3.27.98 \(\int \frac {-1+e^4+4 e^6-\log (5)+\log (x)}{e^8+16 e^{12}+e^4 (8 e^6-2 x)-8 e^6 x+x^2+(-2 e^4-8 e^6+2 x) \log (5)+\log ^2(5)+(2 e^4+8 e^6-2 x-2 \log (5)) \log (x)+\log ^2(x)} \, dx\)

Optimal. Leaf size=22 \[ \frac {x}{e^4+4 e^6-x-\log (5)+\log (x)} \]

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Rubi [F]  time = 0.41, 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 {-1+e^4+4 e^6-\log (5)+\log (x)}{e^8+16 e^{12}+e^4 \left (8 e^6-2 x\right )-8 e^6 x+x^2+\left (-2 e^4-8 e^6+2 x\right ) \log (5)+\log ^2(5)+\left (2 e^4+8 e^6-2 x-2 \log (5)\right ) \log (x)+\log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-1 + E^4 + 4*E^6 - Log[5] + Log[x])/(E^8 + 16*E^12 + E^4*(8*E^6 - 2*x) - 8*E^6*x + x^2 + (-2*E^4 - 8*E^6
+ 2*x)*Log[5] + Log[5]^2 + (2*E^4 + 8*E^6 - 2*x - 2*Log[5])*Log[x] + Log[x]^2),x]

[Out]

-5*Defer[Subst][Defer[Int][(E^4*(1 + 4*E^2) - 5*x + Log[x])^(-2), x], x, x/5] + 25*Defer[Subst][Defer[Int][x/(
E^4*(1 + 4*E^2) - 5*x + Log[x])^2, x], x, x/5] + 5*Defer[Subst][Defer[Int][(E^4*(1 + 4*E^2) - 5*x + Log[x])^(-
1), x], x, x/5]

Rubi steps

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

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Mathematica [A]  time = 0.19, size = 22, normalized size = 1.00 \begin {gather*} \frac {x}{e^4+4 e^6-x+\log \left (\frac {x}{5}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + E^4 + 4*E^6 - Log[5] + Log[x])/(E^8 + 16*E^12 + E^4*(8*E^6 - 2*x) - 8*E^6*x + x^2 + (-2*E^4 -
8*E^6 + 2*x)*Log[5] + Log[5]^2 + (2*E^4 + 8*E^6 - 2*x - 2*Log[5])*Log[x] + Log[x]^2),x]

[Out]

x/(E^4 + 4*E^6 - x + Log[x/5])

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fricas [A]  time = 0.57, size = 21, normalized size = 0.95 \begin {gather*} -\frac {x}{x - 4 \, e^{6} - e^{4} + \log \relax (5) - \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(x)-log(5)+exp(4)+4*exp(3)^2-1)/(log(x)^2+(-2*log(5)+2*exp(4)+8*exp(3)^2-2*x)*log(x)+log(5)^2+(-
2*exp(4)-8*exp(3)^2+2*x)*log(5)+exp(4)^2+(8*exp(3)^2-2*x)*exp(4)+16*exp(3)^4-8*x*exp(3)^2+x^2),x, algorithm="f
ricas")

[Out]

-x/(x - 4*e^6 - e^4 + log(5) - log(x))

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giac [A]  time = 0.31, size = 21, normalized size = 0.95 \begin {gather*} -\frac {x}{x - 4 \, e^{6} - e^{4} + \log \relax (5) - \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(x)-log(5)+exp(4)+4*exp(3)^2-1)/(log(x)^2+(-2*log(5)+2*exp(4)+8*exp(3)^2-2*x)*log(x)+log(5)^2+(-
2*exp(4)-8*exp(3)^2+2*x)*log(5)+exp(4)^2+(8*exp(3)^2-2*x)*exp(4)+16*exp(3)^4-8*x*exp(3)^2+x^2),x, algorithm="g
iac")

[Out]

-x/(x - 4*e^6 - e^4 + log(5) - log(x))

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maple [A]  time = 0.12, size = 21, normalized size = 0.95

method	result	size
risch	\(\frac {x}{4 \,{\mathrm e}^{6}+{\mathrm e}^{4}-\ln \relax (5)-x +\ln \relax (x )}\)	\(21\)
norman	\(\frac {\ln \relax (x )+4 \,{\mathrm e}^{6}-\ln \relax (5)+{\mathrm e}^{4}}{4 \,{\mathrm e}^{6}+{\mathrm e}^{4}-\ln \relax (5)-x +\ln \relax (x )}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((ln(x)-ln(5)+exp(4)+4*exp(3)^2-1)/(ln(x)^2+(-2*ln(5)+2*exp(4)+8*exp(3)^2-2*x)*ln(x)+ln(5)^2+(-2*exp(4)-8*e
xp(3)^2+2*x)*ln(5)+exp(4)^2+(8*exp(3)^2-2*x)*exp(4)+16*exp(3)^4-8*x*exp(3)^2+x^2),x,method=_RETURNVERBOSE)

[Out]

x/(4*exp(6)+exp(4)-ln(5)-x+ln(x))

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maxima [A]  time = 0.91, size = 21, normalized size = 0.95 \begin {gather*} -\frac {x}{x - 4 \, e^{6} - e^{4} + \log \relax (5) - \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(x)-log(5)+exp(4)+4*exp(3)^2-1)/(log(x)^2+(-2*log(5)+2*exp(4)+8*exp(3)^2-2*x)*log(x)+log(5)^2+(-
2*exp(4)-8*exp(3)^2+2*x)*log(5)+exp(4)^2+(8*exp(3)^2-2*x)*exp(4)+16*exp(3)^4-8*x*exp(3)^2+x^2),x, algorithm="m
axima")

[Out]

-x/(x - 4*e^6 - e^4 + log(5) - log(x))

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mupad [B]  time = 2.02, size = 18, normalized size = 0.82 \begin {gather*} \frac {x}{\ln \left (\frac {x}{5}\right )-x+{\mathrm {e}}^4+4\,{\mathrm {e}}^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(4) + 4*exp(6) - log(5) + log(x) - 1)/(exp(8) + 16*exp(12) - 8*x*exp(6) + log(x)^2 - log(5)*(2*exp(4)
- 2*x + 8*exp(6)) - exp(4)*(2*x - 8*exp(6)) + log(5)^2 - log(x)*(2*x - 2*exp(4) - 8*exp(6) + 2*log(5)) + x^2),
x)

[Out]

x/(log(x/5) - x + exp(4) + 4*exp(6))

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sympy [A]  time = 0.13, size = 17, normalized size = 0.77 \begin {gather*} \frac {x}{- x + \log {\relax (x )} - \log {\relax (5 )} + e^{4} + 4 e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((ln(x)-ln(5)+exp(4)+4*exp(3)**2-1)/(ln(x)**2+(-2*ln(5)+2*exp(4)+8*exp(3)**2-2*x)*ln(x)+ln(5)**2+(-2*
exp(4)-8*exp(3)**2+2*x)*ln(5)+exp(4)**2+(8*exp(3)**2-2*x)*exp(4)+16*exp(3)**4-8*x*exp(3)**2+x**2),x)

[Out]

x/(-x + log(x) - log(5) + exp(4) + 4*exp(6))

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