3.25.18 \(\int \frac {(-2+2 x) \log (\frac {50 e^{5+x}}{x})-\log ^2(\frac {50 e^{5+x}}{x})}{x^2} \, dx\)

Optimal. Leaf size=19 \[ 2+\frac {\log ^2\left (\frac {50 e^{5+x}}{x}\right )}{x} \]

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Rubi [F]  time = 0.19, 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 {(-2+2 x) \log \left (\frac {50 e^{5+x}}{x}\right )-\log ^2\left (\frac {50 e^{5+x}}{x}\right )}{x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-2 + 2*x)*Log[(50*E^(5 + x))/x] - Log[(50*E^(5 + x))/x]^2)/x^2,x]

[Out]

-2/x + (2*Log[(50*E^(5 + x))/x])/x - 2*Log[x] + 2*Defer[Int][Log[(50*E^(5 + x))/x]/x, x] - Defer[Int][Log[(50*
E^(5 + x))/x]^2/x^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2 (-1+x) \log \left (\frac {50 e^{5+x}}{x}\right )}{x^2}-\frac {\log ^2\left (\frac {50 e^{5+x}}{x}\right )}{x^2}\right ) \, dx\\ &=2 \int \frac {(-1+x) \log \left (\frac {50 e^{5+x}}{x}\right )}{x^2} \, dx-\int \frac {\log ^2\left (\frac {50 e^{5+x}}{x}\right )}{x^2} \, dx\\ &=2 \int \left (-\frac {\log \left (\frac {50 e^{5+x}}{x}\right )}{x^2}+\frac {\log \left (\frac {50 e^{5+x}}{x}\right )}{x}\right ) \, dx-\int \frac {\log ^2\left (\frac {50 e^{5+x}}{x}\right )}{x^2} \, dx\\ &=-\left (2 \int \frac {\log \left (\frac {50 e^{5+x}}{x}\right )}{x^2} \, dx\right )+2 \int \frac {\log \left (\frac {50 e^{5+x}}{x}\right )}{x} \, dx-\int \frac {\log ^2\left (\frac {50 e^{5+x}}{x}\right )}{x^2} \, dx\\ &=\frac {2 \log \left (\frac {50 e^{5+x}}{x}\right )}{x}-2 \int \frac {-1+x}{x^2} \, dx+2 \int \frac {\log \left (\frac {50 e^{5+x}}{x}\right )}{x} \, dx-\int \frac {\log ^2\left (\frac {50 e^{5+x}}{x}\right )}{x^2} \, dx\\ &=\frac {2 \log \left (\frac {50 e^{5+x}}{x}\right )}{x}-2 \int \left (-\frac {1}{x^2}+\frac {1}{x}\right ) \, dx+2 \int \frac {\log \left (\frac {50 e^{5+x}}{x}\right )}{x} \, dx-\int \frac {\log ^2\left (\frac {50 e^{5+x}}{x}\right )}{x^2} \, dx\\ &=-\frac {2}{x}+\frac {2 \log \left (\frac {50 e^{5+x}}{x}\right )}{x}-2 \log (x)+2 \int \frac {\log \left (\frac {50 e^{5+x}}{x}\right )}{x} \, dx-\int \frac {\log ^2\left (\frac {50 e^{5+x}}{x}\right )}{x^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.06, size = 39, normalized size = 2.05 \begin {gather*} 2+2 x-2 \log \left (\frac {50 e^{5+x}}{x}\right )+\frac {\log ^2\left (\frac {50 e^{5+x}}{x}\right )}{x}-2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-2 + 2*x)*Log[(50*E^(5 + x))/x] - Log[(50*E^(5 + x))/x]^2)/x^2,x]

[Out]

2 + 2*x - 2*Log[(50*E^(5 + x))/x] + Log[(50*E^(5 + x))/x]^2/x - 2*Log[x]

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fricas [A]  time = 0.81, size = 30, normalized size = 1.58 \begin {gather*} \frac {x^{2} + \log \left (50\right )^{2} - 2 \, {\left (x + \log \left (50\right ) + 5\right )} \log \relax (x) + \log \relax (x)^{2} + 10 \, \log \left (50\right ) + 25}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(50*exp(x-log(x)+5))^2+(2*x-2)*log(50*exp(x-log(x)+5)))/x^2,x, algorithm="fricas")

[Out]

