3.3.39 \(\int \frac {10+8 x-2 x^2-2 x \log (5)+(16 x-2 x^2-4 x \log (5)) \log (x)}{-20+5 x+5 \log (5)+(-4 x^2+x^3+x^2 \log (5)) \log (x)} \, dx\)

Optimal. Leaf size=20 \[ \log \left (\frac {2 (-4+x+\log (5))^2}{\left (5+x^2 \log (x)\right )^2}\right ) \]

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Rubi [F]  time = 0.62, 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 {10+8 x-2 x^2-2 x \log (5)+\left (16 x-2 x^2-4 x \log (5)\right ) \log (x)}{-20+5 x+5 \log (5)+\left (-4 x^2+x^3+x^2 \log (5)\right ) \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(10 + 8*x - 2*x^2 - 2*x*Log[5] + (16*x - 2*x^2 - 4*x*Log[5])*Log[x])/(-20 + 5*x + 5*Log[5] + (-4*x^2 + x^3
+ x^2*Log[5])*Log[x]),x]

[Out]

(-2*(8 - Log[25])*Log[x])/(4 - Log[5]) + 2*Log[4 - x - Log[5]] + 20*Defer[Int][1/(x*(5 + x^2*Log[x])), x] - 2*
Defer[Int][x/(5 + x^2*Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {10-2 x^2+x (8-2 \log (5))+\left (16 x-2 x^2-4 x \log (5)\right ) \log (x)}{-20+5 x+5 \log (5)+\left (-4 x^2+x^3+x^2 \log (5)\right ) \log (x)} \, dx\\ &=\int \frac {2 \left (-5+x^2+x (-4+\log (5))+x (-8+x+\log (25)) \log (x)\right )}{(4-x-\log (5)) \left (5+x^2 \log (x)\right )} \, dx\\ &=2 \int \frac {-5+x^2+x (-4+\log (5))+x (-8+x+\log (25)) \log (x)}{(4-x-\log (5)) \left (5+x^2 \log (x)\right )} \, dx\\ &=2 \int \left (\frac {8-x-\log (25)}{x (-4+x+\log (5))}+\frac {-10 x+x^3-x^2 (4-\log (5))+5 (8-\log (25))}{x (4-x-\log (5)) \left (5+x^2 \log (x)\right )}\right ) \, dx\\ &=2 \int \frac {8-x-\log (25)}{x (-4+x+\log (5))} \, dx+2 \int \frac {-10 x+x^3-x^2 (4-\log (5))+5 (8-\log (25))}{x (4-x-\log (5)) \left (5+x^2 \log (x)\right )} \, dx\\ &=2 \int \left (\frac {1}{-4+x+\log (5)}+\frac {8-\log (25)}{x (-4+\log (5))}\right ) \, dx+2 \int \frac {10-x^2}{x \left (5+x^2 \log (x)\right )} \, dx\\ &=-\frac {2 (8-\log (25)) \log (x)}{4-\log (5)}+2 \log (4-x-\log (5))+2 \int \left (\frac {10}{x \left (5+x^2 \log (x)\right )}-\frac {x}{5+x^2 \log (x)}\right ) \, dx\\ &=-\frac {2 (8-\log (25)) \log (x)}{4-\log (5)}+2 \log (4-x-\log (5))-2 \int \frac {x}{5+x^2 \log (x)} \, dx+20 \int \frac {1}{x \left (5+x^2 \log (x)\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 0.34, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {10+8 x-2 x^2-2 x \log (5)+\left (16 x-2 x^2-4 x \log (5)\right ) \log (x)}{-20+5 x+5 \log (5)+\left (-4 x^2+x^3+x^2 \log (5)\right ) \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(10 + 8*x - 2*x^2 - 2*x*Log[5] + (16*x - 2*x^2 - 4*x*Log[5])*Log[x])/(-20 + 5*x + 5*Log[5] + (-4*x^2
+ x^3 + x^2*Log[5])*Log[x]),x]

[Out]

