Integrand size = 134, antiderivative size = 20 \[ \int \frac {-24+546 x^2+3380 x^4+\left (-1+47 x^2+260 x^4\right ) \log \left (\frac {1+5 x^2}{x^2}\right )+\left (x^2+5 x^4\right ) \log ^2\left (\frac {1+5 x^2}{x^2}\right )}{e^4 \left (676 x^2+3380 x^4\right )+e^4 \left (52 x^2+260 x^4\right ) \log \left (\frac {1+5 x^2}{x^2}\right )+e^4 \left (x^2+5 x^4\right ) \log ^2\left (\frac {1+5 x^2}{x^2}\right )} \, dx=\frac {x+\frac {1}{x+x \left (25+\log \left (5+\frac {1}{x^2}\right )\right )}}{e^4} \]
Time = 0.35 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {-24+546 x^2+3380 x^4+\left (-1+47 x^2+260 x^4\right ) \log \left (\frac {1+5 x^2}{x^2}\right )+\left (x^2+5 x^4\right ) \log ^2\left (\frac {1+5 x^2}{x^2}\right )}{e^4 \left (676 x^2+3380 x^4\right )+e^4 \left (52 x^2+260 x^4\right ) \log \left (\frac {1+5 x^2}{x^2}\right )+e^4 \left (x^2+5 x^4\right ) \log ^2\left (\frac {1+5 x^2}{x^2}\right )} \, dx=\frac {x+\frac {1}{x \left (26+\log \left (5+\frac {1}{x^2}\right )\right )}}{e^4} \]
Integrate[(-24 + 546*x^2 + 3380*x^4 + (-1 + 47*x^2 + 260*x^4)*Log[(1 + 5*x ^2)/x^2] + (x^2 + 5*x^4)*Log[(1 + 5*x^2)/x^2]^2)/(E^4*(676*x^2 + 3380*x^4) + E^4*(52*x^2 + 260*x^4)*Log[(1 + 5*x^2)/x^2] + E^4*(x^2 + 5*x^4)*Log[(1 + 5*x^2)/x^2]^2),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.
\(\displaystyle \int \frac {3380 x^4+546 x^2+\left (5 x^4+x^2\right ) \log ^2\left (\frac {5 x^2+1}{x^2}\right )+\left (260 x^4+47 x^2-1\right ) \log \left (\frac {5 x^2+1}{x^2}\right )-24}{e^4 \left (3380 x^4+676 x^2\right )+e^4 \left (5 x^4+x^2\right ) \log ^2\left (\frac {5 x^2+1}{x^2}\right )+e^4 \left (260 x^4+52 x^2\right ) \log \left (\frac {5 x^2+1}{x^2}\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {3380 x^4+546 x^2+\left (5 x^4+x^2\right ) \log ^2\left (\frac {1}{x^2}+5\right )+\left (260 x^4+47 x^2-1\right ) \log \left (\frac {1}{x^2}+5\right )-24}{e^4 x^2 \left (5 x^2+1\right ) \left (\log \left (\frac {1}{x^2}+5\right )+26\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {-3380 x^4-546 x^2-\left (5 x^4+x^2\right ) \log ^2\left (5+\frac {1}{x^2}\right )+\left (-260 x^4-47 x^2+1\right ) \log \left (5+\frac {1}{x^2}\right )+24}{x^2 \left (5 x^2+1\right ) \left (\log \left (5+\frac {1}{x^2}\right )+26\right )^2}dx}{e^4}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {-3380 x^4-546 x^2-\left (5 x^4+x^2\right ) \log ^2\left (5+\frac {1}{x^2}\right )+\left (-260 x^4-47 x^2+1\right ) \log \left (5+\frac {1}{x^2}\right )+24}{x^2 \left (5 x^2+1\right ) \left (\log \left (5+\frac {1}{x^2}\right )+26\right )^2}dx}{e^4}\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle -\frac {\int \left (\frac {1}{x^2 \left (\log \left (5+\frac {1}{x^2}\right )+26\right )}-1-\frac {2}{x^2 \left (\log \left (5+\frac {1}{x^2}\right )+26\right )^2 \left (5 x^2+1\right )}\right )dx}{e^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-2 \int \frac {1}{x^2 \left (5 x^2+1\right ) \left (\log \left (5+\frac {1}{x^2}\right )+26\right )^2}dx+\int \frac {1}{x^2 \left (\log \left (5+\frac {1}{x^2}\right )+26\right )}dx-x}{e^4}\) |
