3.14.87 \(\int \frac {(64-16 x) \log (x)+(-64 x+16 x^2) \log (x) \log (2 x)+(64-16 x+(64-48 x) \log (x)) \log (2 x) \log (\frac {e^x}{\log (2 x)})}{(-64 x^2+48 x^3-12 x^4+x^5) \log ^2(x) \log (2 x)} \, dx\)

Optimal. Leaf size=27 \[ \frac {16 \log \left (\frac {e^x}{\log (2 x)}\right )}{(4-x)^2 x \log (x)} \]

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Rubi [F]  time = 8.30, 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 {(64-16 x) \log (x)+\left (-64 x+16 x^2\right ) \log (x) \log (2 x)+(64-16 x+(64-48 x) \log (x)) \log (2 x) \log \left (\frac {e^x}{\log (2 x)}\right )}{\left (-64 x^2+48 x^3-12 x^4+x^5\right ) \log ^2(x) \log (2 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((64 - 16*x)*Log[x] + (-64*x + 16*x^2)*Log[x]*Log[2*x] + (64 - 16*x + (64 - 48*x)*Log[x])*Log[2*x]*Log[E^x
/Log[2*x]])/((-64*x^2 + 48*x^3 - 12*x^4 + x^5)*Log[x]^2*Log[2*x]),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-((64-16 x) \log (x))-\left (-64 x+16 x^2\right ) \log (x) \log (2 x)-(64-16 x+(64-48 x) \log (x)) \log (2 x) \log \left (\frac {e^x}{\log (2 x)}\right )}{(4-x)^3 x^2 \log ^2(x) \log (2 x)} \, dx\\ &=\int \left (\frac {16 (-1+x \log (2 x))}{(-4+x)^2 x^2 \log (x) \log (2 x)}-\frac {16 (-4+x-4 \log (x)+3 x \log (x)) \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^3 x^2 \log ^2(x)}\right ) \, dx\\ &=16 \int \frac {-1+x \log (2 x)}{(-4+x)^2 x^2 \log (x) \log (2 x)} \, dx-16 \int \frac {(-4+x-4 \log (x)+3 x \log (x)) \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^3 x^2 \log ^2(x)} \, dx\\ &=16 \int \left (\frac {1}{(-4+x)^2 x \log (x)}-\frac {1}{(-4+x)^2 x^2 \log (x) \log (2 x)}\right ) \, dx-16 \int \left (\frac {(-4+x-4 \log (x)+3 x \log (x)) \log \left (\frac {e^x}{\log (2 x)}\right )}{16 (-4+x)^3 \log ^2(x)}-\frac {(-4+x-4 \log (x)+3 x \log (x)) \log \left (\frac {e^x}{\log (2 x)}\right )}{32 (-4+x)^2 \log ^2(x)}+\frac {3 (-4+x-4 \log (x)+3 x \log (x)) \log \left (\frac {e^x}{\log (2 x)}\right )}{256 (-4+x) \log ^2(x)}-\frac {(-4+x-4 \log (x)+3 x \log (x)) \log \left (\frac {e^x}{\log (2 x)}\right )}{64 x^2 \log ^2(x)}-\frac {3 (-4+x-4 \log (x)+3 x \log (x)) \log \left (\frac {e^x}{\log (2 x)}\right )}{256 x \log ^2(x)}\right ) \, dx\\ &=-\left (\frac {3}{16} \int \frac {(-4+x-4 \log (x)+3 x \log (x)) \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x) \log ^2(x)} \, dx\right )+\frac {3}{16} \int \frac {(-4+x-4 \log (x)+3 x \log (x)) \log \left (\frac {e^x}{\log (2 x)}\right )}{x \log ^2(x)} \, dx+\frac {1}{4} \int \frac {(-4+x-4 \log (x)+3 x \log (x)) \log \left (\frac {e^x}{\log (2 x)}\right )}{x^2 \log ^2(x)} \, dx+\frac {1}{2} \int \frac {(-4+x-4 \log (x)+3 x \log (x)) \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^2 \log ^2(x)} \, dx+16 \int \frac {1}{(-4+x)^2 x \log (x)} \, dx-16 \int \frac {1}{(-4+x)^2 x^2 \log (x) \log (2 x)} \, dx-\int \frac {(-4+x-4 \log (x)+3 x \log (x)) \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^3 \log ^2(x)} \, dx\\ &=\frac {3}{16} \int \left (\frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{\log ^2(x)}-\frac {4 \log \left (\frac {e^x}{\log (2 x)}\right )}{x \log ^2(x)}+\frac {3 \log \left (\frac {e^x}{\log (2 x)}\right )}{\log (x)}-\frac {4 \log \left (\frac {e^x}{\log (2 x)}\right )}{x \log (x)}\right ) \, dx-\frac {3}{16} \int \left (-\frac {4 \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x) \log ^2(x)}+\frac {x \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x) \log ^2(x)}-\frac {4 \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x) \log (x)}+\frac {3 x \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x) \log (x)}\right ) \, dx+\frac {1}{4} \int \left (-\frac {4 \log \left (\frac {e^x}{\log (2 x)}\right )}{x^2 \log ^2(x)}+\frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{x \log ^2(x)}-\frac {4 \log \left (\frac {e^x}{\log (2 x)}\right )}{x^2 \log (x)}+\frac {3 \log \left (\frac {e^x}{\log (2 x)}\right )}{x \log (x)}\right ) \, dx+\frac {1}{2} \int \left (-\frac {4 \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^2 \log ^2(x)}+\frac {x \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^2 \log ^2(x)}-\frac {4 \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^2 \log (x)}+\frac {3 x \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^2 \log (x)}\right ) \, dx+16 \int \frac {1}{(-4+x)^2 x \log (x)} \, dx-16 \int \left (\frac {1}{16 (-4+x)^2 \log (x) \log (2 x)}-\frac {1}{32 (-4+x) \log (x) \log (2 x)}+\frac {1}{16 x^2 \log (x) \log (2 x)}+\frac {1}{32 x \log (x) \log (2 x)}\right ) \, dx-\int \left (-\frac {4 \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^3 \log ^2(x)}+\frac {x \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^3 \log ^2(x)}-\frac {4 \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^3 \log (x)}+\frac {3 x \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^3 \log (x)}\right ) \, dx\\ &=\frac {3}{16} \int \frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{\log ^2(x)} \, dx-\frac {3}{16} \int \frac {x \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x) \log ^2(x)} \, dx+\frac {1}{4} \int \frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{x \log ^2(x)} \, dx+\frac {1}{2} \int \frac {1}{(-4+x) \log (x) \log (2 x)} \, dx-\frac {1}{2} \int \frac {1}{x \log (x) \log (2 x)} \, dx+\frac {1}{2} \int \frac {x \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^2 \log ^2(x)} \, dx+\frac {9}{16} \int \frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{\log (x)} \, dx-\frac {9}{16} \int \frac {x \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x) \log (x)} \, dx+\frac {3}{4} \int \frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x) \log ^2(x)} \, dx-\frac {3}{4} \int \frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{x \log ^2(x)} \, dx+\frac {3}{4} \int \frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x) \log (x)} \, dx+\frac {3}{2} \int \frac {x \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^2 \log (x)} \, dx-2 \int \frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^2 \log ^2(x)} \, dx-2 \int \frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^2 \log (x)} \, dx-3 \int \frac {x \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^3 \log (x)} \, dx+4 \int \frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^3 \log ^2(x)} \, dx+4 \int \frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^3 \log (x)} \, dx+16 \int \frac {1}{(-4+x)^2 x \log (x)} \, dx-\int \frac {1}{(-4+x)^2 \log (x) \log (2 x)} \, dx-\int \frac {1}{x^2 \log (x) \log (2 x)} \, dx-\int \frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{x^2 \log ^2(x)} \, dx-\int \frac {x \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^3 \log ^2(x)} \, dx-\int \frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{x^2 \log (x)} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.55, size = 25, normalized size = 0.93 \begin {gather*} \frac {16 \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^2 x \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((64 - 16*x)*Log[x] + (-64*x + 16*x^2)*Log[x]*Log[2*x] + (64 - 16*x + (64 - 48*x)*Log[x])*Log[2*x]*L
og[E^x/Log[2*x]])/((-64*x^2 + 48*x^3 - 12*x^4 + x^5)*Log[x]^2*Log[2*x]),x]

