3.25.71 \(\int \frac {100 x^2+(-8+100 x^3) \log (x)+((4-500 x^2+100 x^3) \log (x)+100 x^2 \log (x) \log (\log (x))) \log (\frac {1-125 x^2+25 x^3+25 x^2 \log (\log (x))}{25 x^2})}{(1-125 x^2+25 x^3) \log (x)+25 x^2 \log (x) \log (\log (x))} \, dx\)

Optimal. Leaf size=19 \[ 25+4 x \log \left (-5+\frac {1}{25 x^2}+x+\log (\log (x))\right ) \]

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Rubi [F]  time = 1.08, 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 {100 x^2+\left (-8+100 x^3\right ) \log (x)+\left (\left (4-500 x^2+100 x^3\right ) \log (x)+100 x^2 \log (x) \log (\log (x))\right ) \log \left (\frac {1-125 x^2+25 x^3+25 x^2 \log (\log (x))}{25 x^2}\right )}{\left (1-125 x^2+25 x^3\right ) \log (x)+25 x^2 \log (x) \log (\log (x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(100*x^2 + (-8 + 100*x^3)*Log[x] + ((4 - 500*x^2 + 100*x^3)*Log[x] + 100*x^2*Log[x]*Log[Log[x]])*Log[(1 -
125*x^2 + 25*x^3 + 25*x^2*Log[Log[x]])/(25*x^2)])/((1 - 125*x^2 + 25*x^3)*Log[x] + 25*x^2*Log[x]*Log[Log[x]]),
x]

[Out]

-8*Defer[Int][(1 - 125*x^2 + 25*x^3 + 25*x^2*Log[Log[x]])^(-1), x] + 100*Defer[Int][x^3/(1 - 125*x^2 + 25*x^3
+ 25*x^2*Log[Log[x]]), x] + 100*Defer[Int][x^2/(Log[x]*(1 - 125*x^2 + 25*x^3 + 25*x^2*Log[Log[x]])), x] + 4*De
fer[Int][Log[-5 + 1/(25*x^2) + x + Log[Log[x]]], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {100 x^2+\left (-8+100 x^3\right ) \log (x)+\left (\left (4-500 x^2+100 x^3\right ) \log (x)+100 x^2 \log (x) \log (\log (x))\right ) \log \left (\frac {1-125 x^2+25 x^3+25 x^2 \log (\log (x))}{25 x^2}\right )}{\log (x) \left (1-125 x^2+25 x^3+25 x^2 \log (\log (x))\right )} \, dx\\ &=\int \frac {4 \left (25 x^2+\log (x) \left (-2+25 x^3+\left (1-125 x^2+25 x^3+25 x^2 \log (\log (x))\right ) \log \left (-5+\frac {1}{25 x^2}+x+\log (\log (x))\right )\right )\right )}{\log (x) \left (1-125 x^2+25 x^3+25 x^2 \log (\log (x))\right )} \, dx\\ &=4 \int \frac {25 x^2+\log (x) \left (-2+25 x^3+\left (1-125 x^2+25 x^3+25 x^2 \log (\log (x))\right ) \log \left (-5+\frac {1}{25 x^2}+x+\log (\log (x))\right )\right )}{\log (x) \left (1-125 x^2+25 x^3+25 x^2 \log (\log (x))\right )} \, dx\\ &=4 \int \left (\frac {25 x^2-2 \log (x)+25 x^3 \log (x)}{\log (x) \left (1-125 x^2+25 x^3+25 x^2 \log (\log (x))\right )}+\log \left (-5+\frac {1}{25 x^2}+x+\log (\log (x))\right )\right ) \, dx\\ &=4 \int \frac {25 x^2-2 \log (x)+25 x^3 \log (x)}{\log (x) \left (1-125 x^2+25 x^3+25 x^2 \log (\log (x))\right )} \, dx+4 \int \log \left (-5+\frac {1}{25 x^2}+x+\log (\log (x))\right ) \, dx\\ &=4 \int \left (-\frac {2}{1-125 x^2+25 x^3+25 x^2 \log (\log (x))}+\frac {25 x^3}{1-125 x^2+25 x^3+25 x^2 \log (\log (x))}+\frac {25 x^2}{\log (x) \left (1-125 x^2+25 x^3+25 x^2 \log (\log (x))\right )}\right ) \, dx+4 \int \log \left (-5+\frac {1}{25 x^2}+x+\log (\log (x))\right ) \, dx\\ &=4 \int \log \left (-5+\frac {1}{25 x^2}+x+\log (\log (x))\right ) \, dx-8 \int \frac {1}{1-125 x^2+25 x^3+25 x^2 \log (\log (x))} \, dx+100 \int \frac {x^3}{1-125 x^2+25 x^3+25 x^2 \log (\log (x))} \, dx+100 \int \frac {x^2}{\log (x) \left (1-125 x^2+25 x^3+25 x^2 \log (\log (x))\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.27, size = 17, normalized size = 0.89 \begin {gather*} 4 x \log \left (-5+\frac {1}{25 x^2}+x+\log (\log (x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(100*x^2 + (-8 + 100*x^3)*Log[x] + ((4 - 500*x^2 + 100*x^3)*Log[x] + 100*x^2*Log[x]*Log[Log[x]])*Log
[(1 - 125*x^2 + 25*x^3 + 25*x^2*Log[Log[x]])/(25*x^2)])/((1 - 125*x^2 + 25*x^3)*Log[x] + 25*x^2*Log[x]*Log[Log
[x]]),x]

