3.64.43 \(\int \frac {-4 e x^2+16 x^3+4 x^4+e^{1+x} (-48 x^2+16 x^3)+12 e x^2 \log (x)}{16 e^{2+2 x}+4 x^2+4 x^3+x^4+e^{1+x} (-16 x-8 x^2)+(-8 e^{2+x}+e (4 x+2 x^2)) \log (x)+e^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=26 \[ \frac {4 x^2}{2+x \left (1+\frac {e \left (-4 e^x+\log (x)\right )}{x^2}\right )} \]

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Rubi [F]  time = 3.43, 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 e x^2+16 x^3+4 x^4+e^{1+x} \left (-48 x^2+16 x^3\right )+12 e x^2 \log (x)}{16 e^{2+2 x}+4 x^2+4 x^3+x^4+e^{1+x} \left (-16 x-8 x^2\right )+\left (-8 e^{2+x}+e \left (4 x+2 x^2\right )\right ) \log (x)+e^2 \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-4*E*x^2 + 16*x^3 + 4*x^4 + E^(1 + x)*(-48*x^2 + 16*x^3) + 12*E*x^2*Log[x])/(16*E^(2 + 2*x) + 4*x^2 + 4*x
^3 + x^4 + E^(1 + x)*(-16*x - 8*x^2) + (-8*E^(2 + x) + E*(4*x + 2*x^2))*Log[x] + E^2*Log[x]^2),x]

[Out]

-4*E*Defer[Int][x^2/(4*E^(1 + x) - 2*x - x^2 - E*Log[x])^2, x] - 8*Defer[Int][x^3/(-4*E^(1 + x) + 2*x + x^2 +
E*Log[x])^2, x] + 4*Defer[Int][x^5/(-4*E^(1 + x) + 2*x + x^2 + E*Log[x])^2, x] + 4*E*Defer[Int][(x^3*Log[x])/(
-4*E^(1 + x) + 2*x + x^2 + E*Log[x])^2, x] + 12*Defer[Int][x^2/(-4*E^(1 + x) + 2*x + x^2 + E*Log[x]), x] - 4*D
efer[Int][x^3/(-4*E^(1 + x) + 2*x + x^2 + E*Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 x^2 \left (-e+4 e^{1+x} (-3+x)+x (4+x)+3 e \log (x)\right )}{\left (4 e^{1+x}-x (2+x)-e \log (x)\right )^2} \, dx\\ &=4 \int \frac {x^2 \left (-e+4 e^{1+x} (-3+x)+x (4+x)+3 e \log (x)\right )}{\left (4 e^{1+x}-x (2+x)-e \log (x)\right )^2} \, dx\\ &=4 \int \left (-\frac {(-3+x) x^2}{-4 e^{1+x}+2 x+x^2+e \log (x)}+\frac {x^2 \left (-e-2 x+x^3+e x \log (x)\right )}{\left (-4 e^{1+x}+2 x+x^2+e \log (x)\right )^2}\right ) \, dx\\ &=-\left (4 \int \frac {(-3+x) x^2}{-4 e^{1+x}+2 x+x^2+e \log (x)} \, dx\right )+4 \int \frac {x^2 \left (-e-2 x+x^3+e x \log (x)\right )}{\left (-4 e^{1+x}+2 x+x^2+e \log (x)\right )^2} \, dx\\ &=4 \int \left (-\frac {e x^2}{\left (4 e^{1+x}-2 x-x^2-e \log (x)\right )^2}-\frac {2 x^3}{\left (-4 e^{1+x}+2 x+x^2+e \log (x)\right )^2}+\frac {x^5}{\left (-4 e^{1+x}+2 x+x^2+e \log (x)\right )^2}+\frac {e x^3 \log (x)}{\left (-4 e^{1+x}+2 x+x^2+e \log (x)\right )^2}\right ) \, dx-4 \int \left (-\frac {3 x^2}{-4 e^{1+x}+2 x+x^2+e \log (x)}+\frac {x^3}{-4 e^{1+x}+2 x+x^2+e \log (x)}\right ) \, dx\\ &=4 \int \frac {x^5}{\left (-4 e^{1+x}+2 x+x^2+e \log (x)\right )^2} \, dx-4 \int \frac {x^3}{-4 e^{1+x}+2 x+x^2+e \log (x)} \, dx-8 \int \frac {x^3}{\left (-4 e^{1+x}+2 x+x^2+e \log (x)\right )^2} \, dx+12 \int \frac {x^2}{-4 e^{1+x}+2 x+x^2+e \log (x)} \, dx-(4 e) \int \frac {x^2}{\left (4 e^{1+x}-2 x-x^2-e \log (x)\right )^2} \, dx+(4 e) \int \frac {x^3 \log (x)}{\left (-4 e^{1+x}+2 x+x^2+e \log (x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.58, size = 25, normalized size = 0.96 \begin {gather*} \frac {4 x^3}{-4 e^{1+x}+2 x+x^2+e \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4*E*x^2 + 16*x^3 + 4*x^4 + E^(1 + x)*(-48*x^2 + 16*x^3) + 12*E*x^2*Log[x])/(16*E^(2 + 2*x) + 4*x^2
+ 4*x^3 + x^4 + E^(1 + x)*(-16*x - 8*x^2) + (-8*E^(2 + x) + E*(4*x + 2*x^2))*Log[x] + E^2*Log[x]^2),x]

