Integrand size = 96, antiderivative size = 27 \[ \int \frac {e^3 \left (x^2-2 x^3+x^4\right ) \log ^3(x)+e^{\frac {1}{x^2 \log ^2(x)}} \left (-6+6 x+(-8+8 x) \log (x)+(-2+2 x) \log ^2(x)+\left (x^2+2 x^3\right ) \log ^3(x)+x^3 \log ^4(x)\right )}{\left (x^3-2 x^4+x^5\right ) \log ^3(x)} \, dx=\left (e^3+\frac {e^{\frac {1}{x^2 \log ^2(x)}}}{1-x}\right ) (3+\log (x)) \] Output:
(3+ln(x))*(exp(1/x^2/ln(x)^2)/(1-x)+exp(3))
Time = 0.12 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.56 \[ \int \frac {e^3 \left (x^2-2 x^3+x^4\right ) \log ^3(x)+e^{\frac {1}{x^2 \log ^2(x)}} \left (-6+6 x+(-8+8 x) \log (x)+(-2+2 x) \log ^2(x)+\left (x^2+2 x^3\right ) \log ^3(x)+x^3 \log ^4(x)\right )}{\left (x^3-2 x^4+x^5\right ) \log ^3(x)} \, dx=\frac {-3 e^{\frac {1}{x^2 \log ^2(x)}}-e^{\frac {1}{x^2 \log ^2(x)}} \log (x)+e^3 (-1+x) \log (x)}{-1+x} \] Input:
Integrate[(E^3*(x^2 - 2*x^3 + x^4)*Log[x]^3 + E^(1/(x^2*Log[x]^2))*(-6 + 6 *x + (-8 + 8*x)*Log[x] + (-2 + 2*x)*Log[x]^2 + (x^2 + 2*x^3)*Log[x]^3 + x^ 3*Log[x]^4))/((x^3 - 2*x^4 + x^5)*Log[x]^3),x]
Output:
(-3*E^(1/(x^2*Log[x]^2)) - E^(1/(x^2*Log[x]^2))*Log[x] + E^3*(-1 + x)*Log[ x])/(-1 + x)
Leaf count is larger than twice the leaf count of optimal. \(76\) vs. \(2(27)=54\).
Time = 3.51 (sec) , antiderivative size = 76, normalized size of antiderivative = 2.81, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.052, Rules used = {2026, 7277, 27, 7293, 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.
\(\displaystyle \int \frac {e^{\frac {1}{x^2 \log ^2(x)}} \left (x^3 \log ^4(x)+\left (2 x^3+x^2\right ) \log ^3(x)+6 x+(2 x-2) \log ^2(x)+(8 x-8) \log (x)-6\right )+e^3 \left (x^4-2 x^3+x^2\right ) \log ^3(x)}{\left (x^5-2 x^4+x^3\right ) \log ^3(x)} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{\frac {1}{x^2 \log ^2(x)}} \left (x^3 \log ^4(x)+\left (2 x^3+x^2\right ) \log ^3(x)+6 x+(2 x-2) \log ^2(x)+(8 x-8) \log (x)-6\right )+e^3 \left (x^4-2 x^3+x^2\right ) \log ^3(x)}{x^3 \left (x^2-2 x+1\right ) \log ^3(x)}dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 4 \int \frac {e^3 \left (x^4-2 x^3+x^2\right ) \log ^3(x)-e^{\frac {1}{x^2 \log ^2(x)}} \left (-x^3 \log ^4(x)-\left (2 x^3+x^2\right ) \log ^3(x)+2 (1-x) \log ^2(x)+8 (1-x) \log (x)-6 x+6\right )}{4 (1-x)^2 x^3 \log ^3(x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {e^3 \left (x^4-2 x^3+x^2\right ) \log ^3(x)-e^{\frac {1}{x^2 \log ^2(x)}} \left (-x^3 \log ^4(x)-\left (2 x^3+x^2\right ) \log ^3(x)-6 x+2 (1-x) \log ^2(x)+8 (1-x) \log (x)+6\right )}{(1-x)^2 x^3 \log ^3(x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{\frac {1}{x^2 \log ^2(x)}} \left (x^3 \log ^4(x)+2 x^3 \log ^3(x)+x^2 \log ^3(x)+6 x+2 x \log ^2(x)-2 \log ^2(x)+8 x \log (x)-8 \log (x)-6\right )}{(x-1)^2 x^3 \log ^3(x)}+\frac {e^3}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^{\frac {1}{x^2 \log ^2(x)}} \left (-3 x-x \log ^2(x)+\log ^2(x)-4 x \log (x)+4 \log (x)+3\right )}{(1-x)^2 x^3 \left (\frac {1}{x^3 \log ^3(x)}+\frac {1}{x^3 \log ^2(x)}\right ) \log ^3(x)}+e^3 \log (x)\) |
Input:
Int[(E^3*(x^2 - 2*x^3 + x^4)*Log[x]^3 + E^(1/(x^2*Log[x]^2))*(-6 + 6*x + ( -8 + 8*x)*Log[x] + (-2 + 2*x)*Log[x]^2 + (x^2 + 2*x^3)*Log[x]^3 + x^3*Log[ x]^4))/((x^3 - 2*x^4 + x^5)*Log[x]^3),x]
Output:
E^3*Log[x] + (E^(1/(x^2*Log[x]^2))*(3 - 3*x + 4*Log[x] - 4*x*Log[x] + Log[ x]^2 - x*Log[x]^2))/((1 - x)^2*x^3*(1/(x^3*Log[x]^3) + 1/(x^3*Log[x]^2))*L og[x]^3)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_)*((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Simp[1/(4^p*c^p) Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n} , x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p] && !AlgebraicFu nctionQ[u, x]
Time = 13.47 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00
method | result | size |
risch | \({\mathrm e}^{3} \ln \left (x \right )-\frac {\left (3+\ln \left (x \right )\right ) {\mathrm e}^{\frac {1}{x^{2} \ln \left (x \right )^{2}}}}{-1+x}\) | \(27\) |
parallelrisch | \(\frac {x \,{\mathrm e}^{3} \ln \left (x \right )-{\mathrm e}^{3} \ln \left (x \right )-\ln \left (x \right ) {\mathrm e}^{\frac {1}{x^{2} \ln \left (x \right )^{2}}}-3 \,{\mathrm e}^{\frac {1}{x^{2} \ln \left (x \right )^{2}}}}{-1+x}\) | \(44\) |
Input:
int(((x^3*ln(x)^4+(2*x^3+x^2)*ln(x)^3+(-2+2*x)*ln(x)^2+(8*x-8)*ln(x)+6*x-6 )*exp(1/x^2/ln(x)^2)+(x^4-2*x^3+x^2)*exp(3)*ln(x)^3)/(x^5-2*x^4+x^3)/ln(x) ^3,x,method=_RETURNVERBOSE)
Output:
exp(3)*ln(x)-(3+ln(x))/(-1+x)*exp(1/x^2/ln(x)^2)
Time = 0.12 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {e^3 \left (x^2-2 x^3+x^4\right ) \log ^3(x)+e^{\frac {1}{x^2 \log ^2(x)}} \left (-6+6 x+(-8+8 x) \log (x)+(-2+2 x) \log ^2(x)+\left (x^2+2 x^3\right ) \log ^3(x)+x^3 \log ^4(x)\right )}{\left (x^3-2 x^4+x^5\right ) \log ^3(x)} \, dx=\frac {{\left (x - 1\right )} e^{3} \log \left (x\right ) - {\left (\log \left (x\right ) + 3\right )} e^{\left (\frac {1}{x^{2} \log \left (x\right )^{2}}\right )}}{x - 1} \] Input:
integrate(((x^3*log(x)^4+(2*x^3+x^2)*log(x)^3+(2*x-2)*log(x)^2+(8*x-8)*log (x)+6*x-6)*exp(1/x^2/log(x)^2)+(x^4-2*x^3+x^2)*exp(3)*log(x)^3)/(x^5-2*x^4 +x^3)/log(x)^3,x, algorithm="fricas")
Output:
((x - 1)*e^3*log(x) - (log(x) + 3)*e^(1/(x^2*log(x)^2)))/(x - 1)
Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {e^3 \left (x^2-2 x^3+x^4\right ) \log ^3(x)+e^{\frac {1}{x^2 \log ^2(x)}} \left (-6+6 x+(-8+8 x) \log (x)+(-2+2 x) \log ^2(x)+\left (x^2+2 x^3\right ) \log ^3(x)+x^3 \log ^4(x)\right )}{\left (x^3-2 x^4+x^5\right ) \log ^3(x)} \, dx=e^{3} \log {\left (x \right )} + \frac {\left (- \log {\left (x \right )} - 3\right ) e^{\frac {1}{x^{2} \log {\left (x \right )}^{2}}}}{x - 1} \] Input:
integrate(((x**3*ln(x)**4+(2*x**3+x**2)*ln(x)**3+(2*x-2)*ln(x)**2+(8*x-8)* ln(x)+6*x-6)*exp(1/x**2/ln(x)**2)+(x**4-2*x**3+x**2)*exp(3)*ln(x)**3)/(x** 5-2*x**4+x**3)/ln(x)**3,x)
Output:
exp(3)*log(x) + (-log(x) - 3)*exp(1/(x**2*log(x)**2))/(x - 1)
\[ \int \frac {e^3 \left (x^2-2 x^3+x^4\right ) \log ^3(x)+e^{\frac {1}{x^2 \log ^2(x)}} \left (-6+6 x+(-8+8 x) \log (x)+(-2+2 x) \log ^2(x)+\left (x^2+2 x^3\right ) \log ^3(x)+x^3 \log ^4(x)\right )}{\left (x^3-2 x^4+x^5\right ) \log ^3(x)} \, dx=\int { \frac {{\left (x^{4} - 2 \, x^{3} + x^{2}\right )} e^{3} \log \left (x\right )^{3} + {\left (x^{3} \log \left (x\right )^{4} + {\left (2 \, x^{3} + x^{2}\right )} \log \left (x\right )^{3} + 2 \, {\left (x - 1\right )} \log \left (x\right )^{2} + 8 \, {\left (x - 1\right )} \log \left (x\right ) + 6 \, x - 6\right )} e^{\left (\frac {1}{x^{2} \log \left (x\right )^{2}}\right )}}{{\left (x^{5} - 2 \, x^{4} + x^{3}\right )} \log \left (x\right )^{3}} \,d x } \] Input:
integrate(((x^3*log(x)^4+(2*x^3+x^2)*log(x)^3+(2*x-2)*log(x)^2+(8*x-8)*log (x)+6*x-6)*exp(1/x^2/log(x)^2)+(x^4-2*x^3+x^2)*exp(3)*log(x)^3)/(x^5-2*x^4 +x^3)/log(x)^3,x, algorithm="maxima")
Output:
e^3*log(x) + integrate((x^3*log(x)^4 + (2*x^3 + x^2)*log(x)^3 + 2*(x - 1)* log(x)^2 + 8*(x - 1)*log(x) + 6*x - 6)*e^(1/(x^2*log(x)^2))/((x^5 - 2*x^4 + x^3)*log(x)^3), x)
\[ \int \frac {e^3 \left (x^2-2 x^3+x^4\right ) \log ^3(x)+e^{\frac {1}{x^2 \log ^2(x)}} \left (-6+6 x+(-8+8 x) \log (x)+(-2+2 x) \log ^2(x)+\left (x^2+2 x^3\right ) \log ^3(x)+x^3 \log ^4(x)\right )}{\left (x^3-2 x^4+x^5\right ) \log ^3(x)} \, dx=\int { \frac {{\left (x^{4} - 2 \, x^{3} + x^{2}\right )} e^{3} \log \left (x\right )^{3} + {\left (x^{3} \log \left (x\right )^{4} + {\left (2 \, x^{3} + x^{2}\right )} \log \left (x\right )^{3} + 2 \, {\left (x - 1\right )} \log \left (x\right )^{2} + 8 \, {\left (x - 1\right )} \log \left (x\right ) + 6 \, x - 6\right )} e^{\left (\frac {1}{x^{2} \log \left (x\right )^{2}}\right )}}{{\left (x^{5} - 2 \, x^{4} + x^{3}\right )} \log \left (x\right )^{3}} \,d x } \] Input:
integrate(((x^3*log(x)^4+(2*x^3+x^2)*log(x)^3+(2*x-2)*log(x)^2+(8*x-8)*log (x)+6*x-6)*exp(1/x^2/log(x)^2)+(x^4-2*x^3+x^2)*exp(3)*log(x)^3)/(x^5-2*x^4 +x^3)/log(x)^3,x, algorithm="giac")
Output:
undef
Timed out. \[ \int \frac {e^3 \left (x^2-2 x^3+x^4\right ) \log ^3(x)+e^{\frac {1}{x^2 \log ^2(x)}} \left (-6+6 x+(-8+8 x) \log (x)+(-2+2 x) \log ^2(x)+\left (x^2+2 x^3\right ) \log ^3(x)+x^3 \log ^4(x)\right )}{\left (x^3-2 x^4+x^5\right ) \log ^3(x)} \, dx=\int \frac {{\mathrm {e}}^{\frac {1}{x^2\,{\ln \left (x\right )}^2}}\,\left (6\,x+{\ln \left (x\right )}^3\,\left (2\,x^3+x^2\right )+\ln \left (x\right )\,\left (8\,x-8\right )+x^3\,{\ln \left (x\right )}^4+{\ln \left (x\right )}^2\,\left (2\,x-2\right )-6\right )+{\mathrm {e}}^3\,{\ln \left (x\right )}^3\,\left (x^4-2\,x^3+x^2\right )}{{\ln \left (x\right )}^3\,\left (x^5-2\,x^4+x^3\right )} \,d x \] Input:
int((exp(1/(x^2*log(x)^2))*(6*x + log(x)^3*(x^2 + 2*x^3) + log(x)*(8*x - 8 ) + x^3*log(x)^4 + log(x)^2*(2*x - 2) - 6) + exp(3)*log(x)^3*(x^2 - 2*x^3 + x^4))/(log(x)^3*(x^3 - 2*x^4 + x^5)),x)
Output:
int((exp(1/(x^2*log(x)^2))*(6*x + log(x)^3*(x^2 + 2*x^3) + log(x)*(8*x - 8 ) + x^3*log(x)^4 + log(x)^2*(2*x - 2) - 6) + exp(3)*log(x)^3*(x^2 - 2*x^3 + x^4))/(log(x)^3*(x^3 - 2*x^4 + x^5)), x)
Time = 0.18 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.74 \[ \int \frac {e^3 \left (x^2-2 x^3+x^4\right ) \log ^3(x)+e^{\frac {1}{x^2 \log ^2(x)}} \left (-6+6 x+(-8+8 x) \log (x)+(-2+2 x) \log ^2(x)+\left (x^2+2 x^3\right ) \log ^3(x)+x^3 \log ^4(x)\right )}{\left (x^3-2 x^4+x^5\right ) \log ^3(x)} \, dx=\frac {-e^{\frac {1}{\mathrm {log}\left (x \right )^{2} x^{2}}} \mathrm {log}\left (x \right )-3 e^{\frac {1}{\mathrm {log}\left (x \right )^{2} x^{2}}}+\mathrm {log}\left (x \right ) e^{3} x -\mathrm {log}\left (x \right ) e^{3}}{x -1} \] Input:
int(((x^3*log(x)^4+(2*x^3+x^2)*log(x)^3+(2*x-2)*log(x)^2+(8*x-8)*log(x)+6* x-6)*exp(1/x^2/log(x)^2)+(x^4-2*x^3+x^2)*exp(3)*log(x)^3)/(x^5-2*x^4+x^3)/ log(x)^3,x)
Output:
( - e**(1/(log(x)**2*x**2))*log(x) - 3*e**(1/(log(x)**2*x**2)) + log(x)*e* *3*x - log(x)*e**3)/(x - 1)