3.96.55 \(\int \frac {e^{\frac {20+29 x-20 x^2-5 x^3-5 \log (4)}{225 x+45 x^2}} (-100-40 x-129 x^2-50 x^3-5 x^4+(25+10 x) \log (4))}{3375 x^2+1350 x^3+135 x^4} \, dx\)

Optimal. Leaf size=34 \[ -5+\frac {1}{3} e^{\frac {\frac {4}{5}+x-x^2-\frac {\log (4)}{5+x}}{9 x}} \]

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Rubi [F]  time = 2.97, 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 {20+29 x-20 x^2-5 x^3-5 \log (4)}{225 x+45 x^2}\right ) \left (-100-40 x-129 x^2-50 x^3-5 x^4+(25+10 x) \log (4)\right )}{3375 x^2+1350 x^3+135 x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((20 + 29*x - 20*x^2 - 5*x^3 - 5*Log[4])/(225*x + 45*x^2))*(-100 - 40*x - 129*x^2 - 50*x^3 - 5*x^4 + (2
5 + 10*x)*Log[4]))/(3375*x^2 + 1350*x^3 + 135*x^4),x]

[Out]

-1/27*Defer[Int][E^((29*x - 20*x^2 - 5*x^3 + 5*(4 - Log[4]))/(x*(225 + 45*x))), x] - ((4 - Log[4])*Defer[Int][
E^((29*x - 20*x^2 - 5*x^3 + 5*(4 - Log[4]))/(x*(225 + 45*x)))/x^2, x])/135 - (Log[4]*Defer[Int][E^((29*x - 20*
x^2 - 5*x^3 + 5*(4 - Log[4]))/(x*(225 + 45*x)))/(5 + x)^2, x])/135

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {20+29 x-20 x^2-5 x^3-5 \log (4)}{225 x+45 x^2}\right ) \left (-100-40 x-129 x^2-50 x^3-5 x^4+(25+10 x) \log (4)\right )}{x^2 \left (3375+1350 x+135 x^2\right )} \, dx\\ &=\int \frac {\exp \left (\frac {20+29 x-20 x^2-5 x^3-5 \log (4)}{225 x+45 x^2}\right ) \left (-100-40 x-129 x^2-50 x^3-5 x^4+(25+10 x) \log (4)\right )}{135 x^2 (5+x)^2} \, dx\\ &=\frac {1}{135} \int \frac {\exp \left (\frac {20+29 x-20 x^2-5 x^3-5 \log (4)}{225 x+45 x^2}\right ) \left (-100-40 x-129 x^2-50 x^3-5 x^4+(25+10 x) \log (4)\right )}{x^2 (5+x)^2} \, dx\\ &=\frac {1}{135} \int \frac {\exp \left (\frac {29 x-20 x^2-5 x^3+5 (4-\log (4))}{x (225+45 x)}\right ) \left (-129 x^2-50 x^3-5 x^4-25 (4-\log (4))-10 x (4-\log (4))\right )}{x^2 (5+x)^2} \, dx\\ &=\frac {1}{135} \int \left (-5 \exp \left (\frac {29 x-20 x^2-5 x^3+5 (4-\log (4))}{x (225+45 x)}\right )+\frac {\exp \left (\frac {29 x-20 x^2-5 x^3+5 (4-\log (4))}{x (225+45 x)}\right ) (-4+\log (4))}{x^2}-\frac {\exp \left (\frac {29 x-20 x^2-5 x^3+5 (4-\log (4))}{x (225+45 x)}\right ) \log (4)}{(5+x)^2}\right ) \, dx\\ &=-\left (\frac {1}{27} \int \exp \left (\frac {29 x-20 x^2-5 x^3+5 (4-\log (4))}{x (225+45 x)}\right ) \, dx\right )+\frac {1}{135} (-4+\log (4)) \int \frac {\exp \left (\frac {29 x-20 x^2-5 x^3+5 (4-\log (4))}{x (225+45 x)}\right )}{x^2} \, dx-\frac {1}{135} \log (4) \int \frac {\exp \left (\frac {29 x-20 x^2-5 x^3+5 (4-\log (4))}{x (225+45 x)}\right )}{(5+x)^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 2.23, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{\frac {20+29 x-20 x^2-5 x^3-5 \log (4)}{225 x+45 x^2}} \left (-100-40 x-129 x^2-50 x^3-5 x^4+(25+10 x) \log (4)\right )}{3375 x^2+1350 x^3+135 x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(E^((20 + 29*x - 20*x^2 - 5*x^3 - 5*Log[4])/(225*x + 45*x^2))*(-100 - 40*x - 129*x^2 - 50*x^3 - 5*x^
4 + (25 + 10*x)*Log[4]))/(3375*x^2 + 1350*x^3 + 135*x^4),x]

[Out]

Integrate[(E^((20 + 29*x - 20*x^2 - 5*x^3 - 5*Log[4])/(225*x + 45*x^2))*(-100 - 40*x - 129*x^2 - 50*x^3 - 5*x^
4 + (25 + 10*x)*Log[4]))/(3375*x^2 + 1350*x^3 + 135*x^4), x]

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fricas [A]  time = 0.61, size = 33, normalized size = 0.97 \begin {gather*} \frac {1}{3} \, e^{\left (-\frac {5 \, x^{3} + 20 \, x^{2} - 29 \, x + 10 \, \log \relax (2) - 20}{45 \, {\left (x^{2} + 5 \, x\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(10*x+25)*log(2)-5*x^4-50*x^3-129*x^2-40*x-100)*exp((-10*log(2)-5*x^3-20*x^2+29*x+20)/(45*x^2+225
*x))/(135*x^4+1350*x^3+3375*x^2),x, algorithm="fricas")

