Integrand size = 89, antiderivative size = 35 \begin {dmath*} \int \frac {-240 e^4-120 e^2 x-15 x^2+e^{\frac {4+3 x+2 x^2}{4 e^2+x}} \left (-4 x^2+2 x^4+e^2 \left (12 x^2+16 x^3\right )\right )}{16 e^4 x^2+8 e^2 x^3+x^4} \, dx=e^{\frac {-x+x^2+(2+x)^2}{4 e^2+x}}+\frac {3 (5-x)}{x} \end {dmath*}
Time = 1.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.80 \begin {dmath*} \int \frac {-240 e^4-120 e^2 x-15 x^2+e^{\frac {4+3 x+2 x^2}{4 e^2+x}} \left (-4 x^2+2 x^4+e^2 \left (12 x^2+16 x^3\right )\right )}{16 e^4 x^2+8 e^2 x^3+x^4} \, dx=e^{\frac {4+3 x+2 x^2}{4 e^2+x}}+\frac {15}{x} \end {dmath*}
Integrate[(-240*E^4 - 120*E^2*x - 15*x^2 + E^((4 + 3*x + 2*x^2)/(4*E^2 + x ))*(-4*x^2 + 2*x^4 + E^2*(12*x^2 + 16*x^3)))/(16*E^4*x^2 + 8*E^2*x^3 + x^4 ),x]
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.
\(\displaystyle \int \frac {-15 x^2+e^{\frac {2 x^2+3 x+4}{x+4 e^2}} \left (2 x^4-4 x^2+e^2 \left (16 x^3+12 x^2\right )\right )-120 e^2 x-240 e^4}{x^4+8 e^2 x^3+16 e^4 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {-15 x^2+e^{\frac {2 x^2+3 x+4}{x+4 e^2}} \left (2 x^4-4 x^2+e^2 \left (16 x^3+12 x^2\right )\right )-120 e^2 x-240 e^4}{x^2 \left (x^2+8 e^2 x+16 e^4\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {-15 x^2+e^{\frac {2 x^2+3 x+4}{x+4 e^2}} \left (2 x^4-4 x^2+e^2 \left (16 x^3+12 x^2\right )\right )-120 e^2 x-240 e^4}{x^2 \left (x+4 e^2\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 e^{\frac {2 x^2+3 x+4}{x+4 e^2}} \left (x^2+8 e^2 x-2 \left (1-3 e^2\right )\right )}{\left (x+4 e^2\right )^2}-\frac {15}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \int e^{\frac {2 x^2+3 x+4}{x+4 e^2}}dx-4 \left (1-3 e^2+8 e^4\right ) \int \frac {e^{\frac {2 x^2+3 x+4}{x+4 e^2}}}{\left (x+4 e^2\right )^2}dx+\frac {15}{x}\) |
Int[(-240*E^4 - 120*E^2*x - 15*x^2 + E^((4 + 3*x + 2*x^2)/(4*E^2 + x))*(-4 *x^2 + 2*x^4 + E^2*(12*x^2 + 16*x^3)))/(16*E^4*x^2 + 8*E^2*x^3 + x^4),x]
3.2.42.3.1 Defintions of rubi rules used
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 = 1.13 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.77
method | result | size |
risch | \(\frac {15}{x}+{\mathrm e}^{\frac {2 x^{2}+3 x +4}{4 \,{\mathrm e}^{2}+x}}\) | \(27\) |
parallelrisch | \(\frac {{\mathrm e}^{\frac {2 x^{2}+3 x +4}{4 \,{\mathrm e}^{2}+x}} x +15}{x}\) | \(31\) |
parts | \(\frac {15}{x}+\frac {{\mathrm e}^{\frac {2 x^{2}+3 x +4}{4 \,{\mathrm e}^{2}+x}} x +4 \,{\mathrm e}^{2} {\mathrm e}^{\frac {2 x^{2}+3 x +4}{4 \,{\mathrm e}^{2}+x}}}{4 \,{\mathrm e}^{2}+x}\) | \(71\) |
norman | \(\frac {{\mathrm e}^{\frac {2 x^{2}+3 x +4}{4 \,{\mathrm e}^{2}+x}} x^{2}+15 x +60 \,{\mathrm e}^{2}+4 x \,{\mathrm e}^{2} {\mathrm e}^{\frac {2 x^{2}+3 x +4}{4 \,{\mathrm e}^{2}+x}}}{x \left (4 \,{\mathrm e}^{2}+x \right )}\) | \(80\) |
int((((16*x^3+12*x^2)*exp(1)^2+2*x^4-4*x^2)*exp((2*x^2+3*x+4)/(4*exp(1)^2+ x))-240*exp(1)^4-120*x*exp(1)^2-15*x^2)/(16*x^2*exp(1)^4+8*x^3*exp(1)^2+x^ 4),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.80 \begin {dmath*} \int \frac {-240 e^4-120 e^2 x-15 x^2+e^{\frac {4+3 x+2 x^2}{4 e^2+x}} \left (-4 x^2+2 x^4+e^2 \left (12 x^2+16 x^3\right )\right )}{16 e^4 x^2+8 e^2 x^3+x^4} \, dx=\frac {x e^{\left (\frac {2 \, x^{2} + 3 \, x + 4}{x + 4 \, e^{2}}\right )} + 15}{x} \end {dmath*}
integrate((((16*x^3+12*x^2)*exp(1)^2+2*x^4-4*x^2)*exp((2*x^2+3*x+4)/(4*exp (1)^2+x))-240*exp(1)^4-120*x*exp(1)^2-15*x^2)/(16*x^2*exp(1)^4+8*x^3*exp(1 )^2+x^4),x, algorithm=\
Time = 0.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.57 \begin {dmath*} \int \frac {-240 e^4-120 e^2 x-15 x^2+e^{\frac {4+3 x+2 x^2}{4 e^2+x}} \left (-4 x^2+2 x^4+e^2 \left (12 x^2+16 x^3\right )\right )}{16 e^4 x^2+8 e^2 x^3+x^4} \, dx=e^{\frac {2 x^{2} + 3 x + 4}{x + 4 e^{2}}} + \frac {15}{x} \end {dmath*}
integrate((((16*x**3+12*x**2)*exp(1)**2+2*x**4-4*x**2)*exp((2*x**2+3*x+4)/ (4*exp(1)**2+x))-240*exp(1)**4-120*x*exp(1)**2-15*x**2)/(16*x**2*exp(1)**4 +8*x**3*exp(1)**2+x**4),x)
Leaf count of result is larger than twice the leaf count of optimal. 132 vs. \(2 (31) = 62\).