[Out]

(16*Log[E^x/Log[2*x]])/((-4 + x)^2*x*Log[x])

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fricas [A]  time = 0.90, size = 31, normalized size = 1.15 \begin {gather*} \frac {16 \, \log \left (\frac {e^{x}}{\log \relax (2) + \log \relax (x)}\right )}{{\left (x^{3} - 8 \, x^{2} + 16 \, x\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-48*x+64)*log(x)-16*x+64)*log(2*x)*log(exp(x)/log(2*x))+(16*x^2-64*x)*log(x)*log(2*x)+(-16*x+64)*
log(x))/(x^5-12*x^4+48*x^3-64*x^2)/log(x)^2/log(2*x),x, algorithm="fricas")

[Out]

16*log(e^x/(log(2) + log(x)))/((x^3 - 8*x^2 + 16*x)*log(x))

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giac [B]  time = 0.37, size = 50, normalized size = 1.85 \begin {gather*} -\frac {16 \, \log \left (\log \relax (2) + \log \relax (x)\right )}{x^{3} \log \relax (x) - 8 \, x^{2} \log \relax (x) + 16 \, x \log \relax (x)} + \frac {16}{x^{2} \log \relax (x) - 8 \, x \log \relax (x) + 16 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-48*x+64)*log(x)-16*x+64)*log(2*x)*log(exp(x)/log(2*x))+(16*x^2-64*x)*log(x)*log(2*x)+(-16*x+64)*
log(x))/(x^5-12*x^4+48*x^3-64*x^2)/log(x)^2/log(2*x),x, algorithm="giac")

[Out]

-16*log(log(2) + log(x))/(x^3*log(x) - 8*x^2*log(x) + 16*x*log(x)) + 16/(x^2*log(x) - 8*x*log(x) + 16*log(x))

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maple [C]  time = 0.62, size = 333, normalized size = 12.33

method	result	size
risch	\(\frac {16 \ln \left ({\mathrm e}^{x}\right )}{x \left (x^{2}-8 x +16\right ) \ln \relax (x )}+\frac {-8 i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (\frac {i}{2 i \ln \relax (2)+2 i \ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{2 i \ln \relax (2)+2 i \ln \relax (x )}\right )+8 i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{2 i \ln \relax (2)+2 i \ln \relax (x )}\right )^{2}-8 i \pi \,\mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{2 i \ln \relax (2)+2 i \ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {{\mathrm e}^{x}}{2 i \ln \relax (2)+2 i \ln \relax (x )}\right )-8 i \pi \mathrm {csgn}\left (\frac {{\mathrm e}^{x}}{2 i \ln \relax (2)+2 i \ln \relax (x )}\right )^{2}+8 i \pi \,\mathrm {csgn}\left (\frac {i}{2 i \ln \relax (2)+2 i \ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{2 i \ln \relax (2)+2 i \ln \relax (x )}\right )^{2}-8 i \pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{2 i \ln \relax (2)+2 i \ln \relax (x )}\right )^{3}+8 i \pi \,\mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{2 i \ln \relax (2)+2 i \ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {{\mathrm e}^{x}}{2 i \ln \relax (2)+2 i \ln \relax (x )}\right )^{2}+8 i \pi \mathrm {csgn}\left (\frac {{\mathrm e}^{x}}{2 i \ln \relax (2)+2 i \ln \relax (x )}\right )^{3}+8 i \pi +16 \ln \relax (2)-16 \ln \left (2 i \ln \relax (2)+2 i \ln \relax (x )\right )}{x \left (x -4\right )^{2} \ln \relax (x )}\)	\(333\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-48*x+64)*ln(x)-16*x+64)*ln(2*x)*ln(exp(x)/ln(2*x))+(16*x^2-64*x)*ln(x)*ln(2*x)+(-16*x+64)*ln(x))/(x^5-
12*x^4+48*x^3-64*x^2)/ln(x)^2/ln(2*x),x,method=_RETURNVERBOSE)

