\(\int \frac {-10-x+x \log (x)+(-1-2 x-x^2+(2 x+2 x^2) \log (x)) \log (\frac {\log (5)}{25})}{2 x \log ^2(x)} \, dx\) [439]

Optimal result

Integrand size = 50, antiderivative size = 28 \[ \int \frac {-10-x+x \log (x)+\left (-1-2 x-x^2+\left (2 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {\log (5)}{25}\right )}{2 x \log ^2(x)} \, dx=\frac {5+\frac {1}{2} \left (x+\log (x)+(1+x)^2 \log \left (\frac {\log (5)}{25}\right )\right )}{\log (x)} \]

[Out]

(5+1/2*x+1/2*ln(1/25*ln(5))*(1+x)^2+1/2*ln(x))/ln(x)

Rubi [F]

\[ \int \frac {-10-x+x \log (x)+\left (-1-2 x-x^2+\left (2 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {\log (5)}{25}\right )}{2 x \log ^2(x)} \, dx=\int \frac {-10-x+x \log (x)+\left (-1-2 x-x^2+\left (2 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {\log (5)}{25}\right )}{2 x \log ^2(x)} \, dx \]

[In]

Int[(-10 - x + x*Log[x] + (-1 - 2*x - x^2 + (2*x + 2*x^2)*Log[x])*Log[Log[5]/25])/(2*x*Log[x]^2),x]

[Out]

ExpIntegralEi[2*Log[x]]*Log[Log[5]/25] + ((1 + 2*Log[Log[5]/25])*LogIntegral[x])/2 + Defer[Int][(-10 + Log[25/
Log[5]] - x^2*Log[Log[5]/25] - x*(1 - Log[625] + 2*Log[Log[5]]))/(x*Log[x]^2), x]/2

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {-10-x+x \log (x)+\left (-1-2 x-x^2+\left (2 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {\log (5)}{25}\right )}{x \log ^2(x)} \, dx \\ & = \frac {1}{2} \int \left (\frac {1+2 \log \left (\frac {\log (5)}{25}\right )+2 x \log \left (\frac {\log (5)}{25}\right )}{\log (x)}+\frac {-10+\log \left (\frac {25}{\log (5)}\right )-x^2 \log \left (\frac {\log (5)}{25}\right )-x (1-\log (625)+2 \log (\log (5)))}{x \log ^2(x)}\right ) \, dx \\ & = \frac {1}{2} \int \frac {1+2 \log \left (\frac {\log (5)}{25}\right )+2 x \log \left (\frac {\log (5)}{25}\right )}{\log (x)} \, dx+\frac {1}{2} \int \frac {-10+\log \left (\frac {25}{\log (5)}\right )-x^2 \log \left (\frac {\log (5)}{25}\right )-x (1-\log (625)+2 \log (\log (5)))}{x \log ^2(x)} \, dx \\ & = \frac {1}{2} \int \left (\frac {2 x \log \left (\frac {\log (5)}{25}\right )}{\log (x)}+\frac {1+2 \log \left (\frac {\log (5)}{25}\right )}{\log (x)}\right ) \, dx+\frac {1}{2} \int \frac {-10+\log \left (\frac {25}{\log (5)}\right )-x^2 \log \left (\frac {\log (5)}{25}\right )-x (1-\log (625)+2 \log (\log (5)))}{x \log ^2(x)} \, dx \\ & = \frac {1}{2} \int \frac {-10+\log \left (\frac {25}{\log (5)}\right )-x^2 \log \left (\frac {\log (5)}{25}\right )-x (1-\log (625)+2 \log (\log (5)))}{x \log ^2(x)} \, dx+\log \left (\frac {\log (5)}{25}\right ) \int \frac {x}{\log (x)} \, dx+\frac {1}{2} \left (1+2 \log \left (\frac {\log (5)}{25}\right )\right ) \int \frac {1}{\log (x)} \, dx \\ & = \frac {1}{2} \left (1+2 \log \left (\frac {\log (5)}{25}\right )\right ) \operatorname {LogIntegral}(x)+\frac {1}{2} \int \frac {-10+\log \left (\frac {25}{\log (5)}\right )-x^2 \log \left (\frac {\log (5)}{25}\right )-x (1-\log (625)+2 \log (\log (5)))}{x \log ^2(x)} \, dx+\log \left (\frac {\log (5)}{25}\right ) \text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right ) \\ & = \operatorname {ExpIntegralEi}(2 \log (x)) \log \left (\frac {\log (5)}{25}\right )+\frac {1}{2} \left (1+2 \log \left (\frac {\log (5)}{25}\right )\right ) \operatorname {LogIntegral}(x)+\frac {1}{2} \int \frac {-10+\log \left (\frac {25}{\log (5)}\right )-x^2 \log \left (\frac {\log (5)}{25}\right )-x (1-\log (625)+2 \log (\log (5)))}{x \log ^2(x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {-10-x+x \log (x)+\left (-1-2 x-x^2+\left (2 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {\log (5)}{25}\right )}{2 x \log ^2(x)} \, dx=\frac {10+x+\log \left (\frac {\log (5)}{25}\right )+2 x \log \left (\frac {\log (5)}{25}\right )+x^2 \log \left (\frac {\log (5)}{25}\right )}{2 \log (x)} \]

