Integrand size = 57, antiderivative size = 21 \[ \int \frac {4 x+30 x^2+16 x^3+2 x^4+e^x \left (-4+2 x+x^2\right )-x^2 \log (x)}{32 x^2+16 x^3+2 x^4} \, dx=x+\frac {3+\frac {e^x}{x}+\log (x)}{8+2 x} \] Output:
x+(3+ln(x)+exp(x)/x)/(2*x+8)
Time = 0.17 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.52 \[ \int \frac {4 x+30 x^2+16 x^3+2 x^4+e^x \left (-4+2 x+x^2\right )-x^2 \log (x)}{32 x^2+16 x^3+2 x^4} \, dx=\frac {e^x+x \left (3+8 x+2 x^2\right )+x \log (x)}{2 x (4+x)} \] Input:
Integrate[(4*x + 30*x^2 + 16*x^3 + 2*x^4 + E^x*(-4 + 2*x + x^2) - x^2*Log[ x])/(32*x^2 + 16*x^3 + 2*x^4),x]
Output:
(E^x + x*(3 + 8*x + 2*x^2) + x*Log[x])/(2*x*(4 + x))
Leaf count is larger than twice the leaf count of optimal. \(51\) vs. \(2(21)=42\).
Time = 1.19 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.43, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, 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.
\(\displaystyle \int \frac {2 x^4+16 x^3+30 x^2+e^x \left (x^2+2 x-4\right )-x^2 \log (x)+4 x}{2 x^4+16 x^3+32 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {2 x^4+16 x^3+30 x^2+e^x \left (x^2+2 x-4\right )-x^2 \log (x)+4 x}{x^2 \left (2 x^2+16 x+32\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {2 x^4+16 x^3+30 x^2+e^x \left (x^2+2 x-4\right )-x^2 \log (x)+4 x}{x^2 \left (\sqrt {2} x+4 \sqrt {2}\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x^2}{(x+4)^2}+\frac {e^x \left (x^2+2 x-4\right )}{2 (x+4)^2 x^2}+\frac {8 x}{(x+4)^2}+\frac {15}{(x+4)^2}+\frac {2}{(x+4)^2 x}-\frac {\log (x)}{2 (x+4)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x-\frac {e^x}{8 (x+4)}+\frac {3}{2 (x+4)}+\frac {e^x}{8 x}-\frac {x \log (x)}{8 (x+4)}+\frac {\log (x)}{8}\) |
Input:
Int[(4*x + 30*x^2 + 16*x^3 + 2*x^4 + E^x*(-4 + 2*x + x^2) - x^2*Log[x])/(3 2*x^2 + 16*x^3 + 2*x^4),x]
Output:
E^x/(8*x) + x + 3/(2*(4 + x)) - E^x/(8*(4 + x)) + Log[x]/8 - (x*Log[x])/(8 *(4 + 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 = 0.58 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24
method | result | size |
norman | \(\frac {x^{3}-\frac {29 x}{2}+\frac {x \ln \left (x \right )}{2}+\frac {{\mathrm e}^{x}}{2}}{\left (4+x \right ) x}\) | \(26\) |
parallelrisch | \(-\frac {-8 x^{3}-4 x \ln \left (x \right )+116 x -4 \,{\mathrm e}^{x}}{8 x \left (4+x \right )}\) | \(29\) |
risch | \(\frac {\ln \left (x \right )}{2 x +8}+\frac {2 x^{3}+8 x^{2}+3 x +{\mathrm e}^{x}}{2 \left (4+x \right ) x}\) | \(37\) |
