3.64.13 \(\int \frac {16-4 x-\log (3)+(-16+2 x+\log (3)) \log (-16 x+2 x^2+x \log (3))}{(-16+2 x+\log (3)) \log ^2(-16 x+2 x^2+x \log (3))} \, dx\)

Optimal. Leaf size=14 \[ \frac {x}{\log (x (-16+2 x+\log (3)))} \]

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Rubi [F]  time = 0.40, 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 {16-4 x-\log (3)+(-16+2 x+\log (3)) \log \left (-16 x+2 x^2+x \log (3)\right )}{(-16+2 x+\log (3)) \log ^2\left (-16 x+2 x^2+x \log (3)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

Defer[Int][(16 - 4*x - Log[3])/((-16 + 2*x + Log[3])*Log[x*(-16 + 2*x + Log[3])]^2), x] + Defer[Int][Log[x*(-1
6 + 2*x + Log[3])]^(-1), x]

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 14, normalized size = 1.00 \begin {gather*} \frac {x}{\log (x (-16+2 x+\log (3)))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x/Log[x*(-16 + 2*x + Log[3])]

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fricas [A]  time = 0.68, size = 18, normalized size = 1.29 \begin {gather*} \frac {x}{\log \left (2 \, x^{2} + x \log \relax (3) - 16 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/log(2*x^2 + x*log(3) - 16*x)

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giac [A]  time = 0.15, size = 18, normalized size = 1.29 \begin {gather*} \frac {x}{\log \left (2 \, x^{2} + x \log \relax (3) - 16 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/log(2*x^2 + x*log(3) - 16*x)

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maple [A]  time = 0.41, size = 19, normalized size = 1.36

method	result	size
norman	\(\frac {x}{\ln \left (x \ln \relax (3)+2 x^{2}-16 x \right )}\)	\(19\)
risch	\(\frac {x}{\ln \left (x \ln \relax (3)+2 x^{2}-16 x \right )}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((ln(3)+2*x-16)*ln(x*ln(3)+2*x^2-16*x)-ln(3)-4*x+16)/(ln(3)+2*x-16)/ln(x*ln(3)+2*x^2-16*x)^2,x,method=_RET
URNVERBOSE)

[Out]

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

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maxima [A]  time = 0.48, size = 15, normalized size = 1.07 \begin {gather*} \frac {x}{\log \left (2 \, x + \log \relax (3) - 16\right ) + \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(log(2*x + log(3) - 16) + log(x))

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mupad [B]  time = 4.40, size = 18, normalized size = 1.29 \begin {gather*} \frac {x}{\ln \left (x\,\ln \relax (3)-16\,x+2\,x^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/log(x*log(3) - 16*x + 2*x^2)

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sympy [A]  time = 0.15, size = 15, normalized size = 1.07 \begin {gather*} \frac {x}{\log {\left (2 x^{2} - 16 x + x \log {\relax (3 )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/log(2*x**2 - 16*x + x*log(3))

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