3.6.44 \(\int \frac {80 x+80 x^2+e^x (120+180 x-25 x^2)+(20 x+20 x^2+e^x (40+35 x-5 x^2)) \log (\frac {e^x (8-x)+4 x}{x})}{-64 e^x x^3+e^{2 x} (-128 x^2+16 x^3)+(-32 e^x x^3+e^{2 x} (-64 x^2+8 x^3)) \log (\frac {e^x (8-x)+4 x}{x})+(-4 e^x x^3+e^{2 x} (-8 x^2+x^3)) \log ^2(\frac {e^x (8-x)+4 x}{x})} \, dx\)

Optimal. Leaf size=28 \[ \frac {5 e^{-x}}{x \left (4+\log \left (4-\frac {e^x (-8+x)}{x}\right )\right )} \]

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Rubi [F]  time = 9.55, 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 {80 x+80 x^2+e^x \left (120+180 x-25 x^2\right )+\left (20 x+20 x^2+e^x \left (40+35 x-5 x^2\right )\right ) \log \left (\frac {e^x (8-x)+4 x}{x}\right )}{-64 e^x x^3+e^{2 x} \left (-128 x^2+16 x^3\right )+\left (-32 e^x x^3+e^{2 x} \left (-64 x^2+8 x^3\right )\right ) \log \left (\frac {e^x (8-x)+4 x}{x}\right )+\left (-4 e^x x^3+e^{2 x} \left (-8 x^2+x^3\right )\right ) \log ^2\left (\frac {e^x (8-x)+4 x}{x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(80*x + 80*x^2 + E^x*(120 + 180*x - 25*x^2) + (20*x + 20*x^2 + E^x*(40 + 35*x - 5*x^2))*Log[(E^x*(8 - x) +
4*x)/x])/(-64*E^x*x^3 + E^(2*x)*(-128*x^2 + 16*x^3) + (-32*E^x*x^3 + E^(2*x)*(-64*x^2 + 8*x^3))*Log[(E^x*(8 -
x) + 4*x)/x] + (-4*E^x*x^3 + E^(2*x)*(-8*x^2 + x^3))*Log[(E^x*(8 - x) + 4*x)/x]^2),x]

[Out]

