\(\int \frac {e^{\frac {81+25 x^2+25 x^2 \log (1-2 x+x^2)-25 x^2 \log (\log (x^2))}{25 x^2}} (50 x^2-50 x^3+(162-162 x+25 x^2+25 x^3) \log (x^2))}{(-25 x^4+25 x^5) \log (x^2)} \, dx\) [395]

   Optimal result
   Rubi [F]
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F(-1)]
   Maxima [F]
   Giac [F]
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 91, antiderivative size = 26 \[ \int \frac {e^{\frac {81+25 x^2+25 x^2 \log \left (1-2 x+x^2\right )-25 x^2 \log \left (\log \left (x^2\right )\right )}{25 x^2}} \left (50 x^2-50 x^3+\left (162-162 x+25 x^2+25 x^3\right ) \log \left (x^2\right )\right )}{\left (-25 x^4+25 x^5\right ) \log \left (x^2\right )} \, dx=\frac {e^{1+\frac {81}{25 x^2}} (-1+x)^2}{x \log \left (x^2\right )} \]

[Out]

exp(1-ln(ln(x^2))+81/25/x^2+ln((-1+x)^2))/x

Rubi [F]

\[ \int \frac {e^{\frac {81+25 x^2+25 x^2 \log \left (1-2 x+x^2\right )-25 x^2 \log \left (\log \left (x^2\right )\right )}{25 x^2}} \left (50 x^2-50 x^3+\left (162-162 x+25 x^2+25 x^3\right ) \log \left (x^2\right )\right )}{\left (-25 x^4+25 x^5\right ) \log \left (x^2\right )} \, dx=\int \frac {\exp \left (\frac {81+25 x^2+25 x^2 \log \left (1-2 x+x^2\right )-25 x^2 \log \left (\log \left (x^2\right )\right )}{25 x^2}\right ) \left (50 x^2-50 x^3+\left (162-162 x+25 x^2+25 x^3\right ) \log \left (x^2\right )\right )}{\left (-25 x^4+25 x^5\right ) \log \left (x^2\right )} \, dx \]

[In]

Int[(E^((81 + 25*x^2 + 25*x^2*Log[1 - 2*x + x^2] - 25*x^2*Log[Log[x^2]])/(25*x^2))*(50*x^2 - 50*x^3 + (162 - 1
62*x + 25*x^2 + 25*x^3)*Log[x^2]))/((-25*x^4 + 25*x^5)*Log[x^2]),x]

[Out]

-2*Defer[Int][E^(1 + 81/(25*x^2))/Log[x^2]^2, x] - 2*Defer[Int][E^(1 + 81/(25*x^2))/(x^2*Log[x^2]^2), x] + Def
er[Int][E^(1 + 81/(25*x^2))/Log[x^2], x] - (162*Defer[Int][E^(1 + 81/(25*x^2))/(x^4*Log[x^2]), x])/25 - (187*D
efer[Int][E^(1 + 81/(25*x^2))/(x^2*Log[x^2]), x])/25 + 2*Defer[Subst][Defer[Int][E^(1 + 81/(25*x))/(x*Log[x]^2
), x], x, x^2] + (162*Defer[Subst][Defer[Int][E^(1 + 81/(25*x))/(x^2*Log[x]), x], x, x^2])/25

Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (\frac {81+25 x^2+25 x^2 \log \left (1-2 x+x^2\right )-25 x^2 \log \left (\log \left (x^2\right )\right )}{25 x^2}\right ) \left (50 x^2-50 x^3+\left (162-162 x+25 x^2+25 x^3\right ) \log \left (x^2\right )\right )}{x^4 (-25+25 x) \log \left (x^2\right )} \, dx \\ & = \int \frac {e^{1+\frac {81}{25 x^2}} (1-x) \left (50 (-1+x) x^2-\left (162-162 x+25 x^2+25 x^3\right ) \log \left (x^2\right )\right )}{25 x^4 \log ^2\left (x^2\right )} \, dx \\ & = \frac {1}{25} \int \frac {e^{1+\frac {81}{25 x^2}} (1-x) \left (50 (-1+x) x^2-\left (162-162 x+25 x^2+25 x^3\right ) \log \left (x^2\right )\right )}{x^4 \log ^2\left (x^2\right )} \, dx \\ & = \frac {1}{25} \int \left (-\frac {50 e^{1+\frac {81}{25 x^2}} (-1+x)^2}{x^2 \log ^2\left (x^2\right )}+\frac {e^{1+\frac {81}{25 x^2}} \left (-162+324 x-187 x^2+25 x^4\right )}{x^4 \log \left (x^2\right )}\right ) \, dx \\ & = \frac {1}{25} \int \frac {e^{1+\frac {81}{25 x^2}} \left (-162+324 x-187 x^2+25 x^4\right )}{x^4 \log \left (x^2\right )} \, dx-2 \int \frac {e^{1+\frac {81}{25 x^2}} (-1+x)^2}{x^2 \log ^2\left (x^2\right )} \, dx \\ & = \frac {1}{25} \int \left (\frac {25 e^{1+\frac {81}{25 x^2}}}{\log \left (x^2\right )}-\frac {162 e^{1+\frac {81}{25 x^2}}}{x^4 \log \left (x^2\right )}+\frac {324 e^{1+\frac {81}{25 x^2}}}{x^3 \log \left (x^2\right )}-\frac {187 e^{1+\frac {81}{25 x^2}}}{x^2 \log \left (x^2\right )}\right ) \, dx-2 \int \left (\frac {e^{1+\frac {81}{25 x^2}}}{\log ^2\left (x^2\right )}+\frac {e^{1+\frac {81}{25 x^2}}}{x^2 \log ^2\left (x^2\right )}-\frac {2 e^{1+\frac {81}{25 x^2}}}{x \log ^2\left (x^2\right )}\right ) \, dx \\ & = -\left (2 \int \frac {e^{1+\frac {81}{25 x^2}}}{\log ^2\left (x^2\right )} \, dx\right )-2 \int \frac {e^{1+\frac {81}{25 x^2}}}{x^2 \log ^2\left (x^2\right )} \, dx+4 \int \frac {e^{1+\frac {81}{25 x^2}}}{x \log ^2\left (x^2\right )} \, dx-\frac {162}{25} \int \frac {e^{1+\frac {81}{25 x^2}}}{x^4 \log \left (x^2\right )} \, dx-\frac {187}{25} \int \frac {e^{1+\frac {81}{25 x^2}}}{x^2 \log \left (x^2\right )} \, dx+\frac {324}{25} \int \frac {e^{1+\frac {81}{25 x^2}}}{x^3 \log \left (x^2\right )} \, dx+\int \frac {e^{1+\frac {81}{25 x^2}}}{\log \left (x^2\right )} \, dx \\ & = -\left (2 \int \frac {e^{1+\frac {81}{25 x^2}}}{\log ^2\left (x^2\right )} \, dx\right )-2 \int \frac {e^{1+\frac {81}{25 x^2}}}{x^2 \log ^2\left (x^2\right )} \, dx+2 \text {Subst}\left (\int \frac {e^{1+\frac {81}{25 x}}}{x \log ^2(x)} \, dx,x,x^2\right )-\frac {162}{25} \int \frac {e^{1+\frac {81}{25 x^2}}}{x^4 \log \left (x^2\right )} \, dx+\frac {162}{25} \text {Subst}\left (\int \frac {e^{1+\frac {81}{25 x}}}{x^2 \log (x)} \, dx,x,x^2\right )-\frac {187}{25} \int \frac {e^{1+\frac {81}{25 x^2}}}{x^2 \log \left (x^2\right )} \, dx+\int \frac {e^{1+\frac {81}{25 x^2}}}{\log \left (x^2\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {81+25 x^2+25 x^2 \log \left (1-2 x+x^2\right )-25 x^2 \log \left (\log \left (x^2\right )\right )}{25 x^2}} \left (50 x^2-50 x^3+\left (162-162 x+25 x^2+25 x^3\right ) \log \left (x^2\right )\right )}{\left (-25 x^4+25 x^5\right ) \log \left (x^2\right )} \, dx=\frac {e^{1+\frac {81}{25 x^2}} (-1+x)^2}{x \log \left (x^2\right )} \]

