3.57.30 \(\int \frac {5^{\frac {5}{(x+x^2 \log (5)) \log (x)}} e^{-4+5^{\frac {5}{(x+x^2 \log (5)) \log (x)}}} (-5 \log (5)-5 x \log ^2(5)+(-5 \log (5)-10 x \log ^2(5)) \log (x))}{(x^2+2 x^3 \log (5)+x^4 \log ^2(5)) \log ^2(x)} \, dx\)

Optimal. Leaf size=24 \[ e^{-4+e^{\frac {5}{\left (x^2+\frac {x}{\log (5)}\right ) \log (x)}}} \]

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Rubi [F]  time = 12.51, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[In]

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]

[Out]

-(Log[5]*Defer[Int][(5^(1 + 5/(x*(1 + x*Log[5])*Log[x]))*E^(-4 + 5^(5/((x + x^2*Log[5])*Log[x]))))/(x^2*Log[x]
^2), x]) + Log[5]^2*Defer[Int][(5^(1 + 5/(x*(1 + x*Log[5])*Log[x]))*E^(-4 + 5^(5/((x + x^2*Log[5])*Log[x]))))/
(x*Log[x]^2), x] - Log[5]^3*Defer[Int][(5^(1 + 5/(x*(1 + x*Log[5])*Log[x]))*E^(-4 + 5^(5/((x + x^2*Log[5])*Log
[x]))))/((1 + x*Log[5])*Log[x]^2), x] - Log[5]*Defer[Int][(5^(1 + 5/(x*(1 + x*Log[5])*Log[x]))*E^(-4 + 5^(5/((
x + x^2*Log[5])*Log[x]))))/(x^2*Log[x]), x] + Log[5]^3*Defer[Int][(5^(1 + 5/(x*(1 + x*Log[5])*Log[x]))*E^(-4 +
5^(5/((x + x^2*Log[5])*Log[x]))))/((1 + x*Log[5])^2*Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\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 )}{x^2 \left (1+2 x \log (5)+x^2 \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 )}{x^2 (1+x \log (5))^2 \log ^2(x)} \, dx\\ &=\int \frac {5^{1+\frac {5}{x (1+x \log (5)) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} \log (5) (-1-x \log (5)-\log (x)-x \log (25) \log (x))}{x^2 (1+x \log (5))^2 \log ^2(x)} \, dx\\ &=\log (5) \int \frac {5^{1+\frac {5}{x (1+x \log (5)) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} (-1-x \log (5)-\log (x)-x \log (25) \log (x))}{x^2 (1+x \log (5))^2 \log ^2(x)} \, dx\\ &=\log (5) \int \left (-\frac {5^{1+\frac {5}{x (1+x \log (5)) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}}}{x^2 (1+x \log (5)) \log ^2(x)}+\frac {5^{1+\frac {5}{x (1+x \log (5)) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} (-1-x \log (25))}{x^2 (1+x \log (5))^2 \log (x)}\right ) \, dx\\ &=-\left (\log (5) \int \frac {5^{1+\frac {5}{x (1+x \log (5)) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}}}{x^2 (1+x \log (5)) \log ^2(x)} \, dx\right )+\log (5) \int \frac {5^{1+\frac {5}{x (1+x \log (5)) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} (-1-x \log (25))}{x^2 (1+x \log (5))^2 \log (x)} \, dx\\ &=-\left (\log (5) \int \left (\frac {5^{1+\frac {5}{x (1+x \log (5)) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}}}{x^2 \log ^2(x)}-\frac {5^{1+\frac {5}{x (1+x \log (5)) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} \log (5)}{x \log ^2(x)}+\frac {5^{1+\frac {5}{x (1+x \log (5)) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} \log ^2(5)}{(1+x \log (5)) \log ^2(x)}\right ) \, dx\right )+\log (5) \int \left (-\frac {5^{1+\frac {5}{x (1+x \log (5)) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}}}{x^2 \log (x)}+\frac {5^{1+\frac {5}{x (1+x \log (5)) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} \log ^2(5)}{(1+x \log (5))^2 \log (x)}\right ) \, dx\\ &=-\left (\log (5) \int \frac {5^{1+\frac {5}{x (1+x \log (5)) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}}}{x^2 \log ^2(x)} \, dx\right )-\log (5) \int \frac {5^{1+\frac {5}{x (1+x \log (5)) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}}}{x^2 \log (x)} \, dx+\log ^2(5) \int \frac {5^{1+\frac {5}{x (1+x \log (5)) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}}}{x \log ^2(x)} \, dx-\log ^3(5) \int \frac {5^{1+\frac {5}{x (1+x \log (5)) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}}}{(1+x \log (5)) \log ^2(x)} \, dx+\log ^3(5) \int \frac {5^{1+\frac {5}{x (1+x \log (5)) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}}}{(1+x \log (5))^2 \log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 3.46, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[In]

