3.98.50 \(\int \frac {e^{-\frac {111+160 x+40 x^2}{110+160 x+40 x^2}} (8+4 x) \log (2)}{605+1760 x+1720 x^2+640 x^3+80 x^4} \, dx\)

Optimal. Leaf size=24 \[ 4+e^{-1-\frac {1}{10 \left (-5+(4+2 x)^2\right )}} \log (2) \]

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Rubi [F]  time = 1.78, 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 {111+160 x+40 x^2}{110+160 x+40 x^2}} (8+4 x) \log (2)}{605+1760 x+1720 x^2+640 x^3+80 x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((8 + 4*x)*Log[2])/(E^((111 + 160*x + 40*x^2)/(110 + 160*x + 40*x^2))*(605 + 1760*x + 1720*x^2 + 640*x^3 +
80*x^4)),x]

[Out]

(32*Log[2]*Defer[Int][1/(E^((111 + 160*x + 40*x^2)/(110 + 160*x + 40*x^2))*(-16 + 4*Sqrt[5] - 8*x)^2), x])/25
- (8*(4 - Sqrt[5])*Log[2]*Defer[Int][1/(E^((111 + 160*x + 40*x^2)/(110 + 160*x + 40*x^2))*(-16 + 4*Sqrt[5] - 8
*x)^2), x])/25 + (32*Log[2]*Defer[Int][1/(E^((111 + 160*x + 40*x^2)/(110 + 160*x + 40*x^2))*(16 + 4*Sqrt[5] +
8*x)^2), x])/25 - (8*(4 + Sqrt[5])*Log[2]*Defer[Int][1/(E^((111 + 160*x + 40*x^2)/(110 + 160*x + 40*x^2))*(16
+ 4*Sqrt[5] + 8*x)^2), x])/25

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log (2) \int \frac {e^{-\frac {111+160 x+40 x^2}{110+160 x+40 x^2}} (8+4 x)}{605+1760 x+1720 x^2+640 x^3+80 x^4} \, dx\\ &=\log (2) \int \frac {4 e^{-\frac {111+160 x+40 x^2}{110+160 x+40 x^2}} (2+x)}{5 \left (11+16 x+4 x^2\right )^2} \, dx\\ &=\frac {1}{5} (4 \log (2)) \int \frac {e^{-\frac {111+160 x+40 x^2}{110+160 x+40 x^2}} (2+x)}{\left (11+16 x+4 x^2\right )^2} \, dx\\ &=\frac {1}{5} (4 \log (2)) \int \left (\frac {2 e^{-\frac {111+160 x+40 x^2}{110+160 x+40 x^2}}}{\left (11+16 x+4 x^2\right )^2}+\frac {e^{-\frac {111+160 x+40 x^2}{110+160 x+40 x^2}} x}{\left (11+16 x+4 x^2\right )^2}\right ) \, dx\\ &=\frac {1}{5} (4 \log (2)) \int \frac {e^{-\frac {111+160 x+40 x^2}{110+160 x+40 x^2}} x}{\left (11+16 x+4 x^2\right )^2} \, dx+\frac {1}{5} (8 \log (2)) \int \frac {e^{-\frac {111+160 x+40 x^2}{110+160 x+40 x^2}}}{\left (11+16 x+4 x^2\right )^2} \, dx\\ &=\frac {1}{5} (4 \log (2)) \int \left (\frac {\left (-16+4 \sqrt {5}\right ) e^{-\frac {111+160 x+40 x^2}{110+160 x+40 x^2}}}{10 \left (-16+4 \sqrt {5}-8 x\right )^2}-\frac {2 e^{-\frac {111+160 x+40 x^2}{110+160 x+40 x^2}}}{5 \sqrt {5} \left (-16+4 \sqrt {5}-8 x\right )}+\frac {\left (-16-4 \sqrt {5}\right ) e^{-\frac {111+160 x+40 x^2}{110+160 x+40 x^2}}}{10 \left (16+4 \sqrt {5}+8 x\right )^2}-\frac {2 e^{-\frac {111+160 x+40 x^2}{110+160 x+40 x^2}}}{5 \sqrt {5} \left (16+4 \sqrt {5}+8 x\right )}\right ) \, dx+\frac {1}{5} (8 \log (2)) \int \left (\frac {4 e^{-\frac {111+160 x+40 x^2}{110+160 x+40 x^2}}}{5 \left (-16+4 \sqrt {5}-8 x\right )^2}+\frac {e^{-\frac {111+160 x+40 x^2}{110+160 x+40 x^2}}}{5 \sqrt {5} \left (-16+4 \sqrt {5}-8 x\right )}+\frac {4 e^{-\frac {111+160 x+40 x^2}{110+160 x+40 x^2}}}{5 \left (16+4 \sqrt {5}+8 x\right )^2}+\frac {e^{-\frac {111+160 x+40 x^2}{110+160 x+40 x^2}}}{5 \sqrt {5} \left (16+4 \sqrt {5}+8 x\right )}\right ) \, dx\\ &=\frac {1}{25} (32 \log (2)) \int \frac {e^{-\frac {111+160 x+40 x^2}{110+160 x+40 x^2}}}{\left (-16+4 \sqrt {5}-8 x\right )^2} \, dx+\frac {1}{25} (32 \log (2)) \int \frac {e^{-\frac {111+160 x+40 x^2}{110+160 x+40 x^2}}}{\left (16+4 \sqrt {5}+8 x\right )^2} \, dx-\frac {1}{25} \left (8 \left (4-\sqrt {5}\right ) \log (2)\right ) \int \frac {e^{-\frac {111+160 x+40 x^2}{110+160 x+40 x^2}}}{\left (-16+4 \sqrt {5}-8 x\right )^2} \, dx-\frac {1}{25} \left (8 \left (4+\sqrt {5}\right ) \log (2)\right ) \int \frac {e^{-\frac {111+160 x+40 x^2}{110+160 x+40 x^2}}}{\left (16+4 \sqrt {5}+8 x\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.25, size = 21, normalized size = 0.88 \begin {gather*} e^{-1-\frac {1}{110+160 x+40 x^2}} \log (2) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((8 + 4*x)*Log[2])/(E^((111 + 160*x + 40*x^2)/(110 + 160*x + 40*x^2))*(605 + 1760*x + 1720*x^2 + 640
*x^3 + 80*x^4)),x]