[In]

Integrate[(E^((81 + 25*x^2 + 25*x^2*Log[1 - 2*x + x^2] - 25*x^2*Log[Log[x^2]])/(25*x^2))*(50*x^2 - 50*x^3 + (1
62 - 162*x + 25*x^2 + 25*x^3)*Log[x^2]))/((-25*x^4 + 25*x^5)*Log[x^2]),x]

[Out]

(E^(1 + 81/(25*x^2))*(-1 + x)^2)/(x*Log[x^2])

Maple [A] (verified)

Time = 35.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62

method result size
parallelrisch \(\frac {{\mathrm e}^{-\frac {25 x^{2} \ln \left (\ln \left (x^{2}\right )\right )-25 x^{2} \ln \left (x^{2}-2 x +1\right )-25 x^{2}-81}{25 x^{2}}}}{x}\) \(42\)

[In]

int(((25*x^3+25*x^2-162*x+162)*ln(x^2)-50*x^3+50*x^2)*exp(1/25*(-25*x^2*ln(ln(x^2))+25*x^2*ln(x^2-2*x+1)+25*x^
2+81)/x^2)/(25*x^5-25*x^4)/ln(x^2),x,method=_RETURNVERBOSE)

[Out]

exp(-1/25*(25*x^2*ln(ln(x^2))-25*x^2*ln(x^2-2*x+1)-25*x^2-81)/x^2)/x

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \[ \int \frac {e^{\frac {81+25 x^2+25 x^2 \log \left (1-2 x+x^2\right )-25 x^2 \log \left (\log \left (x^2\right )\right )}{25 x^2}} \left (50 x^2-50 x^3+\left (162-162 x+25 x^2+25 x^3\right ) \log \left (x^2\right )\right )}{\left (-25 x^4+25 x^5\right ) \log \left (x^2\right )} \, dx=\frac {e^{\left (\frac {25 \, x^{2} \log \left (x^{2} - 2 \, x + 1\right ) - 25 \, x^{2} \log \left (\log \left (x^{2}\right )\right ) + 25 \, x^{2} + 81}{25 \, x^{2}}\right )}}{x} \]

[In]

integrate(((25*x^3+25*x^2-162*x+162)*log(x^2)-50*x^3+50*x^2)*exp(1/25*(-25*x^2*log(log(x^2))+25*x^2*log(x^2-2*
x+1)+25*x^2+81)/x^2)/(25*x^5-25*x^4)/log(x^2),x, algorithm="fricas")

[Out]

e^(1/25*(25*x^2*log(x^2 - 2*x + 1) - 25*x^2*log(log(x^2)) + 25*x^2 + 81)/x^2)/x

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{\frac {81+25 x^2+25 x^2 \log \left (1-2 x+x^2\right )-25 x^2 \log \left (\log \left (x^2\right )\right )}{25 x^2}} \left (50 x^2-50 x^3+\left (162-162 x+25 x^2+25 x^3\right ) \log \left (x^2\right )\right )}{\left (-25 x^4+25 x^5\right ) \log \left (x^2\right )} \, dx=\text {Timed out} \]

[In]

integrate(((25*x**3+25*x**2-162*x+162)*ln(x**2)-50*x**3+50*x**2)*exp(1/25*(-25*x**2*ln(ln(x**2))+25*x**2*ln(x*
*2-2*x+1)+25*x**2+81)/x**2)/(25*x**5-25*x**4)/ln(x**2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {e^{\frac {81+25 x^2+25 x^2 \log \left (1-2 x+x^2\right )-25 x^2 \log \left (\log \left (x^2\right )\right )}{25 x^2}} \left (50 x^2-50 x^3+\left (162-162 x+25 x^2+25 x^3\right ) \log \left (x^2\right )\right )}{\left (-25 x^4+25 x^5\right ) \log \left (x^2\right )} \, dx=\int { -\frac {{\left (50 \, x^{3} - 50 \, x^{2} - {\left (25 \, x^{3} + 25 \, x^{2} - 162 \, x + 162\right )} \log \left (x^{2}\right )\right )} e^{\left (\frac {25 \, x^{2} \log \left (x^{2} - 2 \, x + 1\right ) - 25 \, x^{2} \log \left (\log \left (x^{2}\right )\right ) + 25 \, x^{2} + 81}{25 \, x^{2}}\right )}}{25 \, {\left (x^{5} - x^{4}\right )} \log \left (x^{2}\right )} \,d x } \]

