3.2.52 \(\int \frac {e^{\frac {4 x^3}{11 x-2 e^2 x-2 x^2+2 \log (x)}} (-8 x^2+88 x^3-16 e^2 x^3-8 x^4+24 x^2 \log (x))}{121 x^2+4 e^4 x^2-44 x^3+4 x^4+e^2 (-44 x^2+8 x^3)+(44 x-8 e^2 x-8 x^2) \log (x)+4 \log ^2(x)} \, dx\)

Optimal. Leaf size=33 \[ -\frac {4}{e^2}+e^{\frac {2 x^2}{\frac {11}{2}-e^2-x+\frac {\log (x)}{x}}} \]

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Rubi [A]  time = 1.36, antiderivative size = 26, normalized size of antiderivative = 0.79, number of steps used = 5, number of rules used = 4, integrand size = 125, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {6, 6688, 12, 6706} \begin {gather*} \exp \left (\frac {4 x^3}{\left (-2 x-2 e^2+11\right ) x+2 \log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((4*x^3)/(11*x - 2*E^2*x - 2*x^2 + 2*Log[x]))*(-8*x^2 + 88*x^3 - 16*E^2*x^3 - 8*x^4 + 24*x^2*Log[x]))/(
121*x^2 + 4*E^4*x^2 - 44*x^3 + 4*x^4 + E^2*(-44*x^2 + 8*x^3) + (44*x - 8*E^2*x - 8*x^2)*Log[x] + 4*Log[x]^2),x
]

[Out]

E^((4*x^3)/((11 - 2*E^2 - 2*x)*x + 2*Log[x]))

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

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 {e^{\frac {4 x^3}{11 x-2 e^2 x-2 x^2+2 \log (x)}} \left (-8 x^2+\left (88-16 e^2\right ) x^3-8 x^4+24 x^2 \log (x)\right )}{121 x^2+4 e^4 x^2-44 x^3+4 x^4+e^2 \left (-44 x^2+8 x^3\right )+\left (44 x-8 e^2 x-8 x^2\right ) \log (x)+4 \log ^2(x)} \, dx\\ &=\int \frac {e^{\frac {4 x^3}{11 x-2 e^2 x-2 x^2+2 \log (x)}} \left (-8 x^2+\left (88-16 e^2\right ) x^3-8 x^4+24 x^2 \log (x)\right )}{\left (121+4 e^4\right ) x^2-44 x^3+4 x^4+e^2 \left (-44 x^2+8 x^3\right )+\left (44 x-8 e^2 x-8 x^2\right ) \log (x)+4 \log ^2(x)} \, dx\\ &=\int \frac {8 \exp \left (-\frac {4 x^3}{x \left (-11+2 e^2+2 x\right )-2 \log (x)}\right ) x^2 \left (-1-\left (-11+2 e^2\right ) x-x^2+3 \log (x)\right )}{\left (x \left (-11+2 e^2+2 x\right )-2 \log (x)\right )^2} \, dx\\ &=8 \int \frac {\exp \left (-\frac {4 x^3}{x \left (-11+2 e^2+2 x\right )-2 \log (x)}\right ) x^2 \left (-1-\left (-11+2 e^2\right ) x-x^2+3 \log (x)\right )}{\left (x \left (-11+2 e^2+2 x\right )-2 \log (x)\right )^2} \, dx\\ &=e^{\frac {4 x^3}{\left (11-2 e^2-2 x\right ) x+2 \log (x)}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.97, size = 28, normalized size = 0.85 \begin {gather*} e^{\frac {4 x^3}{11 x-2 e^2 x-2 x^2+2 \log (x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((4*x^3)/(11*x - 2*E^2*x - 2*x^2 + 2*Log[x]))*(-8*x^2 + 88*x^3 - 16*E^2*x^3 - 8*x^4 + 24*x^2*Log[
x]))/(121*x^2 + 4*E^4*x^2 - 44*x^3 + 4*x^4 + E^2*(-44*x^2 + 8*x^3) + (44*x - 8*E^2*x - 8*x^2)*Log[x] + 4*Log[x
]^2),x]

[Out]

E^((4*x^3)/(11*x - 2*E^2*x - 2*x^2 + 2*Log[x]))