[Out]

4*x*Log[-5 + 1/(25*x^2) + x + Log[Log[x]]]

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fricas [A]  time = 0.71, size = 29, normalized size = 1.53 \begin {gather*} 4 \, x \log \left (\frac {25 \, x^{3} + 25 \, x^{2} \log \left (\log \relax (x)\right ) - 125 \, x^{2} + 1}{25 \, x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((100*x^2*log(x)*log(log(x))+(100*x^3-500*x^2+4)*log(x))*log(1/25*(25*x^2*log(log(x))+25*x^3-125*x^2
+1)/x^2)+(100*x^3-8)*log(x)+100*x^2)/(25*x^2*log(x)*log(log(x))+(25*x^3-125*x^2+1)*log(x)),x, algorithm="frica
s")

[Out]

4*x*log(1/25*(25*x^3 + 25*x^2*log(log(x)) - 125*x^2 + 1)/x^2)

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giac [B]  time = 0.37, size = 35, normalized size = 1.84 \begin {gather*} -8 \, x \log \relax (5) + 4 \, x \log \left (25 \, x^{3} + 25 \, x^{2} \log \left (\log \relax (x)\right ) - 125 \, x^{2} + 1\right ) - 8 \, x \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((100*x^2*log(x)*log(log(x))+(100*x^3-500*x^2+4)*log(x))*log(1/25*(25*x^2*log(log(x))+25*x^3-125*x^2
+1)/x^2)+(100*x^3-8)*log(x)+100*x^2)/(25*x^2*log(x)*log(log(x))+(25*x^3-125*x^2+1)*log(x)),x, algorithm="giac"
)

[Out]

-8*x*log(5) + 4*x*log(25*x^3 + 25*x^2*log(log(x)) - 125*x^2 + 1) - 8*x*log(x)

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maple [C]  time = 0.12, size = 237, normalized size = 12.47

method	result	size
risch	\(4 x \ln \left (\frac {1}{25}+x^{3}+\left (\ln \left (\ln \relax (x )\right )-5\right ) x^{2}\right )+2 i \pi x \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-4 i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-2 i \pi x \,\mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (i \left (\frac {1}{25}+x^{3}+\left (\ln \left (\ln \relax (x )\right )-5\right ) x^{2}\right )\right ) \mathrm {csgn}\left (\frac {i \left (\frac {1}{25}+x^{3}+\left (\ln \left (\ln \relax (x )\right )-5\right ) x^{2}\right )}{x^{2}}\right )+2 i \pi x \,\mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (\frac {i \left (\frac {1}{25}+x^{3}+\left (\ln \left (\ln \relax (x )\right )-5\right ) x^{2}\right )}{x^{2}}\right )^{2}+2 i \pi x \mathrm {csgn}\left (i x^{2}\right )^{3}+2 i \pi x \,\mathrm {csgn}\left (i \left (\frac {1}{25}+x^{3}+\left (\ln \left (\ln \relax (x )\right )-5\right ) x^{2}\right )\right ) \mathrm {csgn}\left (\frac {i \left (\frac {1}{25}+x^{3}+\left (\ln \left (\ln \relax (x )\right )-5\right ) x^{2}\right )}{x^{2}}\right )^{2}-2 i \pi x \mathrm {csgn}\left (\frac {i \left (\frac {1}{25}+x^{3}+\left (\ln \left (\ln \relax (x )\right )-5\right ) x^{2}\right )}{x^{2}}\right )^{3}-8 x \ln \relax (x )\)	\(237\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((100*x^2*ln(x)*ln(ln(x))+(100*x^3-500*x^2+4)*ln(x))*ln(1/25*(25*x^2*ln(ln(x))+25*x^3-125*x^2+1)/x^2)+(100
*x^3-8)*ln(x)+100*x^2)/(25*x^2*ln(x)*ln(ln(x))+(25*x^3-125*x^2+1)*ln(x)),x,method=_RETURNVERBOSE)

