Integrand size = 93, antiderivative size = 24 \[ \int \frac {5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} \left (-5 \log (5)-5 x \log ^2(5)+\left (-5 \log (5)-10 x \log ^2(5)\right ) \log (x)\right )}{\left (x^2+2 x^3 \log (5)+x^4 \log ^2(5)\right ) \log ^2(x)} \, dx=e^{-4+e^{\frac {5}{\left (x^2+\frac {x}{\log (5)}\right ) \log (x)}}} \]
\[ \int \frac {5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} \left (-5 \log (5)-5 x \log ^2(5)+\left (-5 \log (5)-10 x \log ^2(5)\right ) \log (x)\right )}{\left (x^2+2 x^3 \log (5)+x^4 \log ^2(5)\right ) \log ^2(x)} \, dx=\int \frac {5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} \left (-5 \log (5)-5 x \log ^2(5)+\left (-5 \log (5)-10 x \log ^2(5)\right ) \log (x)\right )}{\left (x^2+2 x^3 \log (5)+x^4 \log ^2(5)\right ) \log ^2(x)} \, dx \]
Integrate[(5^(5/((x + x^2*Log[5])*Log[x]))*E^(-4 + 5^(5/((x + x^2*Log[5])* Log[x])))*(-5*Log[5] - 5*x*Log[5]^2 + (-5*Log[5] - 10*x*Log[5]^2)*Log[x])) /((x^2 + 2*x^3*Log[5] + x^4*Log[5]^2)*Log[x]^2),x]
Integrate[(5^(5/((x + x^2*Log[5])*Log[x]))*E^(-4 + 5^(5/((x + x^2*Log[5])* Log[x])))*(-5*Log[5] - 5*x*Log[5]^2 + (-5*Log[5] - 10*x*Log[5]^2)*Log[x])) /((x^2 + 2*x^3*Log[5] + x^4*Log[5]^2)*Log[x]^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.
\(\displaystyle \int \frac {5^{\frac {5}{\left (x^2 \log (5)+x\right ) \log (x)}} e^{5^{\frac {5}{\left (x^2 \log (5)+x\right ) \log (x)}}-4} \left (-5 x \log ^2(5)+\left (-10 x \log ^2(5)-5 \log (5)\right ) \log (x)-5 \log (5)\right )}{\left (x^4 \log ^2(5)+2 x^3 \log (5)+x^2\right ) \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {5^{\frac {5}{\left (x^2 \log (5)+x\right ) \log (x)}} e^{5^{\frac {5}{\left (x^2 \log (5)+x\right ) \log (x)}}-4} \left (-5 x \log ^2(5)+\left (-10 x \log ^2(5)-5 \log (5)\right ) \log (x)-5 \log (5)\right )}{x^2 \left (x^2 \log ^2(5)+2 x \log (5)+1\right ) \log ^2(x)}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {5^{\frac {5}{\left (x^2 \log (5)+x\right ) \log (x)}} e^{5^{\frac {5}{\left (x^2 \log (5)+x\right ) \log (x)}}-4} \left (-5 x \log ^2(5)+\left (-10 x \log ^2(5)-5 \log (5)\right ) \log (x)-5 \log (5)\right )}{x^2 (x \log (5)+1)^2 \log ^2(x)}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\log (5) 5^{\frac {5}{x (x \log (5)+1) \log (x)}+1} e^{5^{\frac {5}{\left (x^2 \log (5)+x\right ) \log (x)}}-4} (x (-\log (25)) \log (x)-x \log (5)-\log (x)-1)}{x^2 (x \log (5)+1)^2 \log ^2(x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \log (5) \int -\frac {5^{1+\frac {5}{x (\log (5) x+1) \log (x)}} e^{-4+5^{\frac {5}{\left (\log (5) x^2+x\right ) \log (x)}}} (\log (25) \log (x) x+\log (5) x+\log (x)+1)}{x^2 (\log (5) x+1)^2 \log ^2(x)}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\log (5) \int \frac {5^{1+\frac {5}{x (\log (5) x+1) \log (x)}} e^{-4+5^{\frac {5}{\left (\log (5) x^2+x\right ) \log (x)}}} (\log (25) \log (x) x+\log (5) x+\log (x)+1)}{x^2 (\log (5) x+1)^2 \log ^2(x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\log (5) \int \left (\frac {5^{1+\frac {5}{x (\log (5) x+1) \log (x)}} e^{-4+5^{\frac {5}{\left (\log (5) x^2+x\right ) \log (x)}}} (\log (25) x+1)}{x^2 (\log (5) x+1)^2 \log (x)}+\frac {5^{1+\frac {5}{x (\log (5) x+1) \log (x)}} e^{-4+5^{\frac {5}{\left (\log (5) x^2+x\right ) \log (x)}}}}{x^2 (\log (5) x+1) \log ^2(x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\log (5) \left (\int \frac {5^{1+\frac {5}{x (\log (5) x+1) \log (x)}} e^{-4+5^{\frac {5}{\left (\log (5) x^2+x\right ) \log (x)}}}}{x^2 \log ^2(x)}dx-\log (5) \int \frac {5^{1+\frac {5}{x (\log (5) x+1) \log (x)}} e^{-4+5^{\frac {5}{\left (\log (5) x^2+x\right ) \log (x)}}}}{x \log ^2(x)}dx+\log ^2(5) \int \frac {5^{1+\frac {5}{x (\log (5) x+1) \log (x)}} e^{-4+5^{\frac {5}{\left (\log (5) x^2+x\right ) \log (x)}}}}{(\log (5) x+1) \log ^2(x)}dx-\log ^2(5) \int \frac {5^{1+\frac {5}{x (\log (5) x+1) \log (x)}} e^{-4+5^{\frac {5}{\left (\log (5) x^2+x\right ) \log (x)}}}}{(\log (5) x+1)^2 \log (x)}dx+\int \frac {5^{1+\frac {5}{x (\log (5) x+1) \log (x)}} e^{-4+5^{\frac {5}{\left (\log (5) x^2+x\right ) \log (x)}}}}{x^2 \log (x)}dx\right )\) |
