3.40.90 \(\int \frac {\log (x)+(8 e-16 e \log (x)+(1-2 \log (x)) \log (2 x)) \log (\frac {8 e+\log (2 x)}{4 e})}{72 e x^3+9 x^3 \log (2 x)} \, dx\)

Optimal. Leaf size=23 \[ \frac {\log (x) \log \left (2+\frac {\log (2 x)}{4 e}\right )}{9 x^2} \]

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Rubi [F]  time = 1.01, 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 {\log (x)+(8 e-16 e \log (x)+(1-2 \log (x)) \log (2 x)) \log \left (\frac {8 e+\log (2 x)}{4 e}\right )}{72 e x^3+9 x^3 \log (2 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(Log[x] + (8*E - 16*E*Log[x] + (1 - 2*Log[x])*Log[2*x])*Log[(8*E + Log[2*x])/(4*E)])/(72*E*x^3 + 9*x^3*Log
[2*x]),x]

[Out]

-1/18*1/x^2 + (4*E^(16*E)*ExpIntegralEi[-2*(8*E + Log[2*x])]*Log[x])/9 - ((1 + Log[4])*Log[x])/(9*x^2) - (4*E^
(16*E)*ExpIntegralEi[-2*(8*E + Log[2*x])]*(8*E + Log[2*x]))/9 + Defer[Int][Log[8*E + Log[2*x]]/x^3, x]/9 - (2*
Defer[Int][(Log[x]*Log[8*E + Log[2*x]])/x^3, x])/9

