3.66.51 \(\int \frac {-4 x+32 x^2+e^x (2 x-16 x^2)+(2-16 x+e^x (-1+8 x)) \log (-2+e^x)+(-4 x+16 x^2+e^x (x-4 x^2)) \log (-x+4 x^2)}{2 x-8 x^2+e^x (-x+4 x^2)} \, dx\)

Optimal. Leaf size=21 \[ \left (-2 x+\log \left (-2+e^x\right )\right ) \log \left (-x+4 x^2\right ) \]

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Rubi [F]  time = 2.15, 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 {-4 x+32 x^2+e^x \left (2 x-16 x^2\right )+\left (2-16 x+e^x (-1+8 x)\right ) \log \left (-2+e^x\right )+\left (-4 x+16 x^2+e^x \left (x-4 x^2\right )\right ) \log \left (-x+4 x^2\right )}{2 x-8 x^2+e^x \left (-x+4 x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-4*x + 32*x^2 + E^x*(2*x - 16*x^2) + (2 - 16*x + E^x*(-1 + 8*x))*Log[-2 + E^x] + (-4*x + 16*x^2 + E^x*(x
- 4*x^2))*Log[-x + 4*x^2])/(2*x - 8*x^2 + E^x*(-x + 4*x^2)),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (-2+e^x\right ) (-1+8 x) \log \left (-2+e^x\right )-x \left (2 \left (-2+e^x\right ) (-1+8 x)+\left (-4+e^x\right ) (-1+4 x) \log (x (-1+4 x))\right )}{\left (2-e^x\right ) (1-4 x) x} \, dx\\ &=\int \left (\frac {2 \log (x (-1+4 x))}{-2+e^x}+\frac {2 x-16 x^2-\log \left (-2+e^x\right )+8 x \log \left (-2+e^x\right )+x \log (x (-1+4 x))-4 x^2 \log (x (-1+4 x))}{x (-1+4 x)}\right ) \, dx\\ &=2 \int \frac {\log (x (-1+4 x))}{-2+e^x} \, dx+\int \frac {2 x-16 x^2-\log \left (-2+e^x\right )+8 x \log \left (-2+e^x\right )+x \log (x (-1+4 x))-4 x^2 \log (x (-1+4 x))}{x (-1+4 x)} \, dx\\ &=2 \int \frac {\log (x (-1+4 x))}{-2+e^x} \, dx+\int \left (-\frac {(-1+8 x) \left (2 x-\log \left (-2+e^x\right )\right )}{x (-1+4 x)}-\log (x (-1+4 x))\right ) \, dx\\ &=2 \int \frac {\log (x (-1+4 x))}{-2+e^x} \, dx-\int \frac {(-1+8 x) \left (2 x-\log \left (-2+e^x\right )\right )}{x (-1+4 x)} \, dx-\int \log (x (-1+4 x)) \, dx\\ &=-x \log (-((1-4 x) x))+2 \int 1 \, dx+2 \int \frac {\log (x (-1+4 x))}{-2+e^x} \, dx+\int \frac {1}{-1+4 x} \, dx-\int \left (\frac {2 (-1+8 x)}{-1+4 x}-\frac {(-1+8 x) \log \left (-2+e^x\right )}{x (-1+4 x)}\right ) \, dx\\ &=2 x+\frac {1}{4} \log (1-4 x)-x \log (-((1-4 x) x))-2 \int \frac {-1+8 x}{-1+4 x} \, dx+2 \int \frac {\log (x (-1+4 x))}{-2+e^x} \, dx+\int \frac {(-1+8 x) \log \left (-2+e^x\right )}{x (-1+4 x)} \, dx\\ &=2 x+\frac {1}{4} \log (1-4 x)-x \log (-((1-4 x) x))-2 \int \left (2+\frac {1}{-1+4 x}\right ) \, dx+2 \int \frac {\log (x (-1+4 x))}{-2+e^x} \, dx+\int \left (\frac {\log \left (-2+e^x\right )}{x}+\frac {4 \log \left (-2+e^x\right )}{-1+4 x}\right ) \, dx\\ &=-2 x-\frac {1}{4} \log (1-4 x)-x \log (-((1-4 x) x))+2 \int \frac {\log (x (-1+4 x))}{-2+e^x} \, dx+4 \int \frac {\log \left (-2+e^x\right )}{-1+4 x} \, dx+\int \frac {\log \left (-2+e^x\right )}{x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.43, size = 27, normalized size = 1.29 \begin {gather*} -2 x \log (x (-1+4 x))+\log \left (-2+e^x\right ) \log (x (-1+4 x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4*x + 32*x^2 + E^x*(2*x - 16*x^2) + (2 - 16*x + E^x*(-1 + 8*x))*Log[-2 + E^x] + (-4*x + 16*x^2 + E
^x*(x - 4*x^2))*Log[-x + 4*x^2])/(2*x - 8*x^2 + E^x*(-x + 4*x^2)),x]