[Out]

4*x*ln(1/25+x^3+(ln(ln(x))-5)*x^2)+2*I*Pi*x*csgn(I*x)^2*csgn(I*x^2)-4*I*Pi*x*csgn(I*x)*csgn(I*x^2)^2-2*I*Pi*x*
csgn(I/x^2)*csgn(I*(1/25+x^3+(ln(ln(x))-5)*x^2))*csgn(I/x^2*(1/25+x^3+(ln(ln(x))-5)*x^2))+2*I*Pi*x*csgn(I/x^2)
*csgn(I/x^2*(1/25+x^3+(ln(ln(x))-5)*x^2))^2+2*I*Pi*x*csgn(I*x^2)^3+2*I*Pi*x*csgn(I*(1/25+x^3+(ln(ln(x))-5)*x^2
))*csgn(I/x^2*(1/25+x^3+(ln(ln(x))-5)*x^2))^2-2*I*Pi*x*csgn(I/x^2*(1/25+x^3+(ln(ln(x))-5)*x^2))^3-8*x*ln(x)

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maxima [B]  time = 0.90, size = 35, normalized size = 1.84 \begin {gather*} -8 \, x \log \relax (5) + 4 \, x \log \left (25 \, x^{3} + 25 \, x^{2} \log \left (\log \relax (x)\right ) - 125 \, x^{2} + 1\right ) - 8 \, x \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((100*x^2*log(x)*log(log(x))+(100*x^3-500*x^2+4)*log(x))*log(1/25*(25*x^2*log(log(x))+25*x^3-125*x^2
+1)/x^2)+(100*x^3-8)*log(x)+100*x^2)/(25*x^2*log(x)*log(log(x))+(25*x^3-125*x^2+1)*log(x)),x, algorithm="maxim
a")

[Out]

-8*x*log(5) + 4*x*log(25*x^3 + 25*x^2*log(log(x)) - 125*x^2 + 1) - 8*x*log(x)

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mupad [B]  time = 1.70, size = 25, normalized size = 1.32 \begin {gather*} 4\,x\,\ln \left (\frac {x^2\,\ln \left (\ln \relax (x)\right )-5\,x^2+x^3+\frac {1}{25}}{x^2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log((x^2*log(log(x)) - 5*x^2 + x^3 + 1/25)/x^2)*(log(x)*(100*x^3 - 500*x^2 + 4) + 100*x^2*log(log(x))*log
(x)) + 100*x^2 + log(x)*(100*x^3 - 8))/(log(x)*(25*x^3 - 125*x^2 + 1) + 25*x^2*log(log(x))*log(x)),x)

[Out]

4*x*log((x^2*log(log(x)) - 5*x^2 + x^3 + 1/25)/x^2)

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sympy [A]  time = 2.20, size = 27, normalized size = 1.42 \begin {gather*} 4 x \log {\left (\frac {x^{3} + x^{2} \log {\left (\log {\relax (x )} \right )} - 5 x^{2} + \frac {1}{25}}{x^{2}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((100*x**2*ln(x)*ln(ln(x))+(100*x**3-500*x**2+4)*ln(x))*ln(1/25*(25*x**2*ln(ln(x))+25*x**3-125*x**2+
1)/x**2)+(100*x**3-8)*ln(x)+100*x**2)/(25*x**2*ln(x)*ln(ln(x))+(25*x**3-125*x**2+1)*ln(x)),x)

[Out]

4*x*log((x**3 + x**2*log(log(x)) - 5*x**2 + 1/25)/x**2)

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