3.16.96 \(\int \frac {e^{\frac {2}{5} (5 e^{\frac {-2 e^2+x \log ^2(\frac {2+3 x}{x})}{x^2}}-x)} (-4 x^3-6 x^4+e^{\frac {-2 e^2+x \log ^2(\frac {2+3 x}{x})}{x^2}} (e^2 (80+120 x)-40 x \log (\frac {2+3 x}{x})+(-20 x-30 x^2) \log ^2(\frac {2+3 x}{x})))}{10 x^3+15 x^4} \, dx\)

Optimal. Leaf size=38 \[ e^{2 e^{\frac {-\frac {2 e^2}{x}+\log ^2\left (4-\frac {-2+x}{x}\right )}{x}}-\frac {2 x}{5}} \]

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Rubi [A]  time = 3.06, antiderivative size = 40, normalized size of antiderivative = 1.05, number of steps used = 2, number of rules used = 2, integrand size = 135, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.015, Rules used = {1593, 6706} \begin {gather*} \exp \left (\frac {2}{5} \left (5 e^{-\frac {2 e^2-x \log ^2\left (\frac {3 x+2}{x}\right )}{x^2}}-x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((2*(5*E^((-2*E^2 + x*Log[(2 + 3*x)/x]^2)/x^2) - x))/5)*(-4*x^3 - 6*x^4 + E^((-2*E^2 + x*Log[(2 + 3*x)/
x]^2)/x^2)*(E^2*(80 + 120*x) - 40*x*Log[(2 + 3*x)/x] + (-20*x - 30*x^2)*Log[(2 + 3*x)/x]^2)))/(10*x^3 + 15*x^4
),x]

[Out]

E^((2*(5/E^((2*E^2 - x*Log[(2 + 3*x)/x]^2)/x^2) - x))/5)

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - 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 {\exp \left (\frac {2}{5} \left (5 e^{\frac {-2 e^2+x \log ^2\left (\frac {2+3 x}{x}\right )}{x^2}}-x\right )\right ) \left (-4 x^3-6 x^4+e^{\frac {-2 e^2+x \log ^2\left (\frac {2+3 x}{x}\right )}{x^2}} \left (e^2 (80+120 x)-40 x \log \left (\frac {2+3 x}{x}\right )+\left (-20 x-30 x^2\right ) \log ^2\left (\frac {2+3 x}{x}\right )\right )\right )}{x^3 (10+15 x)} \, dx\\ &=\exp \left (\frac {2}{5} \left (5 e^{-\frac {2 e^2-x \log ^2\left (\frac {2+3 x}{x}\right )}{x^2}}-x\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.27, size = 35, normalized size = 0.92 \begin {gather*} e^{2 e^{-\frac {2 e^2}{x^2}+\frac {\log ^2\left (3+\frac {2}{x}\right )}{x}}-\frac {2 x}{5}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((2*(5*E^((-2*E^2 + x*Log[(2 + 3*x)/x]^2)/x^2) - x))/5)*(-4*x^3 - 6*x^4 + E^((-2*E^2 + x*Log[(2 +
3*x)/x]^2)/x^2)*(E^2*(80 + 120*x) - 40*x*Log[(2 + 3*x)/x] + (-20*x - 30*x^2)*Log[(2 + 3*x)/x]^2)))/(10*x^3 +
15*x^4),x]

[Out]

E^(2*E^((-2*E^2)/x^2 + Log[3 + 2/x]^2/x) - (2*x)/5)

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fricas [A]  time = 0.91, size = 31, normalized size = 0.82 \begin {gather*} e^{\left (-\frac {2}{5} \, x + 2 \, e^{\left (\frac {x \log \left (\frac {3 \, x + 2}{x}\right )^{2} - 2 \, e^{2}}{x^{2}}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-30*x^2-20*x)*log((3*x+2)/x)^2-40*x*log((3*x+2)/x)+(120*x+80)*exp(2))*exp((x*log((3*x+2)/x)^2-2*e
xp(2))/x^2)-6*x^4-4*x^3)*exp(exp((x*log((3*x+2)/x)^2-2*exp(2))/x^2)-1/5*x)^2/(15*x^4+10*x^3),x, algorithm="fri
cas")

[Out]

e^(-2/5*x + 2*e^((x*log((3*x + 2)/x)^2 - 2*e^2)/x^2))

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giac [A]  time = 23.86, size = 30, normalized size = 0.79 \begin {gather*} e^{\left (-\frac {2}{5} \, x + 2 \, e^{\left (\frac {\log \left (\frac {2}{x} + 3\right )^{2}}{x} - \frac {2 \, e^{2}}{x^{2}}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-30*x^2-20*x)*log((3*x+2)/x)^2-40*x*log((3*x+2)/x)+(120*x+80)*exp(2))*exp((x*log((3*x+2)/x)^2-2*e
xp(2))/x^2)-6*x^4-4*x^3)*exp(exp((x*log((3*x+2)/x)^2-2*exp(2))/x^2)-1/5*x)^2/(15*x^4+10*x^3),x, algorithm="gia
c")

