3.91.11 \(\int \frac {e^{\frac {3+e^{10} (-x^2-x^3)}{-e^{10} x^2+e^{10} x^2 \log (\frac {x}{5 \log (x)})}} (3+e^{10} (-x^2-x^3)+(3+e^{10} (x^2+2 x^3)) \log (x)+(-6-e^{10} x^3) \log (x) \log (\frac {x}{5 \log (x)}))}{e^{10} x^3 \log (x)-2 e^{10} x^3 \log (x) \log (\frac {x}{5 \log (x)})+e^{10} x^3 \log (x) \log ^2(\frac {x}{5 \log (x)})} \, dx\)

Optimal. Leaf size=30 \[ e^{\frac {-1+\frac {3}{e^{10} x^2}-x}{-1+\log \left (\frac {x}{5 \log (x)}\right )}} \]

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Rubi [F]  time = 14.11, 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 {3+e^{10} \left (-x^2-x^3\right )}{-e^{10} x^2+e^{10} x^2 \log \left (\frac {x}{5 \log (x)}\right )}\right ) \left (3+e^{10} \left (-x^2-x^3\right )+\left (3+e^{10} \left (x^2+2 x^3\right )\right ) \log (x)+\left (-6-e^{10} x^3\right ) \log (x) \log \left (\frac {x}{5 \log (x)}\right )\right )}{e^{10} x^3 \log (x)-2 e^{10} x^3 \log (x) \log \left (\frac {x}{5 \log (x)}\right )+e^{10} x^3 \log (x) \log ^2\left (\frac {x}{5 \log (x)}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((3 + E^10*(-x^2 - x^3))/(-(E^10*x^2) + E^10*x^2*Log[x/(5*Log[x])]))*(3 + E^10*(-x^2 - x^3) + (3 + E^10
*(x^2 + 2*x^3))*Log[x] + (-6 - E^10*x^3)*Log[x]*Log[x/(5*Log[x])]))/(E^10*x^3*Log[x] - 2*E^10*x^3*Log[x]*Log[x
/(5*Log[x])] + E^10*x^3*Log[x]*Log[x/(5*Log[x])]^2),x]

[Out]