[Out]

E^(-1 - (110 + 160*x + 40*x^2)^(-1))*Log[2]

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fricas [A]  time = 1.06, size = 28, normalized size = 1.17 \begin {gather*} e^{\left (-\frac {40 \, x^{2} + 160 \, x + 111}{10 \, {\left (4 \, x^{2} + 16 \, x + 11\right )}}\right )} \log \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x+8)*log(2)/(80*x^4+640*x^3+1720*x^2+1760*x+605)/exp((40*x^2+160*x+111)/(40*x^2+160*x+110)),x, al
gorithm="fricas")

[Out]

e^(-1/10*(40*x^2 + 160*x + 111)/(4*x^2 + 16*x + 11))*log(2)

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giac [B]  time = 0.18, size = 51, normalized size = 2.12 \begin {gather*} e^{\left (-\frac {4 \, x^{2}}{4 \, x^{2} + 16 \, x + 11} - \frac {16 \, x}{4 \, x^{2} + 16 \, x + 11} - \frac {111}{10 \, {\left (4 \, x^{2} + 16 \, x + 11\right )}}\right )} \log \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x+8)*log(2)/(80*x^4+640*x^3+1720*x^2+1760*x+605)/exp((40*x^2+160*x+111)/(40*x^2+160*x+110)),x, al
gorithm="giac")

[Out]

e^(-4*x^2/(4*x^2 + 16*x + 11) - 16*x/(4*x^2 + 16*x + 11) - 111/10/(4*x^2 + 16*x + 11))*log(2)

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maple [A]  time = 0.27, size = 21, normalized size = 0.88

method	result	size
default	\(\ln \relax (2) {\mathrm e}^{-\frac {1}{40 x^{2}+160 x +110}-1}\)	\(21\)
risch	\(\ln \relax (2) {\mathrm e}^{-\frac {40 x^{2}+160 x +111}{10 \left (4 x^{2}+16 x +11\right )}}\)	\(29\)
gosper	\(\ln \relax (2) {\mathrm e}^{-\frac {40 x^{2}+160 x +111}{10 \left (4 x^{2}+16 x +11\right )}}\)	\(31\)
norman	\(\frac {\left (16 x \ln \relax (2)+4 x^{2} \ln \relax (2)+11 \ln \relax (2)\right ) {\mathrm e}^{-\frac {40 x^{2}+160 x +111}{40 x^{2}+160 x +110}}}{4 x^{2}+16 x +11}\)	\(57\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x+8)*ln(2)/(80*x^4+640*x^3+1720*x^2+1760*x+605)/exp((40*x^2+160*x+111)/(40*x^2+160*x+110)),x,method=_RE
TURNVERBOSE)

[Out]

ln(2)*exp(-1/(40*x^2+160*x+110)-1)

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maxima [A]  time = 0.53, size = 20, normalized size = 0.83 \begin {gather*} e^{\left (-\frac {1}{10 \, {\left (4 \, x^{2} + 16 \, x + 11\right )}} - 1\right )} \log \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x+8)*log(2)/(80*x^4+640*x^3+1720*x^2+1760*x+605)/exp((40*x^2+160*x+111)/(40*x^2+160*x+110)),x, al
gorithm="maxima")

[Out]

e^(-1/10/(4*x^2 + 16*x + 11) - 1)*log(2)

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mupad [B]  time = 0.31, size = 52, normalized size = 2.17 \begin {gather*} {\mathrm {e}}^{-\frac {16\,x}{4\,x^2+16\,x+11}}\,{\mathrm {e}}^{-\frac {4\,x^2}{4\,x^2+16\,x+11}}\,{\mathrm {e}}^{-\frac {111}{40\,x^2+160\,x+110}}\,\ln \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-(160*x + 40*x^2 + 111)/(160*x + 40*x^2 + 110))*log(2)*(4*x + 8))/(1760*x + 1720*x^2 + 640*x^3 + 80*x
^4 + 605),x)

[Out]

exp(-(16*x)/(16*x + 4*x^2 + 11))*exp(-(4*x^2)/(16*x + 4*x^2 + 11))*exp(-111/(160*x + 40*x^2 + 110))*log(2)

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sympy [A]  time = 0.36, size = 24, normalized size = 1.00 \begin {gather*} e^{- \frac {40 x^{2} + 160 x + 111}{40 x^{2} + 160 x + 110}} \log {\relax (2 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x+8)*ln(2)/(80*x**4+640*x**3+1720*x**2+1760*x+605)/exp((40*x**2+160*x+111)/(40*x**2+160*x+110)),x
)

[Out]

exp(-(40*x**2 + 160*x + 111)/(40*x**2 + 160*x + 110))*log(2)

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