\(\int \frac {(-72+120 x-56 x^2+8 x^3) \log (\frac {e^x}{x})+(36-48 x+12 x^2) \log ^2(\frac {e^x}{x})}{e^2} \, dx\) [1413]

Optimal result

Integrand size = 50, antiderivative size = 23 \[ \int \frac {\left (-72+120 x-56 x^2+8 x^3\right ) \log \left (\frac {e^x}{x}\right )+\left (36-48 x+12 x^2\right ) \log ^2\left (\frac {e^x}{x}\right )}{e^2} \, dx=\frac {4 (3-x)^2 x \log ^2\left (\frac {e^x}{x}\right )}{e^2} \]

[Out]

4*x/exp(1)^2*ln(exp(x)/x)^2*(-x+3)^2

Rubi [F]

\[ \int \frac {\left (-72+120 x-56 x^2+8 x^3\right ) \log \left (\frac {e^x}{x}\right )+\left (36-48 x+12 x^2\right ) \log ^2\left (\frac {e^x}{x}\right )}{e^2} \, dx=\int \frac {\left (-72+120 x-56 x^2+8 x^3\right ) \log \left (\frac {e^x}{x}\right )+\left (36-48 x+12 x^2\right ) \log ^2\left (\frac {e^x}{x}\right )}{e^2} \, dx \]

[In]

Int[((-72 + 120*x - 56*x^2 + 8*x^3)*Log[E^x/x] + (36 - 48*x + 12*x^2)*Log[E^x/x]^2)/E^2,x]

[Out]

(-72*x)/E^2 + (66*x^2)/E^2 - (236*x^3)/(9*E^2) + (31*x^4)/(6*E^2) - (2*x^5)/(5*E^2) - (72*x*Log[E^x/x])/E^2 +
(60*x^2*Log[E^x/x])/E^2 - (56*x^3*Log[E^x/x])/(3*E^2) + (2*x^4*Log[E^x/x])/E^2 + (36*Defer[Int][Log[E^x/x]^2,
x])/E^2 - (48*Defer[Int][x*Log[E^x/x]^2, x])/E^2 + (12*Defer[Int][x^2*Log[E^x/x]^2, x])/E^2

