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

Optimal. Leaf size=18 \[ e^{e^{x \left (8+\frac {1}{x^2}+2 x+\log (-2+x)\right )}} \]

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Rubi [F]  time = 16.39, 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 {\exp \left (e^{\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}}+\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}\right ) \left (2-x-16 x^2+x^3+4 x^4+\left (-2 x^2+x^3\right ) \log (-2+x)\right )}{-2 x^2+x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 5.07, size = 22, normalized size = 1.22 \begin {gather*} e^{e^{\frac {1}{x}+8 x+2 x^2} (-2+x)^x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(E^(x^(-1) + 8*x + 2*x^2)*(-2 + x)^x)

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fricas [B]  time = 1.02, size = 78, normalized size = 4.33 \begin {gather*} e^{\left (\frac {2 \, x^{3} + x^{2} \log \left (x - 2\right ) + 8 \, x^{2} + x e^{\left (\frac {2 \, x^{3} + x^{2} \log \left (x - 2\right ) + 8 \, x^{2} + 1}{x}\right )} + 1}{x} - \frac {2 \, x^{3} + x^{2} \log \left (x - 2\right ) + 8 \, x^{2} + 1}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (4 \, x^{4} + x^{3} - 16 \, x^{2} + {\left (x^{3} - 2 \, x^{2}\right )} \log \left (x - 2\right ) - x + 2\right )} e^{\left (\frac {2 \, x^{3} + x^{2} \log \left (x - 2\right ) + 8 \, x^{2} + 1}{x} + e^{\left (\frac {2 \, x^{3} + x^{2} \log \left (x - 2\right ) + 8 \, x^{2} + 1}{x}\right )}\right )}}{x^{3} - 2 \, x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((4*x^4 + x^3 - 16*x^2 + (x^3 - 2*x^2)*log(x - 2) - x + 2)*e^((2*x^3 + x^2*log(x - 2) + 8*x^2 + 1)/x
+ e^((2*x^3 + x^2*log(x - 2) + 8*x^2 + 1)/x))/(x^3 - 2*x^2), x)

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maple [A]  time = 0.13, size = 25, normalized size = 1.39

method	result	size
risch	\({\mathrm e}^{\left (x -2\right )^{x} {\mathrm e}^{\frac {2 x^{3}+8 x^{2}+1}{x}}}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.47, size = 20, normalized size = 1.11 \begin {gather*} e^{\left (e^{\left (2 \, x^{2} + x \log \left (x - 2\right ) + 8 \, x + \frac {1}{x}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(e^(2*x^2 + x*log(x - 2) + 8*x + 1/x))

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mupad [B]  time = 4.54, size = 21, normalized size = 1.17 \begin {gather*} {\mathrm {e}}^{{\mathrm {e}}^{8\,x}\,{\mathrm {e}}^{1/x}\,{\mathrm {e}}^{2\,x^2}\,{\left (x-2\right )}^x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(exp(8*x)*exp(1/x)*exp(2*x^2)*(x - 2)^x)

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sympy [A]  time = 1.91, size = 24, normalized size = 1.33 \begin {gather*} e^{e^{\frac {2 x^{3} + x^{2} \log {\left (x - 2 \right )} + 8 x^{2} + 1}{x}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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