(-5*Defer[Int][1/(E^x*(-8 + x)*(4 + Log[4 + E^x*(-1 + 8/x)])^2), x])/8 + 5*Defer[Int][1/(E^x*x^2*(4 + Log[4 +
E^x*(-1 + 8/x)])^2), x] - (35*Defer[Int][1/(E^x*x*(4 + Log[4 + E^x*(-1 + 8/x)])^2), x])/8 - 20*Defer[Int][1/(E
^x*(-8*E^x - 4*x + E^x*x)*(4 + Log[4 + E^x*(-1 + 8/x)])^2), x] - 20*Defer[Int][1/(E^x*(-8 + x)*(-8*E^x - 4*x +
E^x*x)*(4 + Log[4 + E^x*(-1 + 8/x)])^2), x] + 20*Defer[Int][1/(E^x*x*(-8*E^x - 4*x + E^x*x)*(4 + Log[4 + E^x*
(-1 + 8/x)])^2), x] - 5*Defer[Int][1/(E^x*x^2*(4 + Log[4 + E^x*(-1 + 8/x)])), x] - 5*Defer[Int][1/(E^x*x*(4 +
Log[4 + E^x*(-1 + 8/x)])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-x} \left (80 x (1+x)+5 e^x \left (24+36 x-5 x^2\right )-5 \left (e^x (-8+x)-4 x\right ) (1+x) \log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )}{\left (e^x (-8+x)-4 x\right ) x^2 \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )^2} \, dx\\ &=\int \left (-\frac {20 e^{-x} \left (8-8 x+x^2\right )}{(-8+x) x \left (-8 e^x-4 x+e^x x\right ) \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )^2}-\frac {5 e^{-x} \left (-24-36 x+5 x^2-8 \log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )-7 x \log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )+x^2 \log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )}{(-8+x) x^2 \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )^2}\right ) \, dx\\ &=-\left (5 \int \frac {e^{-x} \left (-24-36 x+5 x^2-8 \log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )-7 x \log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )+x^2 \log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )}{(-8+x) x^2 \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )^2} \, dx\right )-20 \int \frac {e^{-x} \left (8-8 x+x^2\right )}{(-8+x) x \left (-8 e^x-4 x+e^x x\right ) \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )^2} \, dx\\ &=-\left (5 \int \frac {e^{-x} \left (24+36 x-5 x^2-\left (-8-7 x+x^2\right ) \log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )}{(8-x) x^2 \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )^2} \, dx\right )-20 \int \left (\frac {e^{-x}}{\left (-8 e^x-4 x+e^x x\right ) \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )^2}+\frac {e^{-x}}{(-8+x) \left (-8 e^x-4 x+e^x x\right ) \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )^2}-\frac {e^{-x}}{x \left (-8 e^x-4 x+e^x x\right ) \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )^2}\right ) \, dx\\ &=-\left (5 \int \left (\frac {e^{-x} \left (8-8 x+x^2\right )}{(-8+x) x^2 \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )^2}+\frac {e^{-x} (1+x)}{x^2 \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )}\right ) \, dx\right )-20 \int \frac {e^{-x}}{\left (-8 e^x-4 x+e^x x\right ) \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )^2} \, dx-20 \int \frac {e^{-x}}{(-8+x) \left (-8 e^x-4 x+e^x x\right ) \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )^2} \, dx+20 \int \frac {e^{-x}}{x \left (-8 e^x-4 x+e^x x\right ) \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )^2} \, dx\\ &=-\left (5 \int \frac {e^{-x} \left (8-8 x+x^2\right )}{(-8+x) x^2 \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )^2} \, dx\right )-5 \int \frac {e^{-x} (1+x)}{x^2 \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )} \, dx-20 \int \frac {e^{-x}}{\left (-8 e^x-4 x+e^x x\right ) \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )^2} \, dx-20 \int \frac {e^{-x}}{(-8+x) \left (-8 e^x-4 x+e^x x\right ) \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )^2} \, dx+20 \int \frac {e^{-x}}{x \left (-8 e^x-4 x+e^x x\right ) \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )^2} \, dx\\ &=-\left (5 \int \left (\frac {e^{-x}}{8 (-8+x) \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )^2}-\frac {e^{-x}}{x^2 \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )^2}+\frac {7 e^{-x}}{8 x \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )^2}\right ) \, dx\right )-5 \int \left (\frac {e^{-x}}{x^2 \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )}+\frac {e^{-x}}{x \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )}\right ) \, dx-20 \int \frac {e^{-x}}{\left (-8 e^x-4 x+e^x x\right ) \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )^2} \, dx-20 \int \frac {e^{-x}}{(-8+x) \left (-8 e^x-4 x+e^x x\right ) \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )^2} \, dx+20 \int \frac {e^{-x}}{x \left (-8 e^x-4 x+e^x x\right ) \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )^2} \, dx\\ &=-\left (\frac {5}{8} \int \frac {e^{-x}}{(-8+x) \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )^2} \, dx\right )-\frac {35}{8} \int \frac {e^{-x}}{x \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )^2} \, dx+5 \int \frac {e^{-x}}{x^2 \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )^2} \, dx-5 \int \frac {e^{-x}}{x^2 \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )} \, dx-5 \int \frac {e^{-x}}{x \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )} \, dx-20 \int \frac {e^{-x}}{\left (-8 e^x-4 x+e^x x\right ) \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )^2} \, dx-20 \int \frac {e^{-x}}{(-8+x) \left (-8 e^x-4 x+e^x x\right ) \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )^2} \, dx+20 \int \frac {e^{-x}}{x \left (-8 e^x-4 x+e^x x\right ) \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 28, normalized size = 1.00 \begin {gather*} \frac {5 e^{-x}}{x \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(80*x + 80*x^2 + E^x*(120 + 180*x - 25*x^2) + (20*x + 20*x^2 + E^x*(40 + 35*x - 5*x^2))*Log[(E^x*(8
- x) + 4*x)/x])/(-64*E^x*x^3 + E^(2*x)*(-128*x^2 + 16*x^3) + (-32*E^x*x^3 + E^(2*x)*(-64*x^2 + 8*x^3))*Log[(E^
x*(8 - x) + 4*x)/x] + (-4*E^x*x^3 + E^(2*x)*(-8*x^2 + x^3))*Log[(E^x*(8 - x) + 4*x)/x]^2),x]

