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

Optimal. Leaf size=26 \[ 6+\left (e^{\frac {\left (e^2+x\right ) \left (4+x^2\right )}{x}}-x\right )^2+\log (x) \]

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Rubi [F]  time = 0.37, 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+2 x^3+e^{\frac {2 \left (4 x+x^3+e^2 \left (4+x^2\right )\right )}{x}} \left (4 x^3+e^2 \left (-8+2 x^2\right )\right )+e^{\frac {4 x+x^3+e^2 \left (4+x^2\right )}{x}} \left (-2 x^2-4 x^4+e^2 \left (8 x-2 x^3\right )\right )}{x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 25, normalized size = 0.96 \begin {gather*} \left (e^{\frac {\left (e^2+x\right ) \left (4+x^2\right )}{x}}-x\right )^2+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.07, size = 50, normalized size = 1.92 \begin {gather*} x^{2} - 2 \, x e^{\left (\frac {x^{3} + {\left (x^{2} + 4\right )} e^{2} + 4 \, x}{x}\right )} + e^{\left (\frac {2 \, {\left (x^{3} + {\left (x^{2} + 4\right )} e^{2} + 4 \, x\right )}}{x}\right )} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.23, size = 54, normalized size = 2.08 \begin {gather*} x^{2} - 2 \, x e^{\left (\frac {x^{3} + x^{2} e^{2} + 4 \, x + 4 \, e^{2}}{x}\right )} + e^{\left (\frac {2 \, {\left (x^{3} + x^{2} e^{2} + 4 \, x + 4 \, e^{2}\right )}}{x}\right )} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.13, size = 39, normalized size = 1.50

method	result	size
risch	\(\ln \relax (x )+x^{2}+{\mathrm e}^{\frac {2 \left (x^{2}+4\right ) \left (x +{\mathrm e}^{2}\right )}{x}}-2 \,{\mathrm e}^{\frac {\left (x^{2}+4\right ) \left (x +{\mathrm e}^{2}\right )}{x}} x\)	\(39\)
norman	\(\frac {x^{3}+x \,{\mathrm e}^{\frac {2 \left (x^{2}+4\right ) {\mathrm e}^{2}+2 x^{3}+8 x}{x}}-2 x^{2} {\mathrm e}^{\frac {\left (x^{2}+4\right ) {\mathrm e}^{2}+x^{3}+4 x}{x}}}{x}+\ln \relax (x )\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.45, size = 46, normalized size = 1.77 \begin {gather*} x^{2} - 2 \, x e^{\left (x^{2} + x e^{2} + \frac {4 \, e^{2}}{x} + 4\right )} + e^{\left (2 \, x^{2} + 2 \, x e^{2} + \frac {8 \, e^{2}}{x} + 8\right )} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2 - 2*x*e^(x^2 + x*e^2 + 4*e^2/x + 4) + e^(2*x^2 + 2*x*e^2 + 8*e^2/x + 8) + log(x)

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mupad [B]  time = 5.25, size = 46, normalized size = 1.77 \begin {gather*} {\mathrm {e}}^{2\,x\,{\mathrm {e}}^2+\frac {8\,{\mathrm {e}}^2}{x}+2\,x^2+8}+\ln \relax (x)-2\,x\,{\mathrm {e}}^{x\,{\mathrm {e}}^2+\frac {4\,{\mathrm {e}}^2}{x}+x^2+4}+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 0.30, size = 49, normalized size = 1.88 \begin {gather*} x^{2} - 2 x e^{\frac {x^{3} + 4 x + \left (x^{2} + 4\right ) e^{2}}{x}} + e^{\frac {2 \left (x^{3} + 4 x + \left (x^{2} + 4\right ) e^{2}\right )}{x}} + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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