3.5.33 \(\int \frac {e^{-\frac {-3 x+(1-3 x) \log (x)}{x^2 \log (x)}} (-3 e^5 x-6 x^2+(-3 e^5 x-6 x^2) \log (x)-2 e^{\frac {-3 x+(1-3 x) \log (x)}{x^2 \log (x)}} x^3 \log ^2(x)+(e^5 (2-3 x)+4 x-6 x^2-2 x^3) \log ^2(x))}{(e^{10} x^3+4 e^5 x^4+4 x^5) \log ^2(x)} \, dx\)

Optimal. Leaf size=35 \[ \frac {x+e^{-\frac {-3+\frac {1}{x}-\frac {3}{\log (x)}}{x}} x}{x \left (e^5+2 x\right )} \]

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Rubi [F]  time = 0.00, 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*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-3*E^5*x - 6*x^2 + (-3*E^5*x - 6*x^2)*Log[x] - 2*E^((-3*x + (1 - 3*x)*Log[x])/(x^2*Log[x]))*x^3*Log[x]^2
+ (E^5*(2 - 3*x) + 4*x - 6*x^2 - 2*x^3)*Log[x]^2)/(E^((-3*x + (1 - 3*x)*Log[x])/(x^2*Log[x]))*(E^10*x^3 + 4*E^
5*x^4 + 4*x^5)*Log[x]^2),x]

[Out]

$Aborted

Rubi steps

Aborted

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Mathematica [A]  time = 0.31, size = 34, normalized size = 0.97 \begin {gather*} \frac {1+e^{\frac {3 x+(-1+3 x) \log (x)}{x^2 \log (x)}}}{e^5+2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3*E^5*x - 6*x^2 + (-3*E^5*x - 6*x^2)*Log[x] - 2*E^((-3*x + (1 - 3*x)*Log[x])/(x^2*Log[x]))*x^3*Log
[x]^2 + (E^5*(2 - 3*x) + 4*x - 6*x^2 - 2*x^3)*Log[x]^2)/(E^((-3*x + (1 - 3*x)*Log[x])/(x^2*Log[x]))*(E^10*x^3
+ 4*E^5*x^4 + 4*x^5)*Log[x]^2),x]

[Out]

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

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fricas [A]  time = 0.87, size = 32, normalized size = 0.91 \begin {gather*} \frac {e^{\left (\frac {{\left (3 \, x - 1\right )} \log \relax (x) + 3 \, x}{x^{2} \log \relax (x)}\right )} + 1}{2 \, x + e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^3*log(x)^2*exp(((-3*x+1)*log(x)-3*x)/x^2/log(x))+((-3*x+2)*exp(5)-2*x^3-6*x^2+4*x)*log(x)^2+(-
3*x*exp(5)-6*x^2)*log(x)-3*x*exp(5)-6*x^2)/(x^3*exp(5)^2+4*x^4*exp(5)+4*x^5)/log(x)^2/exp(((-3*x+1)*log(x)-3*x
)/x^2/log(x)),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^3*log(x)^2*exp(((-3*x+1)*log(x)-3*x)/x^2/log(x))+((-3*x+2)*exp(5)-2*x^3-6*x^2+4*x)*log(x)^2+(-
3*x*exp(5)-6*x^2)*log(x)-3*x*exp(5)-6*x^2)/(x^3*exp(5)^2+4*x^4*exp(5)+4*x^5)/log(x)^2/exp(((-3*x+1)*log(x)-3*x
)/x^2/log(x)),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Evaluation time: 0.69Unable to divide, perhaps due to rounding error%%%{-13824,[0,25,0]%%%}+%%%{-48384,[0,2
4,1]%%%}+%%

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maple [A]  time = 0.17, size = 41, normalized size = 1.17

method	result	size
risch	\(\frac {1}{2 x +{\mathrm e}^{5}}+\frac {{\mathrm e}^{\frac {3 x \ln \relax (x )-\ln \relax (x )+3 x}{\ln \relax (x ) x^{2}}}}{2 x +{\mathrm e}^{5}}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x^3*ln(x)^2*exp(((-3*x+1)*ln(x)-3*x)/x^2/ln(x))+((-3*x+2)*exp(5)-2*x^3-6*x^2+4*x)*ln(x)^2+(-3*x*exp(5)
-6*x^2)*ln(x)-3*x*exp(5)-6*x^2)/(x^3*exp(5)^2+4*x^4*exp(5)+4*x^5)/ln(x)^2/exp(((-3*x+1)*ln(x)-3*x)/x^2/ln(x)),
x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.66, size = 36, normalized size = 1.03 \begin {gather*} \frac {{\left (e^{\left (\frac {3}{x} + \frac {3}{x \log \relax (x)}\right )} + e^{\left (\frac {1}{x^{2}}\right )}\right )} e^{\left (-\frac {1}{x^{2}}\right )}}{2 \, x + e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^3*log(x)^2*exp(((-3*x+1)*log(x)-3*x)/x^2/log(x))+((-3*x+2)*exp(5)-2*x^3-6*x^2+4*x)*log(x)^2+(-
3*x*exp(5)-6*x^2)*log(x)-3*x*exp(5)-6*x^2)/(x^3*exp(5)^2+4*x^4*exp(5)+4*x^5)/log(x)^2/exp(((-3*x+1)*log(x)-3*x
)/x^2/log(x)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.89, size = 40, normalized size = 1.14 \begin {gather*} \frac {1}{2\,x+{\mathrm {e}}^5}+\frac {{\mathrm {e}}^{-\frac {1}{x^2}}\,{\mathrm {e}}^{3/x}\,{\mathrm {e}}^{\frac {3}{x\,\ln \relax (x)}}}{2\,x+{\mathrm {e}}^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((3*x + log(x)*(3*x - 1))/(x^2*log(x)))*(log(x)*(3*x*exp(5) + 6*x^2) + 3*x*exp(5) + log(x)^2*(6*x^2 -
4*x + 2*x^3 + exp(5)*(3*x - 2)) + 6*x^2 + 2*x^3*exp(-(3*x + log(x)*(3*x - 1))/(x^2*log(x)))*log(x)^2))/(log(x
)^2*(4*x^4*exp(5) + x^3*exp(10) + 4*x^5)),x)

[Out]

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

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sympy [A]  time = 0.49, size = 37, normalized size = 1.06 \begin {gather*} \frac {2}{4 x + 2 e^{5}} + \frac {e^{- \frac {- 3 x + \left (1 - 3 x\right ) \log {\relax (x )}}{x^{2} \log {\relax (x )}}}}{2 x + e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x**3*ln(x)**2*exp(((-3*x+1)*ln(x)-3*x)/x**2/ln(x))+((-3*x+2)*exp(5)-2*x**3-6*x**2+4*x)*ln(x)**2+
(-3*x*exp(5)-6*x**2)*ln(x)-3*x*exp(5)-6*x**2)/(x**3*exp(5)**2+4*x**4*exp(5)+4*x**5)/ln(x)**2/exp(((-3*x+1)*ln(
x)-3*x)/x**2/ln(x)),x)

[Out]

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

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