Rubi steps \begin{align*} \text {integral}& = \frac {\int \left (\left (-72+120 x-56 x^2+8 x^3\right ) \log \left (\frac {e^x}{x}\right )+\left (36-48 x+12 x^2\right ) \log ^2\left (\frac {e^x}{x}\right )\right ) \, dx}{e^2} \\ & = \frac {\int \left (-72+120 x-56 x^2+8 x^3\right ) \log \left (\frac {e^x}{x}\right ) \, dx}{e^2}+\frac {\int \left (36-48 x+12 x^2\right ) \log ^2\left (\frac {e^x}{x}\right ) \, dx}{e^2} \\ & = -\frac {72 x \log \left (\frac {e^x}{x}\right )}{e^2}+\frac {60 x^2 \log \left (\frac {e^x}{x}\right )}{e^2}-\frac {56 x^3 \log \left (\frac {e^x}{x}\right )}{3 e^2}+\frac {2 x^4 \log \left (\frac {e^x}{x}\right )}{e^2}-\frac {\int \frac {2}{3} (1-x) \left (108-90 x+28 x^2-3 x^3\right ) \, dx}{e^2}+\frac {\int \left (36 \log ^2\left (\frac {e^x}{x}\right )-48 x \log ^2\left (\frac {e^x}{x}\right )+12 x^2 \log ^2\left (\frac {e^x}{x}\right )\right ) \, dx}{e^2} \\ & = -\frac {72 x \log \left (\frac {e^x}{x}\right )}{e^2}+\frac {60 x^2 \log \left (\frac {e^x}{x}\right )}{e^2}-\frac {56 x^3 \log \left (\frac {e^x}{x}\right )}{3 e^2}+\frac {2 x^4 \log \left (\frac {e^x}{x}\right )}{e^2}-\frac {2 \int (1-x) \left (108-90 x+28 x^2-3 x^3\right ) \, dx}{3 e^2}+\frac {12 \int x^2 \log ^2\left (\frac {e^x}{x}\right ) \, dx}{e^2}+\frac {36 \int \log ^2\left (\frac {e^x}{x}\right ) \, dx}{e^2}-\frac {48 \int x \log ^2\left (\frac {e^x}{x}\right ) \, dx}{e^2} \\ & = -\frac {72 x \log \left (\frac {e^x}{x}\right )}{e^2}+\frac {60 x^2 \log \left (\frac {e^x}{x}\right )}{e^2}-\frac {56 x^3 \log \left (\frac {e^x}{x}\right )}{3 e^2}+\frac {2 x^4 \log \left (\frac {e^x}{x}\right )}{e^2}-\frac {2 \int \left (108-198 x+118 x^2-31 x^3+3 x^4\right ) \, dx}{3 e^2}+\frac {12 \int x^2 \log ^2\left (\frac {e^x}{x}\right ) \, dx}{e^2}+\frac {36 \int \log ^2\left (\frac {e^x}{x}\right ) \, dx}{e^2}-\frac {48 \int x \log ^2\left (\frac {e^x}{x}\right ) \, dx}{e^2} \\ & = -\frac {72 x}{e^2}+\frac {66 x^2}{e^2}-\frac {236 x^3}{9 e^2}+\frac {31 x^4}{6 e^2}-\frac {2 x^5}{5 e^2}-\frac {72 x \log \left (\frac {e^x}{x}\right )}{e^2}+\frac {60 x^2 \log \left (\frac {e^x}{x}\right )}{e^2}-\frac {56 x^3 \log \left (\frac {e^x}{x}\right )}{3 e^2}+\frac {2 x^4 \log \left (\frac {e^x}{x}\right )}{e^2}+\frac {12 \int x^2 \log ^2\left (\frac {e^x}{x}\right ) \, dx}{e^2}+\frac {36 \int \log ^2\left (\frac {e^x}{x}\right ) \, dx}{e^2}-\frac {48 \int x \log ^2\left (\frac {e^x}{x}\right ) \, dx}{e^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {\left (-72+120 x-56 x^2+8 x^3\right ) \log \left (\frac {e^x}{x}\right )+\left (36-48 x+12 x^2\right ) \log ^2\left (\frac {e^x}{x}\right )}{e^2} \, dx=\frac {4 (-3+x)^2 x \log ^2\left (\frac {e^x}{x}\right )}{e^2} \]

[In]

Integrate[((-72 + 120*x - 56*x^2 + 8*x^3)*Log[E^x/x] + (36 - 48*x + 12*x^2)*Log[E^x/x]^2)/E^2,x]

[Out]

