3.77.65 \(\int \frac {-1536-400 x+96 x^2+25 x^3+e^x (-768-8 x+146 x^2+25 x^3)+(-8 x-4 e^x x+4 x^2) \log (\frac {4 x^2}{2+e^x-x})}{160 x-30 x^3-5 x^4+e^x (80 x+40 x^2+5 x^3)} \, dx\)

Optimal. Leaf size=28 \[ \left (-5+\frac {4}{5 (4+x)}\right ) \log \left (\frac {4 x^2}{2+e^x-x}\right ) \]

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Rubi [F]  time = 1.74, 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 {-1536-400 x+96 x^2+25 x^3+e^x \left (-768-8 x+146 x^2+25 x^3\right )+\left (-8 x-4 e^x x+4 x^2\right ) \log \left (\frac {4 x^2}{2+e^x-x}\right )}{160 x-30 x^3-5 x^4+e^x \left (80 x+40 x^2+5 x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-1536 - 400*x + 96*x^2 + 25*x^3 + E^x*(-768 - 8*x + 146*x^2 + 25*x^3) + (-8*x - 4*E^x*x + 4*x^2)*Log[(4*x
^2)/(2 + E^x - x)])/(160*x - 30*x^3 - 5*x^4 + E^x*(80*x + 40*x^2 + 5*x^3)),x]

[Out]

5*x - 10*Log[x] + (4*Log[(4*x^2)/(2 + E^x - x)])/(5*(4 + x)) - 15*Defer[Int][(2 + E^x - x)^(-1), x] + 5*Defer[
Int][x/(2 + E^x - x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-\frac {\left (-4+e^x (-2+x)+x\right ) \left (384+196 x+25 x^2\right )}{x \left (-2-e^x+x\right )}-4 \log \left (\frac {4 x^2}{2+e^x-x}\right )}{5 (4+x)^2} \, dx\\ &=\frac {1}{5} \int \frac {-\frac {\left (-4+e^x (-2+x)+x\right ) \left (384+196 x+25 x^2\right )}{x \left (-2-e^x+x\right )}-4 \log \left (\frac {4 x^2}{2+e^x-x}\right )}{(4+x)^2} \, dx\\ &=\frac {1}{5} \int \left (\frac {-288+21 x+25 x^2}{\left (2+e^x-x\right ) (4+x)}+\frac {-768-8 x+146 x^2+25 x^3-4 x \log \left (\frac {4 x^2}{2+e^x-x}\right )}{x (4+x)^2}\right ) \, dx\\ &=\frac {1}{5} \int \frac {-288+21 x+25 x^2}{\left (2+e^x-x\right ) (4+x)} \, dx+\frac {1}{5} \int \frac {-768-8 x+146 x^2+25 x^3-4 x \log \left (\frac {4 x^2}{2+e^x-x}\right )}{x (4+x)^2} \, dx\\ &=\frac {1}{5} \int \left (-\frac {79}{2+e^x-x}+\frac {25 x}{2+e^x-x}+\frac {28}{\left (2+e^x-x\right ) (4+x)}\right ) \, dx+\frac {1}{5} \int \left (-\frac {8}{(4+x)^2}-\frac {768}{x (4+x)^2}+\frac {146 x}{(4+x)^2}+\frac {25 x^2}{(4+x)^2}-\frac {4 \log \left (\frac {4 x^2}{2+e^x-x}\right )}{(4+x)^2}\right ) \, dx\\ &=\frac {8}{5 (4+x)}-\frac {4}{5} \int \frac {\log \left (\frac {4 x^2}{2+e^x-x}\right )}{(4+x)^2} \, dx+5 \int \frac {x}{2+e^x-x} \, dx+5 \int \frac {x^2}{(4+x)^2} \, dx+\frac {28}{5} \int \frac {1}{\left (2+e^x-x\right ) (4+x)} \, dx-\frac {79}{5} \int \frac {1}{2+e^x-x} \, dx+\frac {146}{5} \int \frac {x}{(4+x)^2} \, dx-\frac {768}{5} \int \frac {1}{x (4+x)^2} \, dx\\ &=\frac {8}{5 (4+x)}+\frac {4 \log \left (\frac {4 x^2}{2+e^x-x}\right )}{5 (4+x)}-\frac {4}{5} \int \frac {4-e^x (-2+x)-x}{\left (2+e^x-x\right ) x (4+x)} \, dx+5 \int \frac {x}{2+e^x-x} \, dx+5 \int \left (1+\frac {16}{(4+x)^2}-\frac {8}{4+x}\right ) \, dx+\frac {28}{5} \int \frac {1}{\left (2+e^x-x\right ) (4+x)} \, dx-\frac {79}{5} \int \frac {1}{2+e^x-x} \, dx+\frac {146}{5} \int \left (-\frac {4}{(4+x)^2}+\frac {1}{4+x}\right ) \, dx-\frac {768}{5} \int \left (\frac {1}{16 x}-\frac {1}{4 (4+x)^2}-\frac {1}{16 (4+x)}\right ) \, dx\\ &=5 x-\frac {48 \log (x)}{5}+\frac {4 \log \left (\frac {4 x^2}{2+e^x-x}\right )}{5 (4+x)}-\frac {6}{5} \log (4+x)-\frac {4}{5} \int \left (\frac {2-x}{x (4+x)}+\frac {-3+x}{(4+x) \left (-2-e^x+x\right )}\right ) \, dx+5 \int \frac {x}{2+e^x-x} \, dx+\frac {28}{5} \int \frac {1}{\left (2+e^x-x\right ) (4+x)} \, dx-\frac {79}{5} \int \frac {1}{2+e^x-x} \, dx\\ &=5 x-\frac {48 \log (x)}{5}+\frac {4 \log \left (\frac {4 x^2}{2+e^x-x}\right )}{5 (4+x)}-\frac {6}{5} \log (4+x)-\frac {4}{5} \int \frac {2-x}{x (4+x)} \, dx-\frac {4}{5} \int \frac {-3+x}{(4+x) \left (-2-e^x+x\right )} \, dx+5 \int \frac {x}{2+e^x-x} \, dx+\frac {28}{5} \int \frac {1}{\left (2+e^x-x\right ) (4+x)} \, dx-\frac {79}{5} \int \frac {1}{2+e^x-x} \, dx\\ &=5 x-\frac {48 \log (x)}{5}+\frac {4 \log \left (\frac {4 x^2}{2+e^x-x}\right )}{5 (4+x)}-\frac {6}{5} \log (4+x)-\frac {4}{5} \int \left (\frac {1}{2 x}-\frac {3}{2 (4+x)}\right ) \, dx-\frac {4}{5} \int \left (-\frac {1}{2+e^x-x}+\frac {7}{\left (2+e^x-x\right ) (4+x)}\right ) \, dx+5 \int \frac {x}{2+e^x-x} \, dx+\frac {28}{5} \int \frac {1}{\left (2+e^x-x\right ) (4+x)} \, dx-\frac {79}{5} \int \frac {1}{2+e^x-x} \, dx\\ &=5 x-10 \log (x)+\frac {4 \log \left (\frac {4 x^2}{2+e^x-x}\right )}{5 (4+x)}+\frac {4}{5} \int \frac {1}{2+e^x-x} \, dx+5 \int \frac {x}{2+e^x-x} \, dx-\frac {79}{5} \int \frac {1}{2+e^x-x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.53, size = 43, normalized size = 1.54 \begin {gather*} \frac {1}{5} \left (25 \log \left (2+e^x-x\right )-50 \log (x)+\frac {4 \log \left (\frac {4 x^2}{2+e^x-x}\right )}{4+x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1536 - 400*x + 96*x^2 + 25*x^3 + E^x*(-768 - 8*x + 146*x^2 + 25*x^3) + (-8*x - 4*E^x*x + 4*x^2)*Lo
g[(4*x^2)/(2 + E^x - x)])/(160*x - 30*x^3 - 5*x^4 + E^x*(80*x + 40*x^2 + 5*x^3)),x]

