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

Optimal. Leaf size=21 \[ \frac {1}{e^x-e^{-4 \log ^2(x)}-x}+x \]

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Rubi [F]  time = 11.55, 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 {x+e^{2 x} x+e^{-8 \log ^2(x)} x+x^3+e^x \left (-x-2 x^2\right )+e^{-4 \log ^2(x)} \left (-2 e^x x+2 x^2-8 \log (x)\right )}{e^{2 x} x+e^{-8 \log ^2(x)} x-2 e^x x^2+x^3+e^{-4 \log ^2(x)} \left (-2 e^x x+2 x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.20, size = 36, normalized size = 1.71 \begin {gather*} x+\frac {e^{4 \log ^2(x)}}{-1+e^{x+4 \log ^2(x)}-e^{4 \log ^2(x)} x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.67, size = 35, normalized size = 1.67 \begin {gather*} \frac {x^{2} + x e^{\left (-4 \, \log \relax (x)^{2}\right )} - x e^{x} - 1}{x + e^{\left (-4 \, \log \relax (x)^{2}\right )} - e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 1.52, size = 607, normalized size = 28.90 \begin {gather*} \frac {8 \, x^{4} e^{\left (4 \, \log \relax (x)^{2}\right )} \log \relax (x) + x^{4} e^{\left (4 \, \log \relax (x)^{2}\right )} - x^{4} e^{\left (4 \, \log \relax (x)^{2} + x\right )} - 24 \, x^{3} e^{\left (4 \, \log \relax (x)^{2} + x\right )} \log \relax (x) + 2 \, x^{3} e^{\left (4 \, \log \relax (x)^{2} + 2 \, x\right )} - 2 \, x^{3} e^{\left (4 \, \log \relax (x)^{2} + x\right )} - 2 \, x^{3} e^{x} + 16 \, x^{3} \log \relax (x) - 8 \, x^{2} e^{\left (4 \, \log \relax (x)^{2}\right )} \log \relax (x) + 8 \, x^{2} e^{\left (-4 \, \log \relax (x)^{2}\right )} \log \relax (x) + 24 \, x^{2} e^{\left (4 \, \log \relax (x)^{2} + 2 \, x\right )} \log \relax (x) - 32 \, x^{2} e^{x} \log \relax (x) + 2 \, x^{3} - x^{2} e^{\left (4 \, \log \relax (x)^{2}\right )} + x^{2} e^{\left (-4 \, \log \relax (x)^{2}\right )} - x^{2} e^{\left (4 \, \log \relax (x)^{2} + 3 \, x\right )} + x^{2} e^{\left (4 \, \log \relax (x)^{2} + 2 \, x\right )} + x^{2} e^{\left (4 \, \log \relax (x)^{2} + x\right )} - x^{2} e^{\left (-4 \, \log \relax (x)^{2} + x\right )} + 2 \, x^{2} e^{\left (2 \, x\right )} - 2 \, x^{2} e^{x} - 8 \, x e^{\left (4 \, \log \relax (x)^{2} + 3 \, x\right )} \log \relax (x) + 16 \, x e^{\left (4 \, \log \relax (x)^{2} + x\right )} \log \relax (x) - 8 \, x e^{\left (-4 \, \log \relax (x)^{2} + x\right )} \log \relax (x) + 16 \, x e^{\left (2 \, x\right )} \log \relax (x) - x e^{\left (4 \, \log \relax (x)^{2} + 2 \, x\right )} + x e^{\left (4 \, \log \relax (x)^{2} + x\right )} + x e^{x} - 8 \, x \log \relax (x) - 8 \, e^{\left (4 \, \log \relax (x)^{2} + 2 \, x\right )} \log \relax (x) + 8 \, e^{x} \log \relax (x) - x}{8 \, x^{3} e^{\left (4 \, \log \relax (x)^{2}\right )} \log \relax (x) + x^{3} e^{\left (4 \, \log \relax (x)^{2}\right )} - x^{3} e^{\left (4 \, \log \relax (x)^{2} + x\right )} - 24 \, x^{2} e^{\left (4 \, \log \relax (x)^{2} + x\right )} \log \relax (x) + 2 \, x^{2} e^{\left (4 \, \log \relax (x)^{2} + 2 \, x\right )} - 2 \, x^{2} e^{\left (4 \, \log \relax (x)^{2} + x\right )} - 2 \, x^{2} e^{x} + 16 \, x^{2} \log \relax (x) + 8 \, x e^{\left (-4 \, \log \relax (x)^{2}\right )} \log \relax (x) + 24 \, x e^{\left (4 \, \log \relax (x)^{2} + 2 \, x\right )} \log \relax (x) - 32 \, x e^{x} \log \relax (x) + 2 \, x^{2} + x e^{\left (-4 \, \log \relax (x)^{2}\right )} - x e^{\left (4 \, \log \relax (x)^{2} + 3 \, x\right )} + x e^{\left (4 \, \log \relax (x)^{2} + 2 \, x\right )} - x e^{\left (-4 \, \log \relax (x)^{2} + x\right )} + 2 \, x e^{\left (2 \, x\right )} - 2 \, x e^{x} - 8 \, e^{\left (4 \, \log \relax (x)^{2} + 3 \, x\right )} \log \relax (x) - 8 \, e^{\left (-4 \, \log \relax (x)^{2} + x\right )} \log \relax (x) + 16 \, e^{\left (2 \, x\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(8*x^4*e^(4*log(x)^2)*log(x) + x^4*e^(4*log(x)^2) - x^4*e^(4*log(x)^2 + x) - 24*x^3*e^(4*log(x)^2 + x)*log(x)
+ 2*x^3*e^(4*log(x)^2 + 2*x) - 2*x^3*e^(4*log(x)^2 + x) - 2*x^3*e^x + 16*x^3*log(x) - 8*x^2*e^(4*log(x)^2)*log
(x) + 8*x^2*e^(-4*log(x)^2)*log(x) + 24*x^2*e^(4*log(x)^2 + 2*x)*log(x) - 32*x^2*e^x*log(x) + 2*x^3 - x^2*e^(4
*log(x)^2) + x^2*e^(-4*log(x)^2) - x^2*e^(4*log(x)^2 + 3*x) + x^2*e^(4*log(x)^2 + 2*x) + x^2*e^(4*log(x)^2 + x
) - x^2*e^(-4*log(x)^2 + x) + 2*x^2*e^(2*x) - 2*x^2*e^x - 8*x*e^(4*log(x)^2 + 3*x)*log(x) + 16*x*e^(4*log(x)^2
+ x)*log(x) - 8*x*e^(-4*log(x)^2 + x)*log(x) + 16*x*e^(2*x)*log(x) - x*e^(4*log(x)^2 + 2*x) + x*e^(4*log(x)^2
+ x) + x*e^x - 8*x*log(x) - 8*e^(4*log(x)^2 + 2*x)*log(x) + 8*e^x*log(x) - x)/(8*x^3*e^(4*log(x)^2)*log(x) +
x^3*e^(4*log(x)^2) - x^3*e^(4*log(x)^2 + x) - 24*x^2*e^(4*log(x)^2 + x)*log(x) + 2*x^2*e^(4*log(x)^2 + 2*x) -
2*x^2*e^(4*log(x)^2 + x) - 2*x^2*e^x + 16*x^2*log(x) + 8*x*e^(-4*log(x)^2)*log(x) + 24*x*e^(4*log(x)^2 + 2*x)*
log(x) - 32*x*e^x*log(x) + 2*x^2 + x*e^(-4*log(x)^2) - x*e^(4*log(x)^2 + 3*x) + x*e^(4*log(x)^2 + 2*x) - x*e^(
-4*log(x)^2 + x) + 2*x*e^(2*x) - 2*x*e^x - 8*e^(4*log(x)^2 + 3*x)*log(x) - 8*e^(-4*log(x)^2 + x)*log(x) + 16*e
^(2*x)*log(x))

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maple [A]  time = 0.04, size = 20, normalized size = 0.95

method	result	size
risch	\(x -\frac {1}{{\mathrm e}^{-4 \ln \relax (x )^{2}}-{\mathrm e}^{x}+x}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.64, size = 39, normalized size = 1.86 \begin {gather*} \frac {{\left (x^{2} - x e^{x} - 1\right )} e^{\left (4 \, \log \relax (x)^{2}\right )} + x}{{\left (x - e^{x}\right )} e^{\left (4 \, \log \relax (x)^{2}\right )} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.23, size = 19, normalized size = 0.90 \begin {gather*} x-\frac {1}{x+{\mathrm {e}}^{-4\,{\ln \relax (x)}^2}-{\mathrm {e}}^x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - 1/(x + exp(-4*log(x)^2) - exp(x))

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sympy [A]  time = 0.36, size = 17, normalized size = 0.81 \begin {gather*} x + \frac {1}{- x + e^{x} - e^{- 4 \log {\relax (x )}^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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