3.51.56 \(\int \frac {8-17 x+3 x^2+(8-16 x+x^2) \log (\frac {x}{5})+(2-4 x) \log ^2(\frac {x}{5})}{4 x^2-8 x^3+4 x^4+e^2 (4 x^2-8 x^3+4 x^4)+(4 x^2-8 x^3+4 x^4+e^2 (4 x^2-8 x^3+4 x^4)) \log (\frac {x}{5})+(x^2-2 x^3+x^4+e^2 (x^2-2 x^3+x^4)) \log ^2(\frac {x}{5})} \, dx\)

Optimal. Leaf size=31 \[ \frac {2-\frac {x}{2+\log \left (\frac {x}{5}\right )}}{\left (1+e^2\right ) (-1+x) x} \]

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Rubi [F]  time = 0.63, 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 {8-17 x+3 x^2+\left (8-16 x+x^2\right ) \log \left (\frac {x}{5}\right )+(2-4 x) \log ^2\left (\frac {x}{5}\right )}{4 x^2-8 x^3+4 x^4+e^2 \left (4 x^2-8 x^3+4 x^4\right )+\left (4 x^2-8 x^3+4 x^4+e^2 \left (4 x^2-8 x^3+4 x^4\right )\right ) \log \left (\frac {x}{5}\right )+\left (x^2-2 x^3+x^4+e^2 \left (x^2-2 x^3+x^4\right )\right ) \log ^2\left (\frac {x}{5}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(8 - 17*x + 3*x^2 + (8 - 16*x + x^2)*Log[x/5] + (2 - 4*x)*Log[x/5]^2)/(4*x^2 - 8*x^3 + 4*x^4 + E^2*(4*x^2
- 8*x^3 + 4*x^4) + (4*x^2 - 8*x^3 + 4*x^4 + E^2*(4*x^2 - 8*x^3 + 4*x^4))*Log[x/5] + (x^2 - 2*x^3 + x^4 + E^2*(
x^2 - 2*x^3 + x^4))*Log[x/5]^2),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {8-17 x+3 x^2+\left (8-16 x+x^2\right ) \log \left (\frac {x}{5}\right )+(2-4 x) \log ^2\left (\frac {x}{5}\right )}{\left (1+e^2\right ) (1-x)^2 x^2 \left (2+\log \left (\frac {x}{5}\right )\right )^2} \, dx\\ &=\frac {\int \frac {8-17 x+3 x^2+\left (8-16 x+x^2\right ) \log \left (\frac {x}{5}\right )+(2-4 x) \log ^2\left (\frac {x}{5}\right )}{(1-x)^2 x^2 \left (2+\log \left (\frac {x}{5}\right )\right )^2} \, dx}{1+e^2}\\ &=\frac {\int \left (-\frac {2 (-1+2 x)}{(-1+x)^2 x^2}+\frac {1}{(-1+x) x \left (2+\log \left (\frac {x}{5}\right )\right )^2}+\frac {1}{(-1+x)^2 \left (2+\log \left (\frac {x}{5}\right )\right )}\right ) \, dx}{1+e^2}\\ &=\frac {\int \frac {1}{(-1+x) x \left (2+\log \left (\frac {x}{5}\right )\right )^2} \, dx}{1+e^2}+\frac {\int \frac {1}{(-1+x)^2 \left (2+\log \left (\frac {x}{5}\right )\right )} \, dx}{1+e^2}-\frac {2 \int \frac {-1+2 x}{(-1+x)^2 x^2} \, dx}{1+e^2}\\ &=-\frac {2}{\left (1+e^2\right ) (1-x) x}+\frac {\int \frac {1}{(-1+x) x \left (2+\log \left (\frac {x}{5}\right )\right )^2} \, dx}{1+e^2}+\frac {\int \frac {1}{(-1+x)^2 \left (2+\log \left (\frac {x}{5}\right )\right )} \, dx}{1+e^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 39, normalized size = 1.26 \begin {gather*} \frac {\frac {2}{-1+x}-\frac {2}{x}+\frac {1}{(1-x) \left (2+\log \left (\frac {x}{5}\right )\right )}}{1+e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8 - 17*x + 3*x^2 + (8 - 16*x + x^2)*Log[x/5] + (2 - 4*x)*Log[x/5]^2)/(4*x^2 - 8*x^3 + 4*x^4 + E^2*(
4*x^2 - 8*x^3 + 4*x^4) + (4*x^2 - 8*x^3 + 4*x^4 + E^2*(4*x^2 - 8*x^3 + 4*x^4))*Log[x/5] + (x^2 - 2*x^3 + x^4 +
E^2*(x^2 - 2*x^3 + x^4))*Log[x/5]^2),x]

[Out]