[Out]

5/(E^x*x*(4 + Log[4 + E^x*(-1 + 8/x)]))

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fricas [A]  time = 0.52, size = 30, normalized size = 1.07 \begin {gather*} \frac {5}{x e^{x} \log \left (-\frac {{\left (x - 8\right )} e^{x} - 4 \, x}{x}\right ) + 4 \, x e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x^2+35*x+40)*exp(x)+20*x^2+20*x)*log(((8-x)*exp(x)+4*x)/x)+(-25*x^2+180*x+120)*exp(x)+80*x^2+8
0*x)/(((x^3-8*x^2)*exp(x)^2-4*exp(x)*x^3)*log(((8-x)*exp(x)+4*x)/x)^2+((8*x^3-64*x^2)*exp(x)^2-32*exp(x)*x^3)*
log(((8-x)*exp(x)+4*x)/x)+(16*x^3-128*x^2)*exp(x)^2-64*exp(x)*x^3),x, algorithm="fricas")

[Out]

5/(x*e^x*log(-((x - 8)*e^x - 4*x)/x) + 4*x*e^x)

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giac [A]  time = 1.05, size = 32, normalized size = 1.14 \begin {gather*} \frac {5}{x e^{x} \log \left (-\frac {x e^{x} - 4 \, x - 8 \, e^{x}}{x}\right ) + 4 \, x e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x^2+35*x+40)*exp(x)+20*x^2+20*x)*log(((8-x)*exp(x)+4*x)/x)+(-25*x^2+180*x+120)*exp(x)+80*x^2+8
0*x)/(((x^3-8*x^2)*exp(x)^2-4*exp(x)*x^3)*log(((8-x)*exp(x)+4*x)/x)^2+((8*x^3-64*x^2)*exp(x)^2-32*exp(x)*x^3)*
log(((8-x)*exp(x)+4*x)/x)+(16*x^3-128*x^2)*exp(x)^2-64*exp(x)*x^3),x, algorithm="giac")

[Out]

5/(x*e^x*log(-(x*e^x - 4*x - 8*e^x)/x) + 4*x*e^x)

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maple [C]  time = 0.33, size = 195, normalized size = 6.96

method	result	size
risch	\(\frac {10 i {\mathrm e}^{-x}}{x \left (\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (x \left ({\mathrm e}^{x}-4\right )-8 \,{\mathrm e}^{x}\right )\right ) \mathrm {csgn}\left (\frac {i \left (x \left ({\mathrm e}^{x}-4\right )-8 \,{\mathrm e}^{x}\right )}{x}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (x \left ({\mathrm e}^{x}-4\right )-8 \,{\mathrm e}^{x}\right )}{x}\right )^{2}+2 \pi \mathrm {csgn}\left (\frac {i \left (x \left ({\mathrm e}^{x}-4\right )-8 \,{\mathrm e}^{x}\right )}{x}\right )^{2}-\pi \,\mathrm {csgn}\left (i \left (x \left ({\mathrm e}^{x}-4\right )-8 \,{\mathrm e}^{x}\right )\right ) \mathrm {csgn}\left (\frac {i \left (x \left ({\mathrm e}^{x}-4\right )-8 \,{\mathrm e}^{x}\right )}{x}\right )^{2}-\pi \mathrm {csgn}\left (\frac {i \left (x \left ({\mathrm e}^{x}-4\right )-8 \,{\mathrm e}^{x}\right )}{x}\right )^{3}-2 \pi -2 i \ln \relax (x )+2 i \ln \left (x \left ({\mathrm e}^{x}-4\right )-8 \,{\mathrm e}^{x}\right )+8 i\right )}\)	\(195\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-5*x^2+35*x+40)*exp(x)+20*x^2+20*x)*ln(((8-x)*exp(x)+4*x)/x)+(-25*x^2+180*x+120)*exp(x)+80*x^2+80*x)/((
(x^3-8*x^2)*exp(x)^2-4*exp(x)*x^3)*ln(((8-x)*exp(x)+4*x)/x)^2+((8*x^3-64*x^2)*exp(x)^2-32*exp(x)*x^3)*ln(((8-x
)*exp(x)+4*x)/x)+(16*x^3-128*x^2)*exp(x)^2-64*exp(x)*x^3),x,method=_RETURNVERBOSE)

