3.50.86 \(\int \frac {-x+(32+2 x) \log (16+x)}{-64 x^3-4 x^4+(64 x+4 x^2) \log (16+x)+(-16 x^3-x^4+(16 x+x^2) \log (16+x)) \log (\frac {e^3 x^2}{-x^2 \log (2)+\log (2) \log (16+x)})} \, dx\)

Optimal. Leaf size=27 \[ \log \left (4+\log \left (\frac {e^3 x^2}{\log (2) \left (-x^2+\log (16+x)\right )}\right )\right ) \]

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Rubi [A]  time = 0.24, antiderivative size = 25, normalized size of antiderivative = 0.93, number of steps used = 2, number of rules used = 2, integrand size = 91, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.022, Rules used = {6688, 6684} \begin {gather*} \log \left (\log \left (-\frac {x^2}{\log (2) \left (x^2-\log (x+16)\right )}\right )+7\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-x + (32 + 2*x)*Log[16 + x])/(-64*x^3 - 4*x^4 + (64*x + 4*x^2)*Log[16 + x] + (-16*x^3 - x^4 + (16*x + x^2
)*Log[16 + x])*Log[(E^3*x^2)/(-(x^2*Log[2]) + Log[2]*Log[16 + x])]),x]

[Out]

Log[7 + Log[-(x^2/(Log[2]*(x^2 - Log[16 + x])))]]

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 {x-2 (16+x) \log (16+x)}{x (16+x) \left (x^2-\log (16+x)\right ) \left (7+\log \left (-\frac {x^2}{\log (2) \left (x^2-\log (16+x)\right )}\right )\right )} \, dx\\ &=\log \left (7+\log \left (-\frac {x^2}{\log (2) \left (x^2-\log (16+x)\right )}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.28, size = 26, normalized size = 0.96 \begin {gather*} \log \left (-7+\log (\log (2))-\log \left (-\frac {x^2}{x^2-\log (16+x)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-x + (32 + 2*x)*Log[16 + x])/(-64*x^3 - 4*x^4 + (64*x + 4*x^2)*Log[16 + x] + (-16*x^3 - x^4 + (16*x
+ x^2)*Log[16 + x])*Log[(E^3*x^2)/(-(x^2*Log[2]) + Log[2]*Log[16 + x])]),x]

[Out]

Log[-7 + Log[Log[2]] - Log[-(x^2/(x^2 - Log[16 + x]))]]

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fricas [A]  time = 0.64, size = 28, normalized size = 1.04 \begin {gather*} \log \left (\log \left (-\frac {x^{2} e^{3}}{x^{2} \log \relax (2) - \log \relax (2) \log \left (x + 16\right )}\right ) + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+32)*log(x+16)-x)/(((x^2+16*x)*log(x+16)-x^4-16*x^3)*log(x^2*exp(3)/(log(2)*log(x+16)-x^2*log(2
)))+(4*x^2+64*x)*log(x+16)-4*x^4-64*x^3),x, algorithm="fricas")

[Out]

log(log(-x^2*e^3/(x^2*log(2) - log(2)*log(x + 16))) + 4)

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giac [A]  time = 0.30, size = 27, normalized size = 1.00 \begin {gather*} \log \left (-\log \left (x^{2} \log \relax (2) - \log \relax (2) \log \left (x + 16\right )\right ) + \log \left (-x^{2}\right ) + 7\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+32)*log(x+16)-x)/(((x^2+16*x)*log(x+16)-x^4-16*x^3)*log(x^2*exp(3)/(log(2)*log(x+16)-x^2*log(2
)))+(4*x^2+64*x)*log(x+16)-4*x^4-64*x^3),x, algorithm="giac")

[Out]

log(-log(x^2*log(2) - log(2)*log(x + 16)) + log(-x^2) + 7)

