Integrand size = 94, antiderivative size = 26 \begin {dmath*} \int \frac {24+e^x (-24-12 x)+12 x+e^2 (24+12 x)+e^{-3+2 x} (24+12 x)+\left (-12 x-12 e^2 x+e^x \left (-12 x-12 x^2\right )+e^{-3+2 x} \left (36 x+24 x^2\right )\right ) \log (x)}{4 x+4 x^2+x^3} \, dx=\frac {12 \left (1+e^2-e^x+e^{-3+2 x}\right ) \log (x)}{2+x} \end {dmath*}
Time = 0.72 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \begin {dmath*} \int \frac {24+e^x (-24-12 x)+12 x+e^2 (24+12 x)+e^{-3+2 x} (24+12 x)+\left (-12 x-12 e^2 x+e^x \left (-12 x-12 x^2\right )+e^{-3+2 x} \left (36 x+24 x^2\right )\right ) \log (x)}{4 x+4 x^2+x^3} \, dx=\frac {12 \left (e^3+e^5+e^{2 x}-e^{3+x}\right ) \log (x)}{e^3 (2+x)} \end {dmath*}
Integrate[(24 + E^x*(-24 - 12*x) + 12*x + E^2*(24 + 12*x) + E^(-3 + 2*x)*( 24 + 12*x) + (-12*x - 12*E^2*x + E^x*(-12*x - 12*x^2) + E^(-3 + 2*x)*(36*x + 24*x^2))*Log[x])/(4*x + 4*x^2 + x^3),x]
Leaf count is larger than twice the leaf count of optimal. \(81\) vs. \(2(26)=52\).
Time = 1.96 (sec) , antiderivative size = 81, normalized size of antiderivative = 3.12, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, 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.
\(\displaystyle \int \frac {\left (e^x \left (-12 x^2-12 x\right )+e^{2 x-3} \left (24 x^2+36 x\right )-12 e^2 x-12 x\right ) \log (x)+e^x (-12 x-24)+12 x+e^{2 x-3} (12 x+24)+e^2 (12 x+24)+24}{x^3+4 x^2+4 x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (e^x \left (-12 x^2-12 x\right )+e^{2 x-3} \left (24 x^2+36 x\right )-12 e^2 x-12 x\right ) \log (x)+e^x (-12 x-24)+12 x+e^{2 x-3} (12 x+24)+e^2 (12 x+24)+24}{x \left (x^2+4 x+4\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {\left (e^x \left (-12 x^2-12 x\right )+e^{2 x-3} \left (24 x^2+36 x\right )-12 e^2 x-12 x\right ) \log (x)+e^x (-12 x-24)+12 x+e^{2 x-3} (12 x+24)+e^2 (12 x+24)+24}{x (x+2)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {12 e^x \left (x^2 \log (x)+x+x \log (x)+2\right )}{x (x+2)^2}+\frac {12 e^{2 x-3} \left (2 x^2 \log (x)+x+3 x \log (x)+2\right )}{x (x+2)^2}+\frac {12 e^2}{x (x+2)}+\frac {24}{x (x+2)^2}+\frac {12}{(x+2)^2}-\frac {12 \left (1+e^2\right ) \log (x)}{(x+2)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {12 e^x \log (x)}{x+2}+\frac {12 e^{2 x-3} \log (x)}{x+2}-\frac {6 \left (1+e^2\right ) x \log (x)}{x+2}+6 e^2 \log (x)+6 \log (x)+6 \left (1+e^2\right ) \log (x+2)-6 e^2 \log (x+2)-6 \log (x+2)\) |
Int[(24 + E^x*(-24 - 12*x) + 12*x + E^2*(24 + 12*x) + E^(-3 + 2*x)*(24 + 1 2*x) + (-12*x - 12*E^2*x + E^x*(-12*x - 12*x^2) + E^(-3 + 2*x)*(36*x + 24* x^2))*Log[x])/(4*x + 4*x^2 + x^3),x]
6*Log[x] + 6*E^2*Log[x] - (12*E^x*Log[x])/(2 + x) + (12*E^(-3 + 2*x)*Log[x ])/(2 + x) - (6*(1 + E^2)*x*Log[x])/(2 + x) - 6*Log[2 + x] - 6*E^2*Log[2 + x] + 6*(1 + E^2)*Log[2 + x]
3.9.32.3.1 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 = 1.34 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08
method | result | size |
risch | \(\frac {12 \left ({\mathrm e}^{-5+2 x}-{\mathrm e}^{-2+x}+{\mathrm e}^{-2}+1\right ) {\mathrm e}^{2} \ln \left (x \right )}{2+x}\) | \(28\) |
