3.89.15 $\int \frac{e^{-256x+32x^2-x^3}(2+2x+2e^{2x}{x}^2+e^x(4x+x^2-x^3)+(-512x-128x^2+58x^3-3x^4+e^{2x}(-512x^3+128x^4-6x^5)+e^x(-1024x^2+52x^4-3x^5))\log (\frac{2x+x^2+2e^x{x}^2}{2+2e^{x}x}))}{2x+x^2+2e^{2x}{x}^3+e^x(4x^2+x^3)}dx$

Optimal. Leaf size=29 $$e^{-(-16+x{)}^{2}x}\log (x+\frac{x}{2e^x+\frac{2}{x}})$$

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Rubi [B]  time = 0.47, antiderivative size = 152, normalized size of antiderivative = 5.24, number of steps used = 1, number of rules used = 1, integrand size = 169, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.006, Rules used = {2288} $$\begin{array}{c}\frac{e^{-x^3+32x^2-256x}(3x^4-58x^3+128x^2+2e^{2x}(3x^5-64x^4+256x^3)+e^x(3x^5-52x^4+1024x^2)+512x)\log \left(\frac{2e^x{x}^2+x^2+2x}{2(e^{x}x+1)}\right)}{(3x^2-64x+256)(2e^{2x}{x}^3+x^2+e^x(x^3+4x^2)+2x)}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(E^(-256*x + 32*x^2 - x^3)*(2 + 2*x + 2*E^(2*x)*x^2 + E^x*(4*x + x^2 - x^3) + (-512*x - 128*x^2 + 58*x^3 -
3*x^4 + E^(2*x)*(-512*x^3 + 128*x^4 - 6*x^5) + E^x*(-1024*x^2 + 52*x^4 - 3*x^5))*Log[(2*x + x^2 + 2*E^x*x^2)/
(2 + 2*E^x*x)]))/(2*x + x^2 + 2*E^(2*x)*x^3 + E^x*(4*x^2 + x^3)),x]

[Out]

(E^(-256*x + 32*x^2 - x^3)*(512*x + 128*x^2 - 58*x^3 + 3*x^4 + 2*E^(2*x)*(256*x^3 - 64*x^4 + 3*x^5) + E^x*(102
4*x^2 - 52*x^4 + 3*x^5))*Log[(2*x + x^2 + 2*E^x*x^2)/(2*(1 + E^x*x))])/((256 - 64*x + 3*x^2)*(2*x + x^2 + 2*E^
(2*x)*x^3 + E^x*(4*x^2 + x^3)))

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{e^{-256x+32x^2-x^3}(512x+128x^2-58x^3+3x^4+2e^{2x}(256x^3-64x^4+3x^5)+e^x(1024x^2-52x^4+3x^5))\log \left(\frac{2x+x^2+2e^x{x}^2}{2(1+e^{x}x)}\right)}{(256-64x+3x^2)(2x+x^2+2e^{2x}{x}^3+e^x(4x^2+x^3))}\end{array}\end{array}$$

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Mathematica [A]  time = 0.13, size = 33, normalized size = 1.14 $$\begin{array}{c}{e}^{-(-16+x{)}^{2}x}\log \left(\frac{x(2+x+2e^{x}x)}{2+2e^{x}x}\right)\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(E^(-256*x + 32*x^2 - x^3)*(2 + 2*x + 2*E^(2*x)*x^2 + E^x*(4*x + x^2 - x^3) + (-512*x - 128*x^2 + 58
*x^3 - 3*x^4 + E^(2*x)*(-512*x^3 + 128*x^4 - 6*x^5) + E^x*(-1024*x^2 + 52*x^4 - 3*x^5))*Log[(2*x + x^2 + 2*E^x
*x^2)/(2 + 2*E^x*x)]))/(2*x + x^2 + 2*E^(2*x)*x^3 + E^x*(4*x^2 + x^3)),x]

[Out]

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

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fricas [A]  time = 0.49, size = 41, normalized size = 1.41 $$\begin{array}{c}{e}^{(-x^3+32x^2-256x)}\log \left(\frac{2x^2{e}^x+x^2+2x}{2(x{e}^x+1)}\right)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-6*x^5+128*x^4-512*x^3)*exp(x)^2+(-3*x^5+52*x^4-1024*x^2)*exp(x)-3*x^4+58*x^3-128*x^2-512*x)*log(
(2*exp(x)*x^2+x^2+2*x)/(2*exp(x)*x+2))+2*exp(x)^2*x^2+(-x^3+x^2+4*x)*exp(x)+2*x+2)/(2*exp(x)^2*x^3+(x^3+4*x^2)
*exp(x)+x^2+2*x)/exp(x^3-32*x^2+256*x),x, algorithm="fricas")

