\(\int \frac {(-3+\log (4)) \log (25)+e^{e^x-x} ((-1-x) \log (25)+e^x x \log (25))}{5 x^2} \, dx\) [1762]

Optimal result

Integrand size = 41, antiderivative size = 25 \[ \int \frac {(-3+\log (4)) \log (25)+e^{e^x-x} \left ((-1-x) \log (25)+e^x x \log (25)\right )}{5 x^2} \, dx=\frac {\left (3+e^{e^x-x}+x-\log (4)\right ) \log (25)}{5 x} \]

[Out]

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

Rubi [F]

\[ \int \frac {(-3+\log (4)) \log (25)+e^{e^x-x} \left ((-1-x) \log (25)+e^x x \log (25)\right )}{5 x^2} \, dx=\int \frac {(-3+\log (4)) \log (25)+e^{e^x-x} \left ((-1-x) \log (25)+e^x x \log (25)\right )}{5 x^2} \, dx \]

[In]

Int[((-3 + Log[4])*Log[25] + E^(E^x - x)*((-1 - x)*Log[25] + E^x*x*Log[25]))/(5*x^2),x]

[Out]

((3 - Log[4])*Log[25])/(5*x) - (Log[25]*Defer[Int][E^(E^x - x)/x^2, x])/5 + (Log[25]*Defer[Int][E^E^x/x, x])/5
- (Log[25]*Defer[Int][E^(E^x - x)/x, x])/5

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \frac {(-3+\log (4)) \log (25)+e^{e^x-x} \left ((-1-x) \log (25)+e^x x \log (25)\right )}{x^2} \, dx \\ & = \frac {1}{5} \int \left (-\frac {e^{e^x-x} (1+x) \log (25)}{x^2}+\frac {\left (e^{e^x} x-3 \left (1-\frac {2 \log (2)}{3}\right )\right ) \log (25)}{x^2}\right ) \, dx \\ & = -\left (\frac {1}{5} \log (25) \int \frac {e^{e^x-x} (1+x)}{x^2} \, dx\right )+\frac {1}{5} \log (25) \int \frac {e^{e^x} x-3 \left (1-\frac {2 \log (2)}{3}\right )}{x^2} \, dx \\ & = -\left (\frac {1}{5} \log (25) \int \left (\frac {e^{e^x-x}}{x^2}+\frac {e^{e^x-x}}{x}\right ) \, dx\right )+\frac {1}{5} \log (25) \int \left (\frac {e^{e^x}}{x}+\frac {-3+\log (4)}{x^2}\right ) \, dx \\ & = \frac {(3-\log (4)) \log (25)}{5 x}-\frac {1}{5} \log (25) \int \frac {e^{e^x-x}}{x^2} \, dx+\frac {1}{5} \log (25) \int \frac {e^{e^x}}{x} \, dx-\frac {1}{5} \log (25) \int \frac {e^{e^x-x}}{x} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {(-3+\log (4)) \log (25)+e^{e^x-x} \left ((-1-x) \log (25)+e^x x \log (25)\right )}{5 x^2} \, dx=\frac {\left (3+e^{e^x-x}-\log (4)\right ) \log (25)}{5 x} \]

[In]

Integrate[((-3 + Log[4])*Log[25] + E^(E^x - x)*((-1 - x)*Log[25] + E^x*x*Log[25]))/(5*x^2),x]

[Out]

((3 + E^(E^x - x) - Log[4])*Log[25])/(5*x)

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08

method	result	size
norman	\(\frac {\frac {2 \ln \left (5\right ) {\mathrm e}^{{\mathrm e}^{x}-x}}{5}-\frac {4 \ln \left (2\right ) \ln \left (5\right )}{5}+\frac {6 \ln \left (5\right )}{5}}{x}\)	\(27\)
parallelrisch	\(-\frac {4 \ln \left (2\right ) \ln \left (5\right )-2 \ln \left (5\right ) {\mathrm e}^{{\mathrm e}^{x}-x}-6 \ln \left (5\right )}{5 x}\)	\(28\)
risch	\(-\frac {4 \ln \left (2\right ) \ln \left (5\right )}{5 x}+\frac {6 \ln \left (5\right )}{5 x}+\frac {2 \ln \left (5\right ) {\mathrm e}^{{\mathrm e}^{x}-x}}{5 x}\)	\(32\)

[In]

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

[Out]

(2/5*ln(5)*exp(exp(x)-x)-4/5*ln(2)*ln(5)+6/5*ln(5))/x

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {(-3+\log (4)) \log (25)+e^{e^x-x} \left ((-1-x) \log (25)+e^x x \log (25)\right )}{5 x^2} \, dx=-\frac {2 \, {\left ({\left (2 \, \log \left (2\right ) - 3\right )} \log \left (5\right ) - e^{\left (-x + e^{x}\right )} \log \left (5\right )\right )}}{5 \, x} \]

[In]

integrate(1/5*((2*x*exp(x)*log(5)+2*(-1-x)*log(5))*exp(exp(x)-x)+2*(2*log(2)-3)*log(5))/x^2,x, algorithm="fric
as")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {(-3+\log (4)) \log (25)+e^{e^x-x} \left ((-1-x) \log (25)+e^x x \log (25)\right )}{5 x^2} \, dx=\frac {2 e^{- x + e^{x}} \log {\left (5 \right )}}{5 x} - \frac {- \frac {6 \log {\left (5 \right )}}{5} + \frac {4 \log {\left (2 \right )} \log {\left (5 \right )}}{5}}{x} \]

[In]

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

[Out]

2*exp(-x + exp(x))*log(5)/(5*x) - (-6*log(5)/5 + 4*log(2)*log(5)/5)/x

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {(-3+\log (4)) \log (25)+e^{e^x-x} \left ((-1-x) \log (25)+e^x x \log (25)\right )}{5 x^2} \, dx=\frac {2 \, e^{\left (-x + e^{x}\right )} \log \left (5\right )}{5 \, x} - \frac {4 \, \log \left (5\right ) \log \left (2\right )}{5 \, x} + \frac {6 \, \log \left (5\right )}{5 \, x} \]

[In]

integrate(1/5*((2*x*exp(x)*log(5)+2*(-1-x)*log(5))*exp(exp(x)-x)+2*(2*log(2)-3)*log(5))/x^2,x, algorithm="maxi
ma")

[Out]

2/5*e^(-x + e^x)*log(5)/x - 4/5*log(5)*log(2)/x + 6/5*log(5)/x

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {(-3+\log (4)) \log (25)+e^{e^x-x} \left ((-1-x) \log (25)+e^x x \log (25)\right )}{5 x^2} \, dx=\frac {2 \, {\left (e^{\left (-x + e^{x}\right )} \log \left (5\right ) - 2 \, \log \left (5\right ) \log \left (2\right ) + 3 \, \log \left (5\right )\right )}}{5 \, x} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {(-3+\log (4)) \log (25)+e^{e^x-x} \left ((-1-x) \log (25)+e^x x \log (25)\right )}{5 x^2} \, dx=\frac {{\mathrm {e}}^{{\mathrm {e}}^x-x}\,\ln \left (25\right )-2\,\ln \left (5\right )\,\left (2\,\ln \left (2\right )-3\right )}{5\,x} \]

[In]

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

[Out]

(exp(exp(x) - x)*log(25) - 2*log(5)*(2*log(2) - 3))/(5*x)