3.95.15 \(\int \frac {-16 x^2-16 x^3+16 x^4+e^x (4 x-4 x^3+4 x^4)+e^{e^2} (-16 x^2-32 x^3+e^x (8 x+8 x^2-4 x^3))+(16 x^2+32 x^3+e^x (-8 x-8 x^2+4 x^3)) \log (x)}{e^{2 x} x^2-8 e^x x^3+16 x^4+e^{2 e^2} (e^{2 x}-8 e^x x+16 x^2)+e^{e^2} (-2 e^{2 x} x+16 e^x x^2-32 x^3)+(2 e^{2 x} x-16 e^x x^2+32 x^3+e^{e^2} (-2 e^{2 x}+16 e^x x-32 x^2)) \log (x)+(e^{2 x}-8 e^x x+16 x^2) \log ^2(x)} \, dx\)

Optimal. Leaf size=31 \[ \frac {4 x (1+x)}{\left (4-\frac {e^x}{x}\right ) \left (-e^{e^2}+x+\log (x)\right )} \]

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Rubi [F]  time = 7.77, 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 {-16 x^2-16 x^3+16 x^4+e^x \left (4 x-4 x^3+4 x^4\right )+e^{e^2} \left (-16 x^2-32 x^3+e^x \left (8 x+8 x^2-4 x^3\right )\right )+\left (16 x^2+32 x^3+e^x \left (-8 x-8 x^2+4 x^3\right )\right ) \log (x)}{e^{2 x} x^2-8 e^x x^3+16 x^4+e^{2 e^2} \left (e^{2 x}-8 e^x x+16 x^2\right )+e^{e^2} \left (-2 e^{2 x} x+16 e^x x^2-32 x^3\right )+\left (2 e^{2 x} x-16 e^x x^2+32 x^3+e^{e^2} \left (-2 e^{2 x}+16 e^x x-32 x^2\right )\right ) \log (x)+\left (e^{2 x}-8 e^x x+16 x^2\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-16*x^2 - 16*x^3 + 16*x^4 + E^x*(4*x - 4*x^3 + 4*x^4) + E^E^2*(-16*x^2 - 32*x^3 + E^x*(8*x + 8*x^2 - 4*x^
3)) + (16*x^2 + 32*x^3 + E^x*(-8*x - 8*x^2 + 4*x^3))*Log[x])/(E^(2*x)*x^2 - 8*E^x*x^3 + 16*x^4 + E^(2*E^2)*(E^
(2*x) - 8*E^x*x + 16*x^2) + E^E^2*(-2*E^(2*x)*x + 16*E^x*x^2 - 32*x^3) + (2*E^(2*x)*x - 16*E^x*x^2 + 32*x^3 +
E^E^2*(-2*E^(2*x) + 16*E^x*x - 32*x^2))*Log[x] + (E^(2*x) - 8*E^x*x + 16*x^2)*Log[x]^2),x]

[Out]