Integrate[(10 + 8*x - 2*x^2 - 2*x*Log[5] + (16*x - 2*x^2 - 4*x*Log[5])*Log[x])/(-20 + 5*x + 5*Log[5] + (-4*x^2
+ x^3 + x^2*Log[5])*Log[x]), x]

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fricas [A]  time = 1.05, size = 28, normalized size = 1.40 \begin {gather*} 2 \, \log \left (x + \log \relax (5) - 4\right ) - 4 \, \log \relax (x) - 2 \, \log \left (\frac {x^{2} \log \relax (x) + 5}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*log(5)-2*x^2+16*x)*log(x)-2*x*log(5)-2*x^2+8*x+10)/((x^2*log(5)+x^3-4*x^2)*log(x)+5*log(5)+5*
x-20),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.28, size = 20, normalized size = 1.00 \begin {gather*} -2 \, \log \left (x^{2} \log \relax (x) + 5\right ) + 2 \, \log \left (x + \log \relax (5) - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*log(5)-2*x^2+16*x)*log(x)-2*x*log(5)-2*x^2+8*x+10)/((x^2*log(5)+x^3-4*x^2)*log(x)+5*log(5)+5*
x-20),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.16, size = 21, normalized size = 1.05

method	result	size
norman	\(2 \ln \left (\ln \relax (5)-4+x \right )-2 \ln \left (5+x^{2} \ln \relax (x )\right )\)	\(21\)
risch	\(-4 \ln \relax (x )+2 \ln \left (\ln \relax (5)-4+x \right )-2 \ln \left (\ln \relax (x )+\frac {5}{x^{2}}\right )\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x*ln(5)-2*x^2+16*x)*ln(x)-2*x*ln(5)-2*x^2+8*x+10)/((x^2*ln(5)+x^3-4*x^2)*ln(x)+5*ln(5)+5*x-20),x,meth
od=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.89, size = 28, normalized size = 1.40 \begin {gather*} 2 \, \log \left (x + \log \relax (5) - 4\right ) - 4 \, \log \relax (x) - 2 \, \log \left (\frac {x^{2} \log \relax (x) + 5}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*log(5)-2*x^2+16*x)*log(x)-2*x*log(5)-2*x^2+8*x+10)/((x^2*log(5)+x^3-4*x^2)*log(x)+5*log(5)+5*
x-20),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.55, size = 78, normalized size = 3.90 \begin {gather*} \frac {16\,\ln \relax (x)}{\ln \relax (5)-4}-2\,\ln \left (x+\ln \relax (5)-4\right )-2\,\ln \left (x^2\,\ln \relax (x)+5\right )-\frac {16\,\ln \left (x+\ln \relax (5)-4\right )}{\ln \relax (5)-4}-2\,\ln \left (\frac {1}{x^2}\right )+\frac {4\,\ln \relax (5)\,\ln \left (x+\ln \relax (5)-4\right )}{\ln \relax (5)-4}-\frac {4\,\ln \relax (5)\,\ln \relax (x)}{\ln \relax (5)-4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x*log(5) - 8*x + 2*x^2 + log(x)*(4*x*log(5) - 16*x + 2*x^2) - 10)/(5*x + 5*log(5) + log(x)*(x^2*log(5)
- 4*x^2 + x^3) - 20),x)

[Out]

(16*log(x))/(log(5) - 4) - 2*log(x + log(5) - 4) - 2*log(x^2*log(x) + 5) - (16*log(x + log(5) - 4))/(log(5) -
4) - 2*log(1/x^2) + (4*log(5)*log(x + log(5) - 4))/(log(5) - 4) - (4*log(5)*log(x))/(log(5) - 4)

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sympy [A]  time = 0.62, size = 26, normalized size = 1.30 \begin {gather*} - 4 \log {\relax (x )} - 2 \log {\left (\log {\relax (x )} + \frac {5}{x^{2}} \right )} + 2 \log {\left (x - 4 + \log {\relax (5 )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*ln(5)-2*x**2+16*x)*ln(x)-2*x*ln(5)-2*x**2+8*x+10)/((x**2*ln(5)+x**3-4*x**2)*ln(x)+5*ln(5)+5*x
-20),x)

[Out]

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

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