3.56.42 \(\int \frac {-20-2 x+(-20 x-2 x^2) \log (x)+((-40 x-2 x^2) \log (\frac {16}{60 x+3 x^2})+(-40 x-2 x^2) \log (x) \log (\frac {16}{60 x+3 x^2})) \log (1+x \log (x))}{(20 x+x^2) \log (\frac {16}{60 x+3 x^2})+(20 x^2+x^3) \log (x) \log (\frac {16}{60 x+3 x^2})} \, dx\)

Optimal. Leaf size=30 \[ \log \left (\log \left (\frac {4}{3 \left (5+\frac {x}{4}\right ) x}\right )\right )-\log ^2(1+x \log (x)) \]

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Rubi [F]  time = 1.07, 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 {-20-2 x+\left (-20 x-2 x^2\right ) \log (x)+\left (\left (-40 x-2 x^2\right ) \log \left (\frac {16}{60 x+3 x^2}\right )+\left (-40 x-2 x^2\right ) \log (x) \log \left (\frac {16}{60 x+3 x^2}\right )\right ) \log (1+x \log (x))}{\left (20 x+x^2\right ) \log \left (\frac {16}{60 x+3 x^2}\right )+\left (20 x^2+x^3\right ) \log (x) \log \left (\frac {16}{60 x+3 x^2}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-20 - 2*x + (-20*x - 2*x^2)*Log[x] + ((-40*x - 2*x^2)*Log[16/(60*x + 3*x^2)] + (-40*x - 2*x^2)*Log[x]*Log
[16/(60*x + 3*x^2)])*Log[1 + x*Log[x]])/((20*x + x^2)*Log[16/(60*x + 3*x^2)] + (20*x^2 + x^3)*Log[x]*Log[16/(6
0*x + 3*x^2)]),x]

[Out]

-Log[1 + x*Log[x]]^2 + 2*Defer[Int][(-10 - x)/(x*(20 + x)*Log[16/(x*(60 + 3*x))]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-20-2 x+\left (-20 x-2 x^2\right ) \log (x)+\left (\left (-40 x-2 x^2\right ) \log \left (\frac {16}{60 x+3 x^2}\right )+\left (-40 x-2 x^2\right ) \log (x) \log \left (\frac {16}{60 x+3 x^2}\right )\right ) \log (1+x \log (x))}{x (20+x) (1+x \log (x)) \log \left (\frac {16}{x (60+3 x)}\right )} \, dx\\ &=\int \left (\frac {2 (-10-x)}{x (20+x) \log \left (\frac {16}{x (60+3 x)}\right )}-\frac {2 (1+\log (x)) \log (1+x \log (x))}{1+x \log (x)}\right ) \, dx\\ &=2 \int \frac {-10-x}{x (20+x) \log \left (\frac {16}{x (60+3 x)}\right )} \, dx-2 \int \frac {(1+\log (x)) \log (1+x \log (x))}{1+x \log (x)} \, dx\\ &=-\log ^2(1+x \log (x))+2 \int \frac {-10-x}{x (20+x) \log \left (\frac {16}{x (60+3 x)}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 35, normalized size = 1.17 \begin {gather*} -2 \left (\frac {1}{2} \log ^2(1+x \log (x))-\frac {1}{2} \log \left (\log \left (\frac {16}{60 x+3 x^2}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-20 - 2*x + (-20*x - 2*x^2)*Log[x] + ((-40*x - 2*x^2)*Log[16/(60*x + 3*x^2)] + (-40*x - 2*x^2)*Log[
x]*Log[16/(60*x + 3*x^2)])*Log[1 + x*Log[x]])/((20*x + x^2)*Log[16/(60*x + 3*x^2)] + (20*x^2 + x^3)*Log[x]*Log
[16/(60*x + 3*x^2)]),x]

[Out]

-2*(Log[1 + x*Log[x]]^2/2 - Log[Log[16/(60*x + 3*x^2)]]/2)

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fricas [A]  time = 0.75, size = 25, normalized size = 0.83 \begin {gather*} -\log \left (x \log \relax (x) + 1\right )^{2} + \log \left (\log \left (\frac {16}{3 \, {\left (x^{2} + 20 \, x\right )}}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2-40*x)*log(16/(3*x^2+60*x))*log(x)+(-2*x^2-40*x)*log(16/(3*x^2+60*x)))*log(x*log(x)+1)+(-2*
x^2-20*x)*log(x)-2*x-20)/((x^3+20*x^2)*log(16/(3*x^2+60*x))*log(x)+(x^2+20*x)*log(16/(3*x^2+60*x))),x, algorit
hm="fricas")

[Out]

-log(x*log(x) + 1)^2 + log(log(16/3/(x^2 + 20*x)))

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giac [A]  time = 0.32, size = 26, normalized size = 0.87 \begin {gather*} -\log \left (x \log \relax (x) + 1\right )^{2} + \log \left (-4 \, \log \relax (2) + \log \left (3 \, x + 60\right ) + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2-40*x)*log(16/(3*x^2+60*x))*log(x)+(-2*x^2-40*x)*log(16/(3*x^2+60*x)))*log(x*log(x)+1)+(-2*
x^2-20*x)*log(x)-2*x-20)/((x^3+20*x^2)*log(16/(3*x^2+60*x))*log(x)+(x^2+20*x)*log(16/(3*x^2+60*x))),x, algorit
hm="giac")