parallelrisch | \(\frac {12 \,{\mathrm e}^{2} \ln \left (x \right )-12 \,{\mathrm e}^{x} \ln \left (x \right )+12 \ln \left (x \right ) {\mathrm e}^{-3+2 x}+12 \ln \left (x \right )}{2+x}\) | \(34\) |
default | \(\frac {-24 \,{\mathrm e}^{x} \ln \left (x \right )-12 x \,{\mathrm e}^{x} \ln \left (x \right )}{\left (2+x \right )^{2}}+\frac {24 \ln \left (x \right ) {\mathrm e}^{-3+2 x}+12 \ln \left (x \right ) {\mathrm e}^{-3+2 x} x}{\left (2+x \right )^{2}}+\left (12 \,{\mathrm e}^{2}+12\right ) \left (\frac {\ln \left (x \right )}{2}-\frac {\ln \left (2+x \right )}{2}\right )+\left (-12 \,{\mathrm e}^{2}-12\right ) \left (-\frac {\ln \left (2+x \right )}{2}+\frac {x \ln \left (x \right )}{4+2 x}\right )\) | \(92\) |
parts | \(\frac {-24 \,{\mathrm e}^{x} \ln \left (x \right )-12 x \,{\mathrm e}^{x} \ln \left (x \right )}{\left (2+x \right )^{2}}+\frac {24 \ln \left (x \right ) {\mathrm e}^{-3+2 x}+12 \ln \left (x \right ) {\mathrm e}^{-3+2 x} x}{\left (2+x \right )^{2}}+\left (12 \,{\mathrm e}^{2}+12\right ) \left (\frac {\ln \left (x \right )}{2}-\frac {\ln \left (2+x \right )}{2}\right )+\left (-12 \,{\mathrm e}^{2}-12\right ) \left (-\frac {\ln \left (2+x \right )}{2}+\frac {x \ln \left (x \right )}{4+2 x}\right )\) | \(92\) |
int((((24*x^2+36*x)*exp(-3+2*x)+(-12*x^2-12*x)*exp(x)-12*exp(2)*x-12*x)*ln (x)+(12*x+24)*exp(-3+2*x)+(-12*x-24)*exp(x)+(12*x+24)*exp(2)+12*x+24)/(x^3 +4*x^2+4*x),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \begin {dmath*} \int \frac {24+e^x (-24-12 x)+12 x+e^2 (24+12 x)+e^{-3+2 x} (24+12 x)+\left (-12 x-12 e^2 x+e^x \left (-12 x-12 x^2\right )+e^{-3+2 x} \left (36 x+24 x^2\right )\right ) \log (x)}{4 x+4 x^2+x^3} \, dx=\frac {12 \, {\left (e^{5} + e^{3} + e^{\left (2 \, x\right )} - e^{\left (x + 3\right )}\right )} e^{\left (-3\right )} \log \left (x\right )}{x + 2} \end {dmath*}
integrate((((24*x^2+36*x)*exp(-3+2*x)+(-12*x^2-12*x)*exp(x)-12*exp(2)*x-12 *x)*log(x)+(12*x+24)*exp(-3+2*x)+(-12*x-24)*exp(x)+(12*x+24)*exp(2)+12*x+2 4)/(x^3+4*x^2+4*x),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (22) = 44\).
Time = 0.24 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.73 \begin {dmath*} \int \frac {24+e^x (-24-12 x)+12 x+e^2 (24+12 x)+e^{-3+2 x} (24+12 x)+\left (-12 x-12 e^2 x+e^x \left (-12 x-12 x^2\right )+e^{-3+2 x} \left (36 x+24 x^2\right )\right ) \log (x)}{4 x+4 x^2+x^3} \, dx=\frac {\left (12 x \log {\left (x \right )} + 24 \log {\left (x \right )}\right ) e^{2 x} + \left (- 12 x e^{3} \log {\left (x \right )} - 24 e^{3} \log {\left (x \right )}\right ) e^{x}}{x^{2} e^{3} + 4 x e^{3} + 4 e^{3}} + \frac {\left (12 + 12 e^{2}\right ) \log {\left (x \right )}}{x + 2} \end {dmath*}
integrate((((24*x**2+36*x)*exp(-3+2*x)+(-12*x**2-12*x)*exp(x)-12*exp(2)*x- 12*x)*ln(x)+(12*x+24)*exp(-3+2*x)+(-12*x-24)*exp(x)+(12*x+24)*exp(2)+12*x+ 24)/(x**3+4*x**2+4*x),x)
((12*x*log(x) + 24*log(x))*exp(2*x) + (-12*x*exp(3)*log(x) - 24*exp(3)*log (x))*exp(x))/(x**2*exp(3) + 4*x*exp(3) + 4*exp(3)) + (12 + 12*exp(2))*log( x)/(x + 2)
Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (23) = 46\).