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

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fricas [A]  time = 1.99, size = 55, normalized size = 1.77 \begin {gather*} -\frac {x - 2 \, \log \left (\frac {1}{5} \, x\right ) - 4}{2 \, x^{2} + 2 \, {\left (x^{2} - x\right )} e^{2} + {\left (x^{2} + {\left (x^{2} - x\right )} e^{2} - x\right )} \log \left (\frac {1}{5} \, x\right ) - 2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x+2)*log(1/5*x)^2+(x^2-16*x+8)*log(1/5*x)+3*x^2-17*x+8)/(((x^4-2*x^3+x^2)*exp(2)+x^4-2*x^3+x^2)
*log(1/5*x)^2+((4*x^4-8*x^3+4*x^2)*exp(2)+4*x^4-8*x^3+4*x^2)*log(1/5*x)+(4*x^4-8*x^3+4*x^2)*exp(2)+4*x^4-8*x^3
+4*x^2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.27, size = 68, normalized size = 2.19 \begin {gather*} -\frac {x - 2 \, \log \left (\frac {1}{5} \, x\right ) - 4}{x^{2} e^{2} \log \left (\frac {1}{5} \, x\right ) + 2 \, x^{2} e^{2} + x^{2} \log \left (\frac {1}{5} \, x\right ) - x e^{2} \log \left (\frac {1}{5} \, x\right ) + 2 \, x^{2} - 2 \, x e^{2} - x \log \left (\frac {1}{5} \, x\right ) - 2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x+2)*log(1/5*x)^2+(x^2-16*x+8)*log(1/5*x)+3*x^2-17*x+8)/(((x^4-2*x^3+x^2)*exp(2)+x^4-2*x^3+x^2)
*log(1/5*x)^2+((4*x^4-8*x^3+4*x^2)*exp(2)+4*x^4-8*x^3+4*x^2)*log(1/5*x)+(4*x^4-8*x^3+4*x^2)*exp(2)+4*x^4-8*x^3
+4*x^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.18, size = 43, normalized size = 1.39

method	result	size
risch	\(\frac {2}{x \left ({\mathrm e}^{2} x -{\mathrm e}^{2}+x -1\right )}-\frac {1}{\left ({\mathrm e}^{2} x -{\mathrm e}^{2}+x -1\right ) \left (\ln \left (\frac {x}{5}\right )+2\right )}\)	\(43\)
norman	\(\frac {-\frac {x}{{\mathrm e}^{2}+1}+\frac {4}{{\mathrm e}^{2}+1}+\frac {2 \ln \left (\frac {x}{5}\right )}{{\mathrm e}^{2}+1}}{x \left (\ln \left (\frac {x}{5}\right )+2\right ) \left (x -1\right )}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x+2)*ln(1/5*x)^2+(x^2-16*x+8)*ln(1/5*x)+3*x^2-17*x+8)/(((x^4-2*x^3+x^2)*exp(2)+x^4-2*x^3+x^2)*ln(1/5*
x)^2+((4*x^4-8*x^3+4*x^2)*exp(2)+4*x^4-8*x^3+4*x^2)*ln(1/5*x)+(4*x^4-8*x^3+4*x^2)*exp(2)+4*x^4-8*x^3+4*x^2),x,
method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.48, size = 64, normalized size = 2.06 \begin {gather*} \frac {x + 2 \, \log \relax (5) - 2 \, \log \relax (x) - 4}{{\left ({\left (\log \relax (5) - 2\right )} e^{2} + \log \relax (5) - 2\right )} x^{2} - {\left ({\left (\log \relax (5) - 2\right )} e^{2} + \log \relax (5) - 2\right )} x - {\left (x^{2} {\left (e^{2} + 1\right )} - x {\left (e^{2} + 1\right )}\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x+2)*log(1/5*x)^2+(x^2-16*x+8)*log(1/5*x)+3*x^2-17*x+8)/(((x^4-2*x^3+x^2)*exp(2)+x^4-2*x^3+x^2)
*log(1/5*x)^2+((4*x^4-8*x^3+4*x^2)*exp(2)+4*x^4-8*x^3+4*x^2)*log(1/5*x)+(4*x^4-8*x^3+4*x^2)*exp(2)+4*x^4-8*x^3
+4*x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.85, size = 38, normalized size = 1.23 \begin {gather*} \frac {2}{x\,\left ({\mathrm {e}}^2+1\right )\,\left (x-1\right )}-\frac {1}{\left (\ln \left (\frac {x}{5}\right )+2\right )\,\left ({\mathrm {e}}^2+1\right )\,\left (x-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x/5)*(x^2 - 16*x + 8) - log(x/5)^2*(4*x - 2) - 17*x + 3*x^2 + 8)/(log(x/5)^2*(exp(2)*(x^2 - 2*x^3 + x
^4) + x^2 - 2*x^3 + x^4) + log(x/5)*(exp(2)*(4*x^2 - 8*x^3 + 4*x^4) + 4*x^2 - 8*x^3 + 4*x^4) + exp(2)*(4*x^2 -
8*x^3 + 4*x^4) + 4*x^2 - 8*x^3 + 4*x^4),x)

[Out]

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

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sympy [B]  time = 0.40, size = 53, normalized size = 1.71 \begin {gather*} - \frac {1}{2 x + 2 x e^{2} + \left (x + x e^{2} - e^{2} - 1\right ) \log {\left (\frac {x}{5} \right )} - 2 e^{2} - 2} + \frac {2}{x^{2} \left (1 + e^{2}\right ) + x \left (- e^{2} - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x+2)*ln(1/5*x)**2+(x**2-16*x+8)*ln(1/5*x)+3*x**2-17*x+8)/(((x**4-2*x**3+x**2)*exp(2)+x**4-2*x**
3+x**2)*ln(1/5*x)**2+((4*x**4-8*x**3+4*x**2)*exp(2)+4*x**4-8*x**3+4*x**2)*ln(1/5*x)+(4*x**4-8*x**3+4*x**2)*exp
(2)+4*x**4-8*x**3+4*x**2),x)

[Out]

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

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