Defer[Int][E^((3 - E^10*x^2 - E^10*x^3)/(E^10*x^2*(-1 + Log[x/(5*Log[x])])))/(-1 + Log[x/(5*Log[x])])^2, x] -
3*Defer[Int][E^(-10 + (3 - E^10*x^2 - E^10*x^3)/(E^10*x^2*(-1 + Log[x/(5*Log[x])])))/(x^3*(-1 + Log[x/(5*Log[x
])])^2), x] + Defer[Int][E^((3 - E^10*x^2 - E^10*x^3)/(E^10*x^2*(-1 + Log[x/(5*Log[x])])))/(x*(-1 + Log[x/(5*L
og[x])])^2), x] - Defer[Int][E^((3 - E^10*x^2 - E^10*x^3)/(E^10*x^2*(-1 + Log[x/(5*Log[x])])))/(Log[x]*(-1 + L
og[x/(5*Log[x])])^2), x] + 3*Defer[Int][E^(-10 + (3 - E^10*x^2 - E^10*x^3)/(E^10*x^2*(-1 + Log[x/(5*Log[x])]))
)/(x^3*Log[x]*(-1 + Log[x/(5*Log[x])])^2), x] - Defer[Int][E^((3 - E^10*x^2 - E^10*x^3)/(E^10*x^2*(-1 + Log[x/
(5*Log[x])])))/(x*Log[x]*(-1 + Log[x/(5*Log[x])])^2), x] - Defer[Int][E^((3 - E^10*x^2 - E^10*x^3)/(E^10*x^2*(
-1 + Log[x/(5*Log[x])])))/(-1 + Log[x/(5*Log[x])]), x] - 6*Defer[Int][E^(-10 + (3 - E^10*x^2 - E^10*x^3)/(E^10
*x^2*(-1 + Log[x/(5*Log[x])])))/(x^3*(-1 + Log[x/(5*Log[x])])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (-10+\frac {3-e^{10} x^2-e^{10} x^3}{e^{10} x^2 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}\right ) \left (3+e^{10} \left (-x^2-x^3\right )+\left (3+e^{10} \left (x^2+2 x^3\right )\right ) \log (x)+\left (-6-e^{10} x^3\right ) \log (x) \log \left (\frac {x}{5 \log (x)}\right )\right )}{x^3 \log (x) \left (1-\log \left (\frac {x}{5 \log (x)}\right )\right )^2} \, dx\\ &=\int \left (\frac {\exp \left (-10+\frac {3-e^{10} x^2-e^{10} x^3}{e^{10} x^2 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}\right ) \left (-3+e^{10} x^2+e^{10} x^3\right ) (-1+\log (x))}{x^3 \log (x) \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )^2}+\frac {\exp \left (-10+\frac {3-e^{10} x^2-e^{10} x^3}{e^{10} x^2 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}\right ) \left (-6-e^{10} x^3\right )}{x^3 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}\right ) \, dx\\ &=\int \frac {\exp \left (-10+\frac {3-e^{10} x^2-e^{10} x^3}{e^{10} x^2 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}\right ) \left (-3+e^{10} x^2+e^{10} x^3\right ) (-1+\log (x))}{x^3 \log (x) \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )^2} \, dx+\int \frac {\exp \left (-10+\frac {3-e^{10} x^2-e^{10} x^3}{e^{10} x^2 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}\right ) \left (-6-e^{10} x^3\right )}{x^3 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )} \, dx\\ &=\int \left (\frac {\exp \left (\frac {3-e^{10} x^2-e^{10} x^3}{e^{10} x^2 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}\right ) (-1+\log (x))}{\log (x) \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )^2}-\frac {3 \exp \left (-10+\frac {3-e^{10} x^2-e^{10} x^3}{e^{10} x^2 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}\right ) (-1+\log (x))}{x^3 \log (x) \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )^2}+\frac {\exp \left (\frac {3-e^{10} x^2-e^{10} x^3}{e^{10} x^2 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}\right ) (-1+\log (x))}{x \log (x) \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )^2}\right ) \, dx+\int \left (-\frac {\exp \left (\frac {3-e^{10} x^2-e^{10} x^3}{e^{10} x^2 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}\right )}{-1+\log \left (\frac {x}{5 \log (x)}\right )}-\frac {6 \exp \left (-10+\frac {3-e^{10} x^2-e^{10} x^3}{e^{10} x^2 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}\right )}{x^3 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}\right ) \, dx\\ &=-\left (3 \int \frac {\exp \left (-10+\frac {3-e^{10} x^2-e^{10} x^3}{e^{10} x^2 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}\right ) (-1+\log (x))}{x^3 \log (x) \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )^2} \, dx\right )-6 \int \frac {\exp \left (-10+\frac {3-e^{10} x^2-e^{10} x^3}{e^{10} x^2 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}\right )}{x^3 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )} \, dx+\int \frac {\exp \left (\frac {3-e^{10} x^2-e^{10} x^3}{e^{10} x^2 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}\right ) (-1+\log (x))}{\log (x) \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )^2} \, dx+\int \frac {\exp \left (\frac {3-e^{10} x^2-e^{10} x^3}{e^{10} x^2 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}\right ) (-1+\log (x))}{x \log (x) \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )^2} \, dx-\int \frac {\exp \left (\frac {3-e^{10} x^2-e^{10} x^3}{e^{10} x^2 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}\right )}{-1+\log \left (\frac {x}{5 \log (x)}\right )} \, dx\\ &=-\left (3 \int \left (\frac {\exp \left (-10+\frac {3-e^{10} x^2-e^{10} x^3}{e^{10} x^2 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}\right )}{x^3 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )^2}-\frac {\exp \left (-10+\frac {3-e^{10} x^2-e^{10} x^3}{e^{10} x^2 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}\right )}{x^3 \log (x) \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )^2}\right ) \, dx\right )-6 \int \frac {\exp \left (-10+\frac {3-e^{10} x^2-e^{10} x^3}{e^{10} x^2 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}\right )}{x^3 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )} \, dx+\int \left (\frac {\exp \left (\frac {3-e^{10} x^2-e^{10} x^3}{e^{10} x^2 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}\right )}{\left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )^2}-\frac {\exp \left (\frac {3-e^{10} x^2-e^{10} x^3}{e^{10} x^2 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}\right )}{\log (x) \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )^2}\right ) \, dx+\int \left (\frac {\exp \left (\frac {3-e^{10} x^2-e^{10} x^3}{e^{10} x^2 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}\right )}{x \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )^2}-\frac {\exp \left (\frac {3-e^{10} x^2-e^{10} x^3}{e^{10} x^2 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}\right )}{x \log (x) \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )^2}\right ) \, dx-\int \frac {\exp \left (\frac {3-e^{10} x^2-e^{10} x^3}{e^{10} x^2 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}\right )}{-1+\log \left (\frac {x}{5 \log (x)}\right )} \, dx\\ &=-\left (3 \int \frac {\exp \left (-10+\frac {3-e^{10} x^2-e^{10} x^3}{e^{10} x^2 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}\right )}{x^3 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )^2} \, dx\right )+3 \int \frac {\exp \left (-10+\frac {3-e^{10} x^2-e^{10} x^3}{e^{10} x^2 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}\right )}{x^3 \log (x) \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )^2} \, dx-6 \int \frac {\exp \left (-10+\frac {3-e^{10} x^2-e^{10} x^3}{e^{10} x^2 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}\right )}{x^3 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )} \, dx+\int \frac {\exp \left (\frac {3-e^{10} x^2-e^{10} x^3}{e^{10} x^2 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}\right )}{\left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )^2} \, dx+\int \frac {\exp \left (\frac {3-e^{10} x^2-e^{10} x^3}{e^{10} x^2 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}\right )}{x \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )^2} \, dx-\int \frac {\exp \left (\frac {3-e^{10} x^2-e^{10} x^3}{e^{10} x^2 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}\right )}{\log (x) \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )^2} \, dx-\int \frac {\exp \left (\frac {3-e^{10} x^2-e^{10} x^3}{e^{10} x^2 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}\right )}{x \log (x) \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )^2} \, dx-\int \frac {\exp \left (\frac {3-e^{10} x^2-e^{10} x^3}{e^{10} x^2 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}\right )}{-1+\log \left (\frac {x}{5 \log (x)}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.27, size = 36, normalized size = 1.20 \begin {gather*} e^{\frac {3-e^{10} x^2 (1+x)}{e^{10} x^2 \left (-1+\log \left (\frac {x}{5 \log (x)}\right )\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((3 + E^10*(-x^2 - x^3))/(-(E^10*x^2) + E^10*x^2*Log[x/(5*Log[x])]))*(3 + E^10*(-x^2 - x^3) + (3
+ E^10*(x^2 + 2*x^3))*Log[x] + (-6 - E^10*x^3)*Log[x]*Log[x/(5*Log[x])]))/(E^10*x^3*Log[x] - 2*E^10*x^3*Log[x]
*Log[x/(5*Log[x])] + E^10*x^3*Log[x]*Log[x/(5*Log[x])]^2),x]

