3.4.35 \(\int \frac {e^6-4 x^2+e^x x^2+\log (3)}{e^{12}+x^2+16 e^6 x^2+e^{2 x} x^2-8 x^3+16 x^4+e^6 (-2 x-8 e^3 x+8 x^2)+e^3 (8 x^2-32 x^3)+(2 e^6-2 x-8 e^3 x+8 x^2) \log (3)+\log ^2(3)+e^x (-2 e^6 x+2 x^2+8 e^3 x^2-8 x^3-2 x \log (3))} \, dx\)

Optimal. Leaf size=30 \[ \frac {x}{e^6-x+x \left (-e^x+4 \left (-e^3+x\right )\right )+\log (3)} \]

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Rubi [F]  time = 1.59, 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 {e^6-4 x^2+e^x x^2+\log (3)}{e^{12}+x^2+16 e^6 x^2+e^{2 x} x^2-8 x^3+16 x^4+e^6 \left (-2 x-8 e^3 x+8 x^2\right )+e^3 \left (8 x^2-32 x^3\right )+\left (2 e^6-2 x-8 e^3 x+8 x^2\right ) \log (3)+\log ^2(3)+e^x \left (-2 e^6 x+2 x^2+8 e^3 x^2-8 x^3-2 x \log (3)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^6 - 4*x^2 + E^x*x^2 + Log[3])/(E^12 + x^2 + 16*E^6*x^2 + E^(2*x)*x^2 - 8*x^3 + 16*x^4 + E^6*(-2*x - 8*E
^3*x + 8*x^2) + E^3*(8*x^2 - 32*x^3) + (2*E^6 - 2*x - 8*E^3*x + 8*x^2)*Log[3] + Log[3]^2 + E^x*(-2*E^6*x + 2*x
^2 + 8*E^3*x^2 - 8*x^3 - 2*x*Log[3])),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^6-4 x^2+e^x x^2+\log (3)}{e^{12}+e^{2 x} x^2+\left (1+16 e^6\right ) x^2-8 x^3+16 x^4+e^6 \left (-2 x-8 e^3 x+8 x^2\right )+e^3 \left (8 x^2-32 x^3\right )+\left (2 e^6-2 x-8 e^3 x+8 x^2\right ) \log (3)+\log ^2(3)+e^x \left (-2 e^6 x+2 x^2+8 e^3 x^2-8 x^3-2 x \log (3)\right )} \, dx\\ &=\int \frac {-4 x^2+e^x x^2+e^6 \left (1+\frac {\log (3)}{e^6}\right )}{\left (e^x x+\left (1+4 e^3\right ) x-4 x^2-e^6 \left (1+\frac {\log (3)}{e^6}\right )\right )^2} \, dx\\ &=\int \left (\frac {e^6-\left (5+4 e^3\right ) x^2+4 x^3+\log (3)+x \left (e^6+\log (3)\right )}{\left (e^x x+\left (1+4 e^3\right ) x-4 x^2-e^6 \left (1+\frac {\log (3)}{e^6}\right )\right )^2}+\frac {x}{e^x x+\left (1+4 e^3\right ) x-4 x^2-e^6 \left (1+\frac {\log (3)}{e^6}\right )}\right ) \, dx\\ &=\int \frac {e^6-\left (5+4 e^3\right ) x^2+4 x^3+\log (3)+x \left (e^6+\log (3)\right )}{\left (e^x x+\left (1+4 e^3\right ) x-4 x^2-e^6 \left (1+\frac {\log (3)}{e^6}\right )\right )^2} \, dx+\int \frac {x}{e^x x+\left (1+4 e^3\right ) x-4 x^2-e^6 \left (1+\frac {\log (3)}{e^6}\right )} \, dx\\ &=\int \frac {x}{e^x x+\left (1+4 e^3\right ) x-4 x^2-e^6 \left (1+\frac {\log (3)}{e^6}\right )} \, dx+\int \left (\frac {\left (-5-4 e^3\right ) x^2}{\left (e^x x+\left (1+4 e^3\right ) x-4 x^2-e^6 \left (1+\frac {\log (3)}{e^6}\right )\right )^2}+\frac {4 x^3}{\left (e^x x+\left (1+4 e^3\right ) x-4 x^2-e^6 \left (1+\frac {\log (3)}{e^6}\right )\right )^2}+\frac {x \left (e^6+\log (3)\right )}{\left (e^x x+\left (1+4 e^3\right ) x-4 x^2-e^6 \left (1+\frac {\log (3)}{e^6}\right )\right )^2}+\frac {e^6 \left (1+\frac {\log (3)}{e^6}\right )}{\left (e^x x+\left (1+4 e^3\right ) x-4 x^2-e^6 \left (1+\frac {\log (3)}{e^6}\right )\right )^2}\right ) \, dx\\ &=4 \int \frac {x^3}{\left (e^x x+\left (1+4 e^3\right ) x-4 x^2-e^6 \left (1+\frac {\log (3)}{e^6}\right )\right )^2} \, dx+\left (-5-4 e^3\right ) \int \frac {x^2}{\left (e^x x+\left (1+4 e^3\right ) x-4 x^2-e^6 \left (1+\frac {\log (3)}{e^6}\right )\right )^2} \, dx+\left (e^6+\log (3)\right ) \int \frac {1}{\left (e^x x+\left (1+4 e^3\right ) x-4 x^2-e^6 \left (1+\frac {\log (3)}{e^6}\right )\right )^2} \, dx+\left (e^6+\log (3)\right ) \int \frac {x}{\left (e^x x+\left (1+4 e^3\right ) x-4 x^2-e^6 \left (1+\frac {\log (3)}{e^6}\right )\right )^2} \, dx+\int \frac {x}{e^x x+\left (1+4 e^3\right ) x-4 x^2-e^6 \left (1+\frac {\log (3)}{e^6}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.50, size = 32, normalized size = 1.07 \begin {gather*} -\frac {x}{-e^6+x+4 e^3 x+e^x x-4 x^2-\log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^6 - 4*x^2 + E^x*x^2 + Log[3])/(E^12 + x^2 + 16*E^6*x^2 + E^(2*x)*x^2 - 8*x^3 + 16*x^4 + E^6*(-2*x
- 8*E^3*x + 8*x^2) + E^3*(8*x^2 - 32*x^3) + (2*E^6 - 2*x - 8*E^3*x + 8*x^2)*Log[3] + Log[3]^2 + E^x*(-2*E^6*x
+ 2*x^2 + 8*E^3*x^2 - 8*x^3 - 2*x*Log[3])),x]