[In]

integrate(((25*x^3+25*x^2-162*x+162)*log(x^2)-50*x^3+50*x^2)*exp(1/25*(-25*x^2*log(log(x^2))+25*x^2*log(x^2-2*
x+1)+25*x^2+81)/x^2)/(25*x^5-25*x^4)/log(x^2),x, algorithm="maxima")

[Out]

-1/25*integrate((50*x^3 - 50*x^2 - (25*x^3 + 25*x^2 - 162*x + 162)*log(x^2))*e^(1/25*(25*x^2*log(x^2 - 2*x + 1
) - 25*x^2*log(log(x^2)) + 25*x^2 + 81)/x^2)/((x^5 - x^4)*log(x^2)), x)

Giac [F]

\[ \int \frac {e^{\frac {81+25 x^2+25 x^2 \log \left (1-2 x+x^2\right )-25 x^2 \log \left (\log \left (x^2\right )\right )}{25 x^2}} \left (50 x^2-50 x^3+\left (162-162 x+25 x^2+25 x^3\right ) \log \left (x^2\right )\right )}{\left (-25 x^4+25 x^5\right ) \log \left (x^2\right )} \, dx=\int { -\frac {{\left (50 \, x^{3} - 50 \, x^{2} - {\left (25 \, x^{3} + 25 \, x^{2} - 162 \, x + 162\right )} \log \left (x^{2}\right )\right )} e^{\left (\frac {25 \, x^{2} \log \left (x^{2} - 2 \, x + 1\right ) - 25 \, x^{2} \log \left (\log \left (x^{2}\right )\right ) + 25 \, x^{2} + 81}{25 \, x^{2}}\right )}}{25 \, {\left (x^{5} - x^{4}\right )} \log \left (x^{2}\right )} \,d x } \]

[In]

integrate(((25*x^3+25*x^2-162*x+162)*log(x^2)-50*x^3+50*x^2)*exp(1/25*(-25*x^2*log(log(x^2))+25*x^2*log(x^2-2*
x+1)+25*x^2+81)/x^2)/(25*x^5-25*x^4)/log(x^2),x, algorithm="giac")

[Out]

integrate(-1/25*(50*x^3 - 50*x^2 - (25*x^3 + 25*x^2 - 162*x + 162)*log(x^2))*e^(1/25*(25*x^2*log(x^2 - 2*x + 1
) - 25*x^2*log(log(x^2)) + 25*x^2 + 81)/x^2)/((x^5 - x^4)*log(x^2)), x)

Mupad [B] (verification not implemented)

Time = 7.65 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {81+25 x^2+25 x^2 \log \left (1-2 x+x^2\right )-25 x^2 \log \left (\log \left (x^2\right )\right )}{25 x^2}} \left (50 x^2-50 x^3+\left (162-162 x+25 x^2+25 x^3\right ) \log \left (x^2\right )\right )}{\left (-25 x^4+25 x^5\right ) \log \left (x^2\right )} \, dx=\frac {{\mathrm {e}}^{\frac {81}{25\,x^2}+1}\,{\left (x-1\right )}^2}{x\,\ln \left (x^2\right )} \]

[In]

int(-(exp((x^2*log(x^2 - 2*x + 1) + x^2 - x^2*log(log(x^2)) + 81/25)/x^2)*(log(x^2)*(25*x^2 - 162*x + 25*x^3 +
 162) + 50*x^2 - 50*x^3))/(log(x^2)*(25*x^4 - 25*x^5)),x)

[Out]

(exp(81/(25*x^2) + 1)*(x - 1)^2)/(x*log(x^2))