\(\int \frac {e^5 (3-12 x)+x-4 x^2+(-x+2 x^2) \log (x-2 x^2)}{-x^3+2 x^4+e^{10} (-9 x+18 x^2)+e^5 (-6 x^2+12 x^3)} \, dx\) [5102]

Optimal result

Integrand size = 76, antiderivative size = 21 \[ \int \frac {e^5 (3-12 x)+x-4 x^2+\left (-x+2 x^2\right ) \log \left (x-2 x^2\right )}{-x^3+2 x^4+e^{10} \left (-9 x+18 x^2\right )+e^5 \left (-6 x^2+12 x^3\right )} \, dx=\frac {\log (-x (-1+2 x))}{-3 e^5-x} \]

[Out]

ln(-x*(-1+2*x))/(-3*exp(5)-x)

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(85\) vs. \(2(21)=42\).

Time = 0.19 (sec) , antiderivative size = 85, normalized size of antiderivative = 4.05, number of steps used = 12, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.092, Rules used = {6820, 6874, 153, 2581, 36, 31, 29} \[ \int \frac {e^5 (3-12 x)+x-4 x^2+\left (-x+2 x^2\right ) \log \left (x-2 x^2\right )}{-x^3+2 x^4+e^{10} \left (-9 x+18 x^2\right )+e^5 \left (-6 x^2+12 x^3\right )} \, dx=-\frac {\log ((1-2 x) x)}{x+3 e^5}+\frac {\left (1+12 e^5\right ) \log \left (x+3 e^5\right )}{3 e^5 \left (1+6 e^5\right )}-\frac {2 \log \left (x+3 e^5\right )}{1+6 e^5}-\frac {\log \left (x+3 e^5\right )}{3 e^5} \]

[In]

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]

[Out]

-(Log[(1 - 2*x)*x]/(3*E^5 + x)) - Log[3*E^5 + x]/(3*E^5) - (2*Log[3*E^5 + x])/(1 + 6*E^5) + ((1 + 12*E^5)*Log[
3*E^5 + x])/(3*E^5*(1 + 6*E^5))

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 153

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*(g + h*x), x], x] /; FreeQ[{a, b, c, d, e, f, g
, h, m}, x] && (IntegersQ[m, n, p] || (IGtQ[n, 0] && IGtQ[p, 0]))

Rule 2581

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]*((g_.) + (h_.)*(x_))^(m_.),
x_Symbol] :> Simp[(g + h*x)^(m + 1)*(Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]/(h*(m + 1))), x] + (-Dist[b*p*(r/(h
*(m + 1))), Int[(g + h*x)^(m + 1)/(a + b*x), x], x] - Dist[d*q*(r/(h*(m + 1))), Int[(g + h*x)^(m + 1)/(c + d*x
), x], x]) /; FreeQ[{a, b, c, d, e, f, g, h, m, p, q, r}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-\frac {\left (3 e^5+x\right ) (-1+4 x)}{x (-1+2 x)}+\log \left (x-2 x^2\right )}{\left (3 e^5+x\right )^2} \, dx \\ & = \int \left (\frac {1-4 x}{x \left (3 e^5+x\right ) (-1+2 x)}+\frac {\log ((1-2 x) x)}{\left (3 e^5+x\right )^2}\right ) \, dx \\ & = \int \frac {1-4 x}{x \left (3 e^5+x\right ) (-1+2 x)} \, dx+\int \frac {\log ((1-2 x) x)}{\left (3 e^5+x\right )^2} \, dx \\ & = -\frac {\log ((1-2 x) x)}{3 e^5+x}-2 \int \frac {1}{(1-2 x) \left (3 e^5+x\right )} \, dx+\int \frac {1}{x \left (3 e^5+x\right )} \, dx+\int \left (-\frac {1}{3 e^5 x}+\frac {1+12 e^5}{3 e^5 \left (1+6 e^5\right ) \left (3 e^5+x\right )}-\frac {4}{\left (1+6 e^5\right ) (-1+2 x)}\right ) \, dx \\ & = -\frac {2 \log (1-2 x)}{1+6 e^5}-\frac {\log (x)}{3 e^5}-\frac {\log ((1-2 x) x)}{3 e^5+x}+\frac {\left (1+12 e^5\right ) \log \left (3 e^5+x\right )}{3 e^5 \left (1+6 e^5\right )}+\frac {\int \frac {1}{x} \, dx}{3 e^5}-\frac {\int \frac {1}{3 e^5+x} \, dx}{3 e^5}-\frac {2 \int \frac {1}{3 e^5+x} \, dx}{1+6 e^5}-\frac {4 \int \frac {1}{1-2 x} \, dx}{1+6 e^5} \\ & = -\frac {\log ((1-2 x) x)}{3 e^5+x}-\frac {\log \left (3 e^5+x\right )}{3 e^5}-\frac {2 \log \left (3 e^5+x\right )}{1+6 e^5}+\frac {\left (1+12 e^5\right ) \log \left (3 e^5+x\right )}{3 e^5 \left (1+6 e^5\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {e^5 (3-12 x)+x-4 x^2+\left (-x+2 x^2\right ) \log \left (x-2 x^2\right )}{-x^3+2 x^4+e^{10} \left (-9 x+18 x^2\right )+e^5 \left (-6 x^2+12 x^3\right )} \, dx=-\frac {\log ((1-2 x) x)}{3 e^5+x} \]

