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

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

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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 {gather*} \frac {e^{-x^3+32 x^2-256 x} \left (3 x^4-58 x^3+128 x^2+2 e^{2 x} \left (3 x^5-64 x^4+256 x^3\right )+e^x \left (3 x^5-52 x^4+1024 x^2\right )+512 x\right ) \log \left (\frac {2 e^x x^2+x^2+2 x}{2 \left (e^x x+1\right )}\right )}{\left (3 x^2-64 x+256\right ) \left (2 e^{2 x} x^3+x^2+e^x \left (x^3+4 x^2\right )+2 x\right )} \end {gather*}

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

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

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 {gather*} e^{\left (-x^{3} + 32 \, x^{2} - 256 \, x\right )} \log \left (\frac {2 \, x^{2} e^{x} + x^{2} + 2 \, x}{2 \, {\left (x e^{x} + 1\right )}}\right ) \end {gather*}

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 {gather*} e^{\left (-x^{3} + 32 \, x^{2} - 256 \, x\right )} \log \left (\frac {2 \, x^{2} e^{x} + x^{2} + 2 \, x}{2 \, {\left (x e^{x} + 1\right )}}\right ) \end {gather*}

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	\({\mathrm e}^{-\left (x -16\right )^{2} x} \ln \left (1+\left (\frac {1}{2}+{\mathrm e}^{x}\right ) x \right )+\frac {\left (-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i \left (1+\left (\frac {1}{2}+{\mathrm e}^{x}\right ) x \right )}{{\mathrm e}^{x} x +1}\right ) \mathrm {csgn}\left (\frac {i x \left (1+\left (\frac {1}{2}+{\mathrm e}^{x}\right ) x \right )}{{\mathrm e}^{x} x +1}\right )+i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x \left (1+\left (\frac {1}{2}+{\mathrm e}^{x}\right ) x \right )}{{\mathrm e}^{x} x +1}\right )^{2}-i \pi \,\mathrm {csgn}\left (i \left (1+\left (\frac {1}{2}+{\mathrm e}^{x}\right ) x \right )\right ) \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{x} x +1}\right ) \mathrm {csgn}\left (\frac {i \left (1+\left (\frac {1}{2}+{\mathrm e}^{x}\right ) x \right )}{{\mathrm e}^{x} x +1}\right )+i \pi \,\mathrm {csgn}\left (i \left (1+\left (\frac {1}{2}+{\mathrm e}^{x}\right ) x \right )\right ) \mathrm {csgn}\left (\frac {i \left (1+\left (\frac {1}{2}+{\mathrm e}^{x}\right ) x \right )}{{\mathrm e}^{x} x +1}\right )^{2}+i \pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{x} x +1}\right ) \mathrm {csgn}\left (\frac {i \left (1+\left (\frac {1}{2}+{\mathrm e}^{x}\right ) x \right )}{{\mathrm e}^{x} x +1}\right )^{2}-i \pi \mathrm {csgn}\left (\frac {i \left (1+\left (\frac {1}{2}+{\mathrm e}^{x}\right ) x \right )}{{\mathrm e}^{x} x +1}\right )^{3}+i \pi \,\mathrm {csgn}\left (\frac {i \left (1+\left (\frac {1}{2}+{\mathrm e}^{x}\right ) x \right )}{{\mathrm e}^{x} x +1}\right ) \mathrm {csgn}\left (\frac {i x \left (1+\left (\frac {1}{2}+{\mathrm e}^{x}\right ) x \right )}{{\mathrm e}^{x} x +1}\right )^{2}-i \pi \mathrm {csgn}\left (\frac {i x \left (1+\left (\frac {1}{2}+{\mathrm e}^{x}\right ) x \right )}{{\mathrm e}^{x} x +1}\right )^{3}+2 \ln \relax (x )-2 \ln \left ({\mathrm e}^{x} x +1\right )\right ) {\mathrm e}^{-\left (x -16\right )^{2} x}}{2}\)	\(352\)

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 {gather*} -{\left ({\left (\log \relax (2) - \log \relax (x)\right )} e^{\left (-x^{3} + 32 \, x^{2}\right )} - e^{\left (-x^{3} + 32 \, x^{2}\right )} \log \left (2 \, x e^{x} + x + 2\right ) + e^{\left (-x^{3} + 32 \, x^{2}\right )} \log \left (x e^{x} + 1\right )\right )} e^{\left (-256 \, x\right )} \end {gather*}

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 {gather*} \ln \left (\frac {2\,x+2\,x^2\,{\mathrm {e}}^x+x^2}{2\,x\,{\mathrm {e}}^x+2}\right )\,{\mathrm {e}}^{-256\,x}\,{\mathrm {e}}^{-x^3}\,{\mathrm {e}}^{32\,x^2} \end {gather*}

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 {gather*} \text {Timed out} \end {gather*}

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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