Integrand size = 66, antiderivative size = 25 \[ \int \frac {-256+32 x+255 x^2-32 x^3+x^4+e^{-10+x} \left (32 x^2-17 x^4+x^5\right )+\left (256-32 x+x^2\right ) \log (x)}{256 x^2-32 x^3+x^4} \, dx=-2+x+\frac {e^{-10+x} (-2+x) x}{-16+x}-\frac {\log (x)}{x} \] Output:
exp(x-10)/(x-16)*(-2+x)*x-ln(x)/x+x-2
Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {-256+32 x+255 x^2-32 x^3+x^4+e^{-10+x} \left (32 x^2-17 x^4+x^5\right )+\left (256-32 x+x^2\right ) \log (x)}{256 x^2-32 x^3+x^4} \, dx=x+\frac {e^{-10+x} (-2+x) x}{-16+x}-\frac {\log (x)}{x} \] Input:
Integrate[(-256 + 32*x + 255*x^2 - 32*x^3 + x^4 + E^(-10 + x)*(32*x^2 - 17 *x^4 + x^5) + (256 - 32*x + x^2)*Log[x])/(256*x^2 - 32*x^3 + x^4),x]
Output:
x + (E^(-10 + x)*(-2 + x)*x)/(-16 + x) - Log[x]/x
Time = 0.79 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {2026, 7239, 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.
\(\displaystyle \int \frac {x^4-32 x^3+255 x^2+\left (x^2-32 x+256\right ) \log (x)+e^{x-10} \left (x^5-17 x^4+32 x^2\right )+32 x-256}{x^4-32 x^3+256 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {x^4-32 x^3+255 x^2+\left (x^2-32 x+256\right ) \log (x)+e^{x-10} \left (x^5-17 x^4+32 x^2\right )+32 x-256}{x^2 \left (x^2-32 x+256\right )}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \left (-\frac {1}{x^2}+\frac {\log (x)}{x^2}+\frac {e^{x-10} \left (x^3-17 x^2+32\right )}{(x-16)^2}+1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle e^{x-10} x+x+14 e^{x-10}-\frac {224 e^{x-10}}{16-x}-\frac {\log (x)}{x}\) |
Input:
Int[(-256 + 32*x + 255*x^2 - 32*x^3 + x^4 + E^(-10 + x)*(32*x^2 - 17*x^4 + x^5) + (256 - 32*x + x^2)*Log[x])/(256*x^2 - 32*x^3 + x^4),x]
Output:
14*E^(-10 + x) - (224*E^(-10 + x))/(16 - x) + x + E^(-10 + x)*x - Log[x]/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_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.71 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24
method | result | size |
risch | \(-\frac {\ln \left (x \right )}{x}+\frac {x \left (x \,{\mathrm e}^{x -10}+x -2 \,{\mathrm e}^{x -10}-16\right )}{x -16}\) | \(31\) |
default | \(x -\frac {\ln \left (x \right )}{x}+{\mathrm e}^{x -10} \left (x -10\right )+24 \,{\mathrm e}^{x -10}+\frac {224 \,{\mathrm e}^{x -10}}{x -16}\) | \(35\) |
parts | \(x -\frac {\ln \left (x \right )}{x}+{\mathrm e}^{x -10} \left (x -10\right )+24 \,{\mathrm e}^{x -10}+\frac {224 \,{\mathrm e}^{x -10}}{x -16}\) | \(35\) |
parallelrisch | \(\frac {16 x^{3} {\mathrm e}^{x -10}+16 x^{3}-32 x^{2} {\mathrm e}^{x -10}-16 x \ln \left (x \right )-4096 x +256 \ln \left (x \right )}{16 x \left (x -16\right )}\) | \(47\) |