(4*(-3 + x)^2*x*Log[E^x/x]^2)/E^2

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.04

method	result	size
parallelrisch	\({\mathrm e}^{-2} \left (4 x^{3} \ln \left (\frac {{\mathrm e}^{x}}{x}\right )^{2}-24 x^{2} \ln \left (\frac {{\mathrm e}^{x}}{x}\right )^{2}+36 \ln \left (\frac {{\mathrm e}^{x}}{x}\right )^{2} x \right )\)	\(47\)
parts	\(72 x \,{\mathrm e}^{-2}+\frac {2 \,{\mathrm e}^{-2} x^{5}}{5}-2 \,{\mathrm e}^{-2} \ln \left (\frac {{\mathrm e}^{x}}{x}\right ) x^{4}+\frac {56 \,{\mathrm e}^{-2} \ln \left (\frac {{\mathrm e}^{x}}{x}\right ) x^{3}}{3}-60 \,{\mathrm e}^{-2} \ln \left (\frac {{\mathrm e}^{x}}{x}\right ) x^{2}+72 \,{\mathrm e}^{-2} \ln \left (\frac {{\mathrm e}^{x}}{x}\right ) x -66 \,{\mathrm e}^{-2} x^{2}-\frac {31 \,{\mathrm e}^{-2} x^{4}}{6}+\frac {236 \,{\mathrm e}^{-2} x^{3}}{9}-24 \,{\mathrm e}^{-2} \ln \left (\frac {{\mathrm e}^{x}}{x}\right )^{2} x^{2}+36 \,{\mathrm e}^{-2} \ln \left (\frac {{\mathrm e}^{x}}{x}\right )^{2} x +4 \,{\mathrm e}^{-2} \ln \left (\frac {{\mathrm e}^{x}}{x}\right )^{2} x^{3}+8 \,{\mathrm e}^{-2} \left (\frac {\ln \left (\frac {{\mathrm e}^{x}}{x}\right ) x^{4}}{4}-\frac {7 \ln \left (\frac {{\mathrm e}^{x}}{x}\right ) x^{3}}{3}+\frac {15 x^{2} \ln \left (\frac {{\mathrm e}^{x}}{x}\right )}{2}-9 x \ln \left (\frac {{\mathrm e}^{x}}{x}\right )-\frac {x^{5}}{20}+\frac {31 x^{4}}{48}-\frac {59 x^{3}}{18}+\frac {33 x^{2}}{4}-9 x \right )\)	\(235\)
default	\({\mathrm e}^{-2} \left (-6 x^{4} \ln \left (x \right )+36 x \ln \left (x \right )^{2}-72 x \ln \left (x \right )+60 x^{2} \ln \left (\frac {{\mathrm e}^{x}}{x}\right )+4 x^{3} \ln \left (x \right )^{2}-72 x \ln \left (\frac {{\mathrm e}^{x}}{x}\right )-24 x^{2} \ln \left (x \right )^{2}+2 \ln \left (\frac {{\mathrm e}^{x}}{x}\right ) x^{4}-\frac {56 \ln \left (\frac {{\mathrm e}^{x}}{x}\right ) x^{3}}{3}-\frac {16 x^{4}}{3}-24 x^{3}+72 x^{2}+2 x^{5}+\frac {88 x^{3} \ln \left (x \right )}{3}-12 x^{2} \ln \left (x \right )-\frac {88 x^{3} \left (\ln \left (\frac {{\mathrm e}^{x}}{x}\right )+\ln \left (x \right )-x \right )}{3}+12 x^{2} \left (\ln \left (\frac {{\mathrm e}^{x}}{x}\right )+\ln \left (x \right )-x \right )-24 {\left (\ln \left (\frac {{\mathrm e}^{x}}{x}\right )+\ln \left (x \right )-x \right )}^{2} x^{2}+36 {\left (\ln \left (\frac {{\mathrm e}^{x}}{x}\right )+\ln \left (x \right )-x \right )}^{2} x +48 \ln \left (x \right ) \left (\ln \left (\frac {{\mathrm e}^{x}}{x}\right )+\ln \left (x \right )-x \right ) x^{2}-72 \ln \left (x \right ) \left (\ln \left (\frac {{\mathrm e}^{x}}{x}\right )+\ln \left (x \right )-x \right ) x -8 \left (\ln \left (\frac {{\mathrm e}^{x}}{x}\right )+\ln \left (x \right )-x \right ) x^{3} \ln \left (x \right )+72 x \left (\ln \left (\frac {{\mathrm e}^{x}}{x}\right )+\ln \left (x \right )-x \right )+4 {\left (\ln \left (\frac {{\mathrm e}^{x}}{x}\right )+\ln \left (x \right )-x \right )}^{2} x^{3}+6 \left (\ln \left (\frac {{\mathrm e}^{x}}{x}\right )+\ln \left (x \right )-x \right ) x^{4}\right )\)	\(310\)
risch	\(\text {Expression too large to display}\)	\(1508\)

[In]

int(((12*x^2-48*x+36)*ln(exp(x)/x)^2+(8*x^3-56*x^2+120*x-72)*ln(exp(x)/x))/exp(1)^2,x,method=_RETURNVERBOSE)

[Out]

1/exp(1)^2*(4*x^3*ln(exp(x)/x)^2-24*x^2*ln(exp(x)/x)^2+36*ln(exp(x)/x)^2*x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {\left (-72+120 x-56 x^2+8 x^3\right ) \log \left (\frac {e^x}{x}\right )+\left (36-48 x+12 x^2\right ) \log ^2\left (\frac {e^x}{x}\right )}{e^2} \, dx=4 \, {\left (x^{3} - 6 \, x^{2} + 9 \, x\right )} e^{\left (-2\right )} \log \left (\frac {e^{x}}{x}\right )^{2} \]

