\(\int \frac {(-2+2 x) \log (\frac {50 e^{5+x}}{x})-\log ^2(\frac {50 e^{5+x}}{x})}{x^2} \, dx\) [2418]

Optimal result

Integrand size = 37, antiderivative size = 19 \[ \int \frac {(-2+2 x) \log \left (\frac {50 e^{5+x}}{x}\right )-\log ^2\left (\frac {50 e^{5+x}}{x}\right )}{x^2} \, dx=2+\frac {\log ^2\left (\frac {50 e^{5+x}}{x}\right )}{x} \]

[Out]

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

Rubi [F]

\[ \int \frac {(-2+2 x) \log \left (\frac {50 e^{5+x}}{x}\right )-\log ^2\left (\frac {50 e^{5+x}}{x}\right )}{x^2} \, dx=\int \frac {(-2+2 x) \log \left (\frac {50 e^{5+x}}{x}\right )-\log ^2\left (\frac {50 e^{5+x}}{x}\right )}{x^2} \, dx \]

[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{align*} \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{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(39\) vs. \(2(19)=38\).

Time = 0.11 (sec) , antiderivative size = 39, normalized size of antiderivative = 2.05 \[ \int \frac {(-2+2 x) \log \left (\frac {50 e^{5+x}}{x}\right )-\log ^2\left (\frac {50 e^{5+x}}{x}\right )}{x^2} \, dx=2+2 x-2 \log \left (\frac {50 e^{5+x}}{x}\right )+\frac {\log ^2\left (\frac {50 e^{5+x}}{x}\right )}{x}-2 \log (x) \]

[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]

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.58 \[ \int \frac {(-2+2 x) \log \left (\frac {50 e^{5+x}}{x}\right )-\log ^2\left (\frac {50 e^{5+x}}{x}\right )}{x^2} \, dx=\frac {x^{2} + \log \left (50\right )^{2} - 2 \, {\left (x + \log \left (50\right ) + 5\right )} \log \left (x\right ) + \log \left (x\right )^{2} + 10 \, \log \left (50\right ) + 25}{x} \]

[In]

integrate((-log(50*exp(x-log(x)+5))^2+(-2+2*x)*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

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63 \[ \int \frac {(-2+2 x) \log \left (\frac {50 e^{5+x}}{x}\right )-\log ^2\left (\frac {50 e^{5+x}}{x}\right )}{x^2} \, dx=\frac {\log {\left (\frac {50 e^{x + 5}}{x} \right )}^{2}}{x} \]

[In]

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

[Out]

log(50*exp(x + 5)/x)**2/x

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 119 vs. \(2 (19) = 38\).

Time = 0.20 (sec) , antiderivative size = 119, normalized size of antiderivative = 6.26 \[ \int \frac {(-2+2 x) \log \left (\frac {50 e^{5+x}}{x}\right )-\log ^2\left (\frac {50 e^{5+x}}{x}\right )}{x^2} \, dx=-2 \, {\left (x - \log \left (x\right )\right )} \log \left (x\right ) - \log \left (x\right )^{2} - 2 \, {\left (\frac {1}{x} + \log \left (x\right )\right )} \log \left (\frac {50 \, e^{\left (x + 5\right )}}{x}\right ) + 2 \, \log \left (x\right ) \log \left (\frac {50 \, e^{\left (x + 5\right )}}{x}\right ) + 2 \, x + \frac {\log \left (\frac {50 \, e^{\left (x + 5\right )}}{x}\right )^{2}}{x} - \frac {x \log \left (x\right )^{2} + 2 \, x^{2} - 2 \, {\left (x^{2} + x\right )} \log \left (x\right ) - 2}{x} + \frac {2 \, \log \left (\frac {50 \, e^{\left (x + 5\right )}}{x}\right )}{x} - \frac {2}{x} - 2 \, \log \left (x\right ) \]

[In]

integrate((-log(50*exp(x-log(x)+5))^2+(-2+2*x)*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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (19) = 38\).

Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 2.05 \[ \int \frac {(-2+2 x) \log \left (\frac {50 e^{5+x}}{x}\right )-\log ^2\left (\frac {50 e^{5+x}}{x}\right )}{x^2} \, dx=x - \frac {2 \, {\left (\log \left (50\right ) + 5\right )} \log \left (x\right )}{x} + \frac {\log \left (x\right )^{2}}{x} + \frac {\log \left (50\right )^{2} + 10 \, \log \left (50\right ) + 25}{x} - 2 \, \log \left (x\right ) \]

[In]

integrate((-log(50*exp(x-log(x)+5))^2+(-2+2*x)*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)

Mupad [B] (verification not implemented)

Time = 9.07 (sec) , antiderivative size = 42, normalized size of antiderivative = 2.21 \[ \int \frac {(-2+2 x) \log \left (\frac {50 e^{5+x}}{x}\right )-\log ^2\left (\frac {50 e^{5+x}}{x}\right )}{x^2} \, dx=\frac {10\,\ln \left (\frac {1}{x}\right )+10\,\ln \left (50\right )+2\,\ln \left (\frac {1}{x}\right )\,\ln \left (50\right )+{\ln \left (\frac {1}{x}\right )}^2-2\,x\,\ln \left (x\right )+{\ln \left (50\right )}^2+x^2+25}{x} \]

[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