3.88.28 \(\int \frac {e^{2+\frac {e^2}{3 x+12 x^2}} (1+8 x)}{3 x^2+24 x^3+48 x^4} \, dx\)

Optimal. Leaf size=28 \[ 5 e^{e^3}-e^{\frac {e^2}{3 \left (x+4 x^2\right )}} \]

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Rubi [A]  time = 0.33, antiderivative size = 20, normalized size of antiderivative = 0.71, number of steps used = 4, number of rules used = 4, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {1594, 27, 12, 6706} \begin {gather*} -e^{\frac {e^2}{3 \left (4 x^2+x\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(2 + E^2/(3*x + 12*x^2))*(1 + 8*x))/(3*x^2 + 24*x^3 + 48*x^4),x]

[Out]

-E^(E^2/(3*(x + 4*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 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

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 {e^{2+\frac {e^2}{3 x+12 x^2}} (1+8 x)}{x^2 \left (3+24 x+48 x^2\right )} \, dx\\ &=\int \frac {e^{2+\frac {e^2}{3 x+12 x^2}} (1+8 x)}{3 x^2 (1+4 x)^2} \, dx\\ &=\frac {1}{3} \int \frac {e^{2+\frac {e^2}{3 x+12 x^2}} (1+8 x)}{x^2 (1+4 x)^2} \, dx\\ &=-e^{\frac {e^2}{3 \left (x+4 x^2\right )}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 19, normalized size = 0.68 \begin {gather*} -e^{\frac {e^2}{3 x+12 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2 + E^2/(3*x + 12*x^2))*(1 + 8*x))/(3*x^2 + 24*x^3 + 48*x^4),x]

[Out]

-E^(E^2/(3*x + 12*x^2))

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fricas [A]  time = 0.60, size = 27, normalized size = 0.96 \begin {gather*} -e^{\left (\frac {24 \, x^{2} + 6 \, x + e^{2}}{3 \, {\left (4 \, x^{2} + x\right )}} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x+1)*exp(2)*exp(exp(2)/(12*x^2+3*x))/(48*x^4+24*x^3+3*x^2),x, algorithm="fricas")

[Out]

-e^(1/3*(24*x^2 + 6*x + e^2)/(4*x^2 + x) - 2)

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giac [A]  time = 0.16, size = 44, normalized size = 1.57 \begin {gather*} -e^{\left (\frac {8 \, x^{2}}{4 \, x^{2} + x} + \frac {2 \, x}{4 \, x^{2} + x} + \frac {e^{2}}{3 \, {\left (4 \, x^{2} + x\right )}} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x+1)*exp(2)*exp(exp(2)/(12*x^2+3*x))/(48*x^4+24*x^3+3*x^2),x, algorithm="giac")

[Out]

-e^(8*x^2/(4*x^2 + x) + 2*x/(4*x^2 + x) + 1/3*e^2/(4*x^2 + x) - 2)

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maple [A]  time = 0.14, size = 18, normalized size = 0.64

method	result	size
gosper	\(-{\mathrm e}^{\frac {{\mathrm e}^{2}}{3 x \left (4 x +1\right )}}\)	\(18\)
risch	\(-{\mathrm e}^{\frac {{\mathrm e}^{2}}{3 x \left (4 x +1\right )}}\)	\(18\)
norman	\(\frac {-x \,{\mathrm e}^{\frac {{\mathrm e}^{2}}{12 x^{2}+3 x}}-4 x^{2} {\mathrm e}^{\frac {{\mathrm e}^{2}}{12 x^{2}+3 x}}}{x \left (4 x +1\right )}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x+1)*exp(2)*exp(exp(2)/(12*x^2+3*x))/(48*x^4+24*x^3+3*x^2),x,method=_RETURNVERBOSE)

[Out]

-exp(1/3*exp(2)/x/(4*x+1))

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maxima [A]  time = 0.45, size = 22, normalized size = 0.79 \begin {gather*} -e^{\left (-\frac {4 \, e^{2}}{3 \, {\left (4 \, x + 1\right )}} + \frac {e^{2}}{3 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x+1)*exp(2)*exp(exp(2)/(12*x^2+3*x))/(48*x^4+24*x^3+3*x^2),x, algorithm="maxima")

[Out]

-e^(-4/3*e^2/(4*x + 1) + 1/3*e^2/x)

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mupad [B]  time = 5.63, size = 17, normalized size = 0.61 \begin {gather*} -{\mathrm {e}}^{\frac {{\mathrm {e}}^2}{12\,x^2+3\,x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2)*exp(exp(2)/(3*x + 12*x^2))*(8*x + 1))/(3*x^2 + 24*x^3 + 48*x^4),x)

[Out]

-exp(exp(2)/(3*x + 12*x^2))

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sympy [A]  time = 0.24, size = 14, normalized size = 0.50 \begin {gather*} - e^{\frac {e^{2}}{12 x^{2} + 3 x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x+1)*exp(2)*exp(exp(2)/(12*x**2+3*x))/(48*x**4+24*x**3+3*x**2),x)

[Out]

-exp(exp(2)/(12*x**2 + 3*x))

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