Int[(-24 + 546*x^2 + 3380*x^4 + (-1 + 47*x^2 + 260*x^4)*Log[(1 + 5*x^2)/x^ 2] + (x^2 + 5*x^4)*Log[(1 + 5*x^2)/x^2]^2)/(E^4*(676*x^2 + 3380*x^4) + E^4 *(52*x^2 + 260*x^4)*Log[(1 + 5*x^2)/x^2] + E^4*(x^2 + 5*x^4)*Log[(1 + 5*x^ 2)/x^2]^2),x]
3.10.29.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 4.61 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40
method | result | size |
risch | \(x \,{\mathrm e}^{-4}+\frac {{\mathrm e}^{-4}}{x \left (\ln \left (\frac {5 x^{2}+1}{x^{2}}\right )+26\right )}\) | \(28\) |
parallelrisch | \(-\frac {\left (-1-x^{2} \ln \left (\frac {5 x^{2}+1}{x^{2}}\right )-26 x^{2}\right ) {\mathrm e}^{-4}}{x \left (\ln \left (\frac {5 x^{2}+1}{x^{2}}\right )+26\right )}\) | \(50\) |
norman | \(\frac {{\mathrm e}^{-4}+x^{2} {\mathrm e}^{-4} \ln \left (\frac {5 x^{2}+1}{x^{2}}\right )+26 \,{\mathrm e}^{-4} x^{2}}{x \left (\ln \left (\frac {5 x^{2}+1}{x^{2}}\right )+26\right )}\) | \(55\) |
int(((5*x^4+x^2)*ln((5*x^2+1)/x^2)^2+(260*x^4+47*x^2-1)*ln((5*x^2+1)/x^2)+ 3380*x^4+546*x^2-24)/((5*x^4+x^2)*exp(4)*ln((5*x^2+1)/x^2)^2+(260*x^4+52*x ^2)*exp(4)*ln((5*x^2+1)/x^2)+(3380*x^4+676*x^2)*exp(4)),x,method=_RETURNVE RBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (19) = 38\).
Time = 0.25 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.40 \[ \int \frac {-24+546 x^2+3380 x^4+\left (-1+47 x^2+260 x^4\right ) \log \left (\frac {1+5 x^2}{x^2}\right )+\left (x^2+5 x^4\right ) \log ^2\left (\frac {1+5 x^2}{x^2}\right )}{e^4 \left (676 x^2+3380 x^4\right )+e^4 \left (52 x^2+260 x^4\right ) \log \left (\frac {1+5 x^2}{x^2}\right )+e^4 \left (x^2+5 x^4\right ) \log ^2\left (\frac {1+5 x^2}{x^2}\right )} \, dx=\frac {x^{2} \log \left (\frac {5 \, x^{2} + 1}{x^{2}}\right ) + 26 \, x^{2} + 1}{x e^{4} \log \left (\frac {5 \, x^{2} + 1}{x^{2}}\right ) + 26 \, x e^{4}} \]
integrate(((5*x^4+x^2)*log((5*x^2+1)/x^2)^2+(260*x^4+47*x^2-1)*log((5*x^2+ 1)/x^2)+3380*x^4+546*x^2-24)/((5*x^4+x^2)*exp(4)*log((5*x^2+1)/x^2)^2+(260 *x^4+52*x^2)*exp(4)*log((5*x^2+1)/x^2)+(3380*x^4+676*x^2)*exp(4)),x, algor ithm=\
Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.45 \[ \int \frac {-24+546 x^2+3380 x^4+\left (-1+47 x^2+260 x^4\right ) \log \left (\frac {1+5 x^2}{x^2}\right )+\left (x^2+5 x^4\right ) \log ^2\left (\frac {1+5 x^2}{x^2}\right )}{e^4 \left (676 x^2+3380 x^4\right )+e^4 \left (52 x^2+260 x^4\right ) \log \left (\frac {1+5 x^2}{x^2}\right )+e^4 \left (x^2+5 x^4\right ) \log ^2\left (\frac {1+5 x^2}{x^2}\right )} \, dx=\frac {x}{e^{4}} + \frac {1}{x e^{4} \log {\left (\frac {5 x^{2} + 1}{x^{2}} \right )} + 26 x e^{4}} \]
integrate(((5*x**4+x**2)*ln((5*x**2+1)/x**2)**2+(260*x**4+47*x**2-1)*ln((5 *x**2+1)/x**2)+3380*x**4+546*x**2-24)/((5*x**4+x**2)*exp(4)*ln((5*x**2+1)/ x**2)**2+(260*x**4+52*x**2)*exp(4)*ln((5*x**2+1)/x**2)+(3380*x**4+676*x**2 )*exp(4)),x)
Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (19) = 38\).