Time = 0.24 (sec) , antiderivative size = 95, normalized size of antiderivative = 3.65 \begin {dmath*} \int \frac {24+e^x (-24-12 x)+12 x+e^2 (24+12 x)+e^{-3+2 x} (24+12 x)+\left (-12 x-12 e^2 x+e^x \left (-12 x-12 x^2\right )+e^{-3+2 x} \left (36 x+24 x^2\right )\right ) \log (x)}{4 x+4 x^2+x^3} \, dx=6 \, {\left (\frac {2}{x + 2} - \log \left (x + 2\right ) + \log \left (x\right )\right )} e^{2} + 6 \, {\left (e^{2} + 1\right )} \log \left (x + 2\right ) - 6 \, {\left (e^{2} + 1\right )} \log \left (x\right ) + \frac {12 \, {\left ({\left (e^{5} + e^{3}\right )} \log \left (x\right ) + e^{\left (2 \, x\right )} \log \left (x\right ) - e^{\left (x + 3\right )} \log \left (x\right )\right )}}{x e^{3} + 2 \, e^{3}} - \frac {12 \, e^{2}}{x + 2} - 6 \, \log \left (x + 2\right ) + 6 \, \log \left (x\right ) \end {dmath*}
integrate((((24*x^2+36*x)*exp(-3+2*x)+(-12*x^2-12*x)*exp(x)-12*exp(2)*x-12 *x)*log(x)+(12*x+24)*exp(-3+2*x)+(-12*x-24)*exp(x)+(12*x+24)*exp(2)+12*x+2 4)/(x^3+4*x^2+4*x),x, algorithm=\
6*(2/(x + 2) - log(x + 2) + log(x))*e^2 + 6*(e^2 + 1)*log(x + 2) - 6*(e^2 + 1)*log(x) + 12*((e^5 + e^3)*log(x) + e^(2*x)*log(x) - e^(x + 3)*log(x))/ (x*e^3 + 2*e^3) - 12*e^2/(x + 2) - 6*log(x + 2) + 6*log(x)
Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50 \begin {dmath*} \int \frac {24+e^x (-24-12 x)+12 x+e^2 (24+12 x)+e^{-3+2 x} (24+12 x)+\left (-12 x-12 e^2 x+e^x \left (-12 x-12 x^2\right )+e^{-3+2 x} \left (36 x+24 x^2\right )\right ) \log (x)}{4 x+4 x^2+x^3} \, dx=\frac {12 \, {\left (e^{5} \log \left (x\right ) + e^{3} \log \left (x\right ) + e^{\left (2 \, x\right )} \log \left (x\right ) - e^{\left (x + 3\right )} \log \left (x\right )\right )}}{x e^{3} + 2 \, e^{3}} \end {dmath*}
integrate((((24*x^2+36*x)*exp(-3+2*x)+(-12*x^2-12*x)*exp(x)-12*exp(2)*x-12 *x)*log(x)+(12*x+24)*exp(-3+2*x)+(-12*x-24)*exp(x)+(12*x+24)*exp(2)+12*x+2 4)/(x^3+4*x^2+4*x),x, algorithm=\
Timed out. \begin {dmath*} \int \frac {24+e^x (-24-12 x)+12 x+e^2 (24+12 x)+e^{-3+2 x} (24+12 x)+\left (-12 x-12 e^2 x+e^x \left (-12 x-12 x^2\right )+e^{-3+2 x} \left (36 x+24 x^2\right )\right ) \log (x)}{4 x+4 x^2+x^3} \, dx=\int \frac {12\,x+{\mathrm {e}}^{2\,x-3}\,\left (12\,x+24\right )-{\mathrm {e}}^x\,\left (12\,x+24\right )-\ln \left (x\right )\,\left (12\,x+12\,x\,{\mathrm {e}}^2-{\mathrm {e}}^{2\,x-3}\,\left (24\,x^2+36\,x\right )+{\mathrm {e}}^x\,\left (12\,x^2+12\,x\right )\right )+{\mathrm {e}}^2\,\left (12\,x+24\right )+24}{x^3+4\,x^2+4\,x} \,d x \end {dmath*}
int((12*x + exp(2*x - 3)*(12*x + 24) - exp(x)*(12*x + 24) - log(x)*(12*x + 12*x*exp(2) - exp(2*x - 3)*(36*x + 24*x^2) + exp(x)*(12*x + 12*x^2)) + ex p(2)*(12*x + 24) + 24)/(4*x + 4*x^2 + x^3),x)