\(\int \frac {16-12 x+e x-4 x^2+(12 x-e x+8 x^2) \log (x)}{80 x \log ^2(x)} \, dx\) [2228]

Optimal result

Integrand size = 40, antiderivative size = 24 \[ \int \frac {16-12 x+e x-4 x^2+\left (12 x-e x+8 x^2\right ) \log (x)}{80 x \log ^2(x)} \, dx=\frac {x \left (2-\frac {e}{4}+\frac {-4+x}{x}+x\right )}{20 \log (x)} \]

[Out]

1/20*x*(2-1/4*exp(1)+x+(x-4)/x)/ln(x)

Rubi [F]

\[ \int \frac {16-12 x+e x-4 x^2+\left (12 x-e x+8 x^2\right ) \log (x)}{80 x \log ^2(x)} \, dx=\int \frac {16-12 x+e x-4 x^2+\left (12 x-e x+8 x^2\right ) \log (x)}{80 x \log ^2(x)} \, dx \]

[In]

Int[(16 - 12*x + E*x - 4*x^2 + (12*x - E*x + 8*x^2)*Log[x])/(80*x*Log[x]^2),x]

[Out]

ExpIntegralEi[2*Log[x]]/10 + ((12 - E)*LogIntegral[x])/80 + Defer[Int][(16 - (12 - E)*x - 4*x^2)/(x*Log[x]^2),
x]/80

Rubi steps \begin{align*} \text {integral}& = \int \frac {16+(-12+e) x-4 x^2+\left (12 x-e x+8 x^2\right ) \log (x)}{80 x \log ^2(x)} \, dx \\ & = \frac {1}{80} \int \frac {16+(-12+e) x-4 x^2+\left (12 x-e x+8 x^2\right ) \log (x)}{x \log ^2(x)} \, dx \\ & = \frac {1}{80} \int \left (\frac {16-(12-e) x-4 x^2}{x \log ^2(x)}+\frac {12-e+8 x}{\log (x)}\right ) \, dx \\ & = \frac {1}{80} \int \frac {16-(12-e) x-4 x^2}{x \log ^2(x)} \, dx+\frac {1}{80} \int \frac {12-e+8 x}{\log (x)} \, dx \\ & = \frac {1}{80} \int \left (\frac {12 \left (1-\frac {e}{12}\right )}{\log (x)}+\frac {8 x}{\log (x)}\right ) \, dx+\frac {1}{80} \int \frac {16-(12-e) x-4 x^2}{x \log ^2(x)} \, dx \\ & = \frac {1}{80} \int \frac {16-(12-e) x-4 x^2}{x \log ^2(x)} \, dx+\frac {1}{10} \int \frac {x}{\log (x)} \, dx+\frac {1}{80} (12-e) \int \frac {1}{\log (x)} \, dx \\ & = \frac {1}{80} (12-e) \operatorname {LogIntegral}(x)+\frac {1}{80} \int \frac {16-(12-e) x-4 x^2}{x \log ^2(x)} \, dx+\frac {1}{10} \text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right ) \\ & = \frac {1}{10} \operatorname {ExpIntegralEi}(2 \log (x))+\frac {1}{80} (12-e) \operatorname {LogIntegral}(x)+\frac {1}{80} \int \frac {16-(12-e) x-4 x^2}{x \log ^2(x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {16-12 x+e x-4 x^2+\left (12 x-e x+8 x^2\right ) \log (x)}{80 x \log ^2(x)} \, dx=\frac {-16+12 x-e x+4 x^2}{80 \log (x)} \]

[In]

Integrate[(16 - 12*x + E*x - 4*x^2 + (12*x - E*x + 8*x^2)*Log[x])/(80*x*Log[x]^2),x]

[Out]

(-16 + 12*x - E*x + 4*x^2)/(80*Log[x])

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88

method	result	size
norman	\(\frac {-\frac {1}{5}+\left (-\frac {{\mathrm e}}{80}+\frac {3}{20}\right ) x +\frac {x^{2}}{20}}{\ln \left (x \right )}\)	\(21\)
risch	\(-\frac {x \,{\mathrm e}-4 x^{2}-12 x +16}{80 \ln \left (x \right )}\)	\(21\)
parallelrisch	\(-\frac {x \,{\mathrm e}-4 x^{2}-12 x +16}{80 \ln \left (x \right )}\)	\(21\)
default	\(\frac {{\mathrm e} \,\operatorname {Ei}_{1}\left (-\ln \left (x \right )\right )}{80}+\frac {{\mathrm e} \left (-\frac {x}{\ln \left (x \right )}-\operatorname {Ei}_{1}\left (-\ln \left (x \right )\right )\right )}{80}+\frac {x^{2}}{20 \ln \left (x \right )}+\frac {3 x}{20 \ln \left (x \right )}-\frac {1}{5 \ln \left (x \right )}\)	\(54\)
parts	\(\frac {{\mathrm e} \,\operatorname {Ei}_{1}\left (-\ln \left (x \right )\right )}{80}+\frac {{\mathrm e} \left (-\frac {x}{\ln \left (x \right )}-\operatorname {Ei}_{1}\left (-\ln \left (x \right )\right )\right )}{80}+\frac {x^{2}}{20 \ln \left (x \right )}+\frac {3 x}{20 \ln \left (x \right )}-\frac {1}{5 \ln \left (x \right )}\)	\(54\)