[Out]

E^((3 - E^10*x^2*(1 + x))/(E^10*x^2*(-1 + Log[x/(5*Log[x])])))

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fricas [A]  time = 0.56, size = 39, normalized size = 1.30 \begin {gather*} e^{\left (-\frac {{\left (x^{3} + x^{2}\right )} e^{10} - 3}{x^{2} e^{10} \log \left (\frac {x}{5 \, \log \relax (x)}\right ) - x^{2} e^{10}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3*exp(5)^2-6)*log(x)*log(1/5*x/log(x))+((2*x^3+x^2)*exp(5)^2+3)*log(x)+(-x^3-x^2)*exp(5)^2+3)*e
xp(((-x^3-x^2)*exp(5)^2+3)/(x^2*exp(5)^2*log(1/5*x/log(x))-x^2*exp(5)^2))/(x^3*exp(5)^2*log(x)*log(1/5*x/log(x
))^2-2*x^3*exp(5)^2*log(x)*log(1/5*x/log(x))+x^3*exp(5)^2*log(x)),x, algorithm="fricas")

[Out]

e^(-((x^3 + x^2)*e^10 - 3)/(x^2*e^10*log(1/5*x/log(x)) - x^2*e^10))

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giac [B]  time = 29.40, size = 90, normalized size = 3.00 \begin {gather*} e^{\left (-\frac {x^{3} e^{10}}{x^{2} e^{10} \log \left (\frac {x}{5 \, \log \relax (x)}\right ) - x^{2} e^{10}} - \frac {x^{2} e^{10}}{x^{2} e^{10} \log \left (\frac {x}{5 \, \log \relax (x)}\right ) - x^{2} e^{10}} + \frac {3}{x^{2} e^{10} \log \left (\frac {x}{5 \, \log \relax (x)}\right ) - x^{2} e^{10}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3*exp(5)^2-6)*log(x)*log(1/5*x/log(x))+((2*x^3+x^2)*exp(5)^2+3)*log(x)+(-x^3-x^2)*exp(5)^2+3)*e
xp(((-x^3-x^2)*exp(5)^2+3)/(x^2*exp(5)^2*log(1/5*x/log(x))-x^2*exp(5)^2))/(x^3*exp(5)^2*log(x)*log(1/5*x/log(x
))^2-2*x^3*exp(5)^2*log(x)*log(1/5*x/log(x))+x^3*exp(5)^2*log(x)),x, algorithm="giac")

[Out]

e^(-x^3*e^10/(x^2*e^10*log(1/5*x/log(x)) - x^2*e^10) - x^2*e^10/(x^2*e^10*log(1/5*x/log(x)) - x^2*e^10) + 3/(x
^2*e^10*log(1/5*x/log(x)) - x^2*e^10))