4*(1 + 2*E^E^2)*Defer[Int][x/((E^x - 4*x)*(E^E^2 - x - Log[x])^2), x] + 8*E^E^2*Defer[Int][x^2/((E^x - 4*x)*(E
^E^2 - x - Log[x])^2), x] - 4*(1 + E^E^2)*Defer[Int][x^3/((E^x - 4*x)*(E^E^2 - x - Log[x])^2), x] + 4*Defer[In
t][x^4/((E^x - 4*x)*(-E^E^2 + x + Log[x])^2), x] - 8*Defer[Int][(x*Log[x])/((E^x - 4*x)*(-E^E^2 + x + Log[x])^
2), x] - 8*Defer[Int][(x^2*Log[x])/((E^x - 4*x)*(-E^E^2 + x + Log[x])^2), x] + 4*Defer[Int][(x^3*Log[x])/((E^x
- 4*x)*(-E^E^2 + x + Log[x])^2), x] - 16*Defer[Int][x^2/((E^x - 4*x)^2*(-E^E^2 + x + Log[x])), x] + 16*Defer[
Int][x^4/((E^x - 4*x)^2*(-E^E^2 + x + Log[x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 x \left (-4 e^{e^2} x (1+2 x)+e^{e^2+x} \left (2+2 x-x^2\right )+4 x \left (-1-x+x^2\right )+e^x \left (1-x^2+x^3\right )+\left (4 x (1+2 x)+e^x \left (-2-2 x+x^2\right )\right ) \log (x)\right )}{\left (e^x-4 x\right )^2 \left (e^{e^2}-x-\log (x)\right )^2} \, dx\\ &=4 \int \frac {x \left (-4 e^{e^2} x (1+2 x)+e^{e^2+x} \left (2+2 x-x^2\right )+4 x \left (-1-x+x^2\right )+e^x \left (1-x^2+x^3\right )+\left (4 x (1+2 x)+e^x \left (-2-2 x+x^2\right )\right ) \log (x)\right )}{\left (e^x-4 x\right )^2 \left (e^{e^2}-x-\log (x)\right )^2} \, dx\\ &=4 \int \left (\frac {4 x^2 \left (-1+x^2\right )}{\left (e^x-4 x\right )^2 \left (-e^{e^2}+x+\log (x)\right )}+\frac {x \left (1+2 e^{e^2}+2 e^{e^2} x-\left (1+e^{e^2}\right ) x^2+x^3-2 \log (x)-2 x \log (x)+x^2 \log (x)\right )}{\left (e^x-4 x\right ) \left (e^{e^2}-x-\log (x)\right )^2}\right ) \, dx\\ &=4 \int \frac {x \left (1+2 e^{e^2}+2 e^{e^2} x-\left (1+e^{e^2}\right ) x^2+x^3-2 \log (x)-2 x \log (x)+x^2 \log (x)\right )}{\left (e^x-4 x\right ) \left (e^{e^2}-x-\log (x)\right )^2} \, dx+16 \int \frac {x^2 \left (-1+x^2\right )}{\left (e^x-4 x\right )^2 \left (-e^{e^2}+x+\log (x)\right )} \, dx\\ &=4 \int \left (\frac {2 e^{e^2} \left (1+\frac {e^{-e^2}}{2}\right ) x}{\left (e^x-4 x\right ) \left (e^{e^2}-x-\log (x)\right )^2}+\frac {2 e^{e^2} x^2}{\left (e^x-4 x\right ) \left (e^{e^2}-x-\log (x)\right )^2}-\frac {\left (1+e^{e^2}\right ) x^3}{\left (e^x-4 x\right ) \left (e^{e^2}-x-\log (x)\right )^2}+\frac {x^4}{\left (e^x-4 x\right ) \left (-e^{e^2}+x+\log (x)\right )^2}-\frac {2 x \log (x)}{\left (e^x-4 x\right ) \left (-e^{e^2}+x+\log (x)\right )^2}-\frac {2 x^2 \log (x)}{\left (e^x-4 x\right ) \left (-e^{e^2}+x+\log (x)\right )^2}+\frac {x^3 \log (x)}{\left (e^x-4 x\right ) \left (-e^{e^2}+x+\log (x)\right )^2}\right ) \, dx+16 \int \left (-\frac {x^2}{\left (e^x-4 x\right )^2 \left (-e^{e^2}+x+\log (x)\right )}+\frac {x^4}{\left (e^x-4 x\right )^2 \left (-e^{e^2}+x+\log (x)\right )}\right ) \, dx\\ &=4 \int \frac {x^4}{\left (e^x-4 x\right ) \left (-e^{e^2}+x+\log (x)\right )^2} \, dx+4 \int \frac {x^3 \log (x)}{\left (e^x-4 x\right ) \left (-e^{e^2}+x+\log (x)\right )^2} \, dx-8 \int \frac {x \log (x)}{\left (e^x-4 x\right ) \left (-e^{e^2}+x+\log (x)\right )^2} \, dx-8 \int \frac {x^2 \log (x)}{\left (e^x-4 x\right ) \left (-e^{e^2}+x+\log (x)\right )^2} \, dx-16 \int \frac {x^2}{\left (e^x-4 x\right )^2 \left (-e^{e^2}+x+\log (x)\right )} \, dx+16 \int \frac {x^4}{\left (e^x-4 x\right )^2 \left (-e^{e^2}+x+\log (x)\right )} \, dx+\left (8 e^{e^2}\right ) \int \frac {x^2}{\left (e^x-4 x\right ) \left (e^{e^2}-x-\log (x)\right )^2} \, dx-\left (4 \left (1+e^{e^2}\right )\right ) \int \frac {x^3}{\left (e^x-4 x\right ) \left (e^{e^2}-x-\log (x)\right )^2} \, dx+\left (4 \left (1+2 e^{e^2}\right )\right ) \int \frac {x}{\left (e^x-4 x\right ) \left (e^{e^2}-x-\log (x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 32, normalized size = 1.03 \begin {gather*} \frac {4 x^2 (1+x)}{\left (-e^x+4 x\right ) \left (-e^{e^2}+x+\log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16*x^2 - 16*x^3 + 16*x^4 + E^x*(4*x - 4*x^3 + 4*x^4) + E^E^2*(-16*x^2 - 32*x^3 + E^x*(8*x + 8*x^2
- 4*x^3)) + (16*x^2 + 32*x^3 + E^x*(-8*x - 8*x^2 + 4*x^3))*Log[x])/(E^(2*x)*x^2 - 8*E^x*x^3 + 16*x^4 + E^(2*E^
2)*(E^(2*x) - 8*E^x*x + 16*x^2) + E^E^2*(-2*E^(2*x)*x + 16*E^x*x^2 - 32*x^3) + (2*E^(2*x)*x - 16*E^x*x^2 + 32*
x^3 + E^E^2*(-2*E^(2*x) + 16*E^x*x - 32*x^2))*Log[x] + (E^(2*x) - 8*E^x*x + 16*x^2)*Log[x]^2),x]

