3.98.3 \(\int \frac {e^{-\frac {x^2}{1-2 x+x^2}} (20-60 x+20 x^2-30 x^3+e^{\frac {x^2}{1-2 x+x^2}} (-16+40 x-25 x^2-5 x^3+5 x^4+x^5) \log (5))}{(-16+40 x-25 x^2-5 x^3+5 x^4+x^5) \log (5)} \, dx\)

Optimal. Leaf size=28 \[ x-\frac {5 e^{-\frac {x^2}{(1-x)^2}} x}{(4+x) \log (5)} \]

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Rubi [F]  time = 4.65, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {e^{-\frac {x^2}{1-2 x+x^2}} \left (20-60 x+20 x^2-30 x^3+e^{\frac {x^2}{1-2 x+x^2}} \left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)\right )}{\left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(20 - 60*x + 20*x^2 - 30*x^3 + E^(x^2/(1 - 2*x + x^2))*(-16 + 40*x - 25*x^2 - 5*x^3 + 5*x^4 + x^5)*Log[5])
/(E^(x^2/(1 - 2*x + x^2))*(-16 + 40*x - 25*x^2 - 5*x^3 + 5*x^4 + x^5)*Log[5]),x]

[Out]

x - (2*Defer[Int][1/(E^(x^2/(-1 + x)^2)*(-1 + x)^3), x])/Log[5] - (18*Defer[Int][1/(E^(x^2/(-1 + x)^2)*(-1 + x
)^2), x])/(5*Log[5]) - (32*Defer[Int][1/(E^(x^2/(-1 + x)^2)*(-1 + x)), x])/(25*Log[5]) - (20*Defer[Int][1/(E^(
x^2/(-1 + x)^2)*(4 + x)^2), x])/Log[5] + (32*Defer[Int][1/(E^(x^2/(-1 + x)^2)*(4 + x)), x])/(25*Log[5])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {e^{-\frac {x^2}{1-2 x+x^2}} \left (20-60 x+20 x^2-30 x^3+e^{\frac {x^2}{1-2 x+x^2}} \left (-16+40 x-25 x^2-5 x^3+5 x^4+x^5\right ) \log (5)\right )}{-16+40 x-25 x^2-5 x^3+5 x^4+x^5} \, dx}{\log (5)}\\ &=\frac {\int \frac {e^{-\frac {x^2}{(-1+x)^2}} \left (-20+60 x-20 x^2+30 x^3-e^{\frac {x^2}{(-1+x)^2}} (-1+x)^3 (4+x)^2 \log (5)\right )}{(1-x)^3 (4+x)^2} \, dx}{\log (5)}\\ &=\frac {\int \left (\frac {20 e^{-\frac {x^2}{(-1+x)^2}}}{(-1+x)^3 (4+x)^2}-\frac {60 e^{-\frac {x^2}{(-1+x)^2}} x}{(-1+x)^3 (4+x)^2}+\frac {20 e^{-\frac {x^2}{(-1+x)^2}} x^2}{(-1+x)^3 (4+x)^2}-\frac {30 e^{-\frac {x^2}{(-1+x)^2}} x^3}{(-1+x)^3 (4+x)^2}+\log (5)\right ) \, dx}{\log (5)}\\ &=x+\frac {20 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{(-1+x)^3 (4+x)^2} \, dx}{\log (5)}+\frac {20 \int \frac {e^{-\frac {x^2}{(-1+x)^2}} x^2}{(-1+x)^3 (4+x)^2} \, dx}{\log (5)}-\frac {30 \int \frac {e^{-\frac {x^2}{(-1+x)^2}} x^3}{(-1+x)^3 (4+x)^2} \, dx}{\log (5)}-\frac {60 \int \frac {e^{-\frac {x^2}{(-1+x)^2}} x}{(-1+x)^3 (4+x)^2} \, dx}{\log (5)}\\ &=x+\frac {20 \int \left (\frac {e^{-\frac {x^2}{(-1+x)^2}}}{25 (-1+x)^3}+\frac {8 e^{-\frac {x^2}{(-1+x)^2}}}{125 (-1+x)^2}+\frac {8 e^{-\frac {x^2}{(-1+x)^2}}}{625 (-1+x)}-\frac {16 e^{-\frac {x^2}{(-1+x)^2}}}{125 (4+x)^2}-\frac {8 e^{-\frac {x^2}{(-1+x)^2}}}{625 (4+x)}\right ) \, dx}{\log (5)}+\frac {20 \int \left (\frac {e^{-\frac {x^2}{(-1+x)^2}}}{25 (-1+x)^3}-\frac {2 e^{-\frac {x^2}{(-1+x)^2}}}{125 (-1+x)^2}+\frac {3 e^{-\frac {x^2}{(-1+x)^2}}}{625 (-1+x)}-\frac {e^{-\frac {x^2}{(-1+x)^2}}}{125 (4+x)^2}-\frac {3 e^{-\frac {x^2}{(-1+x)^2}}}{625 (4+x)}\right ) \, dx}{\log (5)}-\frac {30 \int \left (\frac {e^{-\frac {x^2}{(-1+x)^2}}}{25 (-1+x)^3}+\frac {13 e^{-\frac {x^2}{(-1+x)^2}}}{125 (-1+x)^2}+\frac {48 e^{-\frac {x^2}{(-1+x)^2}}}{625 (-1+x)}+\frac {64 e^{-\frac {x^2}{(-1+x)^2}}}{125 (4+x)^2}-\frac {48 e^{-\frac {x^2}{(-1+x)^2}}}{625 (4+x)}\right ) \, dx}{\log (5)}-\frac {60 \int \left (\frac {e^{-\frac {x^2}{(-1+x)^2}}}{25 (-1+x)^3}+\frac {3 e^{-\frac {x^2}{(-1+x)^2}}}{125 (-1+x)^2}-\frac {7 e^{-\frac {x^2}{(-1+x)^2}}}{625 (-1+x)}+\frac {4 e^{-\frac {x^2}{(-1+x)^2}}}{125 (4+x)^2}+\frac {7 e^{-\frac {x^2}{(-1+x)^2}}}{625 (4+x)}\right ) \, dx}{\log (5)}\\ &=x+\frac {12 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{-1+x} \, dx}{125 \log (5)}-\frac {12 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{4+x} \, dx}{125 \log (5)}-\frac {4 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{(4+x)^2} \, dx}{25 \log (5)}+\frac {32 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{-1+x} \, dx}{125 \log (5)}-\frac {32 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{4+x} \, dx}{125 \log (5)}-\frac {8 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{(-1+x)^2} \, dx}{25 \log (5)}+\frac {84 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{-1+x} \, dx}{125 \log (5)}-\frac {84 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{4+x} \, dx}{125 \log (5)}+2 \frac {4 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{(-1+x)^3} \, dx}{5 \log (5)}-\frac {6 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{(-1+x)^3} \, dx}{5 \log (5)}+\frac {32 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{(-1+x)^2} \, dx}{25 \log (5)}-\frac {36 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{(-1+x)^2} \, dx}{25 \log (5)}-\frac {48 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{(4+x)^2} \, dx}{25 \log (5)}-\frac {288 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{-1+x} \, dx}{125 \log (5)}+\frac {288 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{4+x} \, dx}{125 \log (5)}-\frac {12 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{(-1+x)^3} \, dx}{5 \log (5)}-\frac {64 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{(4+x)^2} \, dx}{25 \log (5)}-\frac {78 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{(-1+x)^2} \, dx}{25 \log (5)}-\frac {384 \int \frac {e^{-\frac {x^2}{(-1+x)^2}}}{(4+x)^2} \, dx}{25 \log (5)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 2.20, size = 32, normalized size = 1.14 \begin {gather*} \frac {-\frac {5 e^{-\frac {x^2}{(-1+x)^2}} x}{4+x}+(-1+x) \log (5)}{\log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(20 - 60*x + 20*x^2 - 30*x^3 + E^(x^2/(1 - 2*x + x^2))*(-16 + 40*x - 25*x^2 - 5*x^3 + 5*x^4 + x^5)*L
og[5])/(E^(x^2/(1 - 2*x + x^2))*(-16 + 40*x - 25*x^2 - 5*x^3 + 5*x^4 + x^5)*Log[5]),x]

[Out]

((-5*x)/(E^(x^2/(-1 + x)^2)*(4 + x)) + (-1 + x)*Log[5])/Log[5]

