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

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

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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 = {} $$\begin{array}{c}\int \frac{\exp (e^{\frac{1+8x^2+2x^3+x^2\log (-2+x)}{x}}+\frac{1+8x^2+2x^3+x^2\log (-2+x)}{x})(2-x-16x^2+x^3+4x^4+(-2x^2+x^3)\log (-2+x))}{-2x^2+x^3}dx\end{array}$$

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{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{\exp (e^{\frac{1+8x^2+2x^3+x^2\log (-2+x)}{x}}+\frac{1+8x^2+2x^3+x^2\log (-2+x)}{x})(2-x-16x^2+x^3+4x^4+(-2x^2+x^3)\log (-2+x))}{(-2+x)x^2}dx\\ & =\int (\frac{\exp (e^{\frac{1+8x^2+2x^3+x^2\log (-2+x)}{x}}+\frac{1+8x^2+2x^3+x^2\log (-2+x)}{x})(2-x-16x^2+x^3+4x^4)}{(-2+x)x^2}+\exp (e^{\frac{1+8x^2+2x^3+x^2\log (-2+x)}{x}}+\frac{1+8x^2+2x^3+x^2\log (-2+x)}{x})\log (-2+x))dx\\ & =\int \frac{\exp (e^{\frac{1+8x^2+2x^3+x^2\log (-2+x)}{x}}+\frac{1+8x^2+2x^3+x^2\log (-2+x)}{x})(2-x-16x^2+x^3+4x^4)}{(-2+x)x^2}dx+\int \exp (e^{\frac{1+8x^2+2x^3+x^2\log (-2+x)}{x}}+\frac{1+8x^2+2x^3+x^2\log (-2+x)}{x})\log (-2+x)dx\\ & =\int \frac{\exp (e^{\frac{1}{x}+8x+2x^2}(-2+x{)}^x+\frac{1}{x}+8x+2x^2)(-2+x{)}^{-1+x}(2-x-16x^2+x^3+4x^4)}{x^2}dx+\int \exp (e^{\frac{1+8x^2+2x^3+x^2\log (-2+x)}{x}}+\frac{1+8x^2+2x^3+x^2\log (-2+x)}{x})\log (-2+x)dx\\ & =\int (-16\exp (e^{\frac{1}{x}+8x+2x^2}(-2+x{)}^x+\frac{1}{x}+8x+2x^2)(-2+x{)}^{-1+x}+\frac{2\exp (e^{\frac{1}{x}+8x+2x^2}(-2+x{)}^x+\frac{1}{x}+8x+2x^2)(-2+x{)}^{-1+x}}{x^2}-\frac{\exp (e^{\frac{1}{x}+8x+2x^2}(-2+x{)}^x+\frac{1}{x}+8x+2x^2)(-2+x{)}^{-1+x}}{x}+\exp (e^{\frac{1}{x}+8x+2x^2}(-2+x{)}^x+\frac{1}{x}+8x+2x^2)(-2+x{)}^{-1+x}x+4\exp (e^{\frac{1}{x}+8x+2x^2}(-2+x{)}^x+\frac{1}{x}+8x+2x^2)(-2+x{)}^{-1+x}{x}^2)dx+\int \exp (e^{\frac{1+8x^2+2x^3+x^2\log (-2+x)}{x}}+\frac{1+8x^2+2x^3+x^2\log (-2+x)}{x})\log (-2+x)dx\\ & =2\int \frac{\exp (e^{\frac{1}{x}+8x+2x^2}(-2+x{)}^x+\frac{1}{x}+8x+2x^2)(-2+x{)}^{-1+x}}{x^2}dx+4\int \exp (e^{\frac{1}{x}+8x+2x^2}(-2+x{)}^x+\frac{1}{x}+8x+2x^2)(-2+x{)}^{-1+x}{x}^{2}dx-16\int \exp (e^{\frac{1}{x}+8x+2x^2}(-2+x{)}^x+\frac{1}{x}+8x+2x^2)(-2+x{)}^{-1+x}dx-\int \frac{\exp (e^{\frac{1}{x}+8x+2x^2}(-2+x{)}^x+\frac{1}{x}+8x+2x^2)(-2+x{)}^{-1+x}}{x}dx+\int \exp (e^{\frac{1}{x}+8x+2x^2}(-2+x{)}^x+\frac{1}{x}+8x+2x^2)(-2+x{)}^{-1+x}xdx+\int \exp (e^{\frac{1+8x^2+2x^3+x^2\log (-2+x)}{x}}+\frac{1+8x^2+2x^3+x^2\log (-2+x)}{x})\log (-2+x)dx\end{array}\end{array}$$

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

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 $$\begin{array}{c}{e}^{(\frac{2x^3+x^2\log (x-2)+8x^2+x{e}^{\left(\frac{2x^3+x^2\log (x-2)+8x^2+1}{x}\right)}+1}{x}-\frac{2x^3+x^2\log (x-2)+8x^2+1}{x})}\end{array}$$

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 $$\begin{array}{c}\int \frac{(4x^4+x^3-16x^2+(x^3-2x^2)\log (x-2)-x+2)e^{(\frac{2x^3+x^2\log (x-2)+8x^2+1}{x}+e^{\left(\frac{2x^3+x^2\log (x-2)+8x^2+1}{x}\right)})}}{x^3-2x^2}dx\end{array}$$

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	${\mathrm{e}}^{{(x-2)}^x{\mathrm{e}}^{\frac{2x^3+8x^2+1}{x}}}$	$25$

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 $$\begin{array}{c}{e}^{\left(e^{(2x^2+x\log (x-2)+8x+\frac{1}{x})}\right)}\end{array}$$

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 $$\begin{array}{c}{\mathrm{e}}^{{\mathrm{e}}^{8x}{\mathrm{e}}^{1/x}{\mathrm{e}}^{2x^2}{(x-2)}^x}\end{array}$$

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 $$\begin{array}{c}{e}^{e^{\frac{2x^3+x^2\log (x-2)+8x^2+1}{x}}}\end{array}$$

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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