Integrand size = 52, antiderivative size = 22 \[ \int \frac {-25 x-10 x^2-x^3+e^{\frac {e^4}{x}} \left (x^2+e^4 (5+x)\right )}{25 x^2+10 x^3+x^4} \, dx=2-\frac {e^{\frac {e^4}{x}}}{5+x}-\log (x) \] Output:
2-ln(x)-exp(exp(4)/x)/(5+x)
Time = 0.11 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {-25 x-10 x^2-x^3+e^{\frac {e^4}{x}} \left (x^2+e^4 (5+x)\right )}{25 x^2+10 x^3+x^4} \, dx=-\frac {e^{\frac {e^4}{x}}}{5+x}-\log (x) \] Input:
Integrate[(-25*x - 10*x^2 - x^3 + E^(E^4/x)*(x^2 + E^4*(5 + x)))/(25*x^2 + 10*x^3 + x^4),x]
Output:
-(E^(E^4/x)/(5 + x)) - Log[x]
Time = 0.65 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2026, 2007, 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.
\(\displaystyle \int \frac {-x^3-10 x^2+e^{\frac {e^4}{x}} \left (x^2+e^4 (x+5)\right )-25 x}{x^4+10 x^3+25 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {-x^3-10 x^2+e^{\frac {e^4}{x}} \left (x^2+e^4 (x+5)\right )-25 x}{x^2 \left (x^2+10 x+25\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {-x^3-10 x^2+e^{\frac {e^4}{x}} \left (x^2+e^4 (x+5)\right )-25 x}{x^2 (x+5)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{\frac {e^4}{x}} \left (x^2+e^4 x+5 e^4\right )}{x^2 (x+5)^2}-\frac {1}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {e^{\frac {e^4}{x}}}{x+5}-\log (x)\) |
Input:
Int[(-25*x - 10*x^2 - x^3 + E^(E^4/x)*(x^2 + E^4*(5 + x)))/(25*x^2 + 10*x^ 3 + x^4),x]
Output:
-(E^(E^4/x)/(5 + x)) - Log[x]
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
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])
Time = 1.35 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91
method | result | size |
norman | \(-\frac {{\mathrm e}^{\frac {{\mathrm e}^{4}}{x}}}{5+x}-\ln \left (x \right )\) | \(20\) |
risch | \(-\frac {{\mathrm e}^{\frac {{\mathrm e}^{4}}{x}}}{5+x}-\ln \left (x \right )\) | \(20\) |
parallelrisch | \(-\frac {x \ln \left (x \right )+5 \ln \left (x \right )+{\mathrm e}^{\frac {{\mathrm e}^{4}}{x}}}{5+x}\) | \(24\) |
parts | \(-{\mathrm e}^{-4} \left ({\mathrm e}^{8} \left (-\frac {{\mathrm e}^{\frac {{\mathrm e}^{4}}{x}}}{5 \left (\frac {5 \,{\mathrm e}^{4}}{x}+{\mathrm e}^{4}\right )}-\frac {{\mathrm e}^{-\frac {{\mathrm e}^{4}}{5}} \operatorname {expIntegral}_{1}\left (-\frac {{\mathrm e}^{4}}{x}-\frac {{\mathrm e}^{4}}{5}\right )}{25}\right )+{\mathrm e}^{8} \left (\frac {{\mathrm e}^{4} {\mathrm e}^{\frac {{\mathrm e}^{4}}{x}}}{\frac {125 \,{\mathrm e}^{4}}{x}+25 \,{\mathrm e}^{4}}-\left (-\frac {{\mathrm e}^{4}}{125}+\frac {1}{25}\right ) {\mathrm e}^{-\frac {{\mathrm e}^{4}}{5}} \operatorname {expIntegral}_{1}\left (-\frac {{\mathrm e}^{4}}{x}-\frac {{\mathrm e}^{4}}{5}\right )\right )+5 \,{\mathrm e}^{4} \left (\frac {{\mathrm e}^{\frac {{\mathrm e}^{4}}{x}}}{25}-\frac {{\mathrm e}^{\frac {{\mathrm e}^{4}}{x}} {\mathrm e}^{8}}{125 \left (\frac {5 \,{\mathrm e}^{4}}{x}+{\mathrm e}^{4}\right )}-\frac {\left (\frac {{\mathrm e}^{8}}{25}-\frac {2 \,{\mathrm e}^{4}}{5}\right ) {\mathrm e}^{-\frac {{\mathrm e}^{4}}{5}} \operatorname {expIntegral}_{1}\left (-\frac {{\mathrm e}^{4}}{x}-\frac {{\mathrm e}^{4}}{5}\right )}{25}\right )\right )-\ln \left (x \right )\) | \(186\) |