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fricas [B]  time = 0.82, size = 55, normalized size = 1.96 \begin {gather*} \frac {{\left ({\left (x^{2} + 4 \, x\right )} e^{\left (\frac {x^{2}}{x^{2} - 2 \, x + 1}\right )} \log \relax (5) - 5 \, x\right )} e^{\left (-\frac {x^{2}}{x^{2} - 2 \, x + 1}\right )}}{{\left (x + 4\right )} \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^5+5*x^4-5*x^3-25*x^2+40*x-16)*log(5)*exp(x^2/(x^2-2*x+1))-30*x^3+20*x^2-60*x+20)/(x^5+5*x^4-5*x^
3-25*x^2+40*x-16)/log(5)/exp(x^2/(x^2-2*x+1)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.14, size = 41, normalized size = 1.46 \begin {gather*} \frac {x^{2} \log \relax (5) - 5 \, x e^{\left (-\frac {x^{2}}{x^{2} - 2 \, x + 1}\right )} + 4 \, x \log \relax (5)}{{\left (x + 4\right )} \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^5+5*x^4-5*x^3-25*x^2+40*x-16)*log(5)*exp(x^2/(x^2-2*x+1))-30*x^3+20*x^2-60*x+20)/(x^5+5*x^4-5*x^
3-25*x^2+40*x-16)/log(5)/exp(x^2/(x^2-2*x+1)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.43, size = 26, normalized size = 0.93

method	result	size
risch	\(x -\frac {5 x \,{\mathrm e}^{-\frac {x^{2}}{\left (x -1\right )^{2}}}}{\ln \relax (5) \left (4+x \right )}\)	\(26\)
norman	\(\frac {\left (x^{4} {\mathrm e}^{\frac {x^{2}}{x^{2}-2 x +1}}-8 \,{\mathrm e}^{\frac {x^{2}}{x^{2}-2 x +1}}+18 x \,{\mathrm e}^{\frac {x^{2}}{x^{2}-2 x +1}}-11 x^{2} {\mathrm e}^{\frac {x^{2}}{x^{2}-2 x +1}}-\frac {5 x}{\ln \relax (5)}+\frac {10 x^{2}}{\ln \relax (5)}-\frac {5 x^{3}}{\ln \relax (5)}\right ) {\mathrm e}^{-\frac {x^{2}}{x^{2}-2 x +1}}}{\left (4+x \right ) \left (x -1\right )^{2}}\)	\(129\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^5+5*x^4-5*x^3-25*x^2+40*x-16)*ln(5)*exp(x^2/(x^2-2*x+1))-30*x^3+20*x^2-60*x+20)/(x^5+5*x^4-5*x^3-25*x^
2+40*x-16)/ln(5)/exp(x^2/(x^2-2*x+1)),x,method=_RETURNVERBOSE)

[Out]

x-5/ln(5)*x/(4+x)*exp(-x^2/(x-1)^2)

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maxima [B]  time = 0.54, size = 68, normalized size = 2.43 \begin {gather*} -\frac {{\left (5 \, x e^{\left (-\frac {1}{x^{2} - 2 \, x + 1}\right )} - {\left (x^{2} e \log \relax (5) + 4 \, x e \log \relax (5)\right )} e^{\left (\frac {2}{x - 1}\right )}\right )} e^{\left (-\frac {2}{x - 1}\right )}}{{\left (x e + 4 \, e\right )} \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^5+5*x^4-5*x^3-25*x^2+40*x-16)*log(5)*exp(x^2/(x^2-2*x+1))-30*x^3+20*x^2-60*x+20)/(x^5+5*x^4-5*x^
3-25*x^2+40*x-16)/log(5)/exp(x^2/(x^2-2*x+1)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 6.18, size = 30, normalized size = 1.07 \begin {gather*} x-\frac {5\,x\,{\mathrm {e}}^{-\frac {x^2}{x^2-2\,x+1}}}{\ln \relax (5)\,\left (x+4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-x^2/(x^2 - 2*x + 1))*(20*x^2 - 60*x - 30*x^3 + exp(x^2/(x^2 - 2*x + 1))*log(5)*(40*x - 25*x^2 - 5*x^
3 + 5*x^4 + x^5 - 16) + 20))/(log(5)*(40*x - 25*x^2 - 5*x^3 + 5*x^4 + x^5 - 16)),x)

[Out]

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

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sympy [A]  time = 0.26, size = 27, normalized size = 0.96 \begin {gather*} x - \frac {5 x e^{- \frac {x^{2}}{x^{2} - 2 x + 1}}}{x \log {\relax (5 )} + 4 \log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**5+5*x**4-5*x**3-25*x**2+40*x-16)*ln(5)*exp(x**2/(x**2-2*x+1))-30*x**3+20*x**2-60*x+20)/(x**5+5*
x**4-5*x**3-25*x**2+40*x-16)/ln(5)/exp(x**2/(x**2-2*x+1)),x)

[Out]

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

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