3.73.76 \(\int \frac {-6+4 x-24 x^2+16 x^3}{(-5+24 x-40 x^2+16 x^3+16 x^4) \log (\frac {5-14 x+12 x^2+8 x^3}{1-4 x+4 x^2})} \, dx\)

Optimal. Leaf size=27 \[ 1+e^2+\log \left (\log \left (5+2 x+\frac {x}{\left (2 \left (\frac {1}{4}-x\right )+x\right )^2}\right )\right ) \]

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Rubi [F]  time = 0.73, 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 {-6+4 x-24 x^2+16 x^3}{\left (-5+24 x-40 x^2+16 x^3+16 x^4\right ) \log \left (\frac {5-14 x+12 x^2+8 x^3}{1-4 x+4 x^2}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-6 + 4*x - 24*x^2 + 16*x^3)/((-5 + 24*x - 40*x^2 + 16*x^3 + 16*x^4)*Log[(5 - 14*x + 12*x^2 + 8*x^3)/(1 -
4*x + 4*x^2)]),x]

[Out]

-4*Defer[Int][1/((-1 + 2*x)*Log[(5 - 14*x + 12*x^2 + 8*x^3)/(-1 + 2*x)^2]), x] - 14*Defer[Int][1/((5 - 14*x +
12*x^2 + 8*x^3)*Log[(5 - 14*x + 12*x^2 + 8*x^3)/(-1 + 2*x)^2]), x] + 24*Defer[Int][x/((5 - 14*x + 12*x^2 + 8*x
^3)*Log[(5 - 14*x + 12*x^2 + 8*x^3)/(-1 + 2*x)^2]), x] + 24*Defer[Int][x^2/((5 - 14*x + 12*x^2 + 8*x^3)*Log[(5
- 14*x + 12*x^2 + 8*x^3)/(-1 + 2*x)^2]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {4}{(-1+2 x) \log \left (\frac {5-14 x+12 x^2+8 x^3}{(-1+2 x)^2}\right )}+\frac {2 \left (-7+12 x+12 x^2\right )}{\left (5-14 x+12 x^2+8 x^3\right ) \log \left (\frac {5-14 x+12 x^2+8 x^3}{(-1+2 x)^2}\right )}\right ) \, dx\\ &=2 \int \frac {-7+12 x+12 x^2}{\left (5-14 x+12 x^2+8 x^3\right ) \log \left (\frac {5-14 x+12 x^2+8 x^3}{(-1+2 x)^2}\right )} \, dx-4 \int \frac {1}{(-1+2 x) \log \left (\frac {5-14 x+12 x^2+8 x^3}{(-1+2 x)^2}\right )} \, dx\\ &=2 \int \left (-\frac {7}{\left (5-14 x+12 x^2+8 x^3\right ) \log \left (\frac {5-14 x+12 x^2+8 x^3}{(-1+2 x)^2}\right )}+\frac {12 x}{\left (5-14 x+12 x^2+8 x^3\right ) \log \left (\frac {5-14 x+12 x^2+8 x^3}{(-1+2 x)^2}\right )}+\frac {12 x^2}{\left (5-14 x+12 x^2+8 x^3\right ) \log \left (\frac {5-14 x+12 x^2+8 x^3}{(-1+2 x)^2}\right )}\right ) \, dx-4 \int \frac {1}{(-1+2 x) \log \left (\frac {5-14 x+12 x^2+8 x^3}{(-1+2 x)^2}\right )} \, dx\\ &=-\left (4 \int \frac {1}{(-1+2 x) \log \left (\frac {5-14 x+12 x^2+8 x^3}{(-1+2 x)^2}\right )} \, dx\right )-14 \int \frac {1}{\left (5-14 x+12 x^2+8 x^3\right ) \log \left (\frac {5-14 x+12 x^2+8 x^3}{(-1+2 x)^2}\right )} \, dx+24 \int \frac {x}{\left (5-14 x+12 x^2+8 x^3\right ) \log \left (\frac {5-14 x+12 x^2+8 x^3}{(-1+2 x)^2}\right )} \, dx+24 \int \frac {x^2}{\left (5-14 x+12 x^2+8 x^3\right ) \log \left (\frac {5-14 x+12 x^2+8 x^3}{(-1+2 x)^2}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 25, normalized size = 0.93 \begin {gather*} \log \left (\log \left (\frac {5-14 x+12 x^2+8 x^3}{(-1+2 x)^2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-6 + 4*x - 24*x^2 + 16*x^3)/((-5 + 24*x - 40*x^2 + 16*x^3 + 16*x^4)*Log[(5 - 14*x + 12*x^2 + 8*x^3)
/(1 - 4*x + 4*x^2)]),x]

