3.12.89 \(\int \frac {-2+(-4 x-2 x^2) \log (5)+((-1+4 x+x^2) \log (5)+\log (x^2)) \log ((-1+4 x+x^2) \log (5)+\log (x^2))}{((-x+4 x^2+x^3) \log (5)+x \log (x^2)) \log ((-1+4 x+x^2) \log (5)+\log (x^2))} \, dx\)

Optimal. Leaf size=22 \[ \log \left (\frac {x}{\log \left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right )}\right ) \]

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Rubi [F]  time = 1.03, 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 {-2+\left (-4 x-2 x^2\right ) \log (5)+\left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right ) \log \left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right )}{\left (\left (-x+4 x^2+x^3\right ) \log (5)+x \log \left (x^2\right )\right ) \log \left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [F]  time = 0.90, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-2+\left (-4 x-2 x^2\right ) \log (5)+\left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right ) \log \left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right )}{\left (\left (-x+4 x^2+x^3\right ) \log (5)+x \log \left (x^2\right )\right ) \log \left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.57, size = 27, normalized size = 1.23 \begin {gather*} \frac {1}{2} \, \log \left (x^{2}\right ) - \log \left (\log \left ({\left (x^{2} + 4 \, x - 1\right )} \log \relax (5) + \log \left (x^{2}\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.69, size = 27, normalized size = 1.23 \begin {gather*} \log \relax (x) - \log \left (\log \left (x^{2} \log \relax (5) + 4 \, x \log \relax (5) - \log \relax (5) + \log \left (x^{2}\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (\ln \left (x^{2}\right )+\left (x^{2}+4 x -1\right ) \ln \relax (5)\right ) \ln \left (\ln \left (x^{2}\right )+\left (x^{2}+4 x -1\right ) \ln \relax (5)\right )+\left (-2 x^{2}-4 x \right ) \ln \relax (5)-2}{\left (x \ln \left (x^{2}\right )+\left (x^{3}+4 x^{2}-x \right ) \ln \relax (5)\right ) \ln \left (\ln \left (x^{2}\right )+\left (x^{2}+4 x -1\right ) \ln \relax (5)\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((ln(x^2)+(x^2+4*x-1)*ln(5))*ln(ln(x^2)+(x^2+4*x-1)*ln(5))+(-2*x^2-4*x)*ln(5)-2)/(x*ln(x^2)+(x^3+4*x^2-x)*
ln(5))/ln(ln(x^2)+(x^2+4*x-1)*ln(5)),x)

[Out]

int(((ln(x^2)+(x^2+4*x-1)*ln(5))*ln(ln(x^2)+(x^2+4*x-1)*ln(5))+(-2*x^2-4*x)*ln(5)-2)/(x*ln(x^2)+(x^3+4*x^2-x)*
ln(5))/ln(ln(x^2)+(x^2+4*x-1)*ln(5)),x)

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maxima [A]  time = 0.54, size = 27, normalized size = 1.23 \begin {gather*} \log \relax (x) - \log \left (\log \left (x^{2} \log \relax (5) + 4 \, x \log \relax (5) - \log \relax (5) + 2 \, \log \relax (x)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.72, size = 23, normalized size = 1.05 \begin {gather*} \ln \relax (x)-\ln \left (\ln \left (\ln \left (x^2\right )+\ln \relax (5)\,\left (x^2+4\,x-1\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(5)*(4*x + 2*x^2) - log(log(x^2) + log(5)*(4*x + x^2 - 1))*(log(x^2) + log(5)*(4*x + x^2 - 1)) + 2)/(
log(log(x^2) + log(5)*(4*x + x^2 - 1))*(log(5)*(4*x^2 - x + x^3) + x*log(x^2))),x)

[Out]

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

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sympy [A]  time = 0.65, size = 22, normalized size = 1.00 \begin {gather*} \log {\relax (x )} - \log {\left (\log {\left (\left (x^{2} + 4 x - 1\right ) \log {\relax (5 )} + \log {\left (x^{2} \right )} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x) - log(log((x**2 + 4*x - 1)*log(5) + log(x**2)))

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