[Out]

1/3*e^(-1/45*(5*x^3 + 20*x^2 - 29*x + 10*log(2) - 20)/(x^2 + 5*x))

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giac [B]  time = 0.14, size = 68, normalized size = 2.00 \begin {gather*} \frac {1}{3} \, e^{\left (-\frac {x^{3}}{9 \, {\left (x^{2} + 5 \, x\right )}} - \frac {4 \, x^{2}}{9 \, {\left (x^{2} + 5 \, x\right )}} + \frac {29 \, x}{45 \, {\left (x^{2} + 5 \, x\right )}} - \frac {2 \, \log \relax (2)}{9 \, {\left (x^{2} + 5 \, x\right )}} + \frac {4}{9 \, {\left (x^{2} + 5 \, x\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(10*x+25)*log(2)-5*x^4-50*x^3-129*x^2-40*x-100)*exp((-10*log(2)-5*x^3-20*x^2+29*x+20)/(45*x^2+225
*x))/(135*x^4+1350*x^3+3375*x^2),x, algorithm="giac")

[Out]

1/3*e^(-1/9*x^3/(x^2 + 5*x) - 4/9*x^2/(x^2 + 5*x) + 29/45*x/(x^2 + 5*x) - 2/9*log(2)/(x^2 + 5*x) + 4/9/(x^2 +
5*x))

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maple [A]  time = 0.14, size = 33, normalized size = 0.97

method	result	size
gosper	\(\frac {{\mathrm e}^{-\frac {5 x^{3}+20 x^{2}+10 \ln \relax (2)-29 x -20}{45 \left (5+x \right ) x}}}{3}\)	\(33\)
risch	\(\frac {{\mathrm e}^{-\frac {5 x^{3}+20 x^{2}+10 \ln \relax (2)-29 x -20}{45 \left (5+x \right ) x}}}{3}\)	\(33\)
norman	\(\frac {\frac {5 x \,{\mathrm e}^{\frac {-10 \ln \relax (2)-5 x^{3}-20 x^{2}+29 x +20}{45 x^{2}+225 x}}}{3}+\frac {x^{2} {\mathrm e}^{\frac {-10 \ln \relax (2)-5 x^{3}-20 x^{2}+29 x +20}{45 x^{2}+225 x}}}{3}}{\left (5+x \right ) x}\)	\(83\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*(10*x+25)*ln(2)-5*x^4-50*x^3-129*x^2-40*x-100)*exp((-10*ln(2)-5*x^3-20*x^2+29*x+20)/(45*x^2+225*x))/(13
5*x^4+1350*x^3+3375*x^2),x,method=_RETURNVERBOSE)

[Out]

1/3*exp(-1/45*(5*x^3+20*x^2+10*ln(2)-29*x-20)/(5+x)/x)

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maxima [A]  time = 0.62, size = 29, normalized size = 0.85 \begin {gather*} \frac {1}{3} \, e^{\left (-\frac {1}{9} \, x + \frac {2 \, \log \relax (2)}{45 \, {\left (x + 5\right )}} - \frac {2 \, \log \relax (2)}{45 \, x} + \frac {4}{45 \, x} + \frac {1}{9}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(10*x+25)*log(2)-5*x^4-50*x^3-129*x^2-40*x-100)*exp((-10*log(2)-5*x^3-20*x^2+29*x+20)/(45*x^2+225
*x))/(135*x^4+1350*x^3+3375*x^2),x, algorithm="maxima")

[Out]

1/3*e^(-1/9*x + 2/45*log(2)/(x + 5) - 2/45*log(2)/x + 4/45/x + 1/9)

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mupad [B]  time = 9.54, size = 82, normalized size = 2.41 \begin {gather*} \frac {{\mathrm {e}}^{\frac {29\,x}{45\,x^2+225\,x}}\,{\mathrm {e}}^{-\frac {5\,x^3}{45\,x^2+225\,x}}\,{\mathrm {e}}^{-\frac {20\,x^2}{45\,x^2+225\,x}}\,{\mathrm {e}}^{\frac {20}{45\,x^2+225\,x}}}{3\,2^{\frac {10}{45\,x^2+225\,x}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(10*log(2) - 29*x + 20*x^2 + 5*x^3 - 20)/(225*x + 45*x^2))*(40*x - 2*log(2)*(10*x + 25) + 129*x^2 +
50*x^3 + 5*x^4 + 100))/(3375*x^2 + 1350*x^3 + 135*x^4),x)

[Out]

(exp((29*x)/(225*x + 45*x^2))*exp(-(5*x^3)/(225*x + 45*x^2))*exp(-(20*x^2)/(225*x + 45*x^2))*exp(20/(225*x + 4
5*x^2)))/(3*2^(10/(225*x + 45*x^2)))

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sympy [A]  time = 0.53, size = 31, normalized size = 0.91 \begin {gather*} \frac {e^{\frac {- 5 x^{3} - 20 x^{2} + 29 x - 10 \log {\relax (2 )} + 20}{45 x^{2} + 225 x}}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(10*x+25)*ln(2)-5*x**4-50*x**3-129*x**2-40*x-100)*exp((-10*ln(2)-5*x**3-20*x**2+29*x+20)/(45*x**2
+225*x))/(135*x**4+1350*x**3+3375*x**2),x)

[Out]

exp((-5*x**3 - 20*x**2 + 29*x - 10*log(2) + 20)/(45*x**2 + 225*x))/3

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