3.10.36 \(\int \frac {e^{\frac {4+4 x^2+4 e^{78} x^2+x^4+e^{39} (8 x+4 x^3)}{x^2}} (-8+2 x^4+e^{39} (-8 x+4 x^3))}{x^3} \, dx\)

Optimal. Leaf size=18 \[ -1+e^{\left (2 e^{39}+\frac {2}{x}+x\right )^2} \]

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Rubi [A]  time = 0.74, antiderivative size = 19, normalized size of antiderivative = 1.06, number of steps used = 3, number of rules used = 3, integrand size = 61, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {6688, 12, 6706} \begin {gather*} e^{\frac {\left (x^2+2 e^{39} x+2\right )^2}{x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((4 + 4*x^2 + 4*E^78*x^2 + x^4 + E^39*(8*x + 4*x^3))/x^2)*(-8 + 2*x^4 + E^39*(-8*x + 4*x^3)))/x^3,x]

[Out]

E^((2 + 2*E^39*x + x^2)^2/x^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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} &=\int \frac {2 e^{\frac {\left (2+2 e^{39} x+x^2\right )^2}{x^2}} \left (-4-4 e^{39} x+2 e^{39} x^3+x^4\right )}{x^3} \, dx\\ &=2 \int \frac {e^{\frac {\left (2+2 e^{39} x+x^2\right )^2}{x^2}} \left (-4-4 e^{39} x+2 e^{39} x^3+x^4\right )}{x^3} \, dx\\ &=e^{\frac {\left (2+2 e^{39} x+x^2\right )^2}{x^2}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 19, normalized size = 1.06 \begin {gather*} e^{\frac {\left (2+2 e^{39} x+x^2\right )^2}{x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((4 + 4*x^2 + 4*E^78*x^2 + x^4 + E^39*(8*x + 4*x^3))/x^2)*(-8 + 2*x^4 + E^39*(-8*x + 4*x^3)))/x^3
,x]

[Out]

E^((2 + 2*E^39*x + x^2)^2/x^2)

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fricas [B]  time = 1.03, size = 33, normalized size = 1.83 \begin {gather*} e^{\left (\frac {x^{4} + 4 \, x^{2} e^{78} + 4 \, x^{2} + 4 \, {\left (x^{3} + 2 \, x\right )} e^{39} + 4}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^3-8*x)*exp(39)+2*x^4-8)*exp((4*x^2*exp(39)^2+(4*x^3+8*x)*exp(39)+x^4+4*x^2+4)/x^2)/x^3,x, algo
rithm="fricas")

[Out]

e^((x^4 + 4*x^2*e^78 + 4*x^2 + 4*(x^3 + 2*x)*e^39 + 4)/x^2)

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giac [A]  time = 0.36, size = 27, normalized size = 1.50 \begin {gather*} e^{\left (x^{2} + 4 \, x e^{39} + \frac {8 \, e^{39}}{x} + \frac {4}{x^{2}} + 4 \, e^{78} + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^3-8*x)*exp(39)+2*x^4-8)*exp((4*x^2*exp(39)^2+(4*x^3+8*x)*exp(39)+x^4+4*x^2+4)/x^2)/x^3,x, algo
rithm="giac")

[Out]

e^(x^2 + 4*x*e^39 + 8*e^39/x + 4/x^2 + 4*e^78 + 4)

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maple [B]  time = 0.09, size = 35, normalized size = 1.94

method	result	size
risch	\({\mathrm e}^{\frac {4 \,{\mathrm e}^{39} x^{3}+x^{4}+4 x^{2} {\mathrm e}^{78}+8 \,{\mathrm e}^{39} x +4 x^{2}+4}{x^{2}}}\)	\(35\)
gosper	\({\mathrm e}^{\frac {4 \,{\mathrm e}^{39} x^{3}+x^{4}+4 x^{2} {\mathrm e}^{78}+8 \,{\mathrm e}^{39} x +4 x^{2}+4}{x^{2}}}\)	\(37\)
norman	\({\mathrm e}^{\frac {4 x^{2} {\mathrm e}^{78}+\left (4 x^{3}+8 x \right ) {\mathrm e}^{39}+x^{4}+4 x^{2}+4}{x^{2}}}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^3-8*x)*exp(39)+2*x^4-8)*exp((4*x^2*exp(39)^2+(4*x^3+8*x)*exp(39)+x^4+4*x^2+4)/x^2)/x^3,x,method=_RET
URNVERBOSE)

[Out]

exp((4*exp(39)*x^3+x^4+4*x^2*exp(78)+8*exp(39)*x+4*x^2+4)/x^2)

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maxima [A]  time = 1.20, size = 27, normalized size = 1.50 \begin {gather*} e^{\left (x^{2} + 4 \, x e^{39} + \frac {8 \, e^{39}}{x} + \frac {4}{x^{2}} + 4 \, e^{78} + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^3-8*x)*exp(39)+2*x^4-8)*exp((4*x^2*exp(39)^2+(4*x^3+8*x)*exp(39)+x^4+4*x^2+4)/x^2)/x^3,x, algo
rithm="maxima")

[Out]

e^(x^2 + 4*x*e^39 + 8*e^39/x + 4/x^2 + 4*e^78 + 4)

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mupad [B]  time = 0.79, size = 32, normalized size = 1.78 \begin {gather*} {\mathrm {e}}^{\frac {8\,{\mathrm {e}}^{39}}{x}}\,{\mathrm {e}}^{4\,{\mathrm {e}}^{78}}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^4\,{\mathrm {e}}^{\frac {4}{x^2}}\,{\mathrm {e}}^{4\,x\,{\mathrm {e}}^{39}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((exp(39)*(8*x + 4*x^3) + 4*x^2*exp(78) + 4*x^2 + x^4 + 4)/x^2)*(exp(39)*(8*x - 4*x^3) - 2*x^4 + 8))/
x^3,x)

[Out]

exp((8*exp(39))/x)*exp(4*exp(78))*exp(x^2)*exp(4)*exp(4/x^2)*exp(4*x*exp(39))

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sympy [B]  time = 0.21, size = 34, normalized size = 1.89 \begin {gather*} e^{\frac {x^{4} + 4 x^{2} + 4 x^{2} e^{78} + \left (4 x^{3} + 8 x\right ) e^{39} + 4}{x^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**3-8*x)*exp(39)+2*x**4-8)*exp((4*x**2*exp(39)**2+(4*x**3+8*x)*exp(39)+x**4+4*x**2+4)/x**2)/x**
3,x)

[Out]

exp((x**4 + 4*x**2 + 4*x**2*exp(78) + (4*x**3 + 8*x)*exp(39) + 4)/x**2)

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