[Out]

e^(-2/5*x + 2*e^(log(2/x + 3)^2/x - 2*e^2/x^2))

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maple [A]  time = 0.29, size = 34, normalized size = 0.89

method	result	size
risch	\({\mathrm e}^{2 \,{\mathrm e}^{-\frac {-x \ln \left (\frac {3 x +2}{x}\right )^{2}+2 \,{\mathrm e}^{2}}{x^{2}}}-\frac {2 x}{5}}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-30*x^2-20*x)*ln((3*x+2)/x)^2-40*x*ln((3*x+2)/x)+(120*x+80)*exp(2))*exp((x*ln((3*x+2)/x)^2-2*exp(2))/x^
2)-6*x^4-4*x^3)*exp(exp((x*ln((3*x+2)/x)^2-2*exp(2))/x^2)-1/5*x)^2/(15*x^4+10*x^3),x,method=_RETURNVERBOSE)

[Out]

exp(2*exp(-(-x*ln((3*x+2)/x)^2+2*exp(2))/x^2)-2/5*x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {2}{5} \, \int \frac {{\left (3 \, x^{4} + 2 \, x^{3} + 5 \, {\left ({\left (3 \, x^{2} + 2 \, x\right )} \log \left (\frac {3 \, x + 2}{x}\right )^{2} - 4 \, {\left (3 \, x + 2\right )} e^{2} + 4 \, x \log \left (\frac {3 \, x + 2}{x}\right )\right )} e^{\left (\frac {x \log \left (\frac {3 \, x + 2}{x}\right )^{2} - 2 \, e^{2}}{x^{2}}\right )}\right )} e^{\left (-\frac {2}{5} \, x + 2 \, e^{\left (\frac {x \log \left (\frac {3 \, x + 2}{x}\right )^{2} - 2 \, e^{2}}{x^{2}}\right )}\right )}}{3 \, x^{4} + 2 \, x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-30*x^2-20*x)*log((3*x+2)/x)^2-40*x*log((3*x+2)/x)+(120*x+80)*exp(2))*exp((x*log((3*x+2)/x)^2-2*e
xp(2))/x^2)-6*x^4-4*x^3)*exp(exp((x*log((3*x+2)/x)^2-2*exp(2))/x^2)-1/5*x)^2/(15*x^4+10*x^3),x, algorithm="max
ima")

[Out]

-2/5*integrate((3*x^4 + 2*x^3 + 5*((3*x^2 + 2*x)*log((3*x + 2)/x)^2 - 4*(3*x + 2)*e^2 + 4*x*log((3*x + 2)/x))*
e^((x*log((3*x + 2)/x)^2 - 2*e^2)/x^2))*e^(-2/5*x + 2*e^((x*log((3*x + 2)/x)^2 - 2*e^2)/x^2))/(3*x^4 + 2*x^3),
x)

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mupad [B]  time = 1.72, size = 33, normalized size = 0.87 \begin {gather*} {\mathrm {e}}^{-\frac {2\,x}{5}}\,{\mathrm {e}}^{2\,{\mathrm {e}}^{-\frac {2\,{\mathrm {e}}^2}{x^2}}\,{\mathrm {e}}^{\frac {{\ln \left (\frac {3\,x+2}{x}\right )}^2}{x}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2*exp(-(2*exp(2) - x*log((3*x + 2)/x)^2)/x^2) - (2*x)/5)*(exp(-(2*exp(2) - x*log((3*x + 2)/x)^2)/x^2
)*(log((3*x + 2)/x)^2*(20*x + 30*x^2) + 40*x*log((3*x + 2)/x) - exp(2)*(120*x + 80)) + 4*x^3 + 6*x^4))/(10*x^3
+ 15*x^4),x)

[Out]

exp(-(2*x)/5)*exp(2*exp(-(2*exp(2))/x^2)*exp(log((3*x + 2)/x)^2/x))

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sympy [A]  time = 3.50, size = 29, normalized size = 0.76 \begin {gather*} e^{- \frac {2 x}{5} + 2 e^{\frac {x \log {\left (\frac {3 x + 2}{x} \right )}^{2} - 2 e^{2}}{x^{2}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-30*x**2-20*x)*ln((3*x+2)/x)**2-40*x*ln((3*x+2)/x)+(120*x+80)*exp(2))*exp((x*ln((3*x+2)/x)**2-2*e
xp(2))/x**2)-6*x**4-4*x**3)*exp(exp((x*ln((3*x+2)/x)**2-2*exp(2))/x**2)-1/5*x)**2/(15*x**4+10*x**3),x)

[Out]

exp(-2*x/5 + 2*exp((x*log((3*x + 2)/x)**2 - 2*exp(2))/x**2))

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