3.84.57 \(\int \frac {-20+20 x-5 x^2+e^x (4-3 x+x^2+x \log (2))+(-5 x^2-5 x^3+e^x (-6 x+x^2+2 x \log (2))) \log (x)}{16 e^x-8 e^x x^2 \log (x)+e^x x^4 \log ^2(x)} \, dx\)

Optimal. Leaf size=28 \[ \frac {3-x+5 e^{-x} x-\log (2)}{-4+x^2 \log (x)} \]

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Rubi [F]  time = 1.48, 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 {-20+20 x-5 x^2+e^x \left (4-3 x+x^2+x \log (2)\right )+\left (-5 x^2-5 x^3+e^x \left (-6 x+x^2+2 x \log (2)\right )\right ) \log (x)}{16 e^x-8 e^x x^2 \log (x)+e^x x^4 \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-20 + 20*x - 5*x^2 + E^x*(4 - 3*x + x^2 + x*Log[2]) + (-5*x^2 - 5*x^3 + E^x*(-6*x + x^2 + 2*x*Log[2]))*Lo
g[x])/(16*E^x - 8*E^x*x^2*Log[x] + E^x*x^4*Log[x]^2),x]

[Out]

(-5*(4*x - x^3*Log[x]))/(E^x*(4 - x^2*Log[x])^2) + 8*Defer[Int][(-4 + x^2*Log[x])^(-2), x] - 4*(6 - Log[4])*De
fer[Int][1/(x*(-4 + x^2*Log[x])^2), x] - (3 - Log[2])*Defer[Int][x/(-4 + x^2*Log[x])^2, x] + Defer[Int][x^2/(-
4 + x^2*Log[x])^2, x] + Defer[Int][(-4 + x^2*Log[x])^(-1), x] - (6 - Log[4])*Defer[Int][1/(x*(-4 + x^2*Log[x])
), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-x} \left (-5 (-2+x)^2+e^x \left (4+x^2+x (-3+\log (2))\right )+x \left (-5 x (1+x)+e^x (-6+x+\log (4))\right ) \log (x)\right )}{\left (4-x^2 \log (x)\right )^2} \, dx\\ &=\int \left (-\frac {5 e^{-x} \left (4-4 x+x^2+x^2 \log (x)+x^3 \log (x)\right )}{\left (-4+x^2 \log (x)\right )^2}+\frac {4+x^2-3 x \left (1-\frac {\log (2)}{3}\right )+x^2 \log (x)-6 x \left (1-\frac {\log (2)}{3}\right ) \log (x)}{\left (4-x^2 \log (x)\right )^2}\right ) \, dx\\ &=-\left (5 \int \frac {e^{-x} \left (4-4 x+x^2+x^2 \log (x)+x^3 \log (x)\right )}{\left (-4+x^2 \log (x)\right )^2} \, dx\right )+\int \frac {4+x^2-3 x \left (1-\frac {\log (2)}{3}\right )+x^2 \log (x)-6 x \left (1-\frac {\log (2)}{3}\right ) \log (x)}{\left (4-x^2 \log (x)\right )^2} \, dx\\ &=-\frac {5 e^{-x} \left (4 x-x^3 \log (x)\right )}{\left (4-x^2 \log (x)\right )^2}+\int \frac {4+x^2+x (-3+\log (2))+x (-6+x+\log (4)) \log (x)}{\left (4-x^2 \log (x)\right )^2} \, dx\\ &=-\frac {5 e^{-x} \left (4 x-x^3 \log (x)\right )}{\left (4-x^2 \log (x)\right )^2}+\int \left (\frac {8 x+x^3-x^2 (3-\log (2))-4 (6-\log (4))}{x \left (4-x^2 \log (x)\right )^2}+\frac {-6+x+\log (4)}{x \left (-4+x^2 \log (x)\right )}\right ) \, dx\\ &=-\frac {5 e^{-x} \left (4 x-x^3 \log (x)\right )}{\left (4-x^2 \log (x)\right )^2}+\int \frac {8 x+x^3-x^2 (3-\log (2))-4 (6-\log (4))}{x \left (4-x^2 \log (x)\right )^2} \, dx+\int \frac {-6+x+\log (4)}{x \left (-4+x^2 \log (x)\right )} \, dx\\ &=-\frac {5 e^{-x} \left (4 x-x^3 \log (x)\right )}{\left (4-x^2 \log (x)\right )^2}+\int \left (\frac {8}{\left (-4+x^2 \log (x)\right )^2}+\frac {x^2}{\left (-4+x^2 \log (x)\right )^2}+\frac {x (-3+\log (2))}{\left (-4+x^2 \log (x)\right )^2}+\frac {4 (-6+\log (4))}{x \left (-4+x^2 \log (x)\right )^2}\right ) \, dx+\int \left (\frac {1}{-4+x^2 \log (x)}+\frac {-6+\log (4)}{x \left (-4+x^2 \log (x)\right )}\right ) \, dx\\ &=-\frac {5 e^{-x} \left (4 x-x^3 \log (x)\right )}{\left (4-x^2 \log (x)\right )^2}+8 \int \frac {1}{\left (-4+x^2 \log (x)\right )^2} \, dx+(-3+\log (2)) \int \frac {x}{\left (-4+x^2 \log (x)\right )^2} \, dx-(4 (6-\log (4))) \int \frac {1}{x \left (-4+x^2 \log (x)\right )^2} \, dx+(-6+\log (4)) \int \frac {1}{x \left (-4+x^2 \log (x)\right )} \, dx+\int \frac {x^2}{\left (-4+x^2 \log (x)\right )^2} \, dx+\int \frac {1}{-4+x^2 \log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.56, size = 58, normalized size = 2.07 \begin {gather*} \frac {e^{-x} \left (5 x \left (8+x^2\right )-e^x \left (8 x+x^3+x^2 (-3+\log (2))+4 (-6+\log (4))\right )\right )}{\left (8+x^2\right ) \left (-4+x^2 \log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-20 + 20*x - 5*x^2 + E^x*(4 - 3*x + x^2 + x*Log[2]) + (-5*x^2 - 5*x^3 + E^x*(-6*x + x^2 + 2*x*Log[2
]))*Log[x])/(16*E^x - 8*E^x*x^2*Log[x] + E^x*x^4*Log[x]^2),x]