[Out]

-log(x*log(x) + 1)^2 + log(-4*log(2) + log(3*x + 60) + log(x))

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maple [A]  time = 5.46, size = 33, normalized size = 1.10

method	result	size
default	\(-\ln \left (x \ln \relax (x )+1\right )^{2}+\ln \left (-4 \ln \relax (2)+\ln \relax (3)-\ln \left (\frac {1}{x \left (20+x \right )}\right )\right )\)	\(33\)
risch	\(-\ln \left (x \ln \relax (x )+1\right )^{2}+\ln \left (\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i}{20+x}\right ) \mathrm {csgn}\left (\frac {i}{x \left (20+x \right )}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i}{x \left (20+x \right )}\right )^{2}-\pi \,\mathrm {csgn}\left (\frac {i}{20+x}\right ) \mathrm {csgn}\left (\frac {i}{x \left (20+x \right )}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i}{x \left (20+x \right )}\right )^{3}-2 i \ln \relax (3)+8 i \ln \relax (2)-2 i \ln \relax (x )-2 i \ln \left (20+x \right )\right )\)	\(133\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-2*x^2-40*x)*ln(16/(3*x^2+60*x))*ln(x)+(-2*x^2-40*x)*ln(16/(3*x^2+60*x)))*ln(x*ln(x)+1)+(-2*x^2-20*x)*l
n(x)-2*x-20)/((x^3+20*x^2)*ln(16/(3*x^2+60*x))*ln(x)+(x^2+20*x)*ln(16/(3*x^2+60*x))),x,method=_RETURNVERBOSE)

[Out]

-ln(x*ln(x)+1)^2+ln(-4*ln(2)+ln(3)-ln(1/x/(20+x)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -2 \, \int \frac {{\left ({\left (x^{2} + 20 \, x\right )} \log \relax (x) \log \left (\frac {16}{3 \, {\left (x^{2} + 20 \, x\right )}}\right ) + {\left (x^{2} + 20 \, x\right )} \log \left (\frac {16}{3 \, {\left (x^{2} + 20 \, x\right )}}\right )\right )} \log \left (x \log \relax (x) + 1\right ) + {\left (x^{2} + 10 \, x\right )} \log \relax (x) + x + 10}{{\left (x^{3} + 20 \, x^{2}\right )} \log \relax (x) \log \left (\frac {16}{3 \, {\left (x^{2} + 20 \, x\right )}}\right ) + {\left (x^{2} + 20 \, x\right )} \log \left (\frac {16}{3 \, {\left (x^{2} + 20 \, x\right )}}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2-40*x)*log(16/(3*x^2+60*x))*log(x)+(-2*x^2-40*x)*log(16/(3*x^2+60*x)))*log(x*log(x)+1)+(-2*
x^2-20*x)*log(x)-2*x-20)/((x^3+20*x^2)*log(16/(3*x^2+60*x))*log(x)+(x^2+20*x)*log(16/(3*x^2+60*x))),x, algorit
hm="maxima")

[Out]

-2*integrate((((x^2 + 20*x)*log(x)*log(16/3/(x^2 + 20*x)) + (x^2 + 20*x)*log(16/3/(x^2 + 20*x)))*log(x*log(x)
+ 1) + (x^2 + 10*x)*log(x) + x + 10)/((x^3 + 20*x^2)*log(x)*log(16/3/(x^2 + 20*x)) + (x^2 + 20*x)*log(16/3/(x^
2 + 20*x))), x)

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mupad [B]  time = 3.77, size = 27, normalized size = 0.90 \begin {gather*} \ln \left (\ln \left (\frac {16}{3\,x^2+60\,x}\right )\right )-{\ln \left (x\,\ln \relax (x)+1\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x + log(x)*(20*x + 2*x^2) + log(x*log(x) + 1)*(log(16/(60*x + 3*x^2))*(40*x + 2*x^2) + log(16/(60*x +
3*x^2))*log(x)*(40*x + 2*x^2)) + 20)/(log(16/(60*x + 3*x^2))*(20*x + x^2) + log(16/(60*x + 3*x^2))*log(x)*(20*
x^2 + x^3)),x)

[Out]

log(log(16/(60*x + 3*x^2))) - log(x*log(x) + 1)^2

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sympy [A]  time = 0.77, size = 22, normalized size = 0.73 \begin {gather*} - \log {\left (x \log {\relax (x )} + 1 \right )}^{2} + \log {\left (\log {\left (\frac {16}{3 x^{2} + 60 x} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x**2-40*x)*ln(16/(3*x**2+60*x))*ln(x)+(-2*x**2-40*x)*ln(16/(3*x**2+60*x)))*ln(x*ln(x)+1)+(-2*x
**2-20*x)*ln(x)-2*x-20)/((x**3+20*x**2)*ln(16/(3*x**2+60*x))*ln(x)+(x**2+20*x)*ln(16/(3*x**2+60*x))),x)

[Out]

-log(x*log(x) + 1)**2 + log(log(16/(3*x**2 + 60*x)))

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