[In]

integrate(((12*x^2-48*x+36)*log(exp(x)/x)^2+(8*x^3-56*x^2+120*x-72)*log(exp(x)/x))/exp(1)^2,x, algorithm="fric
as")

[Out]

4*(x^3 - 6*x^2 + 9*x)*e^(-2)*log(e^x/x)^2

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {\left (-72+120 x-56 x^2+8 x^3\right ) \log \left (\frac {e^x}{x}\right )+\left (36-48 x+12 x^2\right ) \log ^2\left (\frac {e^x}{x}\right )}{e^2} \, dx=\frac {\left (4 x^{3} - 24 x^{2} + 36 x\right ) \log {\left (\frac {e^{x}}{x} \right )}^{2}}{e^{2}} \]

[In]

integrate(((12*x**2-48*x+36)*ln(exp(x)/x)**2+(8*x**3-56*x**2+120*x-72)*ln(exp(x)/x))/exp(1)**2,x)

[Out]

(4*x**3 - 24*x**2 + 36*x)*exp(-2)*log(exp(x)/x)**2

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {\left (-72+120 x-56 x^2+8 x^3\right ) \log \left (\frac {e^x}{x}\right )+\left (36-48 x+12 x^2\right ) \log ^2\left (\frac {e^x}{x}\right )}{e^2} \, dx=4 \, {\left (x^{3} - 6 \, x^{2} + 9 \, x\right )} e^{\left (-2\right )} \log \left (\frac {e^{x}}{x}\right )^{2} \]

[In]

integrate(((12*x^2-48*x+36)*log(exp(x)/x)^2+(8*x^3-56*x^2+120*x-72)*log(exp(x)/x))/exp(1)^2,x, algorithm="maxi
ma")

[Out]

4*(x^3 - 6*x^2 + 9*x)*e^(-2)*log(e^x/x)^2

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (19) = 38\).

Time = 0.26 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.74 \[ \int \frac {\left (-72+120 x-56 x^2+8 x^3\right ) \log \left (\frac {e^x}{x}\right )+\left (36-48 x+12 x^2\right ) \log ^2\left (\frac {e^x}{x}\right )}{e^2} \, dx=4 \, {\left (x^{5} - 2 \, x^{4} \log \left (x\right ) + x^{3} \log \left (x\right )^{2} - 6 \, x^{4} + 12 \, x^{3} \log \left (x\right ) - 6 \, x^{2} \log \left (x\right )^{2} + 9 \, x^{3} - 18 \, x^{2} \log \left (x\right ) + 9 \, x \log \left (x\right )^{2}\right )} e^{\left (-2\right )} \]

[In]

integrate(((12*x^2-48*x+36)*log(exp(x)/x)^2+(8*x^3-56*x^2+120*x-72)*log(exp(x)/x))/exp(1)^2,x, algorithm="giac
")

[Out]

4*(x^5 - 2*x^4*log(x) + x^3*log(x)^2 - 6*x^4 + 12*x^3*log(x) - 6*x^2*log(x)^2 + 9*x^3 - 18*x^2*log(x) + 9*x*lo
g(x)^2)*e^(-2)

Mupad [B] (verification not implemented)

Time = 8.49 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {\left (-72+120 x-56 x^2+8 x^3\right ) \log \left (\frac {e^x}{x}\right )+\left (36-48 x+12 x^2\right ) \log ^2\left (\frac {e^x}{x}\right )}{e^2} \, dx=4\,x\,{\mathrm {e}}^{-2}\,{\ln \left (\frac {{\mathrm {e}}^x}{x}\right )}^2\,{\left (x-3\right )}^2 \]

[In]

int(exp(-2)*(log(exp(x)/x)*(120*x - 56*x^2 + 8*x^3 - 72) + log(exp(x)/x)^2*(12*x^2 - 48*x + 36)),x)

[Out]

4*x*exp(-2)*log(exp(x)/x)^2*(x - 3)^2