[In]

int(1/80*((-x*exp(1)+8*x^2+12*x)*ln(x)+x*exp(1)-4*x^2-12*x+16)/x/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

(-1/5+(-1/80*exp(1)+3/20)*x+1/20*x^2)/ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {16-12 x+e x-4 x^2+\left (12 x-e x+8 x^2\right ) \log (x)}{80 x \log ^2(x)} \, dx=\frac {4 \, x^{2} - x e + 12 \, x - 16}{80 \, \log \left (x\right )} \]

[In]

integrate(1/80*((-x*exp(1)+8*x^2+12*x)*log(x)+x*exp(1)-4*x^2-12*x+16)/x/log(x)^2,x, algorithm="fricas")

[Out]

1/80*(4*x^2 - x*e + 12*x - 16)/log(x)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {16-12 x+e x-4 x^2+\left (12 x-e x+8 x^2\right ) \log (x)}{80 x \log ^2(x)} \, dx=\frac {4 x^{2} - e x + 12 x - 16}{80 \log {\left (x \right )}} \]

[In]

integrate(1/80*((-x*exp(1)+8*x**2+12*x)*ln(x)+x*exp(1)-4*x**2-12*x+16)/x/ln(x)**2,x)

[Out]

(4*x**2 - E*x + 12*x - 16)/(80*log(x))

Maxima [C] (verification not implemented)

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

Time = 0.22 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.17 \[ \int \frac {16-12 x+e x-4 x^2+\left (12 x-e x+8 x^2\right ) \log (x)}{80 x \log ^2(x)} \, dx=-\frac {1}{80} \, {\rm Ei}\left (\log \left (x\right )\right ) e + \frac {1}{80} \, e \Gamma \left (-1, -\log \left (x\right )\right ) - \frac {1}{5 \, \log \left (x\right )} + \frac {1}{10} \, {\rm Ei}\left (2 \, \log \left (x\right )\right ) + \frac {3}{20} \, {\rm Ei}\left (\log \left (x\right )\right ) - \frac {3}{20} \, \Gamma \left (-1, -\log \left (x\right )\right ) - \frac {1}{10} \, \Gamma \left (-1, -2 \, \log \left (x\right )\right ) \]

[In]

integrate(1/80*((-x*exp(1)+8*x^2+12*x)*log(x)+x*exp(1)-4*x^2-12*x+16)/x/log(x)^2,x, algorithm="maxima")

[Out]

-1/80*Ei(log(x))*e + 1/80*e*gamma(-1, -log(x)) - 1/5/log(x) + 1/10*Ei(2*log(x)) + 3/20*Ei(log(x)) - 3/20*gamma
(-1, -log(x)) - 1/10*gamma(-1, -2*log(x))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {16-12 x+e x-4 x^2+\left (12 x-e x+8 x^2\right ) \log (x)}{80 x \log ^2(x)} \, dx=\frac {4 \, x^{2} - x e + 12 \, x - 16}{80 \, \log \left (x\right )} \]

[In]

integrate(1/80*((-x*exp(1)+8*x^2+12*x)*log(x)+x*exp(1)-4*x^2-12*x+16)/x/log(x)^2,x, algorithm="giac")

[Out]

1/80*(4*x^2 - x*e + 12*x - 16)/log(x)

Mupad [B] (verification not implemented)

Time = 9.52 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {16-12 x+e x-4 x^2+\left (12 x-e x+8 x^2\right ) \log (x)}{80 x \log ^2(x)} \, dx=-\frac {-4\,x^4+\left (\mathrm {e}-12\right )\,x^3+16\,x^2}{80\,x^2\,\ln \left (x\right )} \]

[In]

int(((x*exp(1))/80 - (3*x)/20 - x^2/20 + (log(x)*(12*x - x*exp(1) + 8*x^2))/80 + 1/5)/(x*log(x)^2),x)

[Out]

-(16*x^2 - 4*x^4 + x^3*(exp(1) - 12))/(80*x^2*log(x))