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maple [C]  time = 12.67, size = 124, normalized size = 4.13

method	result	size
risch	\({\mathrm e}^{\frac {2 \left (x^{3} {\mathrm e}^{10}+x^{2} {\mathrm e}^{10}-3\right ) {\mathrm e}^{-10}}{x^{2} \left (i \pi \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )^{3}-i \pi \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )^{2} \mathrm {csgn}\left (i x \right )-i \pi \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )^{2} \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right )+i \pi \,\mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right )-2 \ln \relax (x )+2 \ln \relax (5)+2 \ln \left (\ln \relax (x )\right )+2\right )}}\)	\(124\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^3*exp(5)^2-6)*ln(x)*ln(1/5*x/ln(x))+((2*x^3+x^2)*exp(5)^2+3)*ln(x)+(-x^3-x^2)*exp(5)^2+3)*exp(((-x^3-
x^2)*exp(5)^2+3)/(x^2*exp(5)^2*ln(1/5*x/ln(x))-x^2*exp(5)^2))/(x^3*exp(5)^2*ln(x)*ln(1/5*x/ln(x))^2-2*x^3*exp(
5)^2*ln(x)*ln(1/5*x/ln(x))+x^3*exp(5)^2*ln(x)),x,method=_RETURNVERBOSE)

[Out]

exp(2*(x^3*exp(10)+x^2*exp(10)-3)/x^2*exp(-10)/(I*Pi*csgn(I*x/ln(x))^3-I*Pi*csgn(I*x/ln(x))^2*csgn(I*x)-I*Pi*c
sgn(I*x/ln(x))^2*csgn(I/ln(x))+I*Pi*csgn(I*x/ln(x))*csgn(I*x)*csgn(I/ln(x))-2*ln(x)+2*ln(5)+2*ln(ln(x))+2))

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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(((-x^3*exp(5)^2-6)*log(x)*log(1/5*x/log(x))+((2*x^3+x^2)*exp(5)^2+3)*log(x)+(-x^3-x^2)*exp(5)^2+3)*e
xp(((-x^3-x^2)*exp(5)^2+3)/(x^2*exp(5)^2*log(1/5*x/log(x))-x^2*exp(5)^2))/(x^3*exp(5)^2*log(x)*log(1/5*x/log(x
))^2-2*x^3*exp(5)^2*log(x)*log(1/5*x/log(x))+x^3*exp(5)^2*log(x)),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 = 8.64, size = 69, normalized size = 2.30 \begin {gather*} {\mathrm {e}}^{-\frac {3}{x^2\,{\mathrm {e}}^{10}+x^2\,{\mathrm {e}}^{10}\,\ln \relax (5)-x^2\,{\mathrm {e}}^{10}\,\ln \left (\frac {x}{\ln \relax (x)}\right )}}\,{\mathrm {e}}^{\frac {x}{\ln \relax (5)-\ln \left (\frac {x}{\ln \relax (x)}\right )+1}}\,{\mathrm {e}}^{\frac {1}{\ln \relax (5)-\ln \left (\frac {x}{\ln \relax (x)}\right )+1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((exp(10)*(x^2 + x^3) - 3)/(x^2*exp(10) - x^2*exp(10)*log(x/(5*log(x)))))*(exp(10)*(x^2 + x^3) - log(
x)*(exp(10)*(x^2 + 2*x^3) + 3) + log(x/(5*log(x)))*log(x)*(x^3*exp(10) + 6) - 3))/(x^3*exp(10)*log(x) - 2*x^3*
exp(10)*log(x/(5*log(x)))*log(x) + x^3*exp(10)*log(x/(5*log(x)))^2*log(x)),x)

[Out]

exp(-3/(x^2*exp(10) + x^2*exp(10)*log(5) - x^2*exp(10)*log(x/log(x))))*exp(x/(log(5) - log(x/log(x)) + 1))*exp
(1/(log(5) - log(x/log(x)) + 1))

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sympy [A]  time = 1.51, size = 36, normalized size = 1.20 \begin {gather*} e^{\frac {\left (- x^{3} - x^{2}\right ) e^{10} + 3}{x^{2} e^{10} \log {\left (\frac {x}{5 \log {\relax (x )}} \right )} - x^{2} e^{10}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**3*exp(5)**2-6)*ln(x)*ln(1/5*x/ln(x))+((2*x**3+x**2)*exp(5)**2+3)*ln(x)+(-x**3-x**2)*exp(5)**2+
3)*exp(((-x**3-x**2)*exp(5)**2+3)/(x**2*exp(5)**2*ln(1/5*x/ln(x))-x**2*exp(5)**2))/(x**3*exp(5)**2*ln(x)*ln(1/
5*x/ln(x))**2-2*x**3*exp(5)**2*ln(x)*ln(1/5*x/ln(x))+x**3*exp(5)**2*ln(x)),x)

[Out]

exp(((-x**3 - x**2)*exp(10) + 3)/(x**2*exp(10)*log(x/(5*log(x))) - x**2*exp(10)))

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