[Out]

e^(-x^3 + 32*x^2 - 256*x)*log(1/2*(2*x^2*e^x + x^2 + 2*x)/(x*e^x + 1))

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giac [A]  time = 0.84, size = 41, normalized size = 1.41 $$\begin{array}{c}{e}^{(-x^3+32x^2-256x)}\log \left(\frac{2x^2{e}^x+x^2+2x}{2(x{e}^x+1)}\right)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-6*x^5+128*x^4-512*x^3)*exp(x)^2+(-3*x^5+52*x^4-1024*x^2)*exp(x)-3*x^4+58*x^3-128*x^2-512*x)*log(
(2*exp(x)*x^2+x^2+2*x)/(2*exp(x)*x+2))+2*exp(x)^2*x^2+(-x^3+x^2+4*x)*exp(x)+2*x+2)/(2*exp(x)^2*x^3+(x^3+4*x^2)
*exp(x)+x^2+2*x)/exp(x^3-32*x^2+256*x),x, algorithm="giac")

[Out]

e^(-x^3 + 32*x^2 - 256*x)*log(1/2*(2*x^2*e^x + x^2 + 2*x)/(x*e^x + 1))

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maple [C]  time = 0.27, size = 352, normalized size = 12.14

method	result	size
risch	${\mathrm{e}}^{-{(x-16)}^{2}x}\ln (1+(\frac{1}{2}+{\mathrm{e}}^x)x)+\frac{(-i\pi \mathrm{csgn}\left(ix\right)\mathrm{csgn}\left(\frac{i(1+(\frac{1}{2}+{\mathrm{e}}^x)x)}{{\mathrm{e}}^{x}x+1}\right)\mathrm{csgn}\left(\frac{ix(1+(\frac{1}{2}+{\mathrm{e}}^x)x)}{{\mathrm{e}}^{x}x+1}\right)+i\pi \mathrm{csgn}\left(ix\right)\mathrm{csgn}{\left(\frac{ix(1+(\frac{1}{2}+{\mathrm{e}}^x)x)}{{\mathrm{e}}^{x}x+1}\right)}^2-i\pi \mathrm{csgn}\left(i(1+(\frac{1}{2}+{\mathrm{e}}^x)x)\right)\mathrm{csgn}\left(\frac{i}{{\mathrm{e}}^{x}x+1}\right)\mathrm{csgn}\left(\frac{i(1+(\frac{1}{2}+{\mathrm{e}}^x)x)}{{\mathrm{e}}^{x}x+1}\right)+i\pi \mathrm{csgn}\left(i(1+(\frac{1}{2}+{\mathrm{e}}^x)x)\right)\mathrm{csgn}{\left(\frac{i(1+(\frac{1}{2}+{\mathrm{e}}^x)x)}{{\mathrm{e}}^{x}x+1}\right)}^2+i\pi \mathrm{csgn}\left(\frac{i}{{\mathrm{e}}^{x}x+1}\right)\mathrm{csgn}{\left(\frac{i(1+(\frac{1}{2}+{\mathrm{e}}^x)x)}{{\mathrm{e}}^{x}x+1}\right)}^2-i\pi \mathrm{csgn}{\left(\frac{i(1+(\frac{1}{2}+{\mathrm{e}}^x)x)}{{\mathrm{e}}^{x}x+1}\right)}^3+i\pi \mathrm{csgn}\left(\frac{i(1+(\frac{1}{2}+{\mathrm{e}}^x)x)}{{\mathrm{e}}^{x}x+1}\right)\mathrm{csgn}{\left(\frac{ix(1+(\frac{1}{2}+{\mathrm{e}}^x)x)}{{\mathrm{e}}^{x}x+1}\right)}^2-i\pi \mathrm{csgn}{\left(\frac{ix(1+(\frac{1}{2}+{\mathrm{e}}^x)x)}{{\mathrm{e}}^{x}x+1}\right)}^3+2\ln (x)-2\ln ({\mathrm{e}}^{x}x+1)){\mathrm{e}}^{-{(x-16)}^{2}x}}{2}$	$352$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-6*x^5+128*x^4-512*x^3)*exp(x)^2+(-3*x^5+52*x^4-1024*x^2)*exp(x)-3*x^4+58*x^3-128*x^2-512*x)*ln((2*exp(
x)*x^2+x^2+2*x)/(2*exp(x)*x+2))+2*exp(x)^2*x^2+(-x^3+x^2+4*x)*exp(x)+2*x+2)/(2*exp(x)^2*x^3+(x^3+4*x^2)*exp(x)
+x^2+2*x)/exp(x^3-32*x^2+256*x),x,method=_RETURNVERBOSE)