[Out]

(4*x^3)/(-4*E^(1 + x) + 2*x + x^2 + E*Log[x])

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fricas [A]  time = 0.59, size = 31, normalized size = 1.19 \begin {gather*} \frac {4 \, x^{3} e}{{\left (x^{2} + 2 \, x\right )} e + e^{2} \log \relax (x) - 4 \, e^{\left (x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*x^2*exp(1)*log(x)+(16*x^3-48*x^2)*exp(1)*exp(x)-4*x^2*exp(1)+4*x^4+16*x^3)/(exp(1)^2*log(x)^2+(-
8*exp(1)^2*exp(x)+(2*x^2+4*x)*exp(1))*log(x)+16*exp(1)^2*exp(x)^2+(-8*x^2-16*x)*exp(1)*exp(x)+x^4+4*x^3+4*x^2)
,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.19, size = 25, normalized size = 0.96 \begin {gather*} \frac {4 \, x^{3}}{x^{2} + e \log \relax (x) + 2 \, x - 4 \, e^{\left (x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*x^2*exp(1)*log(x)+(16*x^3-48*x^2)*exp(1)*exp(x)-4*x^2*exp(1)+4*x^4+16*x^3)/(exp(1)^2*log(x)^2+(-
8*exp(1)^2*exp(x)+(2*x^2+4*x)*exp(1))*log(x)+16*exp(1)^2*exp(x)^2+(-8*x^2-16*x)*exp(1)*exp(x)+x^4+4*x^3+4*x^2)
,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 29, normalized size = 1.12

method	result	size
risch	\(-\frac {4 x^{3}}{4 \,{\mathrm e}^{x +1}-{\mathrm e} \ln \relax (x )-x^{2}-2 x}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((12*x^2*exp(1)*ln(x)+(16*x^3-48*x^2)*exp(1)*exp(x)-4*x^2*exp(1)+4*x^4+16*x^3)/(exp(1)^2*ln(x)^2+(-8*exp(1)
^2*exp(x)+(2*x^2+4*x)*exp(1))*ln(x)+16*exp(1)^2*exp(x)^2+(-8*x^2-16*x)*exp(1)*exp(x)+x^4+4*x^3+4*x^2),x,method
=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.42, size = 25, normalized size = 0.96 \begin {gather*} \frac {4 \, x^{3}}{x^{2} + e \log \relax (x) + 2 \, x - 4 \, e^{\left (x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*x^2*exp(1)*log(x)+(16*x^3-48*x^2)*exp(1)*exp(x)-4*x^2*exp(1)+4*x^4+16*x^3)/(exp(1)^2*log(x)^2+(-
8*exp(1)^2*exp(x)+(2*x^2+4*x)*exp(1))*log(x)+16*exp(1)^2*exp(x)^2+(-8*x^2-16*x)*exp(1)*exp(x)+x^4+4*x^3+4*x^2)
,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {16\,x^3-4\,x^2\,\mathrm {e}-{\mathrm {e}}^{x+1}\,\left (48\,x^2-16\,x^3\right )+4\,x^4+12\,x^2\,\mathrm {e}\,\ln \relax (x)}{16\,{\mathrm {e}}^{2\,x+2}+{\mathrm {e}}^2\,{\ln \relax (x)}^2-{\mathrm {e}}^{x+1}\,\left (8\,x^2+16\,x\right )-\ln \relax (x)\,\left (8\,{\mathrm {e}}^{x+2}-\mathrm {e}\,\left (2\,x^2+4\,x\right )\right )+4\,x^2+4\,x^3+x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*x^3 - 4*x^2*exp(1) + 4*x^4 - exp(1)*exp(x)*(48*x^2 - 16*x^3) + 12*x^2*exp(1)*log(x))/(16*exp(2*x)*exp(
2) + exp(2)*log(x)^2 + log(x)*(exp(1)*(4*x + 2*x^2) - 8*exp(2)*exp(x)) + 4*x^2 + 4*x^3 + x^4 - exp(1)*exp(x)*(
16*x + 8*x^2)),x)

[Out]

int((16*x^3 - 4*x^2*exp(1) - exp(x + 1)*(48*x^2 - 16*x^3) + 4*x^4 + 12*x^2*exp(1)*log(x))/(16*exp(2*x + 2) + e
xp(2)*log(x)^2 - exp(x + 1)*(16*x + 8*x^2) - log(x)*(8*exp(x + 2) - exp(1)*(4*x + 2*x^2)) + 4*x^2 + 4*x^3 + x^
4), x)

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sympy [A]  time = 0.32, size = 27, normalized size = 1.04 \begin {gather*} - \frac {4 x^{3}}{- x^{2} - 2 x + 4 e e^{x} - e \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*x**2*exp(1)*ln(x)+(16*x**3-48*x**2)*exp(1)*exp(x)-4*x**2*exp(1)+4*x**4+16*x**3)/(exp(1)**2*ln(x)
**2+(-8*exp(1)**2*exp(x)+(2*x**2+4*x)*exp(1))*ln(x)+16*exp(1)**2*exp(x)**2+(-8*x**2-16*x)*exp(1)*exp(x)+x**4+4
*x**3+4*x**2),x)

[Out]

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

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