3.36.68 \(\int \frac {2 x+e^x (-2+4 x)+2 \log (-3+e^{2 e^4})}{-15 e^{3 x}-45 e^{2 x} x-45 e^x x^2-15 x^3+(45 e^{2 x}+90 e^x x+45 x^2) \log (-3+e^{2 e^4})+(-45 e^x-45 x) \log ^2(-3+e^{2 e^4})+15 \log ^3(-3+e^{2 e^4})} \, dx\)

Optimal. Leaf size=24 \[ \frac {2 x}{15 \left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^2} \]

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Rubi [F]  time = 0.83, 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 {2 x+e^x (-2+4 x)+2 \log \left (-3+e^{2 e^4}\right )}{-15 e^{3 x}-45 e^{2 x} x-45 e^x x^2-15 x^3+\left (45 e^{2 x}+90 e^x x+45 x^2\right ) \log \left (-3+e^{2 e^4}\right )+\left (-45 e^x-45 x\right ) \log ^2\left (-3+e^{2 e^4}\right )+15 \log ^3\left (-3+e^{2 e^4}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(2*x + E^x*(-2 + 4*x) + 2*Log[-3 + E^(2*E^4)])/(-15*E^(3*x) - 45*E^(2*x)*x - 45*E^x*x^2 - 15*x^3 + (45*E^(
2*x) + 90*E^x*x + 45*x^2)*Log[-3 + E^(2*E^4)] + (-45*E^x - 45*x)*Log[-3 + E^(2*E^4)]^2 + 15*Log[-3 + E^(2*E^4)
]^3),x]

[Out]

(-4*(1 + Log[-3 + E^(2*E^4)])*Defer[Int][x/(E^x + x - Log[-3 + E^(2*E^4)])^3, x])/15 + (4*Defer[Int][x^2/(E^x
+ x - Log[-3 + E^(2*E^4)])^3, x])/15 + (2*Defer[Int][(E^x + x - Log[-3 + E^(2*E^4)])^(-2), x])/15 - (4*Defer[I
nt][x/(E^x + x - Log[-3 + E^(2*E^4)])^2, x])/15

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \left (-x-e^x (-1+2 x)-\log \left (-3+e^{2 e^4}\right )\right )}{15 \left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^3} \, dx\\ &=\frac {2}{15} \int \frac {-x-e^x (-1+2 x)-\log \left (-3+e^{2 e^4}\right )}{\left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^3} \, dx\\ &=\frac {2}{15} \int \left (\frac {2 x \left (-1+x-\log \left (-3+e^{2 e^4}\right )\right )}{\left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^3}-\frac {-1+2 x}{\left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^2}\right ) \, dx\\ &=-\left (\frac {2}{15} \int \frac {-1+2 x}{\left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^2} \, dx\right )+\frac {4}{15} \int \frac {x \left (-1+x-\log \left (-3+e^{2 e^4}\right )\right )}{\left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^3} \, dx\\ &=-\left (\frac {2}{15} \int \left (-\frac {1}{\left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^2}+\frac {2 x}{\left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^2}\right ) \, dx\right )+\frac {4}{15} \int \left (\frac {x^2}{\left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^3}-\frac {x \left (1+\log \left (-3+e^{2 e^4}\right )\right )}{\left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^3}\right ) \, dx\\ &=\frac {2}{15} \int \frac {1}{\left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^2} \, dx+\frac {4}{15} \int \frac {x^2}{\left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^3} \, dx-\frac {4}{15} \int \frac {x}{\left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^2} \, dx-\frac {1}{15} \left (4 \left (1+\log \left (-3+e^{2 e^4}\right )\right )\right ) \int \frac {x}{\left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^3} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 24, normalized size = 1.00 \begin {gather*} \frac {2 x}{15 \left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*x + E^x*(-2 + 4*x) + 2*Log[-3 + E^(2*E^4)])/(-15*E^(3*x) - 45*E^(2*x)*x - 45*E^x*x^2 - 15*x^3 + (
45*E^(2*x) + 90*E^x*x + 45*x^2)*Log[-3 + E^(2*E^4)] + (-45*E^x - 45*x)*Log[-3 + E^(2*E^4)]^2 + 15*Log[-3 + E^(
2*E^4)]^3),x]

[Out]

(2*x)/(15*(E^x + x - Log[-3 + E^(2*E^4)])^2)

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fricas [B]  time = 0.70, size = 42, normalized size = 1.75 \begin {gather*} \frac {2 \, x}{15 \, {\left (x^{2} + 2 \, x e^{x} - 2 \, {\left (x + e^{x}\right )} \log \left (e^{\left (2 \, e^{4}\right )} - 3\right ) + \log \left (e^{\left (2 \, e^{4}\right )} - 3\right )^{2} + e^{\left (2 \, x\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(exp(exp(4))^2-3)+(4*x-2)*exp(x)+2*x)/(15*log(exp(exp(4))^2-3)^3+(-45*exp(x)-45*x)*log(exp(exp
(4))^2-3)^2+(45*exp(x)^2+90*exp(x)*x+45*x^2)*log(exp(exp(4))^2-3)-15*exp(x)^3-45*x*exp(x)^2-45*exp(x)*x^2-15*x
^3),x, algorithm="fricas")

[Out]

2/15*x/(x^2 + 2*x*e^x - 2*(x + e^x)*log(e^(2*e^4) - 3) + log(e^(2*e^4) - 3)^2 + e^(2*x))

