Integrand size = 117, antiderivative size = 34 \[ \int \frac {-5+2 x+5 x^2+3 x^3+4 x^4-x^5+e^{x^2} \left (10 x^3+8 x^4-12 x^5-8 x^6+2 x^7\right )+\left (5+4 x-11 x^2+2 x^3\right ) \log \left (\frac {5 x-6 x^2+x^3}{e^2}\right )}{5 x^2+4 x^3-6 x^4-4 x^5+x^6} \, dx=e^{x^2}-\log (x)-\frac {\log \left (\frac {(-5+x) \left (-x+x^2\right )}{e^2}\right )}{x+x^2} \] Output:
exp(x^2)-ln(x)-ln((-5+x)/exp(2)*(x^2-x))/(x^2+x)
Time = 0.08 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15 \[ \int \frac {-5+2 x+5 x^2+3 x^3+4 x^4-x^5+e^{x^2} \left (10 x^3+8 x^4-12 x^5-8 x^6+2 x^7\right )+\left (5+4 x-11 x^2+2 x^3\right ) \log \left (\frac {5 x-6 x^2+x^3}{e^2}\right )}{5 x^2+4 x^3-6 x^4-4 x^5+x^6} \, dx=-\log (x)+\frac {2+e^{x^2} x (1+x)-\log \left (x \left (5-6 x+x^2\right )\right )}{x (1+x)} \] Input:
Integrate[(-5 + 2*x + 5*x^2 + 3*x^3 + 4*x^4 - x^5 + E^x^2*(10*x^3 + 8*x^4 - 12*x^5 - 8*x^6 + 2*x^7) + (5 + 4*x - 11*x^2 + 2*x^3)*Log[(5*x - 6*x^2 + x^3)/E^2])/(5*x^2 + 4*x^3 - 6*x^4 - 4*x^5 + x^6),x]
Output:
-Log[x] + (2 + E^x^2*x*(1 + x) - Log[x*(5 - 6*x + x^2)])/(x*(1 + x))
Time = 4.26 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.50, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, 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.
\(\displaystyle \int \frac {-x^5+4 x^4+3 x^3+5 x^2+\left (2 x^3-11 x^2+4 x+5\right ) \log \left (\frac {x^3-6 x^2+5 x}{e^2}\right )+e^{x^2} \left (2 x^7-8 x^6-12 x^5+8 x^4+10 x^3\right )+2 x-5}{x^6-4 x^5-6 x^4+4 x^3+5 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {-x^5+4 x^4+3 x^3+5 x^2+\left (2 x^3-11 x^2+4 x+5\right ) \log \left (\frac {x^3-6 x^2+5 x}{e^2}\right )+e^{x^2} \left (2 x^7-8 x^6-12 x^5+8 x^4+10 x^3\right )+2 x-5}{x^2 \left (x^4-4 x^3-6 x^2+4 x+5\right )}dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (\frac {-x^5+4 x^4+3 x^3+5 x^2+\left (2 x^3-11 x^2+4 x+5\right ) \log \left (\frac {x^3-6 x^2+5 x}{e^2}\right )+e^{x^2} \left (2 x^7-8 x^6-12 x^5+8 x^4+10 x^3\right )+2 x-5}{18 x^2 (x+1)}+\frac {-x^5+4 x^4+3 x^3+5 x^2+\left (2 x^3-11 x^2+4 x+5\right ) \log \left (\frac {x^3-6 x^2+5 x}{e^2}\right )+e^{x^2} \left (2 x^7-8 x^6-12 x^5+8 x^4+10 x^3\right )+2 x-5}{144 (x-5) x^2}-\frac {-x^5+4 x^4+3 x^3+5 x^2+\left (2 x^3-11 x^2+4 x+5\right ) \log \left (\frac {x^3-6 x^2+5 x}{e^2}\right )+e^{x^2} \left (2 x^7-8 x^6-12 x^5+8 x^4+10 x^3\right )+2 x-5}{16 (x-1) x^2}+\frac {-x^5+4 x^4+3 x^3+5 