3.1.64 \(\int \frac {50+25 x-50 x \log (x) \log (\log (x))}{(32 x+48 x^2+24 x^3+4 x^4) \log (25) \log (x)} \, dx\)

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

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Rubi [F]  time = 0.34, 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 {50+25 x-50 x \log (x) \log (\log (x))}{\left (32 x+48 x^2+24 x^3+4 x^4\right ) \log (25) \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(50 + 25*x - 50*x*Log[x]*Log[Log[x]])/((32*x + 48*x^2 + 24*x^3 + 4*x^4)*Log[25]*Log[x]),x]

[Out]

(25*Defer[Int][1/(x*(2 + x)^2*Log[x]), x])/(4*Log[25]) - (25*Defer[Int][Log[Log[x]]/(2 + x)^3, x])/(2*Log[25])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {50+25 x-50 x \log (x) \log (\log (x))}{\left (32 x+48 x^2+24 x^3+4 x^4\right ) \log (x)} \, dx}{\log (25)}\\ &=\frac {\int \frac {25 (2+x-2 x \log (x) \log (\log (x)))}{4 x (2+x)^3 \log (x)} \, dx}{\log (25)}\\ &=\frac {25 \int \frac {2+x-2 x \log (x) \log (\log (x))}{x (2+x)^3 \log (x)} \, dx}{4 \log (25)}\\ &=\frac {25 \int \left (\frac {1}{x (2+x)^2 \log (x)}-\frac {2 \log (\log (x))}{(2+x)^3}\right ) \, dx}{4 \log (25)}\\ &=\frac {25 \int \frac {1}{x (2+x)^2 \log (x)} \, dx}{4 \log (25)}-\frac {25 \int \frac {\log (\log (x))}{(2+x)^3} \, dx}{2 \log (25)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 16, normalized size = 0.84 \begin {gather*} \frac {25 \log (\log (x))}{4 (2+x)^2 \log (25)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(50 + 25*x - 50*x*Log[x]*Log[Log[x]])/((32*x + 48*x^2 + 24*x^3 + 4*x^4)*Log[25]*Log[x]),x]

[Out]

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

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fricas [A]  time = 0.72, size = 19, normalized size = 1.00 \begin {gather*} \frac {25 \, \log \left (\log \relax (x)\right )}{8 \, {\left (x^{2} + 4 \, x + 4\right )} \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-50*x*log(x)*log(log(x))+25*x+50)/(4*x^4+24*x^3+48*x^2+32*x)/log(5)/log(x),x, algorithm="fricas
")

[Out]

25/8*log(log(x))/((x^2 + 4*x + 4)*log(5))

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giac [A]  time = 0.30, size = 19, normalized size = 1.00 \begin {gather*} \frac {25 \, \log \left (\log \relax (x)\right )}{8 \, {\left (x^{2} + 4 \, x + 4\right )} \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-50*x*log(x)*log(log(x))+25*x+50)/(4*x^4+24*x^3+48*x^2+32*x)/log(5)/log(x),x, algorithm="giac")

[Out]

25/8*log(log(x))/((x^2 + 4*x + 4)*log(5))

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maple [A]  time = 0.03, size = 20, normalized size = 1.05

method	result	size
risch	\(\frac {25 \ln \left (\ln \relax (x )\right )}{8 \ln \relax (5) \left (x^{2}+4 x +4\right )}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(-50*x*ln(x)*ln(ln(x))+25*x+50)/(4*x^4+24*x^3+48*x^2+32*x)/ln(5)/ln(x),x,method=_RETURNVERBOSE)

[Out]

25/8/ln(5)/(x^2+4*x+4)*ln(ln(x))

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maxima [A]  time = 0.59, size = 19, normalized size = 1.00 \begin {gather*} \frac {25 \, \log \left (\log \relax (x)\right )}{8 \, {\left (x^{2} + 4 \, x + 4\right )} \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-50*x*log(x)*log(log(x))+25*x+50)/(4*x^4+24*x^3+48*x^2+32*x)/log(5)/log(x),x, algorithm="maxima
")

[Out]

25/8*log(log(x))/((x^2 + 4*x + 4)*log(5))

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mupad [B]  time = 0.67, size = 14, normalized size = 0.74 \begin {gather*} \frac {25\,\ln \left (\ln \relax (x)\right )}{8\,\ln \relax (5)\,{\left (x+2\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((25*x)/2 - 25*x*log(log(x))*log(x) + 25)/(log(5)*log(x)*(32*x + 48*x^2 + 24*x^3 + 4*x^4)),x)

[Out]

(25*log(log(x)))/(8*log(5)*(x + 2)^2)

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sympy [A]  time = 0.36, size = 26, normalized size = 1.37 \begin {gather*} \frac {25 \log {\left (\log {\relax (x )} \right )}}{8 x^{2} \log {\relax (5 )} + 32 x \log {\relax (5 )} + 32 \log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-50*x*ln(x)*ln(ln(x))+25*x+50)/(4*x**4+24*x**3+48*x**2+32*x)/ln(5)/ln(x),x)

[Out]

25*log(log(x))/(8*x**2*log(5) + 32*x*log(5) + 32*log(5))

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