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\log (x)+(8 e-16 e \log (x)+(1-2 \log (x)) \log (2 x)) \log \left (\frac {8 e+\log (2 x)}{4 e}\right )}{x^3 (72 e+9 \log (2 x))} \, dx\\ &=\int \frac {\log (x)+(8 e-16 e \log (x)+(1-2 \log (x)) \log (2 x)) \log \left (\frac {8 e+\log (2 x)}{4 e}\right )}{9 x^3 (8 e+\log (2 x))} \, dx\\ &=\frac {1}{9} \int \frac {\log (x)+(8 e-16 e \log (x)+(1-2 \log (x)) \log (2 x)) \log \left (\frac {8 e+\log (2 x)}{4 e}\right )}{x^3 (8 e+\log (2 x))} \, dx\\ &=\frac {1}{9} \int \frac {\frac {\log (x)}{8 e+\log (2 x)}+(-1+2 \log (x)) (1+\log (4)-\log (8 e+\log (2 x)))}{x^3} \, dx\\ &=\frac {1}{9} \int \left (\frac {-8 e (1+\log (4))+(1+16 e (1+\log (4))) \log (x)-(1+\log (4)) \log (2 x)+2 (1+\log (4)) \log (x) \log (2 x)}{x^3 (8 e+\log (2 x))}-\frac {(-1+2 \log (x)) \log (8 e+\log (2 x))}{x^3}\right ) \, dx\\ &=\frac {1}{9} \int \frac {-8 e (1+\log (4))+(1+16 e (1+\log (4))) \log (x)-(1+\log (4)) \log (2 x)+2 (1+\log (4)) \log (x) \log (2 x)}{x^3 (8 e+\log (2 x))} \, dx-\frac {1}{9} \int \frac {(-1+2 \log (x)) \log (8 e+\log (2 x))}{x^3} \, dx\\ &=\frac {1}{9} \int \frac {-1-\log (4)+\frac {\log (x) (1+16 e (1+\log (4))+2 (1+\log (4)) \log (2 x))}{8 e+\log (2 x)}}{x^3} \, dx-\frac {1}{9} \int \left (-\frac {\log (8 e+\log (2 x))}{x^3}+\frac {2 \log (x) \log (8 e+\log (2 x))}{x^3}\right ) \, dx\\ &=\frac {1}{9} \int \left (\frac {(1+\log (4)) (-1+2 \log (x))}{x^3}+\frac {\log (x)}{x^3 (8 e+\log (2 x))}\right ) \, dx+\frac {1}{9} \int \frac {\log (8 e+\log (2 x))}{x^3} \, dx-\frac {2}{9} \int \frac {\log (x) \log (8 e+\log (2 x))}{x^3} \, dx\\ &=\frac {1}{9} \int \frac {\log (x)}{x^3 (8 e+\log (2 x))} \, dx+\frac {1}{9} \int \frac {\log (8 e+\log (2 x))}{x^3} \, dx-\frac {2}{9} \int \frac {\log (x) \log (8 e+\log (2 x))}{x^3} \, dx+\frac {1}{9} (1+\log (4)) \int \frac {-1+2 \log (x)}{x^3} \, dx\\ &=\frac {4}{9} e^{16 e} \text {Ei}(-2 (8 e+\log (2 x))) \log (x)-\frac {(1+\log (4)) \log (x)}{9 x^2}-\frac {1}{9} \int \frac {4 e^{16 e} \text {Ei}(-2 (8 e+\log (2 x)))}{x} \, dx+\frac {1}{9} \int \frac {\log (8 e+\log (2 x))}{x^3} \, dx-\frac {2}{9} \int \frac {\log (x) \log (8 e+\log (2 x))}{x^3} \, dx\\ &=\frac {4}{9} e^{16 e} \text {Ei}(-2 (8 e+\log (2 x))) \log (x)-\frac {(1+\log (4)) \log (x)}{9 x^2}+\frac {1}{9} \int \frac {\log (8 e+\log (2 x))}{x^3} \, dx-\frac {2}{9} \int \frac {\log (x) \log (8 e+\log (2 x))}{x^3} \, dx-\frac {1}{9} \left (4 e^{16 e}\right ) \int \frac {\text {Ei}(-2 (8 e+\log (2 x)))}{x} \, dx\\ &=\frac {4}{9} e^{16 e} \text {Ei}(-2 (8 e+\log (2 x))) \log (x)-\frac {(1+\log (4)) \log (x)}{9 x^2}+\frac {1}{9} \int \frac {\log (8 e+\log (2 x))}{x^3} \, dx-\frac {2}{9} \int \frac {\log (x) \log (8 e+\log (2 x))}{x^3} \, dx-\frac {1}{9} \left (4 e^{16 e}\right ) \operatorname {Subst}(\int \text {Ei}(-2 (8 e+x)) \, dx,x,\log (2 x))\\ &=\frac {4}{9} e^{16 e} \text {Ei}(-2 (8 e+\log (2 x))) \log (x)-\frac {(1+\log (4)) \log (x)}{9 x^2}+\frac {1}{9} \int \frac {\log (8 e+\log (2 x))}{x^3} \, dx-\frac {2}{9} \int \frac {\log (x) \log (8 e+\log (2 x))}{x^3} \, dx+\frac {1}{9} \left (2 e^{16 e}\right ) \operatorname {Subst}(\int \text {Ei}(x) \, dx,x,-16 e-2 \log (2 x))\\ &=-\frac {1}{18 x^2}+\frac {4}{9} e^{16 e} \text {Ei}(-2 (8 e+\log (2 x))) \log (x)-\frac {(1+\log (4)) \log (x)}{9 x^2}-\frac {4}{9} e^{16 e} \text {Ei}(-16 e-2 \log (2 x)) (8 e+\log (2 x))+\frac {1}{9} \int \frac {\log (8 e+\log (2 x))}{x^3} \, dx-\frac {2}{9} \int \frac {\log (x) \log (8 e+\log (2 x))}{x^3} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 24, normalized size = 1.04 \begin {gather*} \frac {\log (x) \left (-1+\log \left (\frac {1}{4} (8 e+\log (2 x))\right )\right )}{9 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Log[x] + (8*E - 16*E*Log[x] + (1 - 2*Log[x])*Log[2*x])*Log[(8*E + Log[2*x])/(4*E)])/(72*E*x^3 + 9*x
^3*Log[2*x]),x]

[Out]

(Log[x]*(-1 + Log[(8*E + Log[2*x])/4]))/(9*x^2)