[Out]

exp(-(x-16)^2*x)*ln(1+(1/2+exp(x))*x)+1/2*(-I*Pi*csgn(I*x)*csgn(I*(1+(1/2+exp(x))*x)/(exp(x)*x+1))*csgn(I*x/(e
xp(x)*x+1)*(1+(1/2+exp(x))*x))+I*Pi*csgn(I*x)*csgn(I*x/(exp(x)*x+1)*(1+(1/2+exp(x))*x))^2-I*Pi*csgn(I*(1+(1/2+
exp(x))*x))*csgn(I/(exp(x)*x+1))*csgn(I*(1+(1/2+exp(x))*x)/(exp(x)*x+1))+I*Pi*csgn(I*(1+(1/2+exp(x))*x))*csgn(
I*(1+(1/2+exp(x))*x)/(exp(x)*x+1))^2+I*Pi*csgn(I/(exp(x)*x+1))*csgn(I*(1+(1/2+exp(x))*x)/(exp(x)*x+1))^2-I*Pi*
csgn(I*(1+(1/2+exp(x))*x)/(exp(x)*x+1))^3+I*Pi*csgn(I*(1+(1/2+exp(x))*x)/(exp(x)*x+1))*csgn(I*x/(exp(x)*x+1)*(
1+(1/2+exp(x))*x))^2-I*Pi*csgn(I*x/(exp(x)*x+1)*(1+(1/2+exp(x))*x))^3+2*ln(x)-2*ln(exp(x)*x+1))*exp(-(x-16)^2*
x)

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maxima [B]  time = 31.17, size = 70, normalized size = 2.41 $$\begin{array}{c}-((\log (2)-\log (x))e^{(-x^3+32x^2)}-e^{(-x^3+32x^2)}\log (2x{e}^x+x+2)+e^{(-x^3+32x^2)}\log (x{e}^x+1))e^{(-256x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-6*x^5+128*x^4-512*x^3)*exp(x)^2+(-3*x^5+52*x^4-1024*x^2)*exp(x)-3*x^4+58*x^3-128*x^2-512*x)*log(
(2*exp(x)*x^2+x^2+2*x)/(2*exp(x)*x+2))+2*exp(x)^2*x^2+(-x^3+x^2+4*x)*exp(x)+2*x+2)/(2*exp(x)^2*x^3+(x^3+4*x^2)
*exp(x)+x^2+2*x)/exp(x^3-32*x^2+256*x),x, algorithm="maxima")

[Out]

-((log(2) - log(x))*e^(-x^3 + 32*x^2) - e^(-x^3 + 32*x^2)*log(2*x*e^x + x + 2) + e^(-x^3 + 32*x^2)*log(x*e^x +
1))*e^(-256*x)

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mupad [B]  time = 5.80, size = 42, normalized size = 1.45 $$\begin{array}{c}\ln \left(\frac{2x+2x^2{\mathrm{e}}^x+x^2}{2x{\mathrm{e}}^x+2}\right){\mathrm{e}}^{-256x}{\mathrm{e}}^{-x^3}{\mathrm{e}}^{32x^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(32*x^2 - 256*x - x^3)*(2*x + 2*x^2*exp(2*x) - log((2*x + 2*x^2*exp(x) + x^2)/(2*x*exp(x) + 2))*(512*x
+ exp(x)*(1024*x^2 - 52*x^4 + 3*x^5) + exp(2*x)*(512*x^3 - 128*x^4 + 6*x^5) + 128*x^2 - 58*x^3 + 3*x^4) + exp
(x)*(4*x + x^2 - x^3) + 2))/(2*x + 2*x^3*exp(2*x) + exp(x)*(4*x^2 + x^3) + x^2),x)

[Out]

log((2*x + 2*x^2*exp(x) + x^2)/(2*x*exp(x) + 2))*exp(-256*x)*exp(-x^3)*exp(32*x^2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\text{Timed out}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-6*x**5+128*x**4-512*x**3)*exp(x)**2+(-3*x**5+52*x**4-1024*x**2)*exp(x)-3*x**4+58*x**3-128*x**2-5
12*x)*ln((2*exp(x)*x**2+x**2+2*x)/(2*exp(x)*x+2))+2*exp(x)**2*x**2+(-x**3+x**2+4*x)*exp(x)+2*x+2)/(2*exp(x)**2
*x**3+(x**3+4*x**2)*exp(x)+x**2+2*x)/exp(x**3-32*x**2+256*x),x)

[Out]

Timed out

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