3.61.3 \(\int \frac {128-192 x-16 x^3}{(16 x-48 x^2+21 x^3+4 x^4) \log ^2(\frac {-16+48 x-21 x^2-4 x^3}{9 x^2})} \, dx\) [6003]

Optimal. Leaf size=32 \[ \frac {4}{\log \left (\frac {1}{3} \left (5+\frac {1}{3} \left (-\left (5+\frac {-4+x}{x}\right )^2-4 x\right )\right )\right )} \]

[Out]

4/ln(-4/9*x-1/9*((x-4)/x+5)^2+5/3)

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Rubi [A]
time = 0.06, antiderivative size = 27, normalized size of antiderivative = 0.84, number of steps used = 1, number of rules used = 1, integrand size = 57, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.018, Rules used = {6818} \begin {gather*} \frac {4}{\log \left (-\frac {4 x^3+21 x^2-48 x+16}{9 x^2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(128 - 192*x - 16*x^3)/((16*x - 48*x^2 + 21*x^3 + 4*x^4)*Log[(-16 + 48*x - 21*x^2 - 4*x^3)/(9*x^2)]^2),x]

[Out]

4/Log[-1/9*(16 - 48*x + 21*x^2 + 4*x^3)/x^2]

Rule 6818

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*(y^(m + 1)/(m + 1)), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {4}{\log \left (-\frac {16-48 x+21 x^2+4 x^3}{9 x^2}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.02, size = 27, normalized size = 0.84 \begin {gather*} \frac {4}{\log \left (-\frac {16-48 x+21 x^2+4 x^3}{9 x^2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(128 - 192*x - 16*x^3)/((16*x - 48*x^2 + 21*x^3 + 4*x^4)*Log[(-16 + 48*x - 21*x^2 - 4*x^3)/(9*x^2)]^
2),x]

[Out]

4/Log[-1/9*(16 - 48*x + 21*x^2 + 4*x^3)/x^2]

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Maple [A]
time = 0.36, size = 33, normalized size = 1.03

method	result	size
norman	\(\frac {4}{\ln \left (\frac {-4 x^{3}-21 x^{2}+48 x -16}{9 x^{2}}\right )}\)	\(26\)
risch	\(\frac {4}{\ln \left (\frac {-4 x^{3}-21 x^{2}+48 x -16}{9 x^{2}}\right )}\)	\(26\)
default	\(-\frac {4}{2 \ln \left (3\right )-\ln \left (-\frac {4 x^{3}+21 x^{2}-48 x +16}{x^{2}}\right )}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-16*x^3-192*x+128)/(4*x^4+21*x^3-48*x^2+16*x)/ln(1/9*(-4*x^3-21*x^2+48*x-16)/x^2)^2,x,method=_RETURNVERBO
SE)

[Out]

-4/(2*ln(3)-ln(-(4*x^3+21*x^2-48*x+16)/x^2))

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Maxima [A]
time = 0.54, size = 31, normalized size = 0.97 \begin {gather*} -\frac {4}{2 \, \log \left (3\right ) - \log \left (-4 \, x^{3} - 21 \, x^{2} + 48 \, x - 16\right ) + 2 \, \log \left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*x^3-192*x+128)/(4*x^4+21*x^3-48*x^2+16*x)/log(1/9*(-4*x^3-21*x^2+48*x-16)/x^2)^2,x, algorithm="
maxima")

[Out]

-4/(2*log(3) - log(-4*x^3 - 21*x^2 + 48*x - 16) + 2*log(x))

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Fricas [A]
time = 0.36, size = 25, normalized size = 0.78 \begin {gather*} \frac {4}{\log \left (-\frac {4 \, x^{3} + 21 \, x^{2} - 48 \, x + 16}{9 \, x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*x^3-192*x+128)/(4*x^4+21*x^3-48*x^2+16*x)/log(1/9*(-4*x^3-21*x^2+48*x-16)/x^2)^2,x, algorithm="
fricas")

[Out]

4/log(-1/9*(4*x^3 + 21*x^2 - 48*x + 16)/x^2)

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Sympy [A]
time = 0.06, size = 27, normalized size = 0.84 \begin {gather*} \frac {4}{\log {\left (\frac {- \frac {4 x^{3}}{9} - \frac {7 x^{2}}{3} + \frac {16 x}{3} - \frac {16}{9}}{x^{2}} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*x**3-192*x+128)/(4*x**4+21*x**3-48*x**2+16*x)/ln(1/9*(-4*x**3-21*x**2+48*x-16)/x**2)**2,x)

[Out]

4/log((-4*x**3/9 - 7*x**2/3 + 16*x/3 - 16/9)/x**2)

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Giac [A]
time = 0.43, size = 25, normalized size = 0.78 \begin {gather*} \frac {4}{\log \left (-\frac {4 \, x^{3} + 21 \, x^{2} - 48 \, x + 16}{9 \, x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*x^3-192*x+128)/(4*x^4+21*x^3-48*x^2+16*x)/log(1/9*(-4*x^3-21*x^2+48*x-16)/x^2)^2,x, algorithm="
giac")

[Out]

4/log(-1/9*(4*x^3 + 21*x^2 - 48*x + 16)/x^2)

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Mupad [B]
time = 4.52, size = 25, normalized size = 0.78 \begin {gather*} \frac {4}{\ln \left (-\frac {4\,x^3+21\,x^2-48\,x+16}{9\,x^2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(192*x + 16*x^3 - 128)/(log(-((7*x^2)/3 - (16*x)/3 + (4*x^3)/9 + 16/9)/x^2)^2*(16*x - 48*x^2 + 21*x^3 + 4
*x^4)),x)

[Out]

4/log(-(21*x^2 - 48*x + 4*x^3 + 16)/(9*x^2))

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