[In]

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]

[Out]

-(Log[(1 - 2*x)*x]/(3*E^5 + x))

Maple [A] (verified)

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	\(\frac {\left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (9 \,{\mathrm e}^{10}+6 \textit {\_Z} \,{\mathrm e}^{5}+\textit {\_Z}^{2}\right )}{\sum }\left (-\frac {\underline {\hspace {1.25 ex}}\alpha }{-\left ({\mathrm e}^{5}\right )^{2}+{\mathrm e}^{10}}-\frac {3 \,{\mathrm e}^{5}}{-\left ({\mathrm e}^{5}\right )^{2}+{\mathrm e}^{10}}\right ) \left (\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (-2 x^{2}+x \right )-\left (\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right )-\ln \left (-\frac {x -\underline {\hspace {1.25 ex}}\alpha }{\underline {\hspace {1.25 ex}}\alpha }\right )\right ) \ln \left (\frac {x}{\underline {\hspace {1.25 ex}}\alpha }\right )+\operatorname {dilog}\left (-\frac {x -\underline {\hspace {1.25 ex}}\alpha }{\underline {\hspace {1.25 ex}}\alpha }\right )-\left (\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right )-\ln \left (\frac {2 x -2 \underline {\hspace {1.25 ex}}\alpha }{-2 \underline {\hspace {1.25 ex}}\alpha +1}\right )\right ) \ln \left (\frac {1-2 x}{-2 \underline {\hspace {1.25 ex}}\alpha +1}\right )+\operatorname {dilog}\left (\frac {2 x -2 \underline {\hspace {1.25 ex}}\alpha }{-2 \underline {\hspace {1.25 ex}}\alpha +1}\right )\right )\right )}{18}-\frac {2 \ln \left (-1+2 x \right )}{1+6 \,{\mathrm e}^{5}}-\frac {\left (-12 \,{\mathrm e}^{5}-1\right ) \ln \left (3 \,{\mathrm e}^{5}+x \right )}{3 \,{\mathrm e}^{5} \left (1+6 \,{\mathrm e}^{5}\right )}-\frac {\ln \left (x \right )}{3 \,{\mathrm e}^{5}}\)	\(213\)

[In]

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)

[Out]

-ln(-2*x^2+x)/(3*exp(5)+x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {e^5 (3-12 x)+x-4 x^2+\left (-x+2 x^2\right ) \log \left (x-2 x^2\right )}{-x^3+2 x^4+e^{10} \left (-9 x+18 x^2\right )+e^5 \left (-6 x^2+12 x^3\right )} \, dx=-\frac {\log \left (-2 \, x^{2} + x\right )}{x + 3 \, e^{5}} \]

[In]

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="fricas")

[Out]

-log(-2*x^2 + x)/(x + 3*e^5)