[Out]

10*I*exp(-x)/x/(Pi*csgn(I/x)*csgn(I*(x*(exp(x)-4)-8*exp(x)))*csgn(I/x*(x*(exp(x)-4)-8*exp(x)))-Pi*csgn(I/x)*cs
gn(I/x*(x*(exp(x)-4)-8*exp(x)))^2+2*Pi*csgn(I/x*(x*(exp(x)-4)-8*exp(x)))^2-Pi*csgn(I*(x*(exp(x)-4)-8*exp(x)))*
csgn(I/x*(x*(exp(x)-4)-8*exp(x)))^2-Pi*csgn(I/x*(x*(exp(x)-4)-8*exp(x)))^3-2*Pi-2*I*ln(x)+2*I*ln(x*(exp(x)-4)-
8*exp(x))+8*I)

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maxima [A]  time = 0.90, size = 33, normalized size = 1.18 \begin {gather*} \frac {5}{x e^{x} \log \left (-{\left (x - 8\right )} e^{x} + 4 \, x\right ) - {\left (x \log \relax (x) - 4 \, x\right )} e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x^2+35*x+40)*exp(x)+20*x^2+20*x)*log(((8-x)*exp(x)+4*x)/x)+(-25*x^2+180*x+120)*exp(x)+80*x^2+8
0*x)/(((x^3-8*x^2)*exp(x)^2-4*exp(x)*x^3)*log(((8-x)*exp(x)+4*x)/x)^2+((8*x^3-64*x^2)*exp(x)^2-32*exp(x)*x^3)*
log(((8-x)*exp(x)+4*x)/x)+(16*x^3-128*x^2)*exp(x)^2-64*exp(x)*x^3),x, algorithm="maxima")

[Out]

5/(x*e^x*log(-(x - 8)*e^x + 4*x) - (x*log(x) - 4*x)*e^x)

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mupad [B]  time = 0.96, size = 29, normalized size = 1.04 \begin {gather*} \frac {5\,{\mathrm {e}}^{-x}}{x\,\left (\ln \left (\frac {4\,x-{\mathrm {e}}^x\,\left (x-8\right )}{x}\right )+4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(80*x + log((4*x - exp(x)*(x - 8))/x)*(20*x + exp(x)*(35*x - 5*x^2 + 40) + 20*x^2) + exp(x)*(180*x - 25*x
^2 + 120) + 80*x^2)/(64*x^3*exp(x) + log((4*x - exp(x)*(x - 8))/x)*(32*x^3*exp(x) + exp(2*x)*(64*x^2 - 8*x^3))
+ exp(2*x)*(128*x^2 - 16*x^3) + log((4*x - exp(x)*(x - 8))/x)^2*(4*x^3*exp(x) + exp(2*x)*(8*x^2 - x^3))),x)

[Out]

(5*exp(-x))/(x*(log((4*x - exp(x)*(x - 8))/x) + 4))

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sympy [A]  time = 0.31, size = 26, normalized size = 0.93 \begin {gather*} \frac {5}{x e^{x} \log {\left (\frac {4 x + \left (8 - x\right ) e^{x}}{x} \right )} + 4 x e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x**2+35*x+40)*exp(x)+20*x**2+20*x)*ln(((8-x)*exp(x)+4*x)/x)+(-25*x**2+180*x+120)*exp(x)+80*x**
2+80*x)/(((x**3-8*x**2)*exp(x)**2-4*exp(x)*x**3)*ln(((8-x)*exp(x)+4*x)/x)**2+((8*x**3-64*x**2)*exp(x)**2-32*ex
p(x)*x**3)*ln(((8-x)*exp(x)+4*x)/x)+(16*x**3-128*x**2)*exp(x)**2-64*exp(x)*x**3),x)

[Out]

5/(x*exp(x)*log((4*x + (8 - x)*exp(x))/x) + 4*x*exp(x))

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