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fricas [A]  time = 0.55, size = 26, normalized size = 0.79 \begin {gather*} e^{\left (-\frac {4 \, x^{3}}{2 \, x^{2} + 2 \, x e^{2} - 11 \, x - 2 \, \log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((24*x^2*log(x)-16*x^3*exp(2)-8*x^4+88*x^3-8*x^2)*exp(2*x^3/(2*log(x)-2*exp(2)*x-2*x^2+11*x))^2/(4*lo
g(x)^2+(-8*exp(2)*x-8*x^2+44*x)*log(x)+4*x^2*exp(2)^2+(8*x^3-44*x^2)*exp(2)+4*x^4-44*x^3+121*x^2),x, algorithm
="fricas")

[Out]

e^(-4*x^3/(2*x^2 + 2*x*e^2 - 11*x - 2*log(x)))

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giac [A]  time = 0.68, size = 26, normalized size = 0.79 \begin {gather*} e^{\left (-\frac {4 \, x^{3}}{2 \, x^{2} + 2 \, x e^{2} - 11 \, x - 2 \, \log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((24*x^2*log(x)-16*x^3*exp(2)-8*x^4+88*x^3-8*x^2)*exp(2*x^3/(2*log(x)-2*exp(2)*x-2*x^2+11*x))^2/(4*lo
g(x)^2+(-8*exp(2)*x-8*x^2+44*x)*log(x)+4*x^2*exp(2)^2+(8*x^3-44*x^2)*exp(2)+4*x^4-44*x^3+121*x^2),x, algorithm
="giac")

[Out]

e^(-4*x^3/(2*x^2 + 2*x*e^2 - 11*x - 2*log(x)))

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maple [A]  time = 0.04, size = 27, normalized size = 0.82

method	result	size
risch	\({\mathrm e}^{-\frac {4 x^{3}}{2 \,{\mathrm e}^{2} x +2 x^{2}-2 \ln \relax (x )-11 x}}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((24*x^2*ln(x)-16*x^3*exp(2)-8*x^4+88*x^3-8*x^2)*exp(2*x^3/(2*ln(x)-2*exp(2)*x-2*x^2+11*x))^2/(4*ln(x)^2+(-
8*exp(2)*x-8*x^2+44*x)*ln(x)+4*x^2*exp(2)^2+(8*x^3-44*x^2)*exp(2)+4*x^4-44*x^3+121*x^2),x,method=_RETURNVERBOS
E)

[Out]

exp(-4*x^3/(2*exp(2)*x+2*x^2-2*ln(x)-11*x))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((24*x^2*log(x)-16*x^3*exp(2)-8*x^4+88*x^3-8*x^2)*exp(2*x^3/(2*log(x)-2*exp(2)*x-2*x^2+11*x))^2/(4*lo
g(x)^2+(-8*exp(2)*x-8*x^2+44*x)*log(x)+4*x^2*exp(2)^2+(8*x^3-44*x^2)*exp(2)+4*x^4-44*x^3+121*x^2),x, algorithm
="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

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mupad [B]  time = 0.72, size = 26, normalized size = 0.79 \begin {gather*} {\mathrm {e}}^{\frac {4\,x^3}{11\,x+2\,\ln \relax (x)-2\,x\,{\mathrm {e}}^2-2\,x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((4*x^3)/(11*x + 2*log(x) - 2*x*exp(2) - 2*x^2))*(16*x^3*exp(2) - 24*x^2*log(x) + 8*x^2 - 88*x^3 + 8*
x^4))/(4*log(x)^2 - exp(2)*(44*x^2 - 8*x^3) + 4*x^2*exp(4) + 121*x^2 - 44*x^3 + 4*x^4 - log(x)*(8*x*exp(2) - 4
4*x + 8*x^2)),x)

[Out]

exp((4*x^3)/(11*x + 2*log(x) - 2*x*exp(2) - 2*x^2))

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sympy [A]  time = 0.59, size = 26, normalized size = 0.79 \begin {gather*} e^{\frac {4 x^{3}}{- 2 x^{2} - 2 x e^{2} + 11 x + 2 \log {\relax (x )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((24*x**2*ln(x)-16*x**3*exp(2)-8*x**4+88*x**3-8*x**2)*exp(2*x**3/(2*ln(x)-2*exp(2)*x-2*x**2+11*x))**2
/(4*ln(x)**2+(-8*exp(2)*x-8*x**2+44*x)*ln(x)+4*x**2*exp(2)**2+(8*x**3-44*x**2)*exp(2)+4*x**4-44*x**3+121*x**2)
,x)

[Out]

exp(4*x**3/(-2*x**2 - 2*x*exp(2) + 11*x + 2*log(x)))

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