derivativedivides | \(-{\mathrm e}^{4} \left ({\mathrm e}^{8} \left (-\frac {{\mathrm e}^{\frac {{\mathrm e}^{4}}{x}} {\mathrm e}^{-8}}{5 \left (\frac {5 \,{\mathrm e}^{4}}{x}+{\mathrm e}^{4}\right )}-\frac {{\mathrm e}^{-8} {\mathrm e}^{-\frac {{\mathrm e}^{4}}{5}} \operatorname {expIntegral}_{1}\left (-\frac {{\mathrm e}^{4}}{x}-\frac {{\mathrm e}^{4}}{5}\right )}{25}\right )+{\mathrm e}^{8} \left (\frac {{\mathrm e}^{\frac {{\mathrm e}^{4}}{x}} {\mathrm e}^{-4}}{\frac {125 \,{\mathrm e}^{4}}{x}+25 \,{\mathrm e}^{4}}+\frac {\left ({\mathrm e}^{4}-5\right ) {\mathrm e}^{-8} {\mathrm e}^{-\frac {{\mathrm e}^{4}}{5}} \operatorname {expIntegral}_{1}\left (-\frac {{\mathrm e}^{4}}{x}-\frac {{\mathrm e}^{4}}{5}\right )}{125}\right )-{\mathrm e}^{-4} \ln \left (\frac {{\mathrm e}^{4}}{x}\right )+5 \,{\mathrm e}^{4} \left (\frac {{\mathrm e}^{\frac {{\mathrm e}^{4}}{x}} {\mathrm e}^{-8}}{25}-\frac {{\mathrm e}^{\frac {{\mathrm e}^{4}}{x}}}{125 \left (\frac {5 \,{\mathrm e}^{4}}{x}+{\mathrm e}^{4}\right )}-\frac {{\mathrm e}^{-4} \left ({\mathrm e}^{4}-10\right ) {\mathrm e}^{-\frac {{\mathrm e}^{4}}{5}} \operatorname {expIntegral}_{1}\left (-\frac {{\mathrm e}^{4}}{x}-\frac {{\mathrm e}^{4}}{5}\right )}{625}\right )\right )\) | \(199\) |
default | \(-{\mathrm e}^{4} \left ({\mathrm e}^{8} \left (-\frac {{\mathrm e}^{\frac {{\mathrm e}^{4}}{x}} {\mathrm e}^{-8}}{5 \left (\frac {5 \,{\mathrm e}^{4}}{x}+{\mathrm e}^{4}\right )}-\frac {{\mathrm e}^{-8} {\mathrm e}^{-\frac {{\mathrm e}^{4}}{5}} \operatorname {expIntegral}_{1}\left (-\frac {{\mathrm e}^{4}}{x}-\frac {{\mathrm e}^{4}}{5}\right )}{25}\right )+{\mathrm e}^{8} \left (\frac {{\mathrm e}^{\frac {{\mathrm e}^{4}}{x}} {\mathrm e}^{-4}}{\frac {125 \,{\mathrm e}^{4}}{x}+25 \,{\mathrm e}^{4}}+\frac {\left ({\mathrm e}^{4}-5\right ) {\mathrm e}^{-8} {\mathrm e}^{-\frac {{\mathrm e}^{4}}{5}} \operatorname {expIntegral}_{1}\left (-\frac {{\mathrm e}^{4}}{x}-\frac {{\mathrm e}^{4}}{5}\right )}{125}\right )-{\mathrm e}^{-4} \ln \left (\frac {{\mathrm e}^{4}}{x}\right )+5 \,{\mathrm e}^{4} \left (\frac {{\mathrm e}^{\frac {{\mathrm e}^{4}}{x}} {\mathrm e}^{-8}}{25}-\frac {{\mathrm e}^{\frac {{\mathrm e}^{4}}{x}}}{125 \left (\frac {5 \,{\mathrm e}^{4}}{x}+{\mathrm e}^{4}\right )}-\frac {{\mathrm e}^{-4} \left ({\mathrm e}^{4}-10\right ) {\mathrm e}^{-\frac {{\mathrm e}^{4}}{5}} \operatorname {expIntegral}_{1}\left (-\frac {{\mathrm e}^{4}}{x}-\frac {{\mathrm e}^{4}}{5}\right )}{625}\right )\right )\) | \(199\) |
