3.25.20 \(\int \frac {e^{\frac {8-2 x^5-e^3 x^2 \log (5)}{2 x^3+e^3 \log (5)}} (-48 x^2-8 x^7-8 e^3 x^4 \log (5)-2 e^6 x \log ^2(5))}{4 x^6+4 e^3 x^3 \log (5)+e^6 \log ^2(5)} \, dx\) [2420]

Optimal. Leaf size=26 \[ e^{x \left (-x+\frac {4}{x^4+\frac {1}{2} e^3 x \log (5)}\right )} \]

[Out]

exp((4/(x^4+1/2*x*exp(3)*ln(5))-x)*x)

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Rubi [F]
time = 1.67, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {\exp \left (\frac {8-2 x^5-e^3 x^2 \log (5)}{2 x^3+e^3 \log (5)}\right ) \left (-48 x^2-8 x^7-8 e^3 x^4 \log (5)-2 e^6 x \log ^2(5)\right )}{4 x^6+4 e^3 x^3 \log (5)+e^6 \log ^2(5)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((8 - 2*x^5 - E^3*x^2*Log[5])/(2*x^3 + E^3*Log[5]))*(-48*x^2 - 8*x^7 - 8*E^3*x^4*Log[5] - 2*E^6*x*Log[5
]^2))/(4*x^6 + 4*E^3*x^3*Log[5] + E^6*Log[5]^2),x]

[Out]

-2*Defer[Int][E^((8 - 2*x^5 - E^3*x^2*Log[5])/(2*x^3 + E^3*Log[5]))*x, x] - 48*Defer[Int][(E^((8 - 2*x^5 - E^3
*x^2*Log[5])/(2*x^3 + E^3*Log[5]))*x^2)/(2*x^3 + E^3*Log[5])^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=4 \int \frac {\exp \left (\frac {8-2 x^5-e^3 x^2 \log (5)}{2 x^3+e^3 \log (5)}\right ) \left (-48 x^2-8 x^7-8 e^3 x^4 \log (5)-2 e^6 x \log ^2(5)\right )}{\left (4 x^3+2 e^3 \log (5)\right )^2} \, dx\\ &=4 \int \frac {2 \exp \left (\frac {8-2 x^5-e^3 x^2 \log (5)}{2 x^3+e^3 \log (5)}\right ) x \left (-24 x-4 x^6-4 e^3 x^3 \log (5)-e^6 \log ^2(5)\right )}{\left (4 x^3+2 e^3 \log (5)\right )^2} \, dx\\ &=8 \int \frac {\exp \left (\frac {8-2 x^5-e^3 x^2 \log (5)}{2 x^3+e^3 \log (5)}\right ) x \left (-24 x-4 x^6-4 e^3 x^3 \log (5)-e^6 \log ^2(5)\right )}{\left (4 x^3+2 e^3 \log (5)\right )^2} \, dx\\ &=8 \int \left (-\frac {1}{4} \exp \left (\frac {8-2 x^5-e^3 x^2 \log (5)}{2 x^3+e^3 \log (5)}\right ) x-\frac {6 \exp \left (\frac {8-2 x^5-e^3 x^2 \log (5)}{2 x^3+e^3 \log (5)}\right ) x^2}{\left (2 x^3+e^3 \log (5)\right )^2}\right ) \, dx\\ &=-\left (2 \int \exp \left (\frac {8-2 x^5-e^3 x^2 \log (5)}{2 x^3+e^3 \log (5)}\right ) x \, dx\right )-48 \int \frac {\exp \left (\frac {8-2 x^5-e^3 x^2 \log (5)}{2 x^3+e^3 \log (5)}\right ) x^2}{\left (2 x^3+e^3 \log (5)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.04, size = 24, normalized size = 0.92 \begin {gather*} e^{-x^2+\frac {8}{2 x^3+e^3 \log (5)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((8 - 2*x^5 - E^3*x^2*Log[5])/(2*x^3 + E^3*Log[5]))*(-48*x^2 - 8*x^7 - 8*E^3*x^4*Log[5] - 2*E^6*x
*Log[5]^2))/(4*x^6 + 4*E^3*x^3*Log[5] + E^6*Log[5]^2),x]

[Out]

E^(-x^2 + 8/(2*x^3 + E^3*Log[5]))

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Maple [A]
time = 0.26, size = 32, normalized size = 1.23

method	result	size
gosper	\({\mathrm e}^{-\frac {2 x^{5}+x^{2} {\mathrm e}^{3} \ln \left (5\right )-8}{{\mathrm e}^{3} \ln \left (5\right )+2 x^{3}}}\)	\(32\)
risch	\({\mathrm e}^{-\frac {2 x^{5}+x^{2} {\mathrm e}^{3} \ln \left (5\right )-8}{{\mathrm e}^{3} \ln \left (5\right )+2 x^{3}}}\)	\(32\)
norman	\(\frac {{\mathrm e}^{3} \ln \left (5\right ) {\mathrm e}^{\frac {-x^{2} {\mathrm e}^{3} \ln \left (5\right )-2 x^{5}+8}{{\mathrm e}^{3} \ln \left (5\right )+2 x^{3}}}+2 x^{3} {\mathrm e}^{\frac {-x^{2} {\mathrm e}^{3} \ln \left (5\right )-2 x^{5}+8}{{\mathrm e}^{3} \ln \left (5\right )+2 x^{3}}}}{{\mathrm e}^{3} \ln \left (5\right )+2 x^{3}}\)	\(88\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x*exp(3)^2*ln(5)^2-8*x^4*exp(3)*ln(5)-8*x^7-48*x^2)*exp((-x^2*exp(3)*ln(5)-2*x^5+8)/(exp(3)*ln(5)+2*x^
3))/(exp(3)^2*ln(5)^2+4*x^3*exp(3)*ln(5)+4*x^6),x,method=_RETURNVERBOSE)