[In]

Integrate[(-10 - x + x*Log[x] + (-1 - 2*x - x^2 + (2*x + 2*x^2)*Log[x])*Log[Log[5]/25])/(2*x*Log[x]^2),x]

[Out]

(10 + x + Log[Log[5]/25] + 2*x*Log[Log[5]/25] + x^2*Log[Log[5]/25])/(2*Log[x])

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14

method	result	size
parallelrisch	\(\frac {\ln \left (\frac {\ln \left (5\right )}{25}\right ) x^{2}+2 \ln \left (\frac {\ln \left (5\right )}{25}\right ) x +10+\ln \left (\frac {\ln \left (5\right )}{25}\right )+x}{2 \ln \left (x \right )}\)	\(32\)
norman	\(\frac {\left (-\ln \left (5\right )+\frac {\ln \left (\ln \left (5\right )\right )}{2}\right ) x^{2}+\left (-2 \ln \left (5\right )+\ln \left (\ln \left (5\right )\right )+\frac {1}{2}\right ) x +5-\ln \left (5\right )+\frac {\ln \left (\ln \left (5\right )\right )}{2}}{\ln \left (x \right )}\)	\(42\)
risch	\(-\frac {2 x^{2} \ln \left (5\right )-\ln \left (\ln \left (5\right )\right ) x^{2}+4 x \ln \left (5\right )-2 x \ln \left (\ln \left (5\right )\right )+2 \ln \left (5\right )-\ln \left (\ln \left (5\right )\right )-x -10}{2 \ln \left (x \right )}\)	\(47\)
default	\(2 \ln \left (5\right ) \operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )-\ln \left (\ln \left (5\right )\right ) \operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )+2 \ln \left (5\right ) \operatorname {Ei}_{1}\left (-\ln \left (x \right )\right )-\ln \left (\ln \left (5\right )\right ) \operatorname {Ei}_{1}\left (-\ln \left (x \right )\right )+\ln \left (5\right ) \left (-\frac {x^{2}}{\ln \left (x \right )}-2 \,\operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )\right )-\frac {\ln \left (\ln \left (5\right )\right ) \left (-\frac {x^{2}}{\ln \left (x \right )}-2 \,\operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )\right )}{2}+2 \ln \left (5\right ) \left (-\frac {x}{\ln \left (x \right )}-\operatorname {Ei}_{1}\left (-\ln \left (x \right )\right )\right )-\ln \left (\ln \left (5\right )\right ) \left (-\frac {x}{\ln \left (x \right )}-\operatorname {Ei}_{1}\left (-\ln \left (x \right )\right )\right )-\frac {\ln \left (5\right )}{\ln \left (x \right )}+\frac {\ln \left (\ln \left (5\right )\right )}{2 \ln \left (x \right )}+\frac {x}{2 \ln \left (x \right )}+\frac {5}{\ln \left (x \right )}\)	\(159\)
parts	\(2 \ln \left (5\right ) \operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )-\ln \left (\ln \left (5\right )\right ) \operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )+2 \ln \left (5\right ) \operatorname {Ei}_{1}\left (-\ln \left (x \right )\right )-\ln \left (\ln \left (5\right )\right ) \operatorname {Ei}_{1}\left (-\ln \left (x \right )\right )+\ln \left (5\right ) \left (-\frac {x^{2}}{\ln \left (x \right )}-2 \,\operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )\right )-\frac {\ln \left (\ln \left (5\right )\right ) \left (-\frac {x^{2}}{\ln \left (x \right )}-2 \,\operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )\right )}{2}+2 \ln \left (5\right ) \left (-\frac {x}{\ln \left (x \right )}-\operatorname {Ei}_{1}\left (-\ln \left (x \right )\right )\right )-\ln \left (\ln \left (5\right )\right ) \left (-\frac {x}{\ln \left (x \right )}-\operatorname {Ei}_{1}\left (-\ln \left (x \right )\right )\right )-\frac {\ln \left (5\right )}{\ln \left (x \right )}+\frac {\ln \left (\ln \left (5\right )\right )}{2 \ln \left (x \right )}+\frac {x}{2 \ln \left (x \right )}+\frac {5}{\ln \left (x \right )}\)	\(159\)

[In]

int(1/2*(((2*x^2+2*x)*ln(x)-x^2-2*x-1)*ln(1/25*ln(5))+x*ln(x)-x-10)/x/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

1/2*(ln(1/25*ln(5))*x^2+2*ln(1/25*ln(5))*x+10+ln(1/25*ln(5))+x)/ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int \frac {-10-x+x \log (x)+\left (-1-2 x-x^2+\left (2 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {\log (5)}{25}\right )}{2 x \log ^2(x)} \, dx=\frac {{\left (x^{2} + 2 \, x + 1\right )} \log \left (\frac {1}{25} \, \log \left (5\right )\right ) + x + 10}{2 \, \log \left (x\right )} \]