Input:
int((((5+x)*exp(4)+x^2)*exp(exp(4)/x)-x^3-10*x^2-25*x)/(x^4+10*x^3+25*x^2) ,x,method=_RETURNVERBOSE)
Output:
-exp(exp(4)/x)/(5+x)-ln(x)
Time = 0.10 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {-25 x-10 x^2-x^3+e^{\frac {e^4}{x}} \left (x^2+e^4 (5+x)\right )}{25 x^2+10 x^3+x^4} \, dx=-\frac {{\left (x + 5\right )} \log \left (x\right ) + e^{\left (\frac {e^{4}}{x}\right )}}{x + 5} \] Input:
integrate((((5+x)*exp(4)+x^2)*exp(exp(4)/x)-x^3-10*x^2-25*x)/(x^4+10*x^3+2 5*x^2),x, algorithm="fricas")
Output:
-((x + 5)*log(x) + e^(e^4/x))/(x + 5)
Time = 0.09 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.64 \[ \int \frac {-25 x-10 x^2-x^3+e^{\frac {e^4}{x}} \left (x^2+e^4 (5+x)\right )}{25 x^2+10 x^3+x^4} \, dx=- \log {\left (x \right )} - \frac {e^{\frac {e^{4}}{x}}}{x + 5} \] Input:
integrate((((5+x)*exp(4)+x**2)*exp(exp(4)/x)-x**3-10*x**2-25*x)/(x**4+10*x **3+25*x**2),x)
Output:
-log(x) - exp(exp(4)/x)/(x + 5)
Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {-25 x-10 x^2-x^3+e^{\frac {e^4}{x}} \left (x^2+e^4 (5+x)\right )}{25 x^2+10 x^3+x^4} \, dx=-\frac {e^{\left (\frac {e^{4}}{x}\right )}}{x + 5} - \log \left (x\right ) \] Input:
integrate((((5+x)*exp(4)+x^2)*exp(exp(4)/x)-x^3-10*x^2-25*x)/(x^4+10*x^3+2 5*x^2),x, algorithm="maxima")
Output:
-e^(e^4/x)/(x + 5) - log(x)
Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (20) = 40\).
Time = 0.13 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.45 \[ \int \frac {-25 x-10 x^2-x^3+e^{\frac {e^4}{x}} \left (x^2+e^4 (5+x)\right )}{25 x^2+10 x^3+x^4} \, dx=\frac {{\left (e^{8} \log \left (\frac {e^{4}}{x}\right ) + \frac {5 \, e^{8} \log \left (\frac {e^{4}}{x}\right )}{x} - \frac {e^{\left (\frac {e^{4}}{x} + 8\right )}}{x}\right )} e^{\left (-4\right )}}{\frac {5 \, e^{4}}{x} + e^{4}} \] Input:
integrate((((5+x)*exp(4)+x^2)*exp(exp(4)/x)-x^3-10*x^2-25*x)/(x^4+10*x^3+2 5*x^2),x, algorithm="giac")
Output:
(e^8*log(e^4/x) + 5*e^8*log(e^4/x)/x - e^(e^4/x + 8)/x)*e^(-4)/(5*e^4/x + e^4)
Time = 3.47 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {-25 x-10 x^2-x^3+e^{\frac {e^4}{x}} \left (x^2+e^4 (5+x)\right )}{25 x^2+10 x^3+x^4} \, dx=-\ln \left (x\right )-\frac {{\mathrm {e}}^{\frac {{\mathrm {e}}^4}{x}}}{x+5} \] Input:
int(-(25*x - exp(exp(4)/x)*(exp(4)*(x + 5) + x^2) + 10*x^2 + x^3)/(25*x^2 + 10*x^3 + x^4),x)
Output:
- log(x) - exp(exp(4)/x)/(x + 5)
Time = 0.19 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {-25 x-10 x^2-x^3+e^{\frac {e^4}{x}} \left (x^2+e^4 (5+x)\right )}{25 x^2+10 x^3+x^4} \, dx=\frac {-e^{\frac {e^{4}}{x}}-\mathrm {log}\left (x \right ) x -5 \,\mathrm {log}\left (x \right )}{x +5} \] Input:
int((((5+x)*exp(4)+x^2)*exp(exp(4)/x)-x^3-10*x^2-25*x)/(x^4+10*x^3+25*x^2) ,x)
Output:
( - e**(e**4/x) - log(x)*x - 5*log(x))/(x + 5)