3.103.100 \(\int \frac {e^{\frac {80-8 x+16 x^2+(4+8 x) \log (5)}{x+2 x^2}} (-80-320 x+32 x^2+(-4-16 x-16 x^2) \log (5))}{x^2+4 x^3+4 x^4} \, dx\)

Optimal. Leaf size=27 \[ e^{\frac {4 \left (2 x+\frac {2 (5-x)}{\frac {1}{2}+x}+\log (5)\right )}{x}} \]

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Rubi [F]  time = 2.61, 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 {\exp \left (\frac {80-8 x+16 x^2+(4+8 x) \log (5)}{x+2 x^2}\right ) \left (-80-320 x+32 x^2+\left (-4-16 x-16 x^2\right ) \log (5)\right )}{x^2+4 x^3+4 x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((80 - 8*x + 16*x^2 + (4 + 8*x)*Log[5])/(x + 2*x^2))*(-80 - 320*x + 32*x^2 + (-4 - 16*x - 16*x^2)*Log[5
]))/(x^2 + 4*x^3 + 4*x^4),x]

[Out]

-4*(20 + Log[5])*Defer[Int][E^((80 + 16*x^2 - 8*x*(1 - Log[5]) + Log[625])/(x*(1 + 2*x)))/x^2, x] + 352*Defer[
Int][E^((80 + 16*x^2 - 8*x*(1 - Log[5]) + Log[625])/(x*(1 + 2*x)))/(1 + 2*x)^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {80-8 x+16 x^2+(4+8 x) \log (5)}{x+2 x^2}\right ) \left (-80-320 x+32 x^2+\left (-4-16 x-16 x^2\right ) \log (5)\right )}{x^2 \left (1+4 x+4 x^2\right )} \, dx\\ &=\int \frac {\exp \left (\frac {80-8 x+16 x^2+(4+8 x) \log (5)}{x+2 x^2}\right ) \left (-80-320 x+32 x^2+\left (-4-16 x-16 x^2\right ) \log (5)\right )}{x^2 (1+2 x)^2} \, dx\\ &=\int \frac {\exp \left (\frac {80+16 x^2-8 x (1-\log (5))+\log (625)}{x (1+2 x)}\right ) \left (16 x^2 (2-\log (5))-4 (20+\log (5))-16 x (20+\log (5))\right )}{x^2 (1+2 x)^2} \, dx\\ &=\int \left (\frac {352 \exp \left (\frac {80+16 x^2-8 x (1-\log (5))+\log (625)}{x (1+2 x)}\right )}{(1+2 x)^2}-\frac {4 \exp \left (\frac {80+16 x^2-8 x (1-\log (5))+\log (625)}{x (1+2 x)}\right ) (20+\log (5))}{x^2}\right ) \, dx\\ &=352 \int \frac {\exp \left (\frac {80+16 x^2-8 x (1-\log (5))+\log (625)}{x (1+2 x)}\right )}{(1+2 x)^2} \, dx-(4 (20+\log (5))) \int \frac {\exp \left (\frac {80+16 x^2-8 x (1-\log (5))+\log (625)}{x (1+2 x)}\right )}{x^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 2.16, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{\frac {80-8 x+16 x^2+(4+8 x) \log (5)}{x+2 x^2}} \left (-80-320 x+32 x^2+\left (-4-16 x-16 x^2\right ) \log (5)\right )}{x^2+4 x^3+4 x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(E^((80 - 8*x + 16*x^2 + (4 + 8*x)*Log[5])/(x + 2*x^2))*(-80 - 320*x + 32*x^2 + (-4 - 16*x - 16*x^2)
*Log[5]))/(x^2 + 4*x^3 + 4*x^4),x]

[Out]

Integrate[(E^((80 - 8*x + 16*x^2 + (4 + 8*x)*Log[5])/(x + 2*x^2))*(-80 - 320*x + 32*x^2 + (-4 - 16*x - 16*x^2)
*Log[5]))/(x^2 + 4*x^3 + 4*x^4), x]

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fricas [A]  time = 0.93, size = 30, normalized size = 1.11 \begin {gather*} e^{\left (\frac {4 \, {\left (4 \, x^{2} + {\left (2 \, x + 1\right )} \log \relax (5) - 2 \, x + 20\right )}}{2 \, x^{2} + x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x^2-16*x-4)*log(5)+32*x^2-320*x-80)*exp(((8*x+4)*log(5)+16*x^2-8*x+80)/(2*x^2+x))/(4*x^4+4*x^3
+x^2),x, algorithm="fricas")