x^2+\left (2 x^3-11 x^2+4 x+5\right ) \log \left (\frac {x^3-6 x^2+5 x}{e^2}\right )+e^{x^2} \left (2 x^7-8 x^6-12 x^5+8 x^4+10 x^3\right )+2 x-5}{12 x^2 (x+1)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle e^{x^2}+\frac {2-\log \left (x \left (x^2-6 x+5\right )\right )}{x}-\frac {2-\log \left (x \left (x^2-6 x+5\right )\right )}{x+1}-\log (x)\) |
Input:
Int[(-5 + 2*x + 5*x^2 + 3*x^3 + 4*x^4 - x^5 + E^x^2*(10*x^3 + 8*x^4 - 12*x ^5 - 8*x^6 + 2*x^7) + (5 + 4*x - 11*x^2 + 2*x^3)*Log[(5*x - 6*x^2 + x^3)/E ^2])/(5*x^2 + 4*x^3 - 6*x^4 - 4*x^5 + x^6),x]
Output:
E^x^2 - Log[x] + (2 - Log[x*(5 - 6*x + x^2)])/x - (2 - Log[x*(5 - 6*x + x^ 2)])/(1 + 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_.)*(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 = 4.94 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.76
method | result | size |
parallelrisch | \(\frac {-11 x^{2} \ln \left (x \right )+11 x^{2} {\mathrm e}^{x^{2}}-11 x \ln \left (x \right )+11 \,{\mathrm e}^{x^{2}} x -11 \ln \left (\left (x^{3}-6 x^{2}+5 x \right ) {\mathrm e}^{-2}\right )}{11 x \left (1+x \right )}\) | \(60\) |
risch | \(-\frac {\ln \left (x^{2}-6 x +5\right )}{\left (1+x \right ) x}-\frac {i \pi \,\operatorname {csgn}\left (i \left (x^{2}-6 x +5\right )\right ) {\operatorname {csgn}\left (i x \left (x^{2}-6 x +5\right )\right )}^{2}-i \pi \,\operatorname {csgn}\left (i \left (x^{2}-6 x +5\right )\right ) \operatorname {csgn}\left (i x \left (x^{2}-6 x +5\right )\right ) \operatorname {csgn}\left (i x \right )-4+i \pi {\operatorname {csgn}\left (i x \left (x^{2}-6 x +5\right )\right )}^{2} \operatorname {csgn}\left (i x \right )-i \pi {\operatorname {csgn}\left (i x \left (x^{2}-6 x +5\right )\right )}^{3}+2 x^{2} \ln \left (x \right )-2 x^{2} {\mathrm e}^{x^{2}}+2 x \ln \left (x \right )-2 \,{\mathrm e}^{x^{2}} x +2 \ln \left (x \right )}{2 \left (1+x \right ) x}\) | \(173\) |
Input:
int(((2*x^3-11*x^2+4*x+5)*ln((x^3-6*x^2+5*x)/exp(2))+(2*x^7-8*x^6-12*x^5+8 *x^4+10*x^3)*exp(x^2)-x^5+4*x^4+3*x^3+5*x^2+2*x-5)/(x^6-4*x^5-6*x^4+4*x^3+ 5*x^2),x,method=_RETURNVERBOSE)
Output:
1/11*(-11*x^2*ln(x)+11*x^2*exp(x^2)-11*x*ln(x)+11*exp(x^2)*x-11*ln((x^3-6* x^2+5*x)/exp(2)))/x/(1+x)
