3.88.55 \(\int \frac {-10 e^{\frac {1}{x}} x \log (\frac {1}{x^2})+(-20 e^{\frac {1}{x}} x+e^{\frac {1}{x}} (-10-10 x) \log (\frac {1}{x^2})) \log (x)}{x^3 \log ^2(x)} \, dx\)

Optimal. Leaf size=18 \[ \frac {10 e^{\frac {1}{x}} \log \left (\frac {1}{x^2}\right )}{x \log (x)} \]

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Rubi [F]  time = 1.22, 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 {-10 e^{\frac {1}{x}} x \log \left (\frac {1}{x^2}\right )+\left (-20 e^{\frac {1}{x}} x+e^{\frac {1}{x}} (-10-10 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)}{x^3 \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-10*E^x^(-1)*x*Log[x^(-2)] + (-20*E^x^(-1)*x + E^x^(-1)*(-10 - 10*x)*Log[x^(-2)])*Log[x])/(x^3*Log[x]^2),
x]

[Out]

-10*Defer[Int][(E^x^(-1)*Log[x^(-2)])/(x^2*Log[x]^2), x] - 20*Defer[Int][E^x^(-1)/(x^2*Log[x]), x] - 10*Defer[
Int][(E^x^(-1)*Log[x^(-2)])/(x^3*Log[x]), x] - 10*Defer[Int][(E^x^(-1)*Log[x^(-2)])/(x^2*Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {10 e^{\frac {1}{x}} \left (-2 x \log (x)-\log \left (\frac {1}{x^2}\right ) (x+(1+x) \log (x))\right )}{x^3 \log ^2(x)} \, dx\\ &=10 \int \frac {e^{\frac {1}{x}} \left (-2 x \log (x)-\log \left (\frac {1}{x^2}\right ) (x+(1+x) \log (x))\right )}{x^3 \log ^2(x)} \, dx\\ &=10 \int \left (-\frac {e^{\frac {1}{x}} \log \left (\frac {1}{x^2}\right )}{x^2 \log ^2(x)}+\frac {e^{\frac {1}{x}} \left (-2 x-\log \left (\frac {1}{x^2}\right )-x \log \left (\frac {1}{x^2}\right )\right )}{x^3 \log (x)}\right ) \, dx\\ &=-\left (10 \int \frac {e^{\frac {1}{x}} \log \left (\frac {1}{x^2}\right )}{x^2 \log ^2(x)} \, dx\right )+10 \int \frac {e^{\frac {1}{x}} \left (-2 x-\log \left (\frac {1}{x^2}\right )-x \log \left (\frac {1}{x^2}\right )\right )}{x^3 \log (x)} \, dx\\ &=10 \int \left (-\frac {2 e^{\frac {1}{x}}}{x^2 \log (x)}-\frac {e^{\frac {1}{x}} \log \left (\frac {1}{x^2}\right )}{x^3 \log (x)}-\frac {e^{\frac {1}{x}} \log \left (\frac {1}{x^2}\right )}{x^2 \log (x)}\right ) \, dx-10 \int \frac {e^{\frac {1}{x}} \log \left (\frac {1}{x^2}\right )}{x^2 \log ^2(x)} \, dx\\ &=-\left (10 \int \frac {e^{\frac {1}{x}} \log \left (\frac {1}{x^2}\right )}{x^2 \log ^2(x)} \, dx\right )-10 \int \frac {e^{\frac {1}{x}} \log \left (\frac {1}{x^2}\right )}{x^3 \log (x)} \, dx-10 \int \frac {e^{\frac {1}{x}} \log \left (\frac {1}{x^2}\right )}{x^2 \log (x)} \, dx-20 \int \frac {e^{\frac {1}{x}}}{x^2 \log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 18, normalized size = 1.00 \begin {gather*} \frac {10 e^{\frac {1}{x}} \log \left (\frac {1}{x^2}\right )}{x \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10*E^x^(-1)*x*Log[x^(-2)] + (-20*E^x^(-1)*x + E^x^(-1)*(-10 - 10*x)*Log[x^(-2)])*Log[x])/(x^3*Log[
x]^2),x]