[Out]

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

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fricas [A]  time = 0.79, size = 27, normalized size = 0.90 \begin {gather*} \frac {x}{4 \, x^{2} - 4 \, x e^{3} - x e^{x} - x + e^{6} + \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(4*x^2 - 4*x*e^3 - x*e^x - x + e^6 + log(3))

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giac [A]  time = 0.91, size = 27, normalized size = 0.90 \begin {gather*} \frac {x}{4 \, x^{2} - 4 \, x e^{3} - x e^{x} - x + e^{6} + \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(4*x^2 - 4*x*e^3 - x*e^x - x + e^6 + log(3))

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maple [A]  time = 0.38, size = 28, normalized size = 0.93

method	result	size
norman	\(\frac {x}{-4 x \,{\mathrm e}^{3}+4 x^{2}-{\mathrm e}^{x} x +\ln \relax (3)+{\mathrm e}^{6}-x}\)	\(28\)
risch	\(\frac {x}{-4 x \,{\mathrm e}^{3}+4 x^{2}-{\mathrm e}^{x} x +\ln \relax (3)+{\mathrm e}^{6}-x}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*x^2+ln(3)+exp(6)-4*x^2)/(exp(x)^2*x^2+(-2*x*ln(3)-2*x*exp(6)+8*x^2*exp(3)-8*x^3+2*x^2)*exp(x)+ln(3
)^2+(2*exp(6)-8*x*exp(3)+8*x^2-2*x)*ln(3)+exp(6)^2+(-8*x*exp(3)+8*x^2-2*x)*exp(6)+16*x^2*exp(3)^2+(-32*x^3+8*x
^2)*exp(3)+16*x^4-8*x^3+x^2),x,method=_RETURNVERBOSE)