Time = 0.22 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.70 \[ \int \frac {-24+546 x^2+3380 x^4+\left (-1+47 x^2+260 x^4\right ) \log \left (\frac {1+5 x^2}{x^2}\right )+\left (x^2+5 x^4\right ) \log ^2\left (\frac {1+5 x^2}{x^2}\right )}{e^4 \left (676 x^2+3380 x^4\right )+e^4 \left (52 x^2+260 x^4\right ) \log \left (\frac {1+5 x^2}{x^2}\right )+e^4 \left (x^2+5 x^4\right ) \log ^2\left (\frac {1+5 x^2}{x^2}\right )} \, dx=\frac {x^{2} \log \left (5 \, x^{2} + 1\right ) - 2 \, x^{2} \log \left (x\right ) + 26 \, x^{2} + 1}{x e^{4} \log \left (5 \, x^{2} + 1\right ) - 2 \, x e^{4} \log \left (x\right ) + 26 \, x e^{4}} \]
integrate(((5*x^4+x^2)*log((5*x^2+1)/x^2)^2+(260*x^4+47*x^2-1)*log((5*x^2+ 1)/x^2)+3380*x^4+546*x^2-24)/((5*x^4+x^2)*exp(4)*log((5*x^2+1)/x^2)^2+(260 *x^4+52*x^2)*exp(4)*log((5*x^2+1)/x^2)+(3380*x^4+676*x^2)*exp(4)),x, algor ithm=\
(x^2*log(5*x^2 + 1) - 2*x^2*log(x) + 26*x^2 + 1)/(x*e^4*log(5*x^2 + 1) - 2 *x*e^4*log(x) + 26*x*e^4)
Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (19) = 38\).
Time = 0.38 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.40 \[ \int \frac {-24+546 x^2+3380 x^4+\left (-1+47 x^2+260 x^4\right ) \log \left (\frac {1+5 x^2}{x^2}\right )+\left (x^2+5 x^4\right ) \log ^2\left (\frac {1+5 x^2}{x^2}\right )}{e^4 \left (676 x^2+3380 x^4\right )+e^4 \left (52 x^2+260 x^4\right ) \log \left (\frac {1+5 x^2}{x^2}\right )+e^4 \left (x^2+5 x^4\right ) \log ^2\left (\frac {1+5 x^2}{x^2}\right )} \, dx=\frac {x^{2} \log \left (\frac {5 \, x^{2} + 1}{x^{2}}\right ) + 26 \, x^{2} + 1}{x e^{4} \log \left (\frac {5 \, x^{2} + 1}{x^{2}}\right ) + 26 \, x e^{4}} \]
integrate(((5*x^4+x^2)*log((5*x^2+1)/x^2)^2+(260*x^4+47*x^2-1)*log((5*x^2+ 1)/x^2)+3380*x^4+546*x^2-24)/((5*x^4+x^2)*exp(4)*log((5*x^2+1)/x^2)^2+(260 *x^4+52*x^2)*exp(4)*log((5*x^2+1)/x^2)+(3380*x^4+676*x^2)*exp(4)),x, algor ithm=\
Time = 13.25 (sec) , antiderivative size = 45, normalized size of antiderivative = 2.25 \[ \int \frac {-24+546 x^2+3380 x^4+\left (-1+47 x^2+260 x^4\right ) \log \left (\frac {1+5 x^2}{x^2}\right )+\left (x^2+5 x^4\right ) \log ^2\left (\frac {1+5 x^2}{x^2}\right )}{e^4 \left (676 x^2+3380 x^4\right )+e^4 \left (52 x^2+260 x^4\right ) \log \left (\frac {1+5 x^2}{x^2}\right )+e^4 \left (x^2+5 x^4\right ) \log ^2\left (\frac {1+5 x^2}{x^2}\right )} \, dx=\frac {{\mathrm {e}}^{-4}\,\left (x^2\,\ln \left (\frac {5\,x^2+1}{x^2}\right )+26\,x^2+1\right )}{x\,\left (\ln \left (\frac {5\,x^2+1}{x^2}\right )+26\right )} \]