.25 ex}}\alpha +1}\right )\right )\right )}{18}-\frac {2 \ln \left (-1+2 x \right )}{1+6 \,{\mathrm e}^{5}}-\frac {\left (-12 \,{\mathrm e}^{5}-1\right ) \ln \left (3 \,{\mathrm e}^{5}+x \right )}{3 \,{\mathrm e}^{5} \left (1+6 \,{\mathrm e}^{5}\right )}-\frac {\ln \left (x \right )}{3 \,{\mathrm e}^{5}}\) | \(213\) |
int(((2*x^2-x)*ln(-2*x^2+x)+(-12*x+3)*exp(5)-4*x^2+x)/((18*x^2-9*x)*exp(5) ^2+(12*x^3-6*x^2)*exp(5)+2*x^4-x^3),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {e^5 (3-12 x)+x-4 x^2+\left (-x+2 x^2\right ) \log \left (x-2 x^2\right )}{-x^3+2 x^4+e^{10} \left (-9 x+18 x^2\right )+e^5 \left (-6 x^2+12 x^3\right )} \, dx=-\frac {\log \left (-2 \, x^{2} + x\right )}{x + 3 \, e^{5}} \]
integrate(((2*x^2-x)*log(-2*x^2+x)+(-12*x+3)*exp(5)-4*x^2+x)/((18*x^2-9*x) *exp(5)^2+(12*x^3-6*x^2)*exp(5)+2*x^4-x^3),x, algorithm=\
Time = 0.11 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {e^5 (3-12 x)+x-4 x^2+\left (-x+2 x^2\right ) \log \left (x-2 x^2\right )}{-x^3+2 x^4+e^{10} \left (-9 x+18 x^2\right )+e^5 \left (-6 x^2+12 x^3\right )} \, dx=- \frac {\log {\left (- 2 x^{2} + x \right )}}{x + 3 e^{5}} \]
integrate(((2*x**2-x)*ln(-2*x**2+x)+(-12*x+3)*exp(5)-4*x**2+x)/((18*x**2-9 *x)*exp(5)**2+(12*x**3-6*x**2)*exp(5)+2*x**4-x**3),x)
Leaf count of result is larger than twice the leaf count of optimal. 303 vs. \(2 (19) = 38\).
Time = 0.27 (sec) , antiderivative size = 303, normalized size of antiderivative = 14.43 \[ \int \frac {e^5 (3-12 x)+x-4 x^2+\left (-x+2 x^2\right ) \log \left (x-2 x^2\right )}{-x^3+2 x^4+e^{10} \left (-9 x+18 x^2\right )+e^5 \left (-6 x^2+12 x^3\right )} \, dx=-\frac {1}{3} \, {\left (e^{\left (-10\right )} \log \left (x\right ) - \frac {{\left (12 \, e^{5} + 1\right )} \log \left (x + 3 \, e^{5}\right )}{36 \, e^{20} + 12 \, e^{15} + e^{10}} - \frac {36 \, \log \left (2 \, x - 1\right )}{36 \, e^{10} + 12 \, e^{5} + 1} + \frac {3}{x {\left (6 \, e^{10} + e^{5}\right )} + 18 \, e^{15} + 3 \, e^{10}}\right )} e^{5} - 12 \, {\left (\frac {2 \, \log \left (2 \, x - 1\right )}{36 \, e^{10} + 12 \, e^{5} + 1} - \frac {2 \, \log \left (x + 3 \, e^{5}\right )}{36 \, e^{10} + 12 \, e^{5} + 1} + \frac {1}{x {\left (6 \, e^{5} + 1\right )} + 18 \, e^{10} + 3 \, e^{5}}\right )} e^{5} + \frac {1}{3} \, e^{\left (-5\right )} \log \left (x\right ) - \frac {{\left (12 \, e^{5} + 1\right )} \log \left (x + 3 \, e^{5}\right )}{3 \, {\left (6 \, e^{10} + e^{5}\right )}} - \frac {{\left (6 \, e^{5} + 1\right )} \log \left (x\right ) - {\left (2 \, x - 1\right )} \log \left (-2 \, x + 1\right )}{x {\left (6 \, e^{5} + 1\right )} + 18 \, e^{10} + 3 \, e^{5}} + \frac {12 \, e^{5}}{x {\left (6 \, e^{5} + 1\right )} + 18 \, e^{10} + 3 \, e^{5}} - \frac {2 \, \log \left (2 \, x - 1\right )}{36 \, e^{10} + 12 \, e^{5} + 1} + \frac {2 \, \log \left (x + 3 \, e^{5}\right )}{36 \, e^{10} + 12 \, e^{5} + 1} + \frac {1}{x {\left (6 \, e^{5} + 1\right )} + 18 \, e^{10} + 3 \, e^{5}} \]
integrate(((2*x^2-x)*log(-2*x^2+x)+(-12*x+3)*exp(5)-4*x^2+x)/((18*x^2-9*x) *exp(5)^2+(12*x^3-6*x^2)*exp(5)+2*x^4-x^3),x, algorithm=\
-1/3*(e^(-10)*log(x) - (12*e^5 + 1)*log(x + 3*e^5)/(36*e^20 + 12*e^15 + e^ 10) - 36*log(2*x - 1)/(36*e^10 + 12*e^5 + 1) + 3/(x*(6*e^10 + e^5) + 18*e^ 15 + 3*e^10))*e^5 - 12*(2*log(2*x - 1)/(36*e^10 + 12*e^5 + 1) - 2*log(x + 3*e^5)/(36*e^10 + 12*e^5 + 1) + 1/(x*(6*e^5 + 1) + 18*e^10 + 3*e^5))*e^5 + 1/3*e^(-5)*log(x) - 1/3*(12*e^5 + 1)*log(x + 3*e^5)/(6*e^10 + e^5) - ((6* e^5 + 1)*log(x) - (2*x - 1)*log(-2*x + 1))/(x*(6*e^5 + 1) + 18*e^10 + 3*e^ 5) + 12*e^5/(x*(6*e^5 + 1) + 18*e^10 + 3*e^5) - 2*log(2*x - 1)/(36*e^10 + 12*e^5 + 1) + 2*log(x + 3*e^5)/(36*e^10 + 12*e^5 + 1) + 1/(x*(6*e^5 + 1) + 18*e^10 + 3*e^5)
Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {e^5 (3-12 x)+x-4 x^2+\left (-x+2 x^2\right ) \log \left (x-2 x^2\right )}{-x^3+2 x^4+e^{10} \left (-9 x+18 x^2\right )+e^5 \left (-6 x^2+12 x^3\right )} \, dx=-\frac {\log \left (-2 \, x^{2} + x\right )}{x + 3 \, e^{5}} \]
integrate(((2*x^2-x)*log(-2*x^2+x)+(-12*x+3)*exp(5)-4*x^2+x)/((18*x^2-9*x) *exp(5)^2+(12*x^3-6*x^2)*exp(5)+2*x^4-x^3),x, algorithm=\
Time = 13.92 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {e^5 (3-12 x)+x-4 x^2+\left (-x+2 x^2\right ) \log \left (x-2 x^2\right )}{-x^3+2 x^4+e^{10} \left (-9 x+18 x^2\right )+e^5 \left (-6 x^2+12 x^3\right )} \, dx=-\frac {\ln \left (x-2\,x^2\right )}{x+3\,{\mathrm {e}}^5} \]