[Out]

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

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fricas [A]  time = 0.69, size = 9, normalized size = 0.50 \begin {gather*} -\frac {20 \, e^{\frac {1}{x}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-10*x-10)*exp(1/x)*log(1/x^2)-20*x*exp(1/x))*log(x)-10*x*exp(1/x)*log(1/x^2))/x^3/log(x)^2,x, alg
orithm="fricas")

[Out]

-20*e^(1/x)/x

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giac [A]  time = 0.21, size = 9, normalized size = 0.50 \begin {gather*} -\frac {20 \, e^{\frac {1}{x}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-10*x-10)*exp(1/x)*log(1/x^2)-20*x*exp(1/x))*log(x)-10*x*exp(1/x)*log(1/x^2))/x^3/log(x)^2,x, alg
orithm="giac")

[Out]

-20*e^(1/x)/x

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maple [B]  time = 0.21, size = 41, normalized size = 2.28

method	result	size
derivativedivides	\(\frac {\frac {\left (10 \ln \left (\frac {1}{x^{2}}\right )-20 \ln \left (\frac {1}{x}\right )\right ) {\mathrm e}^{\frac {1}{x}}}{x}+\frac {20 \,{\mathrm e}^{\frac {1}{x}} \ln \left (\frac {1}{x}\right )}{x}}{\ln \relax (x )}\)	\(41\)
default	\(\frac {\frac {\left (10 \ln \left (\frac {1}{x^{2}}\right )-20 \ln \left (\frac {1}{x}\right )\right ) {\mathrm e}^{\frac {1}{x}}}{x}+\frac {20 \,{\mathrm e}^{\frac {1}{x}} \ln \left (\frac {1}{x}\right )}{x}}{\ln \relax (x )}\)	\(41\)
risch	\(-\frac {20 \,{\mathrm e}^{\frac {1}{x}}}{x}+\frac {5 i \pi \,\mathrm {csgn}\left (i x^{2}\right ) {\mathrm e}^{\frac {1}{x}} \left (\mathrm {csgn}\left (i x \right )^{2}-2 \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )+\mathrm {csgn}\left (i x^{2}\right )^{2}\right )}{x \ln \relax (x )}\)	\(64\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-10*x-10)*exp(1/x)*ln(1/x^2)-20*x*exp(1/x))*ln(x)-10*x*exp(1/x)*ln(1/x^2))/x^3/ln(x)^2,x,method=_RETURN
VERBOSE)

[Out]

((10*ln(1/x^2)-20*ln(1/x))/x*exp(1/x)+20*exp(1/x)/x*ln(1/x))/ln(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -10 \, \int \frac {x e^{\frac {1}{x}} \log \left (\frac {1}{x^{2}}\right ) + {\left ({\left (x + 1\right )} e^{\frac {1}{x}} \log \left (\frac {1}{x^{2}}\right ) + 2 \, x e^{\frac {1}{x}}\right )} \log \relax (x)}{x^{3} \log \relax (x)^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-10*x-10)*exp(1/x)*log(1/x^2)-20*x*exp(1/x))*log(x)-10*x*exp(1/x)*log(1/x^2))/x^3/log(x)^2,x, alg
orithm="maxima")

[Out]

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

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mupad [B]  time = 5.71, size = 17, normalized size = 0.94 \begin {gather*} \frac {10\,\ln \left (\frac {1}{x^2}\right )\,{\mathrm {e}}^{1/x}}{x\,\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x)*(20*x*exp(1/x) + log(1/x^2)*exp(1/x)*(10*x + 10)) + 10*x*log(1/x^2)*exp(1/x))/(x^3*log(x)^2),x)

[Out]

(10*log(1/x^2)*exp(1/x))/(x*log(x))

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sympy [A]  time = 0.26, size = 8, normalized size = 0.44 \begin {gather*} - \frac {20 e^{\frac {1}{x}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-10*x-10)*exp(1/x)*ln(1/x**2)-20*x*exp(1/x))*ln(x)-10*x*exp(1/x)*ln(1/x**2))/x**3/ln(x)**2,x)

[Out]

-20*exp(1/x)/x

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