3.25.26 \(\int \frac {-16-20 x+x^2-x^3+(3 x^2-2 x^3) \log (x)+(16+x^2-2 x^3) \log (x) \log (\log (x))}{x^2 \log (x)} \, dx\)

Optimal. Leaf size=32 \[ (3-x) x+\left (-4+\left (5-\left (2+\frac {4}{x}\right )^2-x\right ) x\right ) \log (\log (x)) \]

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Rubi [F]  time = 0.53, 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 {-16-20 x+x^2-x^3+\left (3 x^2-2 x^3\right ) \log (x)+\left (16+x^2-2 x^3\right ) \log (x) \log (\log (x))}{x^2 \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

3*x - x^2 + 16*ExpIntegralEi[-Log[x]] + ExpIntegralEi[2*Log[x]] - (16*Log[Log[x]])/x + x*Log[Log[x]] - x^2*Log
[Log[x]] - LogIntegral[x] + Defer[Int][(-16 - 20*x + x^2 - x^3)/(x^2*Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {-16-20 x+x^2-x^3+3 x^2 \log (x)-2 x^3 \log (x)}{x^2 \log (x)}-\frac {\left (-16-x^2+2 x^3\right ) \log (\log (x))}{x^2}\right ) \, dx\\ &=\int \frac {-16-20 x+x^2-x^3+3 x^2 \log (x)-2 x^3 \log (x)}{x^2 \log (x)} \, dx-\int \frac {\left (-16-x^2+2 x^3\right ) \log (\log (x))}{x^2} \, dx\\ &=\int \left (3-2 x+\frac {-16-20 x+x^2-x^3}{x^2 \log (x)}\right ) \, dx-\int \left (-\log (\log (x))-\frac {16 \log (\log (x))}{x^2}+2 x \log (\log (x))\right ) \, dx\\ &=3 x-x^2-2 \int x \log (\log (x)) \, dx+16 \int \frac {\log (\log (x))}{x^2} \, dx+\int \frac {-16-20 x+x^2-x^3}{x^2 \log (x)} \, dx+\int \log (\log (x)) \, dx\\ &=3 x-x^2-\frac {16 \log (\log (x))}{x}+x \log (\log (x))-x^2 \log (\log (x))+16 \int \frac {1}{x^2 \log (x)} \, dx-\int \frac {1}{\log (x)} \, dx+\int \frac {x}{\log (x)} \, dx+\int \frac {-16-20 x+x^2-x^3}{x^2 \log (x)} \, dx\\ &=3 x-x^2-\frac {16 \log (\log (x))}{x}+x \log (\log (x))-x^2 \log (\log (x))-\text {li}(x)+16 \operatorname {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log (x)\right )+\int \frac {-16-20 x+x^2-x^3}{x^2 \log (x)} \, dx+\operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=3 x-x^2+16 \text {Ei}(-\log (x))+\text {Ei}(2 \log (x))-\frac {16 \log (\log (x))}{x}+x \log (\log (x))-x^2 \log (\log (x))-\text {li}(x)+\int \frac {-16-20 x+x^2-x^3}{x^2 \log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 35, normalized size = 1.09 \begin {gather*} 3 x-x^2-20 \log (\log (x))-\frac {16 \log (\log (x))}{x}+x \log (\log (x))-x^2 \log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

3*x - x^2 - 20*Log[Log[x]] - (16*Log[Log[x]])/x + x*Log[Log[x]] - x^2*Log[Log[x]]

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fricas [A]  time = 0.55, size = 31, normalized size = 0.97 \begin {gather*} -\frac {x^{3} - 3 \, x^{2} + {\left (x^{3} - x^{2} + 20 \, x + 16\right )} \log \left (\log \relax (x)\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x^3 - 3*x^2 + (x^3 - x^2 + 20*x + 16)*log(log(x)))/x

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giac [A]  time = 0.33, size = 31, normalized size = 0.97 \begin {gather*} -x^{2} - {\left (x^{2} - x + \frac {16}{x}\right )} \log \left (\log \relax (x)\right ) + 3 \, x - 20 \, \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^2 - (x^2 - x + 16/x)*log(log(x)) + 3*x - 20*log(log(x))

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maple [A]  time = 0.03, size = 33, normalized size = 1.03

method	result	size
risch	\(-\frac {\left (x^{3}-x^{2}+16\right ) \ln \left (\ln \relax (x )\right )}{x}-x^{2}+3 x -20 \ln \left (\ln \relax (x )\right )\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^3+x^2+16)*ln(x)*ln(ln(x))+(-2*x^3+3*x^2)*ln(x)-x^3+x^2-20*x-16)/x^2/ln(x),x,method=_RETURNVERBOSE)

[Out]

-(x^3-x^2+16)/x*ln(ln(x))-x^2+3*x-20*ln(ln(x))

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maxima [A]  time = 0.52, size = 35, normalized size = 1.09 \begin {gather*} -x^{2} \log \left (\log \relax (x)\right ) - x^{2} + x \log \left (\log \relax (x)\right ) + 3 \, x - \frac {16 \, \log \left (\log \relax (x)\right )}{x} - 20 \, \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^2*log(log(x)) - x^2 + x*log(log(x)) + 3*x - 16*log(log(x))/x - 20*log(log(x))

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mupad [B]  time = 1.42, size = 49, normalized size = 1.53 \begin {gather*} 3\,x-20\,\ln \left (\ln \relax (x)\right )+\ln \left (\ln \relax (x)\right )\,\left (\frac {2\,x^2-3\,x^3}{x}-\frac {-2\,x^3+x^2+16}{x}\right )-x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(20*x - log(x)*(3*x^2 - 2*x^3) - x^2 + x^3 - log(log(x))*log(x)*(x^2 - 2*x^3 + 16) + 16)/(x^2*log(x)),x)

[Out]

3*x - 20*log(log(x)) + log(log(x))*((2*x^2 - 3*x^3)/x - (x^2 - 2*x^3 + 16)/x) - x^2

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sympy [A]  time = 0.37, size = 27, normalized size = 0.84 \begin {gather*} - x^{2} + 3 x - 20 \log {\left (\log {\relax (x )} \right )} + \frac {\left (- x^{3} + x^{2} - 16\right ) \log {\left (\log {\relax (x )} \right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**3+x**2+16)*ln(x)*ln(ln(x))+(-2*x**3+3*x**2)*ln(x)-x**3+x**2-20*x-16)/x**2/ln(x),x)

[Out]

-x**2 + 3*x - 20*log(log(x)) + (-x**3 + x**2 - 16)*log(log(x))/x

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