Time = 0.11 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.35 \[ \int \frac {-5+2 x+5 x^2+3 x^3+4 x^4-x^5+e^{x^2} \left (10 x^3+8 x^4-12 x^5-8 x^6+2 x^7\right )+\left (5+4 x-11 x^2+2 x^3\right ) \log \left (\frac {5 x-6 x^2+x^3}{e^2}\right )}{5 x^2+4 x^3-6 x^4-4 x^5+x^6} \, dx=\frac {{\left (x^{2} + x\right )} e^{\left (x^{2}\right )} - {\left (x^{2} + x\right )} \log \left (x\right ) - \log \left ({\left (x^{3} - 6 \, x^{2} + 5 \, x\right )} e^{\left (-2\right )}\right )}{x^{2} + x} \] Input:
integrate(((2*x^3-11*x^2+4*x+5)*log((x^3-6*x^2+5*x)/exp(2))+(2*x^7-8*x^6-1 2*x^5+8*x^4+10*x^3)*exp(x^2)-x^5+4*x^4+3*x^3+5*x^2+2*x-5)/(x^6-4*x^5-6*x^4 +4*x^3+5*x^2),x, algorithm="fricas")
Output:
((x^2 + x)*e^(x^2) - (x^2 + x)*log(x) - log((x^3 - 6*x^2 + 5*x)*e^(-2)))/( x^2 + x)
Time = 0.21 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {-5+2 x+5 x^2+3 x^3+4 x^4-x^5+e^{x^2} \left (10 x^3+8 x^4-12 x^5-8 x^6+2 x^7\right )+\left (5+4 x-11 x^2+2 x^3\right ) \log \left (\frac {5 x-6 x^2+x^3}{e^2}\right )}{5 x^2+4 x^3-6 x^4-4 x^5+x^6} \, dx=e^{x^{2}} - \log {\left (x \right )} - \frac {\log {\left (\frac {x^{3} - 6 x^{2} + 5 x}{e^{2}} \right )}}{x^{2} + x} \] Input:
integrate(((2*x**3-11*x**2+4*x+5)*ln((x**3-6*x**2+5*x)/exp(2))+(2*x**7-8*x **6-12*x**5+8*x**4+10*x**3)*exp(x**2)-x**5+4*x**4+3*x**3+5*x**2+2*x-5)/(x* *6-4*x**5-6*x**4+4*x**3+5*x**2),x)
Output:
exp(x**2) - log(x) - log((x**3 - 6*x**2 + 5*x)*exp(-2))/(x**2 + x)
Leaf count of result is larger than twice the leaf count of optimal. 90 vs. \(2 (32) = 64\).
Time = 0.09 (sec) , antiderivative size = 90, normalized size of antiderivative = 2.65 \[ \int \frac {-5+2 x+5 x^2+3 x^3+4 x^4-x^5+e^{x^2} \left (10 x^3+8 x^4-12 x^5-8 x^6+2 x^7\right )+\left (5+4 x-11 x^2+2 x^3\right ) \log \left (\frac {5 x-6 x^2+x^3}{e^2}\right )}{5 x^2+4 x^3-6 x^4-4 x^5+x^6} \, dx=\frac {30 \, {\left (x^{2} + x\right )} e^{\left (x^{2}\right )} + 15 \, {\left (x^{2} + x - 2\right )} \log \left (x - 1\right ) + {\left (x^{2} + x - 30\right )} \log \left (x - 5\right ) - 30 \, x - 30 \, \log \left (x\right ) + 30}{30 \, {\left (x^{2} + x\right )}} + \frac {17 \, x + 12}{12 \, {\left (x^{2} + x\right )}} - \frac {5}{12 \, {\left (x + 1\right )}} - \frac {1}{2} \, \log \left (x - 1\right ) - \frac {1}{30} \, \log \left (x - 5\right ) - \log \left (x\right ) \] Input:
integrate(((2*x^3-11*x^2+4*x+5)*log((x^3-6*x^2+5*x)/exp(2))+(2*x^7-8*x^6-1 2*x^5+8*x^4+10*x^3)*exp(x^2)-x^5+4*x^4+3*x^3+5*x^2+2*x-5)/(x^6-4*x^5-6*x^4 +4*x^3+5*x^2),x, algorithm="maxima")
Output:
1/30*(30*(x^2 + x)*e^(x^2) + 15*(x^2 + x - 2)*log(x - 1) + (x^2 + x - 30)* log(x - 5) - 30*x - 30*log(x) + 30)/(x^2 + x) + 1/12*(17*x + 12)/(x^2 + x) - 5/12/(x + 1) - 1/2*log(x - 1) - 1/30*log(x - 5) - log(x)
Time = 0.13 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.50 \[ \int \frac {-5+2 x+5 x^2+3 x^3+4 x^4-x^5+e^{x^2} \left (10 x^3+8 x^4-12 x^5-8 x^6+2 x^7\right )+\left (5+4 x-11 x^2+2 x^3\right ) \log \left (\frac {5 x-6 x^2+x^3}{e^2}\right )}{5 x^2+4 x^3-6 x^4-4 x^5+x^6} \, dx=\frac {x^{2} e^{\left (x^{2}\right )} - x^{2} \log \left (x\right ) + x e^{\left (x^{2}\right )} - x \log \left (x\right ) - \log \left (x^{3} - 6 \, x^{2} + 5 \, x\right ) + 2}{x^{2} + x} \] Input:
integrate(((2*x^3-11*x^2+4*x+5)*log((x^3-6*x^2+5*x)/exp(2))+(2*x^7-8*x^6-1 2*x^5+8*x^4+10*x^3)*exp(x^2)-x^5+4*x^4+3*x^3+5*x^2+2*x-5)/(x^6-4*x^5-6*x^4 +4*x^3+5*x^2),x, algorithm="giac")
Output:
(x^2*e^(x^2) - x^2*log(x) + x*e^(x^2) - x*log(x) - log(x^3 - 6*x^2 + 5*x) + 2)/(x^2 + x)
Time = 2.70 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {-5+2 x+5 x^2+3 x^3+4 x^4-x^5+e^{x^2} \left (10 x^3+8 x^4-12 x^5-8 x^6+2 x^7\right )+\left (5+4 x-11 x^2+2 x^3\right ) \log \left (\frac {5 x-6 x^2+x^3}{e^2}\right )}{5 x^2+4 x^3-6 x^4-4 x^5+x^6} \, dx={\mathrm {e}}^{x^2}-\ln \left (x\right )-\frac {\ln \left ({\mathrm {e}}^{-2}\,\left (x^3-6\,x^2+5\,x\right )\right )}{x^2+x} \] Input:
int((2*x + exp(x^2)*(10*x^3 + 8*x^4 - 12*x^5 - 8*x^6 + 2*x^7) + log(exp(-2 )*(5*x - 6*x^2 + x^3))*(4*x - 11*x^2 + 2*x^3 + 5) + 5*x^2 + 3*x^3 + 4*x^4 - x^5 - 5)/(5*x^2 + 4*x^3 - 6*x^4 - 4*x^5 + x^6),x)
Output:
exp(x^2) - log(x) - log(exp(-2)*(5*x - 6*x^2 + x^3))/(x + x^2)
Time = 0.16 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.79 \[ \int \frac {-5+2 x+5 x^2+3 x^3+4 x^4-x^5+e^{x^2} \left (10 x^3+8 x^4-12 x^5-8 x^6+2 x^7\right )+\left (5+4 x-11 x^2+2 x^3\right ) \log \left (\frac {5 x-6 x^2+x^3}{e^2}\right )}{5 x^2+4 x^3-6 x^4-4 x^5+x^6} \, dx=\frac {e^{x^{2}} x^{2}+e^{x^{2}} x -\mathrm {log}\left (\frac {x^{3}-6 x^{2}+5 x}{e^{2}}\right )-\mathrm {log}\left (x \right ) x^{2}-\mathrm {log}\left (x \right ) x +x^{2}+x}{x \left (x +1\right )} \] Input:
int(((2*x^3-11*x^2+4*x+5)*log((x^3-6*x^2+5*x)/exp(2))+(2*x^7-8*x^6-12*x^5+ 8*x^4+10*x^3)*exp(x^2)-x^5+4*x^4+3*x^3+5*x^2+2*x-5)/(x^6-4*x^5-6*x^4+4*x^3 +5*x^2),x)
Output:
(e**(x**2)*x**2 + e**(x**2)*x - log((x**3 - 6*x**2 + 5*x)/e**2) - log(x)*x **2 - log(x)*x + x**2 + x)/(x*(x + 1))