Time = 0.28 (sec) , antiderivative size = 132, normalized size of antiderivative = 3.77 \begin {dmath*} \int \frac {-240 e^4-120 e^2 x-15 x^2+e^{\frac {4+3 x+2 x^2}{4 e^2+x}} \left (-4 x^2+2 x^4+e^2 \left (12 x^2+16 x^3\right )\right )}{16 e^4 x^2+8 e^2 x^3+x^4} \, dx=-\frac {15}{2} \, {\left (e^{\left (-6\right )} \log \left (x + 4 \, e^{2}\right ) - e^{\left (-6\right )} \log \left (x\right ) - \frac {4 \, {\left (x + 2 \, e^{2}\right )}}{x^{2} e^{4} + 4 \, x e^{6}}\right )} e^{4} + \frac {15}{2} \, {\left (e^{\left (-4\right )} \log \left (x + 4 \, e^{2}\right ) - e^{\left (-4\right )} \log \left (x\right ) - \frac {4}{x e^{2} + 4 \, e^{4}}\right )} e^{2} + \frac {15}{x + 4 \, e^{2}} + e^{\left (2 \, x + \frac {32 \, e^{4}}{x + 4 \, e^{2}} - \frac {12 \, e^{2}}{x + 4 \, e^{2}} + \frac {4}{x + 4 \, e^{2}} - 8 \, e^{2} + 3\right )} \end {dmath*}
integrate((((16*x^3+12*x^2)*exp(1)^2+2*x^4-4*x^2)*exp((2*x^2+3*x+4)/(4*exp (1)^2+x))-240*exp(1)^4-120*x*exp(1)^2-15*x^2)/(16*x^2*exp(1)^4+8*x^3*exp(1 )^2+x^4),x, algorithm=\
-15/2*(e^(-6)*log(x + 4*e^2) - e^(-6)*log(x) - 4*(x + 2*e^2)/(x^2*e^4 + 4* x*e^6))*e^4 + 15/2*(e^(-4)*log(x + 4*e^2) - e^(-4)*log(x) - 4/(x*e^2 + 4*e ^4))*e^2 + 15/(x + 4*e^2) + e^(2*x + 32*e^4/(x + 4*e^2) - 12*e^2/(x + 4*e^ 2) + 4/(x + 4*e^2) - 8*e^2 + 3)
Time = 0.40 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.49 \begin {dmath*} \int \frac {-240 e^4-120 e^2 x-15 x^2+e^{\frac {4+3 x+2 x^2}{4 e^2+x}} \left (-4 x^2+2 x^4+e^2 \left (12 x^2+16 x^3\right )\right )}{16 e^4 x^2+8 e^2 x^3+x^4} \, dx=\frac {{\left (x e^{\left ({\left (2 \, e^{2} + 1\right )} e^{\left (-2\right )} + \frac {2 \, x^{2} e^{2} + 3 \, x e^{2} - x}{x e^{2} + 4 \, e^{4}}\right )} + 15 \, e^{2}\right )} e^{\left (-2\right )}}{x} \end {dmath*}
integrate((((16*x^3+12*x^2)*exp(1)^2+2*x^4-4*x^2)*exp((2*x^2+3*x+4)/(4*exp (1)^2+x))-240*exp(1)^4-120*x*exp(1)^2-15*x^2)/(16*x^2*exp(1)^4+8*x^3*exp(1 )^2+x^4),x, algorithm=\
Time = 0.62 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.26 \begin {dmath*} \int \frac {-240 e^4-120 e^2 x-15 x^2+e^{\frac {4+3 x+2 x^2}{4 e^2+x}} \left (-4 x^2+2 x^4+e^2 \left (12 x^2+16 x^3\right )\right )}{16 e^4 x^2+8 e^2 x^3+x^4} \, dx=\frac {15}{x}+{\mathrm {e}}^{\frac {3\,x}{x+4\,{\mathrm {e}}^2}}\,{\mathrm {e}}^{\frac {2\,x^2}{x+4\,{\mathrm {e}}^2}}\,{\mathrm {e}}^{\frac {4}{x+4\,{\mathrm {e}}^2}} \end {dmath*}