Int[(5^(5/((x + x^2*Log[5])*Log[x]))*E^(-4 + 5^(5/((x + x^2*Log[5])*Log[x] )))*(-5*Log[5] - 5*x*Log[5]^2 + (-5*Log[5] - 10*x*Log[5]^2)*Log[x]))/((x^2 + 2*x^3*Log[5] + x^4*Log[5]^2)*Log[x]^2),x]
3.27.30.3.1 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_.)*(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 = 167.87 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92
method | result | size |
risch | \({\mathrm e}^{3125^{\frac {1}{x \left (x \ln \left (5\right )+1\right ) \ln \left (x \right )}}-4}\) | \(22\) |
parallelrisch | \({\mathrm e}^{{\mathrm e}^{\frac {5 \ln \left (5\right )}{x \left (x \ln \left (5\right )+1\right ) \ln \left (x \right )}}-4}\) | \(24\) |
int(((-10*x*ln(5)^2-5*ln(5))*ln(x)-5*x*ln(5)^2-5*ln(5))*exp(5*ln(5)/(x^2*l n(5)+x)/ln(x))*exp(exp(5*ln(5)/(x^2*ln(5)+x)/ln(x))-4)/(x^4*ln(5)^2+2*x^3* ln(5)+x^2)/ln(x)^2,x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} \left (-5 \log (5)-5 x \log ^2(5)+\left (-5 \log (5)-10 x \log ^2(5)\right ) \log (x)\right )}{\left (x^2+2 x^3 \log (5)+x^4 \log ^2(5)\right ) \log ^2(x)} \, dx=e^{\left (5^{\frac {5}{{\left (x^{2} \log \left (5\right ) + x\right )} \log \left (x\right )}} - 4\right )} \]
integrate(((-10*x*log(5)^2-5*log(5))*log(x)-5*x*log(5)^2-5*log(5))*exp(5*l og(5)/(x^2*log(5)+x)/log(x))*exp(exp(5*log(5)/(x^2*log(5)+x)/log(x))-4)/(x ^4*log(5)^2+2*x^3*log(5)+x^2)/log(x)^2,x, algorithm=\
Exception generated. \[ \int \frac {5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} \left (-5 \log (5)-5 x \log ^2(5)+\left (-5 \log (5)-10 x \log ^2(5)\right ) \log (x)\right )}{\left (x^2+2 x^3 \log (5)+x^4 \log ^2(5)\right ) \log ^2(x)} \, dx=\text {Exception raised: TypeError} \]
integrate(((-10*x*ln(5)**2-5*ln(5))*ln(x)-5*x*ln(5)**2-5*ln(5))*exp(5*ln(5 )/(x**2*ln(5)+x)/ln(x))*exp(exp(5*ln(5)/(x**2*ln(5)+x)/ln(x))-4)/(x**4*ln( 5)**2+2*x**3*ln(5)+x**2)/ln(x)**2,x)
Time = 0.47 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} \left (-5 \log (5)-5 x \log ^2(5)+\left (-5 \log (5)-10 x \log ^2(5)\right ) \log (x)\right )}{\left (x^2+2 x^3 \log (5)+x^4 \log ^2(5)\right ) \log ^2(x)} \, dx=e^{\left (e^{\left (-\frac {5 \, \log \left (5\right )^{2}}{{\left (x \log \left (5\right ) + 1\right )} \log \left (x\right )} + \frac {5 \, \log \left (5\right )}{x \log \left (x\right )}\right )} - 4\right )} \]
integrate(((-10*x*log(5)^2-5*log(5))*log(x)-5*x*log(5)^2-5*log(5))*exp(5*l og(5)/(x^2*log(5)+x)/log(x))*exp(exp(5*log(5)/(x^2*log(5)+x)/log(x))-4)/(x ^4*log(5)^2+2*x^3*log(5)+x^2)/log(x)^2,x, algorithm=\
Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} \left (-5 \log (5)-5 x \log ^2(5)+\left (-5 \log (5)-10 x \log ^2(5)\right ) \log (x)\right )}{\left (x^2+2 x^3 \log (5)+x^4 \log ^2(5)\right ) \log ^2(x)} \, dx=e^{\left (5^{\frac {5}{x^{2} \log \left (5\right ) \log \left (x\right ) + x \log \left (x\right )}} - 4\right )} \]
integrate(((-10*x*log(5)^2-5*log(5))*log(x)-5*x*log(5)^2-5*log(5))*exp(5*l og(5)/(x^2*log(5)+x)/log(x))*exp(exp(5*log(5)/(x^2*log(5)+x)/log(x))-4)/(x ^4*log(5)^2+2*x^3*log(5)+x^2)/log(x)^2,x, algorithm=\
Time = 11.89 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} \left (-5 \log (5)-5 x \log ^2(5)+\left (-5 \log (5)-10 x \log ^2(5)\right ) \log (x)\right )}{\left (x^2+2 x^3 \log (5)+x^4 \log ^2(5)\right ) \log ^2(x)} \, dx={\mathrm {e}}^{5^{\frac {5}{x\,\ln \left (x\right )+x^2\,\ln \left (5\right )\,\ln \left (x\right )}}}\,{\mathrm {e}}^{-4} \]