3.96.98 \(\int \frac {1+e^2+2 x^2-\log (x)+(e^2 (-1-2 x)+2 x^2+x \log (5)+\log (x)) \log (\frac {-2 x^2+e^2 (1+2 x)-x \log (5)-\log (x)}{x})}{e^2 (-1-2 x)+2 x^2+x \log (5)+\log (x)} \, dx\)

Optimal. Leaf size=28 \[ x \log \left (2 e^2-2 x-\log (5)+\frac {e^2-\log (x)}{x}\right ) \]

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Rubi [F]  time = 0.73, 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^2+2 x^2-\log (x)+\left (e^2 (-1-2 x)+2 x^2+x \log (5)+\log (x)\right ) \log \left (\frac {-2 x^2+e^2 (1+2 x)-x \log (5)-\log (x)}{x}\right )}{e^2 (-1-2 x)+2 x^2+x \log (5)+\log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(1 + E^2 + 2*x^2 - Log[x] + (E^2*(-1 - 2*x) + 2*x^2 + x*Log[5] + Log[x])*Log[(-2*x^2 + E^2*(1 + 2*x) - x*L
og[5] - Log[x])/x])/(E^2*(-1 - 2*x) + 2*x^2 + x*Log[5] + Log[x]),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 29, normalized size = 1.04 \begin {gather*} x \log \left (-\frac {-e^2 (1+2 x)+x (2 x+\log (5))+\log (x)}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + E^2 + 2*x^2 - Log[x] + (E^2*(-1 - 2*x) + 2*x^2 + x*Log[5] + Log[x])*Log[(-2*x^2 + E^2*(1 + 2*x)
- x*Log[5] - Log[x])/x])/(E^2*(-1 - 2*x) + 2*x^2 + x*Log[5] + Log[x]),x]

[Out]

x*Log[-((-(E^2*(1 + 2*x)) + x*(2*x + Log[5]) + Log[x])/x)]

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fricas [A]  time = 0.60, size = 29, normalized size = 1.04 \begin {gather*} x \log \left (-\frac {2 \, x^{2} - {\left (2 \, x + 1\right )} e^{2} + x \log \relax (5) + \log \relax (x)}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((log(x)+x*log(5)+(-2*x-1)*exp(2)+2*x^2)*log((-log(x)-x*log(5)+(2*x+1)*exp(2)-2*x^2)/x)-log(x)+exp(2
)+2*x^2+1)/(log(x)+x*log(5)+(-2*x-1)*exp(2)+2*x^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.22, size = 31, normalized size = 1.11 \begin {gather*} x \log \left (-2 \, x^{2} + 2 \, x e^{2} - x \log \relax (5) + e^{2} - \log \relax (x)\right ) - x \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((log(x)+x*log(5)+(-2*x-1)*exp(2)+2*x^2)*log((-log(x)-x*log(5)+(2*x+1)*exp(2)-2*x^2)/x)-log(x)+exp(2
)+2*x^2+1)/(log(x)+x*log(5)+(-2*x-1)*exp(2)+2*x^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.31, size = 31, normalized size = 1.11

method	result	size
norman	\(x \ln \left (\frac {-\ln \relax (x )-x \ln \relax (5)+\left (2 x +1\right ) {\mathrm e}^{2}-2 x^{2}}{x}\right )\)	\(31\)
risch	\(\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (\left (\frac {1}{2}+x \right ) {\mathrm e}^{2}-\frac {x \ln \relax (5)}{2}-x^{2}-\frac {\ln \relax (x )}{2}\right )}{x}\right )^{2}}{2}+\frac {i \pi x \,\mathrm {csgn}\left (i \left (\left (\frac {1}{2}+x \right ) {\mathrm e}^{2}-\frac {x \ln \relax (5)}{2}-x^{2}-\frac {\ln \relax (x )}{2}\right )\right ) \mathrm {csgn}\left (\frac {i \left (\left (\frac {1}{2}+x \right ) {\mathrm e}^{2}-\frac {x \ln \relax (5)}{2}-x^{2}-\frac {\ln \relax (x )}{2}\right )}{x}\right )^{2}}{2}-\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (\left (\frac {1}{2}+x \right ) {\mathrm e}^{2}-\frac {x \ln \relax (5)}{2}-x^{2}-\frac {\ln \relax (x )}{2}\right )\right ) \mathrm {csgn}\left (\frac {i \left (\left (\frac {1}{2}+x \right ) {\mathrm e}^{2}-\frac {x \ln \relax (5)}{2}-x^{2}-\frac {\ln \relax (x )}{2}\right )}{x}\right )}{2}-\frac {i \pi x \mathrm {csgn}\left (\frac {i \left (\left (\frac {1}{2}+x \right ) {\mathrm e}^{2}-\frac {x \ln \relax (5)}{2}-x^{2}-\frac {\ln \relax (x )}{2}\right )}{x}\right )^{3}}{2}+x \ln \relax (2)-x \ln \relax (x )+x \ln \left (\left (\frac {1}{2}+x \right ) {\mathrm e}^{2}-\frac {x \ln \relax (5)}{2}-x^{2}-\frac {\ln \relax (x )}{2}\right )\)	\(237\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((ln(x)+x*ln(5)+(-2*x-1)*exp(2)+2*x^2)*ln((-ln(x)-x*ln(5)+(2*x+1)*exp(2)-2*x^2)/x)-ln(x)+exp(2)+2*x^2+1)/(
ln(x)+x*ln(5)+(-2*x-1)*exp(2)+2*x^2),x,method=_RETURNVERBOSE)

[Out]

x*ln((-ln(x)-x*ln(5)+(2*x+1)*exp(2)-2*x^2)/x)

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maxima [A]  time = 0.47, size = 32, normalized size = 1.14 \begin {gather*} x \log \left (-2 \, x^{2} + x {\left (2 \, e^{2} - \log \relax (5)\right )} + e^{2} - \log \relax (x)\right ) - x \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((log(x)+x*log(5)+(-2*x-1)*exp(2)+2*x^2)*log((-log(x)-x*log(5)+(2*x+1)*exp(2)-2*x^2)/x)-log(x)+exp(2
)+2*x^2+1)/(log(x)+x*log(5)+(-2*x-1)*exp(2)+2*x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 7.71, size = 29, normalized size = 1.04 \begin {gather*} x\,\ln \left (-\frac {\ln \relax (x)+x\,\ln \relax (5)+2\,x^2-{\mathrm {e}}^2\,\left (2\,x+1\right )}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2) - log(x) + log(-(log(x) + x*log(5) + 2*x^2 - exp(2)*(2*x + 1))/x)*(log(x) + x*log(5) + 2*x^2 - exp
(2)*(2*x + 1)) + 2*x^2 + 1)/(log(x) + x*log(5) + 2*x^2 - exp(2)*(2*x + 1)),x)

[Out]

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

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sympy [A]  time = 0.51, size = 26, normalized size = 0.93 \begin {gather*} x \log {\left (\frac {- 2 x^{2} - x \log {\relax (5 )} + \left (2 x + 1\right ) e^{2} - \log {\relax (x )}}{x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((ln(x)+x*ln(5)+(-2*x-1)*exp(2)+2*x**2)*ln((-ln(x)-x*ln(5)+(2*x+1)*exp(2)-2*x**2)/x)-ln(x)+exp(2)+2*
x**2+1)/(ln(x)+x*ln(5)+(-2*x-1)*exp(2)+2*x**2),x)

[Out]

x*log((-2*x**2 - x*log(5) + (2*x + 1)*exp(2) - log(x))/x)

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