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fricas [A]  time = 0.77, size = 21, normalized size = 0.91 \begin {gather*} \frac {\log \left (\frac {1}{4} \, {\left (8 \, e + \log \relax (2) + \log \relax (x)\right )} e^{\left (-1\right )}\right ) \log \relax (x)}{9 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*log(x)+1)*log(2*x)-16*exp(1)*log(x)+8*exp(1))*log(1/4*(log(2*x)+8*exp(1))/exp(1))+log(x))/(9*x
^3*log(2*x)+72*x^3*exp(1)),x, algorithm="fricas")

[Out]

1/9*log(1/4*(8*e + log(2) + log(x))*e^(-1))*log(x)/x^2

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giac [A]  time = 0.14, size = 28, normalized size = 1.22 \begin {gather*} -\frac {2 \, \log \relax (2) \log \relax (x) - \log \relax (x) \log \left (8 \, e + \log \relax (2) + \log \relax (x)\right ) + \log \relax (x)}{9 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*log(x)+1)*log(2*x)-16*exp(1)*log(x)+8*exp(1))*log(1/4*(log(2*x)+8*exp(1))/exp(1))+log(x))/(9*x
^3*log(2*x)+72*x^3*exp(1)),x, algorithm="giac")

[Out]

-1/9*(2*log(2)*log(x) - log(x)*log(8*e + log(2) + log(x)) + log(x))/x^2

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maple [A]  time = 0.10, size = 22, normalized size = 0.96

method	result	size
risch	\(\frac {\ln \relax (x ) \ln \left (\frac {\left (\ln \relax (2)+\ln \relax (x )+8 \,{\mathrm e}\right ) {\mathrm e}^{-1}}{4}\right )}{9 x^{2}}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-2*ln(x)+1)*ln(2*x)-16*exp(1)*ln(x)+8*exp(1))*ln(1/4*(ln(2*x)+8*exp(1))/exp(1))+ln(x))/(9*x^3*ln(2*x)+7
2*x^3*exp(1)),x,method=_RETURNVERBOSE)

[Out]

1/9/x^2*ln(x)*ln(1/4*(ln(2)+ln(x)+8*exp(1))*exp(-1))

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maxima [A]  time = 0.50, size = 29, normalized size = 1.26 \begin {gather*} -\frac {{\left (2 \, \log \relax (2) + 1\right )} \log \relax (x) - \log \relax (x) \log \left (8 \, e + \log \relax (2) + \log \relax (x)\right )}{9 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*log(x)+1)*log(2*x)-16*exp(1)*log(x)+8*exp(1))*log(1/4*(log(2*x)+8*exp(1))/exp(1))+log(x))/(9*x
^3*log(2*x)+72*x^3*exp(1)),x, algorithm="maxima")

[Out]

-1/9*((2*log(2) + 1)*log(x) - log(x)*log(8*e + log(2) + log(x)))/x^2

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mupad [B]  time = 2.98, size = 21, normalized size = 0.91 \begin {gather*} \frac {\ln \relax (x)\,\left (\ln \left (\frac {\ln \left (2\,x\right )}{4}+2\,\mathrm {e}\right )-1\right )}{9\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x) - log(exp(-1)*(log(2*x)/4 + 2*exp(1)))*(log(2*x)*(2*log(x) - 1) - 8*exp(1) + 16*exp(1)*log(x)))/(9
*x^3*log(2*x) + 72*x^3*exp(1)),x)

[Out]

(log(x)*(log(log(2*x)/4 + 2*exp(1)) - 1))/(9*x^2)

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sympy [A]  time = 0.48, size = 27, normalized size = 1.17 \begin {gather*} \frac {\log {\relax (x )} \log {\left (\frac {\frac {\log {\relax (x )}}{4} + \frac {\log {\relax (2 )}}{4} + 2 e}{e} \right )}}{9 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*ln(x)+1)*ln(2*x)-16*exp(1)*ln(x)+8*exp(1))*ln(1/4*(ln(2*x)+8*exp(1))/exp(1))+ln(x))/(9*x**3*ln
(2*x)+72*x**3*exp(1)),x)

[Out]

log(x)*log((log(x)/4 + log(2)/4 + 2*E)*exp(-1))/(9*x**2)

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