(x^2 + log(50)^2 - 2*(x + log(50) + 5)*log(x) + log(x)^2 + 10*log(50) + 25)/x

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giac [B]  time = 0.16, size = 39, normalized size = 2.05 \begin {gather*} x - \frac {2 \, {\left (\log \left (50\right ) + 5\right )} \log \relax (x)}{x} + \frac {\log \relax (x)^{2}}{x} + \frac {\log \left (50\right )^{2} + 10 \, \log \left (50\right ) + 25}{x} - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(50*exp(x-log(x)+5))^2+(2*x-2)*log(50*exp(x-log(x)+5)))/x^2,x, algorithm="giac")

[Out]

x - 2*(log(50) + 5)*log(x)/x + log(x)^2/x + (log(50)^2 + 10*log(50) + 25)/x - 2*log(x)

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maple [A]  time = 0.11, size = 18, normalized size = 0.95

method	result	size
default	\(\frac {\ln \left (50 \,{\mathrm e}^{x -\ln \relax (x )+5}\right )^{2}}{x}\)	\(18\)
risch	\(\frac {\ln \left (\frac {{\mathrm e}^{5+x}}{x}\right )^{2}}{x}+\frac {\left (4 \ln \relax (5)+2 \ln \relax (2)\right ) \ln \left (\frac {{\mathrm e}^{5+x}}{x}\right )}{x}+\frac {\ln \relax (2)^{2}}{x}+\frac {4 \ln \relax (2) \ln \relax (5)}{x}+\frac {4 \ln \relax (5)^{2}}{x}\)	\(65\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-ln(50*exp(x-ln(x)+5))^2+(2*x-2)*ln(50*exp(x-ln(x)+5)))/x^2,x,method=_RETURNVERBOSE)

[Out]

ln(50*exp(x-ln(x)+5))^2/x

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maxima [B]  time = 0.48, size = 119, normalized size = 6.26 \begin {gather*} -2 \, {\left (x - \log \relax (x)\right )} \log \relax (x) - \log \relax (x)^{2} - 2 \, {\left (\frac {1}{x} + \log \relax (x)\right )} \log \left (\frac {50 \, e^{\left (x + 5\right )}}{x}\right ) + 2 \, \log \relax (x) \log \left (\frac {50 \, e^{\left (x + 5\right )}}{x}\right ) + 2 \, x + \frac {\log \left (\frac {50 \, e^{\left (x + 5\right )}}{x}\right )^{2}}{x} - \frac {x \log \relax (x)^{2} + 2 \, x^{2} - 2 \, {\left (x^{2} + x\right )} \log \relax (x) - 2}{x} + \frac {2 \, \log \left (\frac {50 \, e^{\left (x + 5\right )}}{x}\right )}{x} - \frac {2}{x} - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(50*exp(x-log(x)+5))^2+(2*x-2)*log(50*exp(x-log(x)+5)))/x^2,x, algorithm="maxima")

[Out]

-2*(x - log(x))*log(x) - log(x)^2 - 2*(1/x + log(x))*log(50*e^(x + 5)/x) + 2*log(x)*log(50*e^(x + 5)/x) + 2*x
+ log(50*e^(x + 5)/x)^2/x - (x*log(x)^2 + 2*x^2 - 2*(x^2 + x)*log(x) - 2)/x + 2*log(50*e^(x + 5)/x)/x - 2/x -
2*log(x)

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mupad [B]  time = 1.47, size = 42, normalized size = 2.21 \begin {gather*} \frac {10\,\ln \left (\frac {1}{x}\right )+10\,\ln \left (50\right )+2\,\ln \left (\frac {1}{x}\right )\,\ln \left (50\right )+{\ln \left (\frac {1}{x}\right )}^2-2\,x\,\ln \relax (x)+{\ln \left (50\right )}^2+x^2+25}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(50*exp(x - log(x) + 5))^2 - log(50*exp(x - log(x) + 5))*(2*x - 2))/x^2,x)

[Out]

(10*log(1/x) + 10*log(50) + 2*log(1/x)*log(50) + log(1/x)^2 - 2*x*log(x) + log(50)^2 + x^2 + 25)/x

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sympy [A]  time = 0.17, size = 12, normalized size = 0.63 \begin {gather*} \frac {\log {\left (\frac {50 e^{x + 5}}{x} \right )}^{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-ln(50*exp(x-ln(x)+5))**2+(2*x-2)*ln(50*exp(x-ln(x)+5)))/x**2,x)

[Out]

log(50*exp(x + 5)/x)**2/x

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