3.96.40 \(\int \frac {e^{4 x-2 x^2}+16 x^2+4 \log (5)+e^{2 x-x^2} (-8 x+(-2+2 x) \log (5))}{e^{4 x-2 x^2}-8 e^{2 x-x^2} x+16 x^2} \, dx\)

Optimal. Leaf size=24 \[ x+\frac {\log (5)}{e^{-x+(3-x) x}-4 x} \]

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Rubi [F]  time = 3.21, 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^{4 x-2 x^2}+16 x^2+4 \log (5)+e^{2 x-x^2} (-8 x+(-2+2 x) \log (5))}{e^{4 x-2 x^2}-8 e^{2 x-x^2} x+16 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(4*x - 2*x^2) + 16*x^2 + 4*Log[5] + E^(2*x - x^2)*(-8*x + (-2 + 2*x)*Log[5]))/(E^(4*x - 2*x^2) - 8*E^(2
*x - x^2)*x + 16*x^2),x]

[Out]

x + E^(-2*x + x^2)*Log[5] + 4*Log[5]*Defer[Int][E^(2*x^2)/(E^(2*x) - 4*E^x^2*x)^2, x] - 8*Log[5]*Defer[Int][(E
^(2*x^2)*x)/(-E^(2*x) + 4*E^x^2*x)^2, x] + 8*Log[5]*Defer[Int][(E^(2*x^2)*x^2)/(-E^(2*x) + 4*E^x^2*x)^2, x] +
8*Log[5]*Defer[Int][(E^(2*x^2)*x)/(-E^(4*x) + 4*E^(x*(2 + x))*x), x] - 8*Log[5]*Defer[Int][(E^(2*x^2)*x^2)/(-E
^(4*x) + 4*E^(x*(2 + x))*x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{2 x^2} \left (e^{4 x-2 x^2}+16 x^2+4 \log (5)+e^{2 x-x^2} (-8 x+(-2+2 x) \log (5))\right )}{\left (e^{2 x}-4 e^{x^2} x\right )^2} \, dx\\ &=\int \left (1+2 e^{-2 x+x^2} (-1+x) \log (5)-\frac {8 e^{2 x^2} (-1+x) x \log (5)}{-e^{4 x}+4 e^{x (2+x)} x}+\frac {4 e^{2 x^2} \left (1-2 x+2 x^2\right ) \log (5)}{\left (e^{2 x}-4 e^{x^2} x\right )^2}\right ) \, dx\\ &=x+(2 \log (5)) \int e^{-2 x+x^2} (-1+x) \, dx+(4 \log (5)) \int \frac {e^{2 x^2} \left (1-2 x+2 x^2\right )}{\left (e^{2 x}-4 e^{x^2} x\right )^2} \, dx-(8 \log (5)) \int \frac {e^{2 x^2} (-1+x) x}{-e^{4 x}+4 e^{x (2+x)} x} \, dx\\ &=x+e^{-2 x+x^2} \log (5)+(4 \log (5)) \int \left (\frac {e^{2 x^2}}{\left (e^{2 x}-4 e^{x^2} x\right )^2}-\frac {2 e^{2 x^2} x}{\left (-e^{2 x}+4 e^{x^2} x\right )^2}+\frac {2 e^{2 x^2} x^2}{\left (-e^{2 x}+4 e^{x^2} x\right )^2}\right ) \, dx-(8 \log (5)) \int \left (-\frac {e^{2 x^2} x}{-e^{4 x}+4 e^{x (2+x)} x}+\frac {e^{2 x^2} x^2}{-e^{4 x}+4 e^{x (2+x)} x}\right ) \, dx\\ &=x+e^{-2 x+x^2} \log (5)+(4 \log (5)) \int \frac {e^{2 x^2}}{\left (e^{2 x}-4 e^{x^2} x\right )^2} \, dx-(8 \log (5)) \int \frac {e^{2 x^2} x}{\left (-e^{2 x}+4 e^{x^2} x\right )^2} \, dx+(8 \log (5)) \int \frac {e^{2 x^2} x^2}{\left (-e^{2 x}+4 e^{x^2} x\right )^2} \, dx+(8 \log (5)) \int \frac {e^{2 x^2} x}{-e^{4 x}+4 e^{x (2+x)} x} \, dx-(8 \log (5)) \int \frac {e^{2 x^2} x^2}{-e^{4 x}+4 e^{x (2+x)} x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.71, size = 39, normalized size = 1.62 \begin {gather*} \frac {e^{2 x} x+e^{x^2} \left (-4 x^2+\log (5)\right )}{e^{2 x}-4 e^{x^2} x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(4*x - 2*x^2) + 16*x^2 + 4*Log[5] + E^(2*x - x^2)*(-8*x + (-2 + 2*x)*Log[5]))/(E^(4*x - 2*x^2) -
8*E^(2*x - x^2)*x + 16*x^2),x]