[Out]

-2*x*Log[x*(-1 + 4*x)] + Log[-2 + E^x]*Log[x*(-1 + 4*x)]

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fricas [A]  time = 0.78, size = 30, normalized size = 1.43 \begin {gather*} -2 \, x \log \left (4 \, x^{2} - x\right ) + \log \left (4 \, x^{2} - x\right ) \log \left (e^{x} - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x-1)*exp(x)-16*x+2)*log(exp(x)-2)+((-4*x^2+x)*exp(x)+16*x^2-4*x)*log(4*x^2-x)+(-16*x^2+2*x)*exp
(x)+32*x^2-4*x)/((4*x^2-x)*exp(x)-8*x^2+2*x),x, algorithm="fricas")

[Out]

-2*x*log(4*x^2 - x) + log(4*x^2 - x)*log(e^x - 2)

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giac [A]  time = 0.28, size = 30, normalized size = 1.43 \begin {gather*} -2 \, x \log \left (4 \, x^{2} - x\right ) + \log \left (4 \, x^{2} - x\right ) \log \left (e^{x} - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x-1)*exp(x)-16*x+2)*log(exp(x)-2)+((-4*x^2+x)*exp(x)+16*x^2-4*x)*log(4*x^2-x)+(-16*x^2+2*x)*exp
(x)+32*x^2-4*x)/((4*x^2-x)*exp(x)-8*x^2+2*x),x, algorithm="giac")

[Out]

-2*x*log(4*x^2 - x) + log(4*x^2 - x)*log(e^x - 2)

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (8 x -1\right ) {\mathrm e}^{x}-16 x +2\right ) \ln \left ({\mathrm e}^{x}-2\right )+\left (\left (-4 x^{2}+x \right ) {\mathrm e}^{x}+16 x^{2}-4 x \right ) \ln \left (4 x^{2}-x \right )+\left (-16 x^{2}+2 x \right ) {\mathrm e}^{x}+32 x^{2}-4 x}{\left (4 x^{2}-x \right ) {\mathrm e}^{x}-8 x^{2}+2 x}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((8*x-1)*exp(x)-16*x+2)*ln(exp(x)-2)+((-4*x^2+x)*exp(x)+16*x^2-4*x)*ln(4*x^2-x)+(-16*x^2+2*x)*exp(x)+32*x
^2-4*x)/((4*x^2-x)*exp(x)-8*x^2+2*x),x)

[Out]

int((((8*x-1)*exp(x)-16*x+2)*ln(exp(x)-2)+((-4*x^2+x)*exp(x)+16*x^2-4*x)*ln(4*x^2-x)+(-16*x^2+2*x)*exp(x)+32*x
^2-4*x)/((4*x^2-x)*exp(x)-8*x^2+2*x),x)

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maxima [A]  time = 0.45, size = 30, normalized size = 1.43 \begin {gather*} -2 \, x \log \left (4 \, x - 1\right ) - 2 \, x \log \relax (x) + {\left (\log \left (4 \, x - 1\right ) + \log \relax (x)\right )} \log \left (e^{x} - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x-1)*exp(x)-16*x+2)*log(exp(x)-2)+((-4*x^2+x)*exp(x)+16*x^2-4*x)*log(4*x^2-x)+(-16*x^2+2*x)*exp
(x)+32*x^2-4*x)/((4*x^2-x)*exp(x)-8*x^2+2*x),x, algorithm="maxima")

[Out]

-2*x*log(4*x - 1) - 2*x*log(x) + (log(4*x - 1) + log(x))*log(e^x - 2)

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mupad [B]  time = 4.32, size = 23, normalized size = 1.10 \begin {gather*} -\ln \left (4\,x^2-x\right )\,\left (2\,x-\ln \left ({\mathrm {e}}^x-2\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*(2*x - 16*x^2) - 4*x + log(exp(x) - 2)*(exp(x)*(8*x - 1) - 16*x + 2) + 32*x^2 + log(4*x^2 - x)*(e
xp(x)*(x - 4*x^2) - 4*x + 16*x^2))/(exp(x)*(x - 4*x^2) - 2*x + 8*x^2),x)

[Out]

-log(4*x^2 - x)*(2*x - log(exp(x) - 2))

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sympy [A]  time = 0.64, size = 26, normalized size = 1.24 \begin {gather*} - 2 x \log {\left (4 x^{2} - x \right )} + \log {\left (4 x^{2} - x \right )} \log {\left (e^{x} - 2 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x-1)*exp(x)-16*x+2)*ln(exp(x)-2)+((-4*x**2+x)*exp(x)+16*x**2-4*x)*ln(4*x**2-x)+(-16*x**2+2*x)*e
xp(x)+32*x**2-4*x)/((4*x**2-x)*exp(x)-8*x**2+2*x),x)

[Out]

-2*x*log(4*x**2 - x) + log(4*x**2 - x)*log(exp(x) - 2)

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