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {e^5 (3-12 x)+x-4 x^2+\left (-x+2 x^2\right ) \log \left (x-2 x^2\right )}{-x^3+2 x^4+e^{10} \left (-9 x+18 x^2\right )+e^5 \left (-6 x^2+12 x^3\right )} \, dx=- \frac {\log {\left (- 2 x^{2} + x \right )}}{x + 3 e^{5}} \]

[In]

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)

[Out]

-log(-2*x**2 + x)/(x + 3*exp(5))

Maxima [B] (verification not implemented)

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 \[ \int \frac {e^5 (3-12 x)+x-4 x^2+\left (-x+2 x^2\right ) \log \left (x-2 x^2\right )}{-x^3+2 x^4+e^{10} \left (-9 x+18 x^2\right )+e^5 \left (-6 x^2+12 x^3\right )} \, dx=-\frac {1}{3} \, {\left (e^{\left (-10\right )} \log \left (x\right ) - \frac {{\left (12 \, e^{5} + 1\right )} \log \left (x + 3 \, e^{5}\right )}{36 \, e^{20} + 12 \, e^{15} + e^{10}} - \frac {36 \, \log \left (2 \, x - 1\right )}{36 \, e^{10} + 12 \, e^{5} + 1} + \frac {3}{x {\left (6 \, e^{10} + e^{5}\right )} + 18 \, e^{15} + 3 \, e^{10}}\right )} e^{5} - 12 \, {\left (\frac {2 \, \log \left (2 \, x - 1\right )}{36 \, e^{10} + 12 \, e^{5} + 1} - \frac {2 \, \log \left (x + 3 \, e^{5}\right )}{36 \, e^{10} + 12 \, e^{5} + 1} + \frac {1}{x {\left (6 \, e^{5} + 1\right )} + 18 \, e^{10} + 3 \, e^{5}}\right )} e^{5} + \frac {1}{3} \, e^{\left (-5\right )} \log \left (x\right ) - \frac {{\left (12 \, e^{5} + 1\right )} \log \left (x + 3 \, e^{5}\right )}{3 \, {\left (6 \, e^{10} + e^{5}\right )}} - \frac {{\left (6 \, e^{5} + 1\right )} \log \left (x\right ) - {\left (2 \, x - 1\right )} \log \left (-2 \, x + 1\right )}{x {\left (6 \, e^{5} + 1\right )} + 18 \, e^{10} + 3 \, e^{5}} + \frac {12 \, e^{5}}{x {\left (6 \, e^{5} + 1\right )} + 18 \, e^{10} + 3 \, e^{5}} - \frac {2 \, \log \left (2 \, x - 1\right )}{36 \, e^{10} + 12 \, e^{5} + 1} + \frac {2 \, \log \left (x + 3 \, e^{5}\right )}{36 \, e^{10} + 12 \, e^{5} + 1} + \frac {1}{x {\left (6 \, e^{5} + 1\right )} + 18 \, e^{10} + 3 \, e^{5}} \]

[In]

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="maxima")

[Out]

-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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {e^5 (3-12 x)+x-4 x^2+\left (-x+2 x^2\right ) \log \left (x-2 x^2\right )}{-x^3+2 x^4+e^{10} \left (-9 x+18 x^2\right )+e^5 \left (-6 x^2+12 x^3\right )} \, dx=-\frac {\log \left (-2 \, x^{2} + x\right )}{x + 3 \, e^{5}} \]

[In]

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="giac")

[Out]

-log(-2*x^2 + x)/(x + 3*e^5)

Mupad [B] (verification not implemented)

Time = 13.92 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {e^5 (3-12 x)+x-4 x^2+\left (-x+2 x^2\right ) \log \left (x-2 x^2\right )}{-x^3+2 x^4+e^{10} \left (-9 x+18 x^2\right )+e^5 \left (-6 x^2+12 x^3\right )} \, dx=-\frac {\ln \left (x-2\,x^2\right )}{x+3\,{\mathrm {e}}^5} \]

[In]

int((4*x^2 - x + log(x - 2*x^2)*(x - 2*x^2) + exp(5)*(12*x - 3))/(exp(10)*(9*x - 18*x^2) + exp(5)*(6*x^2 - 12*
x^3) + x^3 - 2*x^4),x)

[Out]

-log(x - 2*x^2)/(x + 3*exp(5))