3.87.49 \(\int \frac {e^{\frac {20-55 x+e^4 x+5 x^2-5 x^4}{x}} (-20+5 x^2-15 x^4)}{x^2} \, dx\)

Optimal. Leaf size=21 \[ e^{e^4-5 \left (11-\frac {4}{x}-x+x^3\right )} \]

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Rubi [A]  time = 0.25, antiderivative size = 26, normalized size of antiderivative = 1.24, number of steps used = 1, number of rules used = 1, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {6706} \begin {gather*} e^{\frac {-5 x^4+5 x^2+e^4 x-55 x+20}{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((20 - 55*x + E^4*x + 5*x^2 - 5*x^4)/x)*(-20 + 5*x^2 - 15*x^4))/x^2,x]

[Out]

E^((20 - 55*x + E^4*x + 5*x^2 - 5*x^4)/x)

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=e^{\frac {20-55 x+e^4 x+5 x^2-5 x^4}{x}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 20, normalized size = 0.95 \begin {gather*} e^{-55+e^4+\frac {20}{x}+5 x-5 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((20 - 55*x + E^4*x + 5*x^2 - 5*x^4)/x)*(-20 + 5*x^2 - 15*x^4))/x^2,x]

[Out]

E^(-55 + E^4 + 20/x + 5*x - 5*x^3)

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fricas [A]  time = 0.59, size = 26, normalized size = 1.24 \begin {gather*} e^{\left (-\frac {5 \, x^{4} - 5 \, x^{2} - x e^{4} + 55 \, x - 20}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-15*x^4+5*x^2-20)*exp((x*exp(4)-5*x^4+5*x^2-55*x+20)/x)/x^2,x, algorithm="fricas")

[Out]

e^(-(5*x^4 - 5*x^2 - x*e^4 + 55*x - 20)/x)

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giac [A]  time = 0.15, size = 18, normalized size = 0.86 \begin {gather*} e^{\left (-5 \, x^{3} + 5 \, x + \frac {20}{x} + e^{4} - 55\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-15*x^4+5*x^2-20)*exp((x*exp(4)-5*x^4+5*x^2-55*x+20)/x)/x^2,x, algorithm="giac")

[Out]

e^(-5*x^3 + 5*x + 20/x + e^4 - 55)

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maple [A]  time = 0.14, size = 25, normalized size = 1.19

method	result	size
gosper	\({\mathrm e}^{\frac {x \,{\mathrm e}^{4}-5 x^{4}+5 x^{2}-55 x +20}{x}}\)	\(25\)
norman	\({\mathrm e}^{\frac {x \,{\mathrm e}^{4}-5 x^{4}+5 x^{2}-55 x +20}{x}}\)	\(25\)
risch	\({\mathrm e}^{\frac {x \,{\mathrm e}^{4}-5 x^{4}+5 x^{2}-55 x +20}{x}}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-15*x^4+5*x^2-20)*exp((x*exp(4)-5*x^4+5*x^2-55*x+20)/x)/x^2,x,method=_RETURNVERBOSE)

[Out]

exp((x*exp(4)-5*x^4+5*x^2-55*x+20)/x)

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maxima [A]  time = 0.56, size = 18, normalized size = 0.86 \begin {gather*} e^{\left (-5 \, x^{3} + 5 \, x + \frac {20}{x} + e^{4} - 55\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-15*x^4+5*x^2-20)*exp((x*exp(4)-5*x^4+5*x^2-55*x+20)/x)/x^2,x, algorithm="maxima")

[Out]

e^(-5*x^3 + 5*x + 20/x + e^4 - 55)

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mupad [B]  time = 5.30, size = 22, normalized size = 1.05 \begin {gather*} {\mathrm {e}}^{5\,x}\,{\mathrm {e}}^{-55}\,{\mathrm {e}}^{-5\,x^3}\,{\mathrm {e}}^{20/x}\,{\mathrm {e}}^{{\mathrm {e}}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((x*exp(4) - 55*x + 5*x^2 - 5*x^4 + 20)/x)*(15*x^4 - 5*x^2 + 20))/x^2,x)

[Out]

exp(5*x)*exp(-55)*exp(-5*x^3)*exp(20/x)*exp(exp(4))

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sympy [A]  time = 0.14, size = 22, normalized size = 1.05 \begin {gather*} e^{\frac {- 5 x^{4} + 5 x^{2} - 55 x + x e^{4} + 20}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-15*x**4+5*x**2-20)*exp((x*exp(4)-5*x**4+5*x**2-55*x+20)/x)/x**2,x)

[Out]

exp((-5*x**4 + 5*x**2 - 55*x + x*exp(4) + 20)/x)

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