[Out]

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

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fricas [A]  time = 0.63, size = 42, normalized size = 1.75 \begin {gather*} \frac {4 \, x^{2} - x e^{\left (-x^{2} + 2 \, x\right )} - \log \relax (5)}{4 \, x - e^{\left (-x^{2} + 2 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(-x^2+2*x)^2+((2*x-2)*log(5)-8*x)*exp(-x^2+2*x)+4*log(5)+16*x^2)/(exp(-x^2+2*x)^2-8*x*exp(-x^2+2
*x)+16*x^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.22, size = 42, normalized size = 1.75 \begin {gather*} \frac {4 \, x^{2} - x e^{\left (-x^{2} + 2 \, x\right )} - \log \relax (5)}{4 \, x - e^{\left (-x^{2} + 2 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(-x^2+2*x)^2+((2*x-2)*log(5)-8*x)*exp(-x^2+2*x)+4*log(5)+16*x^2)/(exp(-x^2+2*x)^2-8*x*exp(-x^2+2
*x)+16*x^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 22, normalized size = 0.92

method	result	size
risch	\(x -\frac {\ln \relax (5)}{4 x -{\mathrm e}^{-\left (x -2\right ) x}}\)	\(22\)
norman	\(\frac {4 x^{2}-x \,{\mathrm e}^{-x^{2}+2 x}-\ln \relax (5)}{4 x -{\mathrm e}^{-x^{2}+2 x}}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-x^2+2*x)^2+((2*x-2)*ln(5)-8*x)*exp(-x^2+2*x)+4*ln(5)+16*x^2)/(exp(-x^2+2*x)^2-8*x*exp(-x^2+2*x)+16*x
^2),x,method=_RETURNVERBOSE)

[Out]

x-ln(5)/(4*x-exp(-(x-2)*x))

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maxima [A]  time = 0.47, size = 40, normalized size = 1.67 \begin {gather*} \frac {{\left (4 \, x^{2} - \log \relax (5)\right )} e^{\left (x^{2}\right )} - x e^{\left (2 \, x\right )}}{4 \, x e^{\left (x^{2}\right )} - e^{\left (2 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(-x^2+2*x)^2+((2*x-2)*log(5)-8*x)*exp(-x^2+2*x)+4*log(5)+16*x^2)/(exp(-x^2+2*x)^2-8*x*exp(-x^2+2
*x)+16*x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.31, size = 24, normalized size = 1.00 \begin {gather*} x-\frac {\ln \relax (5)}{4\,x-{\mathrm {e}}^{2\,x-x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*log(5) + exp(4*x - 2*x^2) - exp(2*x - x^2)*(8*x - log(5)*(2*x - 2)) + 16*x^2)/(exp(4*x - 2*x^2) - 8*x*e
xp(2*x - x^2) + 16*x^2),x)

[Out]

x - log(5)/(4*x - exp(2*x - x^2))

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sympy [A]  time = 0.13, size = 15, normalized size = 0.62 \begin {gather*} x + \frac {\log {\relax (5 )}}{- 4 x + e^{- x^{2} + 2 x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(-x**2+2*x)**2+((2*x-2)*ln(5)-8*x)*exp(-x**2+2*x)+4*ln(5)+16*x**2)/(exp(-x**2+2*x)**2-8*x*exp(-x
**2+2*x)+16*x**2),x)

[Out]

x + log(5)/(-4*x + exp(-x**2 + 2*x))

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