3.101.62 \(\int \frac {-6-369 x+149 x^2-15 x^3+(12+369 x-15 x^3) \log (x) \log (\log (x))}{x^3 \log (x)} \, dx\)

Optimal. Leaf size=29 \[ \frac {\left (6-3 \left (5 (5-x)^2+\frac {2}{x}\right )-x\right ) \log (\log (x))}{x} \]

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Rubi [F]  time = 0.33, 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 {-6-369 x+149 x^2-15 x^3+\left (12+369 x-15 x^3\right ) \log (x) \log (\log (x))}{x^3 \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-6 - 369*x + 149*x^2 - 15*x^3 + (12 + 369*x - 15*x^3)*Log[x]*Log[Log[x]])/(x^3*Log[x]),x]

[Out]

6*ExpIntegralEi[-2*Log[x]] + 369*ExpIntegralEi[-Log[x]] - (6*Log[Log[x]])/x^2 - (369*Log[Log[x]])/x - 15*x*Log
[Log[x]] + 15*LogIntegral[x] + Defer[Int][(-6 - 369*x + 149*x^2 - 15*x^3)/(x^3*Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {-6-369 x+149 x^2-15 x^3}{x^3 \log (x)}-\frac {3 \left (-4-123 x+5 x^3\right ) \log (\log (x))}{x^3}\right ) \, dx\\ &=-\left (3 \int \frac {\left (-4-123 x+5 x^3\right ) \log (\log (x))}{x^3} \, dx\right )+\int \frac {-6-369 x+149 x^2-15 x^3}{x^3 \log (x)} \, dx\\ &=-\left (3 \int \left (5 \log (\log (x))-\frac {4 \log (\log (x))}{x^3}-\frac {123 \log (\log (x))}{x^2}\right ) \, dx\right )+\int \frac {-6-369 x+149 x^2-15 x^3}{x^3 \log (x)} \, dx\\ &=12 \int \frac {\log (\log (x))}{x^3} \, dx-15 \int \log (\log (x)) \, dx+369 \int \frac {\log (\log (x))}{x^2} \, dx+\int \frac {-6-369 x+149 x^2-15 x^3}{x^3 \log (x)} \, dx\\ &=-\frac {6 \log (\log (x))}{x^2}-\frac {369 \log (\log (x))}{x}-15 x \log (\log (x))+6 \int \frac {1}{x^3 \log (x)} \, dx+15 \int \frac {1}{\log (x)} \, dx+369 \int \frac {1}{x^2 \log (x)} \, dx+\int \frac {-6-369 x+149 x^2-15 x^3}{x^3 \log (x)} \, dx\\ &=-\frac {6 \log (\log (x))}{x^2}-\frac {369 \log (\log (x))}{x}-15 x \log (\log (x))+15 \text {li}(x)+6 \operatorname {Subst}\left (\int \frac {e^{-2 x}}{x} \, dx,x,\log (x)\right )+369 \operatorname {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log (x)\right )+\int \frac {-6-369 x+149 x^2-15 x^3}{x^3 \log (x)} \, dx\\ &=6 \text {Ei}(-2 \log (x))+369 \text {Ei}(-\log (x))-\frac {6 \log (\log (x))}{x^2}-\frac {369 \log (\log (x))}{x}-15 x \log (\log (x))+15 \text {li}(x)+\int \frac {-6-369 x+149 x^2-15 x^3}{x^3 \log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 21, normalized size = 0.72 \begin {gather*} \left (149-\frac {3 \left (2+123 x+5 x^3\right )}{x^2}\right ) \log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-6 - 369*x + 149*x^2 - 15*x^3 + (12 + 369*x - 15*x^3)*Log[x]*Log[Log[x]])/(x^3*Log[x]),x]

[Out]

(149 - (3*(2 + 123*x + 5*x^3))/x^2)*Log[Log[x]]

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fricas [A]  time = 0.60, size = 23, normalized size = 0.79 \begin {gather*} -\frac {{\left (15 \, x^{3} - 149 \, x^{2} + 369 \, x + 6\right )} \log \left (\log \relax (x)\right )}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x^3+369*x+12)*log(x)*log(log(x))-15*x^3+149*x^2-369*x-6)/x^3/log(x),x, algorithm="fricas")

[Out]

-(15*x^3 - 149*x^2 + 369*x + 6)*log(log(x))/x^2

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giac [A]  time = 0.14, size = 24, normalized size = 0.83 \begin {gather*} -3 \, {\left (5 \, x + \frac {123 \, x + 2}{x^{2}}\right )} \log \left (\log \relax (x)\right ) + 149 \, \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x^3+369*x+12)*log(x)*log(log(x))-15*x^3+149*x^2-369*x-6)/x^3/log(x),x, algorithm="giac")

[Out]

-3*(5*x + (123*x + 2)/x^2)*log(log(x)) + 149*log(log(x))

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maple [A]  time = 0.03, size = 25, normalized size = 0.86

method	result	size
risch	\(-\frac {3 \left (5 x^{3}+123 x +2\right ) \ln \left (\ln \relax (x )\right )}{x^{2}}+149 \ln \left (\ln \relax (x )\right )\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-15*x^3+369*x+12)*ln(x)*ln(ln(x))-15*x^3+149*x^2-369*x-6)/x^3/ln(x),x,method=_RETURNVERBOSE)

[Out]

-3*(5*x^3+123*x+2)/x^2*ln(ln(x))+149*ln(ln(x))

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maxima [A]  time = 0.41, size = 28, normalized size = 0.97 \begin {gather*} -15 \, x \log \left (\log \relax (x)\right ) - \frac {369 \, \log \left (\log \relax (x)\right )}{x} - \frac {6 \, \log \left (\log \relax (x)\right )}{x^{2}} + 149 \, \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x^3+369*x+12)*log(x)*log(log(x))-15*x^3+149*x^2-369*x-6)/x^3/log(x),x, algorithm="maxima")

[Out]

-15*x*log(log(x)) - 369*log(log(x))/x - 6*log(log(x))/x^2 + 149*log(log(x))

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mupad [B]  time = 6.00, size = 23, normalized size = 0.79 \begin {gather*} -\frac {\ln \left (\ln \relax (x)\right )\,\left (15\,x^3-149\,x^2+369\,x+6\right )}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(369*x - 149*x^2 + 15*x^3 - log(log(x))*log(x)*(369*x - 15*x^3 + 12) + 6)/(x^3*log(x)),x)

[Out]

-(log(log(x))*(369*x - 149*x^2 + 15*x^3 + 6))/x^2

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sympy [A]  time = 0.35, size = 26, normalized size = 0.90 \begin {gather*} 149 \log {\left (\log {\relax (x )} \right )} + \frac {\left (- 15 x^{3} - 369 x - 6\right ) \log {\left (\log {\relax (x )} \right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x**3+369*x+12)*ln(x)*ln(ln(x))-15*x**3+149*x**2-369*x-6)/x**3/ln(x),x)

[Out]

149*log(log(x)) + (-15*x**3 - 369*x - 6)*log(log(x))/x**2

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