3.1.85 \(\int \frac {20 x+(3+x^2) \log (3+x^2)}{(-96+3 x-32 x^2+x^3-(-30-10 x^2) \log (\frac {4}{3})) \log (3+x^2)+(30+10 x^2) \log (3+x^2) \log (\log (3+x^2))} \, dx\)

Optimal. Leaf size=23 \[ \log \left (-2+x-10 \left (3-\log \left (\frac {4}{3}\right )-\log \left (\log \left (3+x^2\right )\right )\right )\right ) \]

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Rubi [A]  time = 0.19, antiderivative size = 22, normalized size of antiderivative = 0.96, number of steps used = 2, number of rules used = 2, integrand size = 74, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {6688, 6684} \begin {gather*} \log \left (10 \log \left (\log \left (x^2+3\right )\right )+x-2 \left (16-5 \log \left (\frac {4}{3}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(20*x + (3 + x^2)*Log[3 + x^2])/((-96 + 3*x - 32*x^2 + x^3 - (-30 - 10*x^2)*Log[4/3])*Log[3 + x^2] + (30 +
10*x^2)*Log[3 + x^2]*Log[Log[3 + x^2]]),x]

[Out]

Log[x - 2*(16 - 5*Log[4/3]) + 10*Log[Log[3 + x^2]]]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {20 x+\left (3+x^2\right ) \log \left (3+x^2\right )}{\left (3+x^2\right ) \log \left (3+x^2\right ) \left (x-32 \left (1-\frac {5}{16} \log \left (\frac {4}{3}\right )\right )+10 \log \left (\log \left (3+x^2\right )\right )\right )} \, dx\\ &=\log \left (x-2 \left (16-5 \log \left (\frac {4}{3}\right )\right )+10 \log \left (\log \left (3+x^2\right )\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.10, size = 21, normalized size = 0.91 \begin {gather*} \log \left (32-x-10 \log \left (\frac {4}{3}\right )-10 \log \left (\log \left (3+x^2\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(20*x + (3 + x^2)*Log[3 + x^2])/((-96 + 3*x - 32*x^2 + x^3 - (-30 - 10*x^2)*Log[4/3])*Log[3 + x^2] +
(30 + 10*x^2)*Log[3 + x^2]*Log[Log[3 + x^2]]),x]

[Out]

Log[32 - x - 10*Log[4/3] - 10*Log[Log[3 + x^2]]]

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fricas [A]  time = 0.60, size = 17, normalized size = 0.74 \begin {gather*} \log \left (x - 10 \, \log \left (\frac {3}{4}\right ) + 10 \, \log \left (\log \left (x^{2} + 3\right )\right ) - 32\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+3)*log(x^2+3)+20*x)/((10*x^2+30)*log(x^2+3)*log(log(x^2+3))+((-10*x^2-30)*log(3/4)+x^3-32*x^2+
3*x-96)*log(x^2+3)),x, algorithm="fricas")

[Out]

log(x - 10*log(3/4) + 10*log(log(x^2 + 3)) - 32)

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giac [A]  time = 0.50, size = 21, normalized size = 0.91 \begin {gather*} \log \left (x - 10 \, \log \relax (3) + 20 \, \log \relax (2) + 10 \, \log \left (\log \left (x^{2} + 3\right )\right ) - 32\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+3)*log(x^2+3)+20*x)/((10*x^2+30)*log(x^2+3)*log(log(x^2+3))+((-10*x^2-30)*log(3/4)+x^3-32*x^2+
3*x-96)*log(x^2+3)),x, algorithm="giac")

[Out]

log(x - 10*log(3) + 20*log(2) + 10*log(log(x^2 + 3)) - 32)

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maple [A]  time = 0.08, size = 22, normalized size = 0.96

method	result	size
risch	\(\ln \left (-\ln \relax (3)+2 \ln \relax (2)+\frac {x}{10}+\ln \left (\ln \left (x^{2}+3\right )\right )-\frac {16}{5}\right )\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2+3)*ln(x^2+3)+20*x)/((10*x^2+30)*ln(x^2+3)*ln(ln(x^2+3))+((-10*x^2-30)*ln(3/4)+x^3-32*x^2+3*x-96)*ln(
x^2+3)),x,method=_RETURNVERBOSE)

[Out]

ln(-ln(3)+2*ln(2)+1/10*x+ln(ln(x^2+3))-16/5)

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maxima [A]  time = 0.69, size = 21, normalized size = 0.91 \begin {gather*} \log \left (\frac {1}{10} \, x - \log \relax (3) + 2 \, \log \relax (2) + \log \left (\log \left (x^{2} + 3\right )\right ) - \frac {16}{5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+3)*log(x^2+3)+20*x)/((10*x^2+30)*log(x^2+3)*log(log(x^2+3))+((-10*x^2-30)*log(3/4)+x^3-32*x^2+
3*x-96)*log(x^2+3)),x, algorithm="maxima")

[Out]

log(1/10*x - log(3) + 2*log(2) + log(log(x^2 + 3)) - 16/5)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} -\int \frac {20\,x+\ln \left (x^2+3\right )\,\left (x^2+3\right )}{\ln \left (x^2+3\right )\,\left (\ln \left (\frac {3}{4}\right )\,\left (10\,x^2+30\right )-3\,x+32\,x^2-x^3+96\right )-\ln \left (\ln \left (x^2+3\right )\right )\,\ln \left (x^2+3\right )\,\left (10\,x^2+30\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(20*x + log(x^2 + 3)*(x^2 + 3))/(log(x^2 + 3)*(log(3/4)*(10*x^2 + 30) - 3*x + 32*x^2 - x^3 + 96) - log(lo
g(x^2 + 3))*log(x^2 + 3)*(10*x^2 + 30)),x)

[Out]

-int((20*x + log(x^2 + 3)*(x^2 + 3))/(log(x^2 + 3)*(log(3/4)*(10*x^2 + 30) - 3*x + 32*x^2 - x^3 + 96) - log(lo
g(x^2 + 3))*log(x^2 + 3)*(10*x^2 + 30)), x)

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sympy [A]  time = 0.53, size = 24, normalized size = 1.04 \begin {gather*} \log {\left (\frac {x}{10} + \log {\left (\log {\left (x^{2} + 3 \right )} \right )} - \frac {16}{5} - \log {\relax (3 )} + 2 \log {\relax (2 )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**2+3)*ln(x**2+3)+20*x)/((10*x**2+30)*ln(x**2+3)*ln(ln(x**2+3))+((-10*x**2-30)*ln(3/4)+x**3-32*x*
*2+3*x-96)*ln(x**2+3)),x)

[Out]

log(x/10 + log(log(x**2 + 3)) - 16/5 - log(3) + 2*log(2))

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