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giac [B]  time = 0.18, size = 51, normalized size = 2.12 \begin {gather*} \frac {2 \, x}{15 \, {\left (x^{2} + 2 \, x e^{x} - 2 \, x \log \left (e^{\left (2 \, e^{4}\right )} - 3\right ) - 2 \, e^{x} \log \left (e^{\left (2 \, e^{4}\right )} - 3\right ) + \log \left (e^{\left (2 \, e^{4}\right )} - 3\right )^{2} + e^{\left (2 \, x\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(exp(exp(4))^2-3)+(4*x-2)*exp(x)+2*x)/(15*log(exp(exp(4))^2-3)^3+(-45*exp(x)-45*x)*log(exp(exp
(4))^2-3)^2+(45*exp(x)^2+90*exp(x)*x+45*x^2)*log(exp(exp(4))^2-3)-15*exp(x)^3-45*x*exp(x)^2-45*exp(x)*x^2-15*x
^3),x, algorithm="giac")

[Out]

2/15*x/(x^2 + 2*x*e^x - 2*x*log(e^(2*e^4) - 3) - 2*e^x*log(e^(2*e^4) - 3) + log(e^(2*e^4) - 3)^2 + e^(2*x))

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maple [A]  time = 0.28, size = 22, normalized size = 0.92

method	result	size
norman	\(\frac {2 x}{15 \left (\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{4}}-3\right )-x -{\mathrm e}^{x}\right )^{2}}\)	\(22\)
risch	\(\frac {2 x}{15 \left (\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{4}}-3\right )-x -{\mathrm e}^{x}\right )^{2}}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*ln(exp(exp(4))^2-3)+(4*x-2)*exp(x)+2*x)/(15*ln(exp(exp(4))^2-3)^3+(-45*exp(x)-45*x)*ln(exp(exp(4))^2-3)
^2+(45*exp(x)^2+90*exp(x)*x+45*x^2)*ln(exp(exp(4))^2-3)-15*exp(x)^3-45*x*exp(x)^2-45*exp(x)*x^2-15*x^3),x,meth
od=_RETURNVERBOSE)

[Out]

2/15*x/(ln(exp(exp(4))^2-3)-x-exp(x))^2

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maxima [B]  time = 0.53, size = 50, normalized size = 2.08 \begin {gather*} \frac {2 \, x}{15 \, {\left (x^{2} + 2 \, {\left (x - \log \left (e^{\left (2 \, e^{4}\right )} - 3\right )\right )} e^{x} - 2 \, x \log \left (e^{\left (2 \, e^{4}\right )} - 3\right ) + \log \left (e^{\left (2 \, e^{4}\right )} - 3\right )^{2} + e^{\left (2 \, x\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(exp(exp(4))^2-3)+(4*x-2)*exp(x)+2*x)/(15*log(exp(exp(4))^2-3)^3+(-45*exp(x)-45*x)*log(exp(exp
(4))^2-3)^2+(45*exp(x)^2+90*exp(x)*x+45*x^2)*log(exp(exp(4))^2-3)-15*exp(x)^3-45*x*exp(x)^2-45*exp(x)*x^2-15*x
^3),x, algorithm="maxima")

[Out]

2/15*x/(x^2 + 2*(x - log(e^(2*e^4) - 3))*e^x - 2*x*log(e^(2*e^4) - 3) + log(e^(2*e^4) - 3)^2 + e^(2*x))

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mupad [B]  time = 0.48, size = 57, normalized size = 2.38 \begin {gather*} \frac {2\,x}{15\,\left ({\mathrm {e}}^{2\,x}-2\,{\mathrm {e}}^x\,\ln \left ({\mathrm {e}}^{2\,{\mathrm {e}}^4}-3\right )+{\ln \left ({\mathrm {e}}^{2\,{\mathrm {e}}^4}-3\right )}^2+2\,x\,{\mathrm {e}}^x-2\,x\,\ln \left ({\mathrm {e}}^{2\,{\mathrm {e}}^4}-3\right )+x^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x + 2*log(exp(2*exp(4)) - 3) + exp(x)*(4*x - 2))/(15*exp(3*x) + 45*x*exp(2*x) + 45*x^2*exp(x) + log(ex
p(2*exp(4)) - 3)^2*(45*x + 45*exp(x)) - log(exp(2*exp(4)) - 3)*(45*exp(2*x) + 90*x*exp(x) + 45*x^2) - 15*log(e
xp(2*exp(4)) - 3)^3 + 15*x^3),x)

[Out]

(2*x)/(15*(exp(2*x) - 2*exp(x)*log(exp(2*exp(4)) - 3) + log(exp(2*exp(4)) - 3)^2 + 2*x*exp(x) - 2*x*log(exp(2*
exp(4)) - 3) + x^2))

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sympy [B]  time = 0.19, size = 60, normalized size = 2.50 \begin {gather*} \frac {2 x}{15 x^{2} - 30 x \log {\left (-3 + e^{2 e^{4}} \right )} + \left (30 x - 30 \log {\left (-3 + e^{2 e^{4}} \right )}\right ) e^{x} + 15 e^{2 x} + 15 \log {\left (-3 + e^{2 e^{4}} \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*ln(exp(exp(4))**2-3)+(4*x-2)*exp(x)+2*x)/(15*ln(exp(exp(4))**2-3)**3+(-45*exp(x)-45*x)*ln(exp(exp
(4))**2-3)**2+(45*exp(x)**2+90*exp(x)*x+45*x**2)*ln(exp(exp(4))**2-3)-15*exp(x)**3-45*x*exp(x)**2-45*exp(x)*x*
*2-15*x**3),x)

[Out]

2*x/(15*x**2 - 30*x*log(-3 + exp(2*exp(4))) + (30*x - 30*log(-3 + exp(2*exp(4))))*exp(x) + 15*exp(2*x) + 15*lo
g(-3 + exp(2*exp(4)))**2)

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