[Out]

exp(-(2*x^5+x^2*exp(3)*ln(5)-8)/(exp(3)*ln(5)+2*x^3))

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Maxima [A]
time = 0.60, size = 22, normalized size = 0.85 \begin {gather*} e^{\left (-x^{2} + \frac {8}{2 \, x^{3} + e^{3} \log \left (5\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*exp(3)^2*log(5)^2-8*x^4*exp(3)*log(5)-8*x^7-48*x^2)*exp((-x^2*exp(3)*log(5)-2*x^5+8)/(exp(3)*l
og(5)+2*x^3))/(exp(3)^2*log(5)^2+4*x^3*exp(3)*log(5)+4*x^6),x, algorithm="maxima")

[Out]

e^(-x^2 + 8/(2*x^3 + e^3*log(5)))

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Fricas [A]
time = 0.37, size = 31, normalized size = 1.19 \begin {gather*} e^{\left (-\frac {2 \, x^{5} + x^{2} e^{3} \log \left (5\right ) - 8}{2 \, x^{3} + e^{3} \log \left (5\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*exp(3)^2*log(5)^2-8*x^4*exp(3)*log(5)-8*x^7-48*x^2)*exp((-x^2*exp(3)*log(5)-2*x^5+8)/(exp(3)*l
og(5)+2*x^3))/(exp(3)^2*log(5)^2+4*x^3*exp(3)*log(5)+4*x^6),x, algorithm="fricas")

[Out]

e^(-(2*x^5 + x^2*e^3*log(5) - 8)/(2*x^3 + e^3*log(5)))

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Sympy [A]
time = 0.16, size = 29, normalized size = 1.12 \begin {gather*} e^{\frac {- 2 x^{5} - x^{2} e^{3} \log {\left (5 \right )} + 8}{2 x^{3} + e^{3} \log {\left (5 \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*exp(3)**2*ln(5)**2-8*x**4*exp(3)*ln(5)-8*x**7-48*x**2)*exp((-x**2*exp(3)*ln(5)-2*x**5+8)/(exp(
3)*ln(5)+2*x**3))/(exp(3)**2*ln(5)**2+4*x**3*exp(3)*ln(5)+4*x**6),x)

[Out]

exp((-2*x**5 - x**2*exp(3)*log(5) + 8)/(2*x**3 + exp(3)*log(5)))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (22) = 44\).
time = 0.54, size = 56, normalized size = 2.15 \begin {gather*} e^{\left (-\frac {2 \, x^{5} e^{3} \log \left (5\right ) + x^{2} e^{6} \log \left (5\right )^{2} + 16 \, x^{3}}{2 \, x^{3} e^{3} \log \left (5\right ) + e^{6} \log \left (5\right )^{2}} + \frac {8 \, e^{\left (-3\right )}}{\log \left (5\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*exp(3)^2*log(5)^2-8*x^4*exp(3)*log(5)-8*x^7-48*x^2)*exp((-x^2*exp(3)*log(5)-2*x^5+8)/(exp(3)*l
og(5)+2*x^3))/(exp(3)^2*log(5)^2+4*x^3*exp(3)*log(5)+4*x^6),x, algorithm="giac")

[Out]

e^(-(2*x^5*e^3*log(5) + x^2*e^6*log(5)^2 + 16*x^3)/(2*x^3*e^3*log(5) + e^6*log(5)^2) + 8*e^(-3)/log(5))

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Mupad [B]
time = 1.93, size = 59, normalized size = 2.27 \begin {gather*} \frac {{\mathrm {e}}^{-\frac {2\,x^5}{2\,x^3+{\mathrm {e}}^3\,\ln \left (5\right )}}\,{\mathrm {e}}^{\frac {8}{2\,x^3+{\mathrm {e}}^3\,\ln \left (5\right )}}}{5^{\frac {x^2\,{\mathrm {e}}^3}{2\,x^3+{\mathrm {e}}^3\,\ln \left (5\right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(2*x^5 + x^2*exp(3)*log(5) - 8)/(exp(3)*log(5) + 2*x^3))*(48*x^2 + 8*x^7 + 8*x^4*exp(3)*log(5) + 2*
x*exp(6)*log(5)^2))/(exp(6)*log(5)^2 + 4*x^6 + 4*x^3*exp(3)*log(5)),x)

[Out]

(exp(-(2*x^5)/(exp(3)*log(5) + 2*x^3))*exp(8/(exp(3)*log(5) + 2*x^3)))/5^((x^2*exp(3))/(exp(3)*log(5) + 2*x^3)
)

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