[Out]

x/(-4*x*exp(3)+4*x^2-exp(x)*x+ln(3)+exp(6)-x)

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maxima [A]  time = 0.70, size = 28, normalized size = 0.93 \begin {gather*} \frac {x}{4 \, x^{2} - x {\left (4 \, e^{3} + 1\right )} - x e^{x} + e^{6} + \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(4*x^2 - x*(4*e^3 + 1) - x*e^x + e^6 + log(3))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\mathrm {e}}^6+\ln \relax (3)+x^2\,{\mathrm {e}}^x-4\,x^2}{{\mathrm {e}}^{12}-{\mathrm {e}}^x\,\left (2\,x\,{\mathrm {e}}^6+2\,x\,\ln \relax (3)-8\,x^2\,{\mathrm {e}}^3-2\,x^2+8\,x^3\right )-\ln \relax (3)\,\left (2\,x-2\,{\mathrm {e}}^6+8\,x\,{\mathrm {e}}^3-8\,x^2\right )-{\mathrm {e}}^6\,\left (2\,x+8\,x\,{\mathrm {e}}^3-8\,x^2\right )+x^2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^3\,\left (8\,x^2-32\,x^3\right )+16\,x^2\,{\mathrm {e}}^6+{\ln \relax (3)}^2+x^2-8\,x^3+16\,x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(6) + log(3) + x^2*exp(x) - 4*x^2)/(exp(12) - exp(x)*(2*x*exp(6) + 2*x*log(3) - 8*x^2*exp(3) - 2*x^2 +
8*x^3) - log(3)*(2*x - 2*exp(6) + 8*x*exp(3) - 8*x^2) - exp(6)*(2*x + 8*x*exp(3) - 8*x^2) + x^2*exp(2*x) + ex
p(3)*(8*x^2 - 32*x^3) + 16*x^2*exp(6) + log(3)^2 + x^2 - 8*x^3 + 16*x^4),x)

[Out]

int((exp(6) + log(3) + x^2*exp(x) - 4*x^2)/(exp(12) - exp(x)*(2*x*exp(6) + 2*x*log(3) - 8*x^2*exp(3) - 2*x^2 +
8*x^3) - log(3)*(2*x - 2*exp(6) + 8*x*exp(3) - 8*x^2) - exp(6)*(2*x + 8*x*exp(3) - 8*x^2) + x^2*exp(2*x) + ex
p(3)*(8*x^2 - 32*x^3) + 16*x^2*exp(6) + log(3)^2 + x^2 - 8*x^3 + 16*x^4), x)

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sympy [A]  time = 0.25, size = 27, normalized size = 0.90 \begin {gather*} - \frac {x}{- 4 x^{2} + x e^{x} + x + 4 x e^{3} - e^{6} - \log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)*x**2+ln(3)+exp(6)-4*x**2)/(exp(x)**2*x**2+(-2*x*ln(3)-2*x*exp(6)+8*x**2*exp(3)-8*x**3+2*x**2
)*exp(x)+ln(3)**2+(2*exp(6)-8*x*exp(3)+8*x**2-2*x)*ln(3)+exp(6)**2+(-8*x*exp(3)+8*x**2-2*x)*exp(6)+16*x**2*exp
(3)**2+(-32*x**3+8*x**2)*exp(3)+16*x**4-8*x**3+x**2),x)

[Out]

-x/(-4*x**2 + x*exp(x) + x + 4*x*exp(3) - exp(6) - log(3))

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