3.23.33 \(\int \frac {-1+x+(-5+10 x) \log (x)+(1-2 x) \log (x) \log (\log (x))}{(x^2-2 x^3+x^4) \log (x)} \, dx\)

Optimal. Leaf size=17 \[ \frac {5-\log (\log (x))}{x-x^2} \]

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Rubi [F]  time = 0.69, 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+x+(-5+10 x) \log (x)+(1-2 x) \log (x) \log (\log (x))}{\left (x^2-2 x^3+x^4\right ) \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-1 + x + (-5 + 10*x)*Log[x] + (1 - 2*x)*Log[x]*Log[Log[x]])/((x^2 - 2*x^3 + x^4)*Log[x]),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-1+x+(-5+10 x) \log (x)+(1-2 x) \log (x) \log (\log (x))}{x^2 \left (1-2 x+x^2\right ) \log (x)} \, dx\\ &=\int \frac {-1+x+(-5+10 x) \log (x)+(1-2 x) \log (x) \log (\log (x))}{(-1+x)^2 x^2 \log (x)} \, dx\\ &=\int \left (\frac {-1+x-5 \log (x)+10 x \log (x)}{(-1+x)^2 x^2 \log (x)}-\frac {(-1+2 x) \log (\log (x))}{(-1+x)^2 x^2}\right ) \, dx\\ &=\int \frac {-1+x-5 \log (x)+10 x \log (x)}{(-1+x)^2 x^2 \log (x)} \, dx-\int \frac {(-1+2 x) \log (\log (x))}{(-1+x)^2 x^2} \, dx\\ &=\int \left (\frac {5 (-1+2 x)}{(-1+x)^2 x^2}+\frac {1}{(-1+x) x^2 \log (x)}\right ) \, dx-\int \left (\frac {\log (\log (x))}{(-1+x)^2}-\frac {\log (\log (x))}{x^2}\right ) \, dx\\ &=5 \int \frac {-1+2 x}{(-1+x)^2 x^2} \, dx+\int \frac {1}{(-1+x) x^2 \log (x)} \, dx-\int \frac {\log (\log (x))}{(-1+x)^2} \, dx+\int \frac {\log (\log (x))}{x^2} \, dx\\ &=\frac {5}{(1-x) x}-\frac {\log (\log (x))}{x}+\int \frac {1}{x^2 \log (x)} \, dx+\int \frac {1}{(-1+x) x^2 \log (x)} \, dx-\int \frac {\log (\log (x))}{(-1+x)^2} \, dx\\ &=\frac {5}{(1-x) x}-\frac {\log (\log (x))}{x}+\int \frac {1}{(-1+x) x^2 \log (x)} \, dx-\int \frac {\log (\log (x))}{(-1+x)^2} \, dx+\operatorname {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log (x)\right )\\ &=\frac {5}{(1-x) x}+\text {Ei}(-\log (x))-\frac {\log (\log (x))}{x}+\int \frac {1}{(-1+x) x^2 \log (x)} \, dx-\int \frac {\log (\log (x))}{(-1+x)^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 14, normalized size = 0.82 \begin {gather*} \frac {-5+\log (\log (x))}{(-1+x) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + x + (-5 + 10*x)*Log[x] + (1 - 2*x)*Log[x]*Log[Log[x]])/((x^2 - 2*x^3 + x^4)*Log[x]),x]

[Out]

(-5 + Log[Log[x]])/((-1 + x)*x)

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fricas [A]  time = 0.53, size = 15, normalized size = 0.88 \begin {gather*} \frac {\log \left (\log \relax (x)\right ) - 5}{x^{2} - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1-2*x)*log(x)*log(log(x))+(10*x-5)*log(x)+x-1)/(x^4-2*x^3+x^2)/log(x),x, algorithm="fricas")

[Out]

(log(log(x)) - 5)/(x^2 - x)

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giac [A]  time = 0.21, size = 28, normalized size = 1.65 \begin {gather*} {\left (\frac {1}{x - 1} - \frac {1}{x}\right )} \log \left (\log \relax (x)\right ) - \frac {5}{x - 1} + \frac {5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1-2*x)*log(x)*log(log(x))+(10*x-5)*log(x)+x-1)/(x^4-2*x^3+x^2)/log(x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 24, normalized size = 1.41

method	result	size
risch	\(\frac {\ln \left (\ln \relax (x )\right )}{x \left (x -1\right )}-\frac {5}{x \left (x -1\right )}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1-2*x)*ln(x)*ln(ln(x))+(10*x-5)*ln(x)+x-1)/(x^4-2*x^3+x^2)/ln(x),x,method=_RETURNVERBOSE)

[Out]

1/x/(x-1)*ln(ln(x))-5/x/(x-1)

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maxima [A]  time = 0.56, size = 15, normalized size = 0.88 \begin {gather*} \frac {\log \left (\log \relax (x)\right ) - 5}{x^{2} - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1-2*x)*log(x)*log(log(x))+(10*x-5)*log(x)+x-1)/(x^4-2*x^3+x^2)/log(x),x, algorithm="maxima")

[Out]

(log(log(x)) - 5)/(x^2 - x)

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mupad [B]  time = 1.43, size = 14, normalized size = 0.82 \begin {gather*} \frac {\ln \left (\ln \relax (x)\right )-5}{x\,\left (x-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + log(x)*(10*x - 5) - log(log(x))*log(x)*(2*x - 1) - 1)/(log(x)*(x^2 - 2*x^3 + x^4)),x)

[Out]

(log(log(x)) - 5)/(x*(x - 1))

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sympy [A]  time = 0.33, size = 15, normalized size = 0.88 \begin {gather*} \frac {\log {\left (\log {\relax (x )} \right )}}{x^{2} - x} - \frac {5}{x^{2} - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1-2*x)*ln(x)*ln(ln(x))+(10*x-5)*ln(x)+x-1)/(x**4-2*x**3+x**2)/ln(x),x)

[Out]

log(log(x))/(x**2 - x) - 5/(x**2 - x)

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