default | \(-\frac {{\mathrm e}^{x}}{8 \left (4+x \right )}+\frac {{\mathrm e}^{x}}{8 x}-\frac {\ln \left (x \right ) x}{8 \left (4+x \right )}+x +\frac {3}{2 \left (4+x \right )}+\frac {\ln \left (x \right )}{8}\) | \(40\) |
parts | \(-\frac {{\mathrm e}^{x}}{8 \left (4+x \right )}+\frac {{\mathrm e}^{x}}{8 x}-\frac {\ln \left (x \right ) x}{8 \left (4+x \right )}+x +\frac {3}{2 \left (4+x \right )}+\frac {\ln \left (x \right )}{8}\) | \(40\) |
orering | \(\frac {\left (4 x^{6}-4 x^{5}+3 x^{4}+21 x^{3}-42 x^{2}-2 x -8\right ) \left (-x^{2} \ln \left (x \right )+\left (x^{2}+2 x -4\right ) {\mathrm e}^{x}+2 x^{4}+16 x^{3}+30 x^{2}+4 x \right )}{\left (4 x^{5}-12 x^{4}+23 x^{3}-23 x^{2}-2 x +2\right ) \left (2 x^{4}+16 x^{3}+32 x^{2}\right )}+\frac {x \left (4 x^{6}+4 x^{5}-11 x^{4}+34 x^{3}+43 x^{2}-279 x -16\right ) \left (\frac {-2 x \ln \left (x \right )+59 x +\left (2+2 x \right ) {\mathrm e}^{x}+\left (x^{2}+2 x -4\right ) {\mathrm e}^{x}+8 x^{3}+48 x^{2}+4}{2 x^{4}+16 x^{3}+32 x^{2}}-\frac {\left (-x^{2} \ln \left (x \right )+\left (x^{2}+2 x -4\right ) {\mathrm e}^{x}+2 x^{4}+16 x^{3}+30 x^{2}+4 x \right ) \left (8 x^{3}+48 x^{2}+64 x \right )}{\left (2 x^{4}+16 x^{3}+32 x^{2}\right )^{2}}\right )}{4 x^{5}-12 x^{4}+23 x^{3}-23 x^{2}-2 x +2}-\frac {\left (4 x^{4}-11 x^{2}+23 x +1\right ) \left (4+x \right ) x^{2} \left (\frac {-2 \ln \left (x \right )+57+2 \,{\mathrm e}^{x}+2 \left (2+2 x \right ) {\mathrm e}^{x}+\left (x^{2}+2 x -4\right ) {\mathrm e}^{x}+24 x^{2}+96 x}{2 x^{4}+16 x^{3}+32 x^{2}}-\frac {2 \left (-2 x \ln \left (x \right )+59 x +\left (2+2 x \right ) {\mathrm e}^{x}+\left (x^{2}+2 x -4\right ) {\mathrm e}^{x}+8 x^{3}+48 x^{2}+4\right ) \left (8 x^{3}+48 x^{2}+64 x \right )}{\left (2 x^{4}+16 x^{3}+32 x^{2}\right )^{2}}+\frac {2 \left (-x^{2} \ln \left (x \right )+\left (x^{2}+2 x -4\right ) {\mathrm e}^{x}+2 x^{4}+16 x^{3}+30 x^{2}+4 x \right ) \left (8 x^{3}+48 x^{2}+64 x \right )^{2}}{\left (2 x^{4}+16 x^{3}+32 x^{2}\right )^{3}}-\frac {\left (-x^{2} \ln \left (x \right )+\left (x^{2}+2 x -4\right ) {\mathrm e}^{x}+2 x^{4}+16 x^{3}+30 x^{2}+4 x \right ) \left (24 x^{2}+96 x +64\right )}{\left (2 x^{4}+16 x^{3}+32 x^{2}\right )^{2}}\right )}{4 x^{5}-12 x^{4}+23 x^{3}-23 x^{2}-2 x +2}\) | \(625\) |
Input:
int((-x^2*ln(x)+(x^2+2*x-4)*exp(x)+2*x^4+16*x^3+30*x^2+4*x)/(2*x^4+16*x^3+ 32*x^2),x,method=_RETURNVERBOSE)
Output:
(x^3-29/2*x+1/2*x*ln(x)+1/2*exp(x))/(4+x)/x
Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.48 \[ \int \frac {4 x+30 x^2+16 x^3+2 x^4+e^x \left (-4+2 x+x^2\right )-x^2 \log (x)}{32 x^2+16 x^3+2 x^4} \, dx=\frac {2 \, x^{3} + 8 \, x^{2} + x \log \left (x\right ) + 3 \, x + e^{x}}{2 \, {\left (x^{2} + 4 \, x\right )}} \] Input:
integrate((-x^2*log(x)+(x^2+2*x-4)*exp(x)+2*x^4+16*x^3+30*x^2+4*x)/(2*x^4+ 16*x^3+32*x^2),x, algorithm="fricas")
Output:
1/2*(2*x^3 + 8*x^2 + x*log(x) + 3*x + e^x)/(x^2 + 4*x)
Time = 0.13 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29 \[ \int \frac {4 x+30 x^2+16 x^3+2 x^4+e^x \left (-4+2 x+x^2\right )-x^2 \log (x)}{32 x^2+16 x^3+2 x^4} \, dx=x + \frac {e^{x}}{2 x^{2} + 8 x} + \frac {\log {\left (x \right )}}{2 x + 8} + \frac {3}{2 x + 8} \] Input:
integrate((-x**2*ln(x)+(x**2+2*x-4)*exp(x)+2*x**4+16*x**3+30*x**2+4*x)/(2* x**4+16*x**3+32*x**2),x)
Output:
x + exp(x)/(2*x**2 + 8*x) + log(x)/(2*x + 8) + 3/(2*x + 8)
Time = 0.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29 \[ \int \frac {4 x+30 x^2+16 x^3+2 x^4+e^x \left (-4+2 x+x^2\right )-x^2 \log (x)}{32 x^2+16 x^3+2 x^4} \, dx=x + \frac {x \log \left (x\right ) + e^{x}}{2 \, {\left (x^{2} + 4 \, x\right )}} + \frac {3}{2 \, {\left (x + 4\right )}} \] Input:
integrate((-x^2*log(x)+(x^2+2*x-4)*exp(x)+2*x^4+16*x^3+30*x^2+4*x)/(2*x^4+ 16*x^3+32*x^2),x, algorithm="maxima")
Output:
x + 1/2*(x*log(x) + e^x)/(x^2 + 4*x) + 3/2/(x + 4)
Time = 0.12 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.48 \[ \int \frac {4 x+30 x^2+16 x^3+2 x^4+e^x \left (-4+2 x+x^2\right )-x^2 \log (x)}{32 x^2+16 x^3+2 x^4} \, dx=\frac {2 \, x^{3} + 8 \, x^{2} + x \log \left (x\right ) + 3 \, x + e^{x}}{2 \, {\left (x^{2} + 4 \, x\right )}} \] Input:
integrate((-x^2*log(x)+(x^2+2*x-4)*exp(x)+2*x^4+16*x^3+30*x^2+4*x)/(2*x^4+ 16*x^3+32*x^2),x, algorithm="giac")
Output:
1/2*(2*x^3 + 8*x^2 + x*log(x) + 3*x + e^x)/(x^2 + 4*x)
Time = 0.60 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {4 x+30 x^2+16 x^3+2 x^4+e^x \left (-4+2 x+x^2\right )-x^2 \log (x)}{32 x^2+16 x^3+2 x^4} \, dx=x+\frac {\frac {{\mathrm {e}}^x}{2}+x\,\left (\frac {\ln \left (x\right )}{2}+\frac {3}{2}\right )}{x\,\left (x+4\right )} \] Input:
int((4*x - x^2*log(x) + exp(x)*(2*x + x^2 - 4) + 30*x^2 + 16*x^3 + 2*x^4)/ (32*x^2 + 16*x^3 + 2*x^4),x)
Output:
x + (exp(x)/2 + x*(log(x)/2 + 3/2))/(x*(x + 4))
Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.48 \[ \int \frac {4 x+30 x^2+16 x^3+2 x^4+e^x \left (-4+2 x+x^2\right )-x^2 \log (x)}{32 x^2+16 x^3+2 x^4} \, dx=\frac {4 e^{x}+4 \,\mathrm {log}\left (x \right ) x +8 x^{3}+29 x^{2}}{8 x \left (x +4\right )} \] Input:
int((-x^2*log(x)+(x^2+2*x-4)*exp(x)+2*x^4+16*x^3+30*x^2+4*x)/(2*x^4+16*x^3 +32*x^2),x)
Output:
(4*e**x + 4*log(x)*x + 8*x**3 + 29*x**2)/(8*x*(x + 4))