[Out]

(5*x*(8 + x^2) - E^x*(8*x + x^3 + x^2*(-3 + Log[2]) + 4*(-6 + Log[4])))/(E^x*(8 + x^2)*(-4 + x^2*Log[x]))

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fricas [A]  time = 0.69, size = 29, normalized size = 1.04 \begin {gather*} -\frac {{\left (x + \log \relax (2) - 3\right )} e^{x} - 5 \, x}{x^{2} e^{x} \log \relax (x) - 4 \, e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x*log(2)+x^2-6*x)*exp(x)-5*x^3-5*x^2)*log(x)+(x*log(2)+x^2-3*x+4)*exp(x)-5*x^2+20*x-20)/(x^4*ex
p(x)*log(x)^2-8*x^2*exp(x)*log(x)+16*exp(x)),x, algorithm="fricas")

[Out]

-((x + log(2) - 3)*e^x - 5*x)/(x^2*e^x*log(x) - 4*e^x)

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giac [A]  time = 0.16, size = 27, normalized size = 0.96 \begin {gather*} \frac {5 \, x e^{\left (-x\right )} - x - \log \relax (2) + 3}{x^{2} \log \relax (x) - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x*log(2)+x^2-6*x)*exp(x)-5*x^3-5*x^2)*log(x)+(x*log(2)+x^2-3*x+4)*exp(x)-5*x^2+20*x-20)/(x^4*ex
p(x)*log(x)^2-8*x^2*exp(x)*log(x)+16*exp(x)),x, algorithm="giac")

[Out]

(5*x*e^(-x) - x - log(2) + 3)/(x^2*log(x) - 4)

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maple [A]  time = 0.10, size = 34, normalized size = 1.21

method	result	size
risch	\(-\frac {\left ({\mathrm e}^{x} \ln \relax (2)+{\mathrm e}^{x} x -5 x -3 \,{\mathrm e}^{x}\right ) {\mathrm e}^{-x}}{x^{2} \ln \relax (x )-4}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x*ln(2)+x^2-6*x)*exp(x)-5*x^3-5*x^2)*ln(x)+(x*ln(2)+x^2-3*x+4)*exp(x)-5*x^2+20*x-20)/(x^4*exp(x)*ln(x
)^2-8*x^2*exp(x)*ln(x)+16*exp(x)),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.50, size = 27, normalized size = 0.96 \begin {gather*} \frac {5 \, x e^{\left (-x\right )} - x - \log \relax (2) + 3}{x^{2} \log \relax (x) - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x*log(2)+x^2-6*x)*exp(x)-5*x^3-5*x^2)*log(x)+(x*log(2)+x^2-3*x+4)*exp(x)-5*x^2+20*x-20)/(x^4*ex
p(x)*log(x)^2-8*x^2*exp(x)*log(x)+16*exp(x)),x, algorithm="maxima")

[Out]

(5*x*e^(-x) - x - log(2) + 3)/(x^2*log(x) - 4)

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mupad [B]  time = 5.58, size = 86, normalized size = 3.07 \begin {gather*} -\frac {x^2\,{\mathrm {e}}^{-x}\,\left (8\,{\mathrm {e}}^x-40\right )-x\,{\mathrm {e}}^{-x}\,\left (24\,{\mathrm {e}}^x-8\,{\mathrm {e}}^x\,\ln \relax (2)\right )+x^4\,{\mathrm {e}}^{-x}\,\left ({\mathrm {e}}^x-5\right )-x^3\,{\mathrm {e}}^{-x}\,\left (3\,{\mathrm {e}}^x-{\mathrm {e}}^x\,\ln \relax (2)\right )}{\left (x^3+8\,x\right )\,\left (x^2\,\ln \relax (x)-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x)*(5*x^2 - exp(x)*(2*x*log(2) - 6*x + x^2) + 5*x^3) - exp(x)*(x*log(2) - 3*x + x^2 + 4) - 20*x + 5*
x^2 + 20)/(16*exp(x) + x^4*exp(x)*log(x)^2 - 8*x^2*exp(x)*log(x)),x)

[Out]

-(x^2*exp(-x)*(8*exp(x) - 40) - x*exp(-x)*(24*exp(x) - 8*exp(x)*log(2)) + x^4*exp(-x)*(exp(x) - 5) - x^3*exp(-
x)*(3*exp(x) - exp(x)*log(2)))/((8*x + x^3)*(x^2*log(x) - 4))

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sympy [A]  time = 0.37, size = 29, normalized size = 1.04 \begin {gather*} \frac {5 x e^{- x}}{x^{2} \log {\relax (x )} - 4} + \frac {- x - \log {\relax (2 )} + 3}{x^{2} \log {\relax (x )} - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x*ln(2)+x**2-6*x)*exp(x)-5*x**3-5*x**2)*ln(x)+(x*ln(2)+x**2-3*x+4)*exp(x)-5*x**2+20*x-20)/(x**4
*exp(x)*ln(x)**2-8*x**2*exp(x)*ln(x)+16*exp(x)),x)

[Out]

5*x*exp(-x)/(x**2*log(x) - 4) + (-x - log(2) + 3)/(x**2*log(x) - 4)

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