[Out]

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

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fricas [A]  time = 0.69, size = 46, normalized size = 1.48 \begin {gather*} \frac {4 \, {\left (x^{3} + x^{2}\right )}}{4 \, x^{2} - x e^{x} - {\left (4 \, x - e^{x}\right )} e^{\left (e^{2}\right )} + {\left (4 \, x - e^{x}\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x^3-8*x^2-8*x)*exp(x)+32*x^3+16*x^2)*log(x)+((-4*x^3+8*x^2+8*x)*exp(x)-32*x^3-16*x^2)*exp(exp(2
))+(4*x^4-4*x^3+4*x)*exp(x)+16*x^4-16*x^3-16*x^2)/((exp(x)^2-8*exp(x)*x+16*x^2)*log(x)^2+((-2*exp(x)^2+16*exp(
x)*x-32*x^2)*exp(exp(2))+2*x*exp(x)^2-16*exp(x)*x^2+32*x^3)*log(x)+(exp(x)^2-8*exp(x)*x+16*x^2)*exp(exp(2))^2+
(-2*x*exp(x)^2+16*exp(x)*x^2-32*x^3)*exp(exp(2))+exp(x)^2*x^2-8*exp(x)*x^3+16*x^4),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.31, size = 44, normalized size = 1.42 \begin {gather*} \frac {4 \, {\left (x^{3} + x^{2}\right )}}{4 \, x^{2} - x e^{x} - 4 \, x e^{\left (e^{2}\right )} + 4 \, x \log \relax (x) - e^{x} \log \relax (x) + e^{\left (x + e^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x^3-8*x^2-8*x)*exp(x)+32*x^3+16*x^2)*log(x)+((-4*x^3+8*x^2+8*x)*exp(x)-32*x^3-16*x^2)*exp(exp(2
))+(4*x^4-4*x^3+4*x)*exp(x)+16*x^4-16*x^3-16*x^2)/((exp(x)^2-8*exp(x)*x+16*x^2)*log(x)^2+((-2*exp(x)^2+16*exp(
x)*x-32*x^2)*exp(exp(2))+2*x*exp(x)^2-16*exp(x)*x^2+32*x^3)*log(x)+(exp(x)^2-8*exp(x)*x+16*x^2)*exp(exp(2))^2+
(-2*x*exp(x)^2+16*exp(x)*x^2-32*x^3)*exp(exp(2))+exp(x)^2*x^2-8*exp(x)*x^3+16*x^4),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.07, size = 32, normalized size = 1.03

method	result	size
risch	\(-\frac {4 \left (x +1\right ) x^{2}}{\left (4 x -{\mathrm e}^{x}\right ) \left ({\mathrm e}^{{\mathrm e}^{2}}-x -\ln \relax (x )\right )}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((4*x^3-8*x^2-8*x)*exp(x)+32*x^3+16*x^2)*ln(x)+((-4*x^3+8*x^2+8*x)*exp(x)-32*x^3-16*x^2)*exp(exp(2))+(4*x
^4-4*x^3+4*x)*exp(x)+16*x^4-16*x^3-16*x^2)/((exp(x)^2-8*exp(x)*x+16*x^2)*ln(x)^2+((-2*exp(x)^2+16*exp(x)*x-32*
x^2)*exp(exp(2))+2*x*exp(x)^2-16*exp(x)*x^2+32*x^3)*ln(x)+(exp(x)^2-8*exp(x)*x+16*x^2)*exp(exp(2))^2+(-2*x*exp
(x)^2+16*exp(x)*x^2-32*x^3)*exp(exp(2))+exp(x)^2*x^2-8*exp(x)*x^3+16*x^4),x,method=_RETURNVERBOSE)

[Out]

-4*(x+1)*x^2/(4*x-exp(x))/(exp(exp(2))-x-ln(x))