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maple [C]  time = 0.12, size = 244, normalized size = 9.04

method	result	size
risch	\(\ln \left (\frac {i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}}{2}+\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (\frac {i}{-\ln \left (x +16\right )+x^{2}}\right ) \mathrm {csgn}\left (\frac {i x^{2}}{-\ln \left (x +16\right )+x^{2}}\right )}{2}-\frac {i \pi \mathrm {csgn}\left (\frac {i x^{2}}{-\ln \left (x +16\right )+x^{2}}\right )^{3}}{2}+\frac {i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )}{2}-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (\frac {i x^{2}}{-\ln \left (x +16\right )+x^{2}}\right )^{2}}{2}-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+i \pi \mathrm {csgn}\left (\frac {i x^{2}}{-\ln \left (x +16\right )+x^{2}}\right )^{2}-\frac {i \pi \,\mathrm {csgn}\left (\frac {i}{-\ln \left (x +16\right )+x^{2}}\right ) \mathrm {csgn}\left (\frac {i x^{2}}{-\ln \left (x +16\right )+x^{2}}\right )^{2}}{2}-i \pi -7+\ln \left (\ln \relax (2)\right )-2 \ln \relax (x )+\ln \left (-\ln \left (x +16\right )+x^{2}\right )\right )\)	\(244\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x+32)*ln(x+16)-x)/(((x^2+16*x)*ln(x+16)-x^4-16*x^3)*ln(x^2*exp(3)/(ln(2)*ln(x+16)-x^2*ln(2)))+(4*x^2+6
4*x)*ln(x+16)-4*x^4-64*x^3),x,method=_RETURNVERBOSE)

[Out]

ln(1/2*I*Pi*csgn(I*x^2)^3+1/2*I*Pi*csgn(I*x^2)*csgn(I/(-ln(x+16)+x^2))*csgn(I*x^2/(-ln(x+16)+x^2))-1/2*I*Pi*cs
gn(I*x^2/(-ln(x+16)+x^2))^3+1/2*I*Pi*csgn(I*x)^2*csgn(I*x^2)-1/2*I*Pi*csgn(I*x^2)*csgn(I*x^2/(-ln(x+16)+x^2))^
2-I*Pi*csgn(I*x)*csgn(I*x^2)^2+I*Pi*csgn(I*x^2/(-ln(x+16)+x^2))^2-1/2*I*Pi*csgn(I/(-ln(x+16)+x^2))*csgn(I*x^2/
(-ln(x+16)+x^2))^2-I*Pi-7+ln(ln(2))-2*ln(x)+ln(-ln(x+16)+x^2))

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maxima [A]  time = 0.48, size = 21, normalized size = 0.78 \begin {gather*} \log \left (\log \left (-x^{2} + \log \left (x + 16\right )\right ) - 2 \, \log \relax (x) + \log \left (\log \relax (2)\right ) - 7\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+32)*log(x+16)-x)/(((x^2+16*x)*log(x+16)-x^4-16*x^3)*log(x^2*exp(3)/(log(2)*log(x+16)-x^2*log(2
)))+(4*x^2+64*x)*log(x+16)-4*x^4-64*x^3),x, algorithm="maxima")

[Out]

log(log(-x^2 + log(x + 16)) - 2*log(x) + log(log(2)) - 7)

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mupad [B]  time = 4.96, size = 27, normalized size = 1.00 \begin {gather*} \ln \left (\ln \left (\frac {x^2\,{\mathrm {e}}^3}{\ln \left (x+16\right )\,\ln \relax (2)-x^2\,\ln \relax (2)}\right )+4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - log(x + 16)*(2*x + 32))/(64*x^3 - log(x + 16)*(64*x + 4*x^2) + 4*x^4 + log((x^2*exp(3))/(log(x + 16)*
log(2) - x^2*log(2)))*(16*x^3 - log(x + 16)*(16*x + x^2) + x^4)),x)

[Out]

log(log((x^2*exp(3))/(log(x + 16)*log(2) - x^2*log(2))) + 4)

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sympy [A]  time = 0.64, size = 26, normalized size = 0.96 \begin {gather*} \log {\left (\log {\left (\frac {x^{2} e^{3}}{- x^{2} \log {\relax (2 )} + \log {\relax (2 )} \log {\left (x + 16 \right )}} \right )} + 4 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+32)*ln(x+16)-x)/(((x**2+16*x)*ln(x+16)-x**4-16*x**3)*ln(x**2*exp(3)/(ln(2)*ln(x+16)-x**2*ln(2)
))+(4*x**2+64*x)*ln(x+16)-4*x**4-64*x**3),x)

[Out]

log(log(x**2*exp(3)/(-x**2*log(2) + log(2)*log(x + 16))) + 4)

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