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]

[Out]

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]

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fricas [A]  time = 0.61, size = 21, normalized size = 0.88 \begin {gather*} e^{\left (5^{\frac {5}{{\left (x^{2} \log \relax (5) + x\right )} \log \relax (x)}} - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x*log(5)^2-5*log(5))*log(x)-5*x*log(5)^2-5*log(5))*exp(5*log(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="fricas")

[Out]

e^(5^(5/((x^2*log(5) + x)*log(x))) - 4)

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giac [A]  time = 0.23, size = 22, normalized size = 0.92 \begin {gather*} e^{\left (5^{\frac {5}{x^{2} \log \relax (5) \log \relax (x) + x \log \relax (x)}} - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x*log(5)^2-5*log(5))*log(x)-5*x*log(5)^2-5*log(5))*exp(5*log(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="giac")

[Out]

e^(5^(5/(x^2*log(5)*log(x) + x*log(x))) - 4)

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maple [A]  time = 0.13, size = 22, normalized size = 0.92

method	result	size
risch	\({\mathrm e}^{3125^{\frac {1}{x \left (x \ln \relax (5)+1\right ) \ln \relax (x )}}-4}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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*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,method=_RETURNVERBOSE)

[Out]

exp(3125^(1/x/(x*ln(5)+1)/ln(x))-4)

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maxima [A]  time = 0.72, size = 34, normalized size = 1.42 \begin {gather*} e^{\left (e^{\left (-\frac {5 \, \log \relax (5)^{2}}{{\left (x \log \relax (5) + 1\right )} \log \relax (x)} + \frac {5 \, \log \relax (5)}{x \log \relax (x)}\right )} - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x*log(5)^2-5*log(5))*log(x)-5*x*log(5)^2-5*log(5))*exp(5*log(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="maxima")

[Out]

e^(e^(-5*log(5)^2/((x*log(5) + 1)*log(x)) + 5*log(5)/(x*log(x))) - 4)

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mupad [B]  time = 4.13, size = 23, normalized size = 0.96 \begin {gather*} {\mathrm {e}}^{5^{\frac {5}{x\,\ln \relax (x)+x^2\,\ln \relax (5)\,\ln \relax (x)}}}\,{\mathrm {e}}^{-4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp((5*log(5))/(log(x)*(x + x^2*log(5)))) - 4)*exp((5*log(5))/(log(x)*(x + x^2*log(5))))*(5*log(5) +
log(x)*(5*log(5) + 10*x*log(5)^2) + 5*x*log(5)^2))/(log(x)^2*(x^4*log(5)^2 + 2*x^3*log(5) + x^2)),x)

[Out]

exp(5^(5/(x*log(x) + x^2*log(5)*log(x))))*exp(-4)

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sympy [A]  time = 2.42, size = 20, normalized size = 0.83 \begin {gather*} e^{e^{\frac {5 \log {\relax (5 )}}{\left (x^{2} \log {\relax (5 )} + x\right ) \log {\relax (x )}}} - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

exp(exp(5*log(5)/((x**2*log(5) + x)*log(x))) - 4)

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