orering | \(\frac {x \left (2 x^{9}+36 x^{8}-4403 x^{7}+66626 x^{6}+7728 x^{5}-3423164 x^{4}-3285832 x^{3}+3534336 x^{2}-123904 x +32768\right ) \left (\left (x^{2}-32 x +256\right ) \ln \left (x \right )+\left (x^{5}-17 x^{4}+32 x^{2}\right ) {\mathrm e}^{x -10}+x^{4}-32 x^{3}+255 x^{2}+32 x -256\right )}{\left (2 x^{9}-88 x^{8}+1057 x^{7}+982 x^{6}-53836 x^{5}-53508 x^{4}+25344 x^{3}-64512 x^{2}+41984 x +32768\right ) \left (x^{4}-32 x^{3}+256 x^{2}\right )}+\frac {x^{2} \left (2 x^{9}-32 x^{8}-1577 x^{7}+38247 x^{6}-128909 x^{5}-1384822 x^{4}+1713190 x^{3}+336000 x^{2}-3086848 x +131072\right ) \left (\frac {\left (2 x -32\right ) \ln \left (x \right )+\frac {x^{2}-32 x +256}{x}+\left (5 x^{4}-68 x^{3}+64 x \right ) {\mathrm e}^{x -10}+\left (x^{5}-17 x^{4}+32 x^{2}\right ) {\mathrm e}^{x -10}+4 x^{3}-96 x^{2}+510 x +32}{x^{4}-32 x^{3}+256 x^{2}}-\frac {\left (\left (x^{2}-32 x +256\right ) \ln \left (x \right )+\left (x^{5}-17 x^{4}+32 x^{2}\right ) {\mathrm e}^{x -10}+x^{4}-32 x^{3}+255 x^{2}+32 x -256\right ) \left (4 x^{3}-96 x^{2}+512 x \right )}{\left (x^{4}-32 x^{3}+256 x^{2}\right )^{2}}\right )}{2 x^{9}-88 x^{8}+1057 x^{7}+982 x^{6}-53836 x^{5}-53508 x^{4}+25344 x^{3}-64512 x^{2}+41984 x +32768}-\frac {x^{3} \left (2 x^{7}-1481 x^{5}+11589 x^{4}+67586 x^{3}-33102 x^{2}-63328 x +2048\right ) \left (x -16\right ) \left (\frac {2 \ln \left (x \right )+\frac {4 x -64}{x}-\frac {x^{2}-32 x +256}{x^{2}}+\left (20 x^{3}-204 x^{2}+64\right ) {\mathrm e}^{x -10}+2 \left (5 x^{4}-68 x^{3}+64 x \right ) {\mathrm e}^{x -10}+\left (x^{5}-17 x^{4}+32 x^{2}\right ) {\mathrm e}^{x -10}+12 x^{2}-192 x +510}{x^{4}-32 x^{3}+256 x^{2}}-\frac {2 \left (\left (2 x -32\right ) \ln \left (x \right )+\frac {x^{2}-32 x +256}{x}+\left (5 x^{4}-68 x^{3}+64 x \right ) {\mathrm e}^{x -10}+\left (x^{5}-17 x^{4}+32 x^{2}\right ) {\mathrm e}^{x -10}+4 x^{3}-96 x^{2}+510 x +32\right ) \left (4 x^{3}-96 x^{2}+512 x \right )}{\left (x^{4}-32 x^{3}+256 x^{2}\right )^{2}}+\frac {2 \left (\left (x^{2}-32 x +256\right ) \ln \left (x \right )+\left (x^{5}-17 x^{4}+32 x^{2}\right ) {\mathrm e}^{x -10}+x^{4}-32 x^{3}+255 x^{2}+32 x -256\right ) \left (4 x^{3}-96 x^{2}+512 x \right )^{2}}{\left (x^{4}-32 x^{3}+256 x^{2}\right )^{3}}-\frac {\left (\left (x^{2}-32 x +256\right ) \ln \left (x \right )+\left (x^{5}-17 x^{4}+32 x^{2}\right ) {\mathrm e}^{x -10}+x^{4}-32 x^{3}+255 x^{2}+32 x -256\right ) \left (12 x^{2}-192 x +512\right )}{\left (x^{4}-32 x^{3}+256 x^{2}\right )^{2}}\right )}{2 x^{9}-88 x^{8}+1057 x^{7}+982 x^{6}-53836 x^{5}-53508 x^{4}+25344 x^{3}-64512 x^{2}+41984 x +32768}\) | \(886\) |
Input:
int(((x^2-32*x+256)*ln(x)+(x^5-17*x^4+32*x^2)*exp(x-10)+x^4-32*x^3+255*x^2 +32*x-256)/(x^4-32*x^3+256*x^2),x,method=_RETURNVERBOSE)
Output:
-ln(x)/x+x*(x*exp(x-10)+x-2*exp(x-10)-16)/(x-16)
Time = 0.10 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.60 \[ \int \frac {-256+32 x+255 x^2-32 x^3+x^4+e^{-10+x} \left (32 x^2-17 x^4+x^5\right )+\left (256-32 x+x^2\right ) \log (x)}{256 x^2-32 x^3+x^4} \, dx=\frac {x^{3} - 16 \, x^{2} + {\left (x^{3} - 2 \, x^{2}\right )} e^{\left (x - 10\right )} - {\left (x - 16\right )} \log \left (x\right )}{x^{2} - 16 \, x} \] Input:
integrate(((x^2-32*x+256)*log(x)+(x^5-17*x^4+32*x^2)*exp(x-10)+x^4-32*x^3+ 255*x^2+32*x-256)/(x^4-32*x^3+256*x^2),x, algorithm="fricas")
Output:
(x^3 - 16*x^2 + (x^3 - 2*x^2)*e^(x - 10) - (x - 16)*log(x))/(x^2 - 16*x)
Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {-256+32 x+255 x^2-32 x^3+x^4+e^{-10+x} \left (32 x^2-17 x^4+x^5\right )+\left (256-32 x+x^2\right ) \log (x)}{256 x^2-32 x^3+x^4} \, dx=x + \frac {\left (x^{2} - 2 x\right ) e^{x - 10}}{x - 16} - \frac {\log {\left (x \right )}}{x} \] Input:
integrate(((x**2-32*x+256)*ln(x)+(x**5-17*x**4+32*x**2)*exp(x-10)+x**4-32* x**3+255*x**2+32*x-256)/(x**4-32*x**3+256*x**2),x)
Output:
x + (x**2 - 2*x)*exp(x - 10)/(x - 16) - log(x)/x
\[ \int \frac {-256+32 x+255 x^2-32 x^3+x^4+e^{-10+x} \left (32 x^2-17 x^4+x^5\right )+\left (256-32 x+x^2\right ) \log (x)}{256 x^2-32 x^3+x^4} \, dx=\int { \frac {x^{4} - 32 \, x^{3} + 255 \, x^{2} + {\left (x^{5} - 17 \, x^{4} + 32 \, x^{2}\right )} e^{\left (x - 10\right )} + {\left (x^{2} - 32 \, x + 256\right )} \log \left (x\right ) + 32 \, x - 256}{x^{4} - 32 \, x^{3} + 256 \, x^{2}} \,d x } \] Input:
integrate(((x^2-32*x+256)*log(x)+(x^5-17*x^4+32*x^2)*exp(x-10)+x^4-32*x^3+ 255*x^2+32*x-256)/(x^4-32*x^3+256*x^2),x, algorithm="maxima")
Output:
x - 32*e^6*exp_integral_e(2, -x + 16)/(x - 16) + 2*(x - 8)/(x^2 - 16*x) - (log(x) + 1)/x - 1/(x - 16) + integrate((x^3 - 17*x^2)*e^x/(x^2*e^10 - 32* x*e^10 + 256*e^10), x)
Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (24) = 48\).
Time = 0.12 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.20 \[ \int \frac {-256+32 x+255 x^2-32 x^3+x^4+e^{-10+x} \left (32 x^2-17 x^4+x^5\right )+\left (256-32 x+x^2\right ) \log (x)}{256 x^2-32 x^3+x^4} \, dx=\frac {x^{3} e^{10} + x^{3} e^{x} - 16 \, x^{2} e^{10} - 2 \, x^{2} e^{x} - x e^{10} \log \left (x\right ) + 16 \, e^{10} \log \left (x\right )}{x^{2} e^{10} - 16 \, x e^{10}} \] Input:
integrate(((x^2-32*x+256)*log(x)+(x^5-17*x^4+32*x^2)*exp(x-10)+x^4-32*x^3+ 255*x^2+32*x-256)/(x^4-32*x^3+256*x^2),x, algorithm="giac")
Output:
(x^3*e^10 + x^3*e^x - 16*x^2*e^10 - 2*x^2*e^x - x*e^10*log(x) + 16*e^10*lo g(x))/(x^2*e^10 - 16*x*e^10)
Time = 6.98 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {-256+32 x+255 x^2-32 x^3+x^4+e^{-10+x} \left (32 x^2-17 x^4+x^5\right )+\left (256-32 x+x^2\right ) \log (x)}{256 x^2-32 x^3+x^4} \, dx=x-\frac {\ln \left (x\right )}{x}-\frac {{\mathrm {e}}^{x-10}\,\left (2\,x-x^2\right )}{x-16} \] Input:
int((32*x + exp(x - 10)*(32*x^2 - 17*x^4 + x^5) + log(x)*(x^2 - 32*x + 256 ) + 255*x^2 - 32*x^3 + x^4 - 256)/(256*x^2 - 32*x^3 + x^4),x)
Output:
x - log(x)/x - (exp(x - 10)*(2*x - x^2))/(x - 16)
Time = 0.16 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.32 \[ \int \frac {-256+32 x+255 x^2-32 x^3+x^4+e^{-10+x} \left (32 x^2-17 x^4+x^5\right )+\left (256-32 x+x^2\right ) \log (x)}{256 x^2-32 x^3+x^4} \, dx=\frac {e^{x} x^{3}-2 e^{x} x^{2}-\mathrm {log}\left (x \right ) e^{10} x +16 \,\mathrm {log}\left (x \right ) e^{10}+e^{10} x^{3}-16 e^{10} x^{2}}{e^{10} x \left (x -16\right )} \] Input:
int(((x^2-32*x+256)*log(x)+(x^5-17*x^4+32*x^2)*exp(x-10)+x^4-32*x^3+255*x^ 2+32*x-256)/(x^4-32*x^3+256*x^2),x)
Output:
(e**x*x**3 - 2*e**x*x**2 - log(x)*e**10*x + 16*log(x)*e**10 + e**10*x**3 - 16*e**10*x**2)/(e**10*x*(x - 16))