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maxima [A]  time = 0.51, size = 41, normalized size = 1.32 \begin {gather*} \frac {4 \, {\left (x^{3} + x^{2}\right )}}{4 \, x^{2} - {\left (x - e^{\left (e^{2}\right )} + \log \relax (x)\right )} e^{x} - 4 \, x e^{\left (e^{2}\right )} + 4 \, x \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x^3-8*x^2-8*x)*exp(x)+32*x^3+16*x^2)*log(x)+((-4*x^3+8*x^2+8*x)*exp(x)-32*x^3-16*x^2)*exp(exp(2
))+(4*x^4-4*x^3+4*x)*exp(x)+16*x^4-16*x^3-16*x^2)/((exp(x)^2-8*exp(x)*x+16*x^2)*log(x)^2+((-2*exp(x)^2+16*exp(
x)*x-32*x^2)*exp(exp(2))+2*x*exp(x)^2-16*exp(x)*x^2+32*x^3)*log(x)+(exp(x)^2-8*exp(x)*x+16*x^2)*exp(exp(2))^2+
(-2*x*exp(x)^2+16*exp(x)*x^2-32*x^3)*exp(exp(2))+exp(x)^2*x^2-8*exp(x)*x^3+16*x^4),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 9.47, size = 110, normalized size = 3.55 \begin {gather*} \frac {4\,x^2\,\left (4\,x-2\,{\mathrm {e}}^{x+{\mathrm {e}}^2}-{\mathrm {e}}^x+2\,{\mathrm {e}}^{{\mathrm {e}}^2}\,{\mathrm {e}}^x-x^2\,{\mathrm {e}}^x-2\,x\,{\mathrm {e}}^{x+{\mathrm {e}}^2}-2\,x\,{\mathrm {e}}^x+8\,x^2+4\,x^3+x^2\,{\mathrm {e}}^{x+{\mathrm {e}}^2}+2\,x\,{\mathrm {e}}^{{\mathrm {e}}^2}\,{\mathrm {e}}^x-x^2\,{\mathrm {e}}^{{\mathrm {e}}^2}\,{\mathrm {e}}^x\right )}{{\left (4\,x-{\mathrm {e}}^x\right )}^2\,\left (x+1\right )\,\left (x-{\mathrm {e}}^{{\mathrm {e}}^2}+\ln \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(16*x^2 + 16*x^3 - 16*x^4 + exp(exp(2))*(16*x^2 + 32*x^3 - exp(x)*(8*x + 8*x^2 - 4*x^3)) - exp(x)*(4*x -
4*x^3 + 4*x^4) - log(x)*(16*x^2 + 32*x^3 - exp(x)*(8*x + 8*x^2 - 4*x^3)))/(exp(2*exp(2))*(exp(2*x) - 8*x*exp(x
) + 16*x^2) - 8*x^3*exp(x) - log(x)*(exp(exp(2))*(2*exp(2*x) - 16*x*exp(x) + 32*x^2) - 2*x*exp(2*x) + 16*x^2*e
xp(x) - 32*x^3) + log(x)^2*(exp(2*x) - 8*x*exp(x) + 16*x^2) + x^2*exp(2*x) + 16*x^4 - exp(exp(2))*(2*x*exp(2*x
) - 16*x^2*exp(x) + 32*x^3)),x)

[Out]

(4*x^2*(4*x - 2*exp(x + exp(2)) - exp(x) + 2*exp(exp(2))*exp(x) - x^2*exp(x) - 2*x*exp(x + exp(2)) - 2*x*exp(x
) + 8*x^2 + 4*x^3 + x^2*exp(x + exp(2)) + 2*x*exp(exp(2))*exp(x) - x^2*exp(exp(2))*exp(x)))/((4*x - exp(x))^2*
(x + 1)*(x - exp(exp(2)) + log(x)))

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sympy [A]  time = 0.46, size = 44, normalized size = 1.42 \begin {gather*} \frac {- 4 x^{3} - 4 x^{2}}{- 4 x^{2} - 4 x \log {\relax (x )} + 4 x e^{e^{2}} + \left (x + \log {\relax (x )} - e^{e^{2}}\right ) e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x**3-8*x**2-8*x)*exp(x)+32*x**3+16*x**2)*ln(x)+((-4*x**3+8*x**2+8*x)*exp(x)-32*x**3-16*x**2)*ex
p(exp(2))+(4*x**4-4*x**3+4*x)*exp(x)+16*x**4-16*x**3-16*x**2)/((exp(x)**2-8*exp(x)*x+16*x**2)*ln(x)**2+((-2*ex
p(x)**2+16*exp(x)*x-32*x**2)*exp(exp(2))+2*x*exp(x)**2-16*exp(x)*x**2+32*x**3)*ln(x)+(exp(x)**2-8*exp(x)*x+16*
x**2)*exp(exp(2))**2+(-2*x*exp(x)**2+16*exp(x)*x**2-32*x**3)*exp(exp(2))+exp(x)**2*x**2-8*exp(x)*x**3+16*x**4)
,x)

[Out]

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

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