[Out]

(25*Log[2 + E^x - x] - 50*Log[x] + (4*Log[(4*x^2)/(2 + E^x - x)])/(4 + x))/5

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fricas [A]  time = 1.69, size = 27, normalized size = 0.96 \begin {gather*} -\frac {{\left (25 \, x + 96\right )} \log \left (-\frac {4 \, x^{2}}{x - e^{x} - 2}\right )}{5 \, {\left (x + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*exp(x)*x+4*x^2-8*x)*log(4*x^2/(exp(x)+2-x))+(25*x^3+146*x^2-8*x-768)*exp(x)+25*x^3+96*x^2-400*x
-1536)/((5*x^3+40*x^2+80*x)*exp(x)-5*x^4-30*x^3+160*x),x, algorithm="fricas")

[Out]

-1/5*(25*x + 96)*log(-4*x^2/(x - e^x - 2))/(x + 4)

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giac [B]  time = 0.17, size = 55, normalized size = 1.96 \begin {gather*} \frac {25 \, x \log \left (x - e^{x} - 2\right ) - 50 \, x \log \relax (x) + 100 \, \log \left (x - e^{x} - 2\right ) - 200 \, \log \relax (x) + 4 \, \log \left (-\frac {4 \, x^{2}}{x - e^{x} - 2}\right )}{5 \, {\left (x + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*exp(x)*x+4*x^2-8*x)*log(4*x^2/(exp(x)+2-x))+(25*x^3+146*x^2-8*x-768)*exp(x)+25*x^3+96*x^2-400*x
-1536)/((5*x^3+40*x^2+80*x)*exp(x)-5*x^4-30*x^3+160*x),x, algorithm="giac")

[Out]