[In]

integrate(1/2*(((2*x^2+2*x)*log(x)-x^2-2*x-1)*log(1/25*log(5))+x*log(x)-x-10)/x/log(x)^2,x, algorithm="fricas"
)

[Out]

1/2*((x^2 + 2*x + 1)*log(1/25*log(5)) + x + 10)/log(x)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.75 \[ \int \frac {-10-x+x \log (x)+\left (-1-2 x-x^2+\left (2 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {\log (5)}{25}\right )}{2 x \log ^2(x)} \, dx=\frac {- 2 x^{2} \log {\left (5 \right )} + x^{2} \log {\left (\log {\left (5 \right )} \right )} - 4 x \log {\left (5 \right )} + 2 x \log {\left (\log {\left (5 \right )} \right )} + x - 2 \log {\left (5 \right )} + \log {\left (\log {\left (5 \right )} \right )} + 10}{2 \log {\left (x \right )}} \]

[In]

integrate(1/2*(((2*x**2+2*x)*ln(x)-x**2-2*x-1)*ln(1/25*ln(5))+x*ln(x)-x-10)/x/ln(x)**2,x)

[Out]

(-2*x**2*log(5) + x**2*log(log(5)) - 4*x*log(5) + 2*x*log(log(5)) + x - 2*log(5) + log(log(5)) + 10)/(2*log(x)
)

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.23 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.75 \[ \int \frac {-10-x+x \log (x)+\left (-1-2 x-x^2+\left (2 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {\log (5)}{25}\right )}{2 x \log ^2(x)} \, dx={\rm Ei}\left (2 \, \log \left (x\right )\right ) \log \left (\frac {1}{25} \, \log \left (5\right )\right ) + {\rm Ei}\left (\log \left (x\right )\right ) \log \left (\frac {1}{25} \, \log \left (5\right )\right ) - \Gamma \left (-1, -\log \left (x\right )\right ) \log \left (\frac {1}{25} \, \log \left (5\right )\right ) - \Gamma \left (-1, -2 \, \log \left (x\right )\right ) \log \left (\frac {1}{25} \, \log \left (5\right )\right ) + \frac {\log \left (\frac {1}{25} \, \log \left (5\right )\right )}{2 \, \log \left (x\right )} + \frac {5}{\log \left (x\right )} + \frac {1}{2} \, {\rm Ei}\left (\log \left (x\right )\right ) - \frac {1}{2} \, \Gamma \left (-1, -\log \left (x\right )\right ) \]

[In]

integrate(1/2*(((2*x^2+2*x)*log(x)-x^2-2*x-1)*log(1/25*log(5))+x*log(x)-x-10)/x/log(x)^2,x, algorithm="maxima"
)

[Out]

Ei(2*log(x))*log(1/25*log(5)) + Ei(log(x))*log(1/25*log(5)) - gamma(-1, -log(x))*log(1/25*log(5)) - gamma(-1,
-2*log(x))*log(1/25*log(5)) + 1/2*log(1/25*log(5))/log(x) + 5/log(x) + 1/2*Ei(log(x)) - 1/2*gamma(-1, -log(x))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (22) = 44\).

Time = 0.28 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.64 \[ \int \frac {-10-x+x \log (x)+\left (-1-2 x-x^2+\left (2 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {\log (5)}{25}\right )}{2 x \log ^2(x)} \, dx=-\frac {2 \, x^{2} \log \left (5\right ) - x^{2} \log \left (\log \left (5\right )\right ) + 4 \, x \log \left (5\right ) - 2 \, x \log \left (\log \left (5\right )\right ) - x + 2 \, \log \left (5\right ) - \log \left (\log \left (5\right )\right ) - 10}{2 \, \log \left (x\right )} \]

[In]

integrate(1/2*(((2*x^2+2*x)*log(x)-x^2-2*x-1)*log(1/25*log(5))+x*log(x)-x-10)/x/log(x)^2,x, algorithm="giac")

[Out]

-1/2*(2*x^2*log(5) - x^2*log(log(5)) + 4*x*log(5) - 2*x*log(log(5)) - x + 2*log(5) - log(log(5)) - 10)/log(x)

Mupad [B] (verification not implemented)

Time = 8.16 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.54 \[ \int \frac {-10-x+x \log (x)+\left (-1-2 x-x^2+\left (2 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {\log (5)}{25}\right )}{2 x \log ^2(x)} \, dx=\frac {\ln \left (\frac {\ln \left (5\right )}{25}\right )\,x^4+\left (2\,\ln \left (\frac {\ln \left (5\right )}{25}\right )+1\right )\,x^3+\left (\ln \left (\frac {\ln \left (5\right )}{25}\right )+10\right )\,x^2}{2\,x^2\,\ln \left (x\right )} \]

[In]

int(-(x/2 + (log(log(5)/25)*(2*x - log(x)*(2*x + 2*x^2) + x^2 + 1))/2 - (x*log(x))/2 + 5)/(x*log(x)^2),x)

[Out]

(x^3*(2*log(log(5)/25) + 1) + x^4*log(log(5)/25) + x^2*(log(log(5)/25) + 10))/(2*x^2*log(x))