[Out]

e^(4*(4*x^2 + (2*x + 1)*log(5) - 2*x + 20)/(2*x^2 + x))

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giac [B]  time = 0.17, size = 66, normalized size = 2.44 \begin {gather*} e^{\left (\frac {16 \, x^{2}}{2 \, x^{2} + x} + \frac {8 \, x \log \relax (5)}{2 \, x^{2} + x} - \frac {8 \, x}{2 \, x^{2} + x} + \frac {4 \, \log \relax (5)}{2 \, x^{2} + x} + \frac {80}{2 \, x^{2} + x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x^2-16*x-4)*log(5)+32*x^2-320*x-80)*exp(((8*x+4)*log(5)+16*x^2-8*x+80)/(2*x^2+x))/(4*x^4+4*x^3
+x^2),x, algorithm="giac")

[Out]

e^(16*x^2/(2*x^2 + x) + 8*x*log(5)/(2*x^2 + x) - 8*x/(2*x^2 + x) + 4*log(5)/(2*x^2 + x) + 80/(2*x^2 + x))

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maple [A]  time = 0.14, size = 31, normalized size = 1.15

method	result	size
gosper	\({\mathrm e}^{\frac {8 x \ln \relax (5)+16 x^{2}+4 \ln \relax (5)-8 x +80}{\left (2 x +1\right ) x}}\)	\(31\)
risch	\({\mathrm e}^{\frac {8 x \ln \relax (5)+16 x^{2}+4 \ln \relax (5)-8 x +80}{\left (2 x +1\right ) x}}\)	\(31\)
norman	\(\frac {x \,{\mathrm e}^{\frac {\left (8 x +4\right ) \ln \relax (5)+16 x^{2}-8 x +80}{2 x^{2}+x}}+2 x^{2} {\mathrm e}^{\frac {\left (8 x +4\right ) \ln \relax (5)+16 x^{2}-8 x +80}{2 x^{2}+x}}}{\left (2 x +1\right ) x}\)	\(78\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-16*x^2-16*x-4)*ln(5)+32*x^2-320*x-80)*exp(((8*x+4)*ln(5)+16*x^2-8*x+80)/(2*x^2+x))/(4*x^4+4*x^3+x^2),x,
method=_RETURNVERBOSE)

[Out]

exp(4*(2*x*ln(5)+4*x^2+ln(5)-2*x+20)/(2*x+1)/x)

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maxima [A]  time = 0.63, size = 24, normalized size = 0.89 \begin {gather*} e^{\left (\frac {4 \, \log \relax (5)}{x} - \frac {176}{2 \, x + 1} + \frac {80}{x} + 8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x^2-16*x-4)*log(5)+32*x^2-320*x-80)*exp(((8*x+4)*log(5)+16*x^2-8*x+80)/(2*x^2+x))/(4*x^4+4*x^3
+x^2),x, algorithm="maxima")

[Out]

e^(4*log(5)/x - 176/(2*x + 1) + 80/x + 8)

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mupad [B]  time = 7.16, size = 39, normalized size = 1.44 \begin {gather*} {625}^{1/x}\,{\mathrm {e}}^{\frac {80}{2\,x^2+x}}\,{\mathrm {e}}^{-\frac {8}{2\,x+1}}\,{\mathrm {e}}^{\frac {16\,x}{2\,x+1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((log(5)*(8*x + 4) - 8*x + 16*x^2 + 80)/(x + 2*x^2))*(320*x + log(5)*(16*x + 16*x^2 + 4) - 32*x^2 + 8
0))/(x^2 + 4*x^3 + 4*x^4),x)

[Out]

625^(1/x)*exp(80/(x + 2*x^2))*exp(-8/(2*x + 1))*exp((16*x)/(2*x + 1))

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sympy [A]  time = 0.55, size = 26, normalized size = 0.96 \begin {gather*} e^{\frac {16 x^{2} - 8 x + \left (8 x + 4\right ) \log {\relax (5 )} + 80}{2 x^{2} + x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x**2-16*x-4)*ln(5)+32*x**2-320*x-80)*exp(((8*x+4)*ln(5)+16*x**2-8*x+80)/(2*x**2+x))/(4*x**4+4*
x**3+x**2),x)

[Out]

exp((16*x**2 - 8*x + (8*x + 4)*log(5) + 80)/(2*x**2 + x))

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