1/5*(25*x*log(x - e^x - 2) - 50*x*log(x) + 100*log(x - e^x - 2) - 200*log(x) + 4*log(-4*x^2/(x - e^x - 2)))/(x
+ 4)

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maple [A]  time = 0.27, size = 38, normalized size = 1.36

method	result	size
norman	\(\frac {4 \ln \left (\frac {4 x^{2}}{{\mathrm e}^{x}+2-x}\right )}{5 \left (4+x \right )}-10 \ln \relax (x )+5 \ln \left (-{\mathrm e}^{x}+x -2\right )\)	\(38\)
risch	\(-\frac {4 \ln \left (-{\mathrm e}^{x}+x -2\right )}{5 \left (4+x \right )}+\frac {-2 i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}-2 i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2}+2 i \pi \mathrm {csgn}\left (\frac {i x^{2}}{{\mathrm e}^{x}+2-x}\right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 i \pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{x}+2-x}\right ) \mathrm {csgn}\left (\frac {i x^{2}}{{\mathrm e}^{x}+2-x}\right )^{2}-2 i \pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{x}+2-x}\right ) \mathrm {csgn}\left (\frac {i x^{2}}{{\mathrm e}^{x}+2-x}\right ) \mathrm {csgn}\left (i x^{2}\right )-4 i \pi \mathrm {csgn}\left (\frac {i x^{2}}{{\mathrm e}^{x}+2-x}\right )^{2}+4 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-2 i \pi \mathrm {csgn}\left (\frac {i x^{2}}{{\mathrm e}^{x}+2-x}\right )^{3}+4 i \pi -50 x \ln \relax (x )+25 \ln \left ({\mathrm e}^{x}+2-x \right ) x -192 \ln \relax (x )+8 \ln \relax (2)+100 \ln \left ({\mathrm e}^{x}+2-x \right )}{20+5 x}\)	\(260\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*exp(x)*x+4*x^2-8*x)*ln(4*x^2/(exp(x)+2-x))+(25*x^3+146*x^2-8*x-768)*exp(x)+25*x^3+96*x^2-400*x-1536)/
((5*x^3+40*x^2+80*x)*exp(x)-5*x^4-30*x^3+160*x),x,method=_RETURNVERBOSE)

[Out]

4/5*ln(4*x^2/(exp(x)+2-x))/(4+x)-10*ln(x)+5*ln(-exp(x)+x-2)

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maxima [A]  time = 0.49, size = 36, normalized size = 1.29 \begin {gather*} -\frac {2 \, {\left (25 \, x + 96\right )} \log \relax (x) - {\left (25 \, x + 96\right )} \log \left (-x + e^{x} + 2\right ) - 8 \, \log \relax (2)}{5 \, {\left (x + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*exp(x)*x+4*x^2-8*x)*log(4*x^2/(exp(x)+2-x))+(25*x^3+146*x^2-8*x-768)*exp(x)+25*x^3+96*x^2-400*x
-1536)/((5*x^3+40*x^2+80*x)*exp(x)-5*x^4-30*x^3+160*x),x, algorithm="maxima")

[Out]

-1/5*(2*(25*x + 96)*log(x) - (25*x + 96)*log(-x + e^x + 2) - 8*log(2))/(x + 4)

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mupad [B]  time = 5.18, size = 39, normalized size = 1.39 \begin {gather*} 5\,\ln \left ({\mathrm {e}}^x-x+2\right )-10\,\ln \relax (x)+\frac {4\,\ln \left (\frac {4\,x^2}{{\mathrm {e}}^x-x+2}\right )}{5\,\left (x+4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(400*x + log((4*x^2)/(exp(x) - x + 2))*(8*x + 4*x*exp(x) - 4*x^2) - 96*x^2 - 25*x^3 + exp(x)*(8*x - 146*x
^2 - 25*x^3 + 768) + 1536)/(160*x - 30*x^3 - 5*x^4 + exp(x)*(80*x + 40*x^2 + 5*x^3)),x)

[Out]

5*log(exp(x) - x + 2) - 10*log(x) + (4*log((4*x^2)/(exp(x) - x + 2)))/(5*(x + 4))

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sympy [A]  time = 0.50, size = 34, normalized size = 1.21 \begin {gather*} - 10 \log {\relax (x )} + 5 \log {\left (- x + e^{x} + 2 \right )} + \frac {4 \log {\left (\frac {4 x^{2}}{- x + e^{x} + 2} \right )}}{5 x + 20} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*exp(x)*x+4*x**2-8*x)*ln(4*x**2/(exp(x)+2-x))+(25*x**3+146*x**2-8*x-768)*exp(x)+25*x**3+96*x**2-
400*x-1536)/((5*x**3+40*x**2+80*x)*exp(x)-5*x**4-30*x**3+160*x),x)

[Out]

-10*log(x) + 5*log(-x + exp(x) + 2) + 4*log(4*x**2/(-x + exp(x) + 2))/(5*x + 20)

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