[Out]

16/x/(x^2-8*x+16)/ln(x)*ln(exp(x))+8*(-I*Pi*csgn(I*exp(x))*csgn(I/(2*I*ln(2)+2*I*ln(x)))*csgn(I*exp(x)/(2*I*ln
(2)+2*I*ln(x)))+I*Pi*csgn(I*exp(x))*csgn(I*exp(x)/(2*I*ln(2)+2*I*ln(x)))^2-I*Pi*csgn(I*exp(x)/(2*I*ln(2)+2*I*l
n(x)))*csgn(exp(x)/(2*I*ln(2)+2*I*ln(x)))-I*Pi*csgn(exp(x)/(2*I*ln(2)+2*I*ln(x)))^2+I*Pi*csgn(I/(2*I*ln(2)+2*I
*ln(x)))*csgn(I*exp(x)/(2*I*ln(2)+2*I*ln(x)))^2-I*Pi*csgn(I*exp(x)/(2*I*ln(2)+2*I*ln(x)))^3+I*Pi*csgn(I*exp(x)
/(2*I*ln(2)+2*I*ln(x)))*csgn(exp(x)/(2*I*ln(2)+2*I*ln(x)))^2+I*Pi*csgn(exp(x)/(2*I*ln(2)+2*I*ln(x)))^3+I*Pi+2*
ln(2)-2*ln(2*I*ln(2)+2*I*ln(x)))/x/(x-4)^2/ln(x)

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maxima [A]  time = 0.54, size = 30, normalized size = 1.11 \begin {gather*} \frac {16 \, {\left (x - \log \left (\log \relax (2) + \log \relax (x)\right )\right )}}{{\left (x^{3} - 8 \, x^{2} + 16 \, x\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-48*x+64)*log(x)-16*x+64)*log(2*x)*log(exp(x)/log(2*x))+(16*x^2-64*x)*log(x)*log(2*x)+(-16*x+64)*
log(x))/(x^5-12*x^4+48*x^3-64*x^2)/log(x)^2/log(2*x),x, algorithm="maxima")

[Out]

16*(x - log(log(2) + log(x)))/((x^3 - 8*x^2 + 16*x)*log(x))

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mupad [B]  time = 1.39, size = 68, normalized size = 2.52 \begin {gather*} -\frac {\left (\ln \relax (x)\,\left (\frac {48\,x^3-256\,x^2+256\,x}{x^2\,{\left (x-4\right )}^4}-\frac {48\,x-64}{x\,{\left (x-4\right )}^3}\right )-\frac {16}{x\,{\left (x-4\right )}^2}\right )\,\left (x+\ln \left (\frac {1}{\ln \left (2\,x\right )}\right )\right )}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)*(16*x - 64) + log(2*x)*log(exp(x)/log(2*x))*(16*x + log(x)*(48*x - 64) - 64) + log(2*x)*log(x)*(64
*x - 16*x^2))/(log(2*x)*log(x)^2*(64*x^2 - 48*x^3 + 12*x^4 - x^5)),x)

[Out]

-((log(x)*((256*x - 256*x^2 + 48*x^3)/(x^2*(x - 4)^4) - (48*x - 64)/(x*(x - 4)^3)) - 16/(x*(x - 4)^2))*(x + lo
g(1/log(2*x))))/log(x)

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sympy [A]  time = 0.68, size = 34, normalized size = 1.26 \begin {gather*} \frac {16 \log {\left (\frac {e^{x}}{\log {\relax (x )} + \log {\relax (2 )}} \right )}}{x^{3} \log {\relax (x )} - 8 x^{2} \log {\relax (x )} + 16 x \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-48*x+64)*ln(x)-16*x+64)*ln(2*x)*ln(exp(x)/ln(2*x))+(16*x**2-64*x)*ln(x)*ln(2*x)+(-16*x+64)*ln(x)
)/(x**5-12*x**4+48*x**3-64*x**2)/ln(x)**2/ln(2*x),x)

[Out]

16*log(exp(x)/(log(x) + log(2)))/(x**3*log(x) - 8*x**2*log(x) + 16*x*log(x))

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