[Out]

Log[Log[(5 - 14*x + 12*x^2 + 8*x^3)/(-1 + 2*x)^2]]

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fricas [A]  time = 0.58, size = 30, normalized size = 1.11 \begin {gather*} \log \left (\log \left (\frac {8 \, x^{3} + 12 \, x^{2} - 14 \, x + 5}{4 \, x^{2} - 4 \, x + 1}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*x^3-24*x^2+4*x-6)/(16*x^4+16*x^3-40*x^2+24*x-5)/log((8*x^3+12*x^2-14*x+5)/(4*x^2-4*x+1)),x, algo
rithm="fricas")

[Out]

log(log((8*x^3 + 12*x^2 - 14*x + 5)/(4*x^2 - 4*x + 1)))

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giac [A]  time = 0.21, size = 30, normalized size = 1.11 \begin {gather*} \log \left (\log \left (\frac {8 \, x^{3} + 12 \, x^{2} - 14 \, x + 5}{4 \, x^{2} - 4 \, x + 1}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*x^3-24*x^2+4*x-6)/(16*x^4+16*x^3-40*x^2+24*x-5)/log((8*x^3+12*x^2-14*x+5)/(4*x^2-4*x+1)),x, algo
rithm="giac")

[Out]

log(log((8*x^3 + 12*x^2 - 14*x + 5)/(4*x^2 - 4*x + 1)))

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maple [A]  time = 0.04, size = 31, normalized size = 1.15

method	result	size
norman	\(\ln \left (\ln \left (\frac {8 x^{3}+12 x^{2}-14 x +5}{4 x^{2}-4 x +1}\right )\right )\)	\(31\)
risch	\(\ln \left (\ln \left (\frac {8 x^{3}+12 x^{2}-14 x +5}{4 x^{2}-4 x +1}\right )\right )\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*x^3-24*x^2+4*x-6)/(16*x^4+16*x^3-40*x^2+24*x-5)/ln((8*x^3+12*x^2-14*x+5)/(4*x^2-4*x+1)),x,method=_RETU
RNVERBOSE)

[Out]

ln(ln((8*x^3+12*x^2-14*x+5)/(4*x^2-4*x+1)))

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maxima [A]  time = 0.41, size = 26, normalized size = 0.96 \begin {gather*} \log \left (\log \left (8 \, x^{3} + 12 \, x^{2} - 14 \, x + 5\right ) - 2 \, \log \left (2 \, x - 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*x^3-24*x^2+4*x-6)/(16*x^4+16*x^3-40*x^2+24*x-5)/log((8*x^3+12*x^2-14*x+5)/(4*x^2-4*x+1)),x, algo
rithm="maxima")

[Out]

log(log(8*x^3 + 12*x^2 - 14*x + 5) - 2*log(2*x - 1))

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mupad [B]  time = 4.88, size = 22, normalized size = 0.81 \begin {gather*} \ln \left (\ln \left (2\,x+\frac {4\,x}{4\,x^2-4\,x+1}+5\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x - 24*x^2 + 16*x^3 - 6)/(log((12*x^2 - 14*x + 8*x^3 + 5)/(4*x^2 - 4*x + 1))*(24*x - 40*x^2 + 16*x^3 +
16*x^4 - 5)),x)

[Out]

log(log(2*x + (4*x)/(4*x^2 - 4*x + 1) + 5))

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sympy [A]  time = 0.34, size = 27, normalized size = 1.00 \begin {gather*} \log {\left (\log {\left (\frac {8 x^{3} + 12 x^{2} - 14 x + 5}{4 x^{2} - 4 x + 1} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*x**3-24*x**2+4*x-6)/(16*x**4+16*x**3-40*x**2+24*x-5)/ln((8*x**3+12*x**2-14*x+5)/(4*x**2-4*x+1)),
x)

[Out]

log(log((8*x**3 + 12*x**2 - 14*x + 5)/(4*x**2 - 4*x + 1)))

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