3.7.37 \(\int \frac {-25+140 x-176 x^2+64 x^3+e^{-\frac {x}{-5+4 x}} (-25+35 x-16 x^2)+(25-40 x+16 x^2) \log (5)}{25-40 x+16 x^2} \, dx\)

Optimal. Leaf size=22 \[ x \left (-1-e^{\frac {1}{-4+\frac {5}{x}}}+2 x+\log (5)\right ) \]

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Rubi [F]  time = 0.42, 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 {-25+140 x-176 x^2+64 x^3+e^{-\frac {x}{-5+4 x}} \left (-25+35 x-16 x^2\right )+\left (25-40 x+16 x^2\right ) \log (5)}{25-40 x+16 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-25 + 140*x - 176*x^2 + 64*x^3 + (-25 + 35*x - 16*x^2)/E^(x/(-5 + 4*x)) + (25 - 40*x + 16*x^2)*Log[5])/(2
5 - 40*x + 16*x^2),x]

[Out]

(-5*E^(-1/4 + 5/(4*(5 - 4*x))))/4 + 2*x^2 + (5*ExpIntegralEi[5/(4*(5 - 4*x))])/(16*E^(1/4)) - x*(1 - Log[5]) -
Defer[Int][E^(x/(5 - 4*x)), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-25+140 x-176 x^2+64 x^3+e^{-\frac {x}{-5+4 x}} \left (-25+35 x-16 x^2\right )+\left (25-40 x+16 x^2\right ) \log (5)}{(-5+4 x)^2} \, dx\\ &=\int \left (-1+4 x+\frac {e^{\frac {x}{5-4 x}} \left (-25+35 x-16 x^2\right )}{(5-4 x)^2}+\log (5)\right ) \, dx\\ &=2 x^2-x (1-\log (5))+\int \frac {e^{\frac {x}{5-4 x}} \left (-25+35 x-16 x^2\right )}{(5-4 x)^2} \, dx\\ &=2 x^2-x (1-\log (5))+\int \left (-e^{\frac {x}{5-4 x}}-\frac {25 e^{\frac {x}{5-4 x}}}{4 (-5+4 x)^2}-\frac {5 e^{\frac {x}{5-4 x}}}{4 (-5+4 x)}\right ) \, dx\\ &=2 x^2-x (1-\log (5))-\frac {5}{4} \int \frac {e^{\frac {x}{5-4 x}}}{-5+4 x} \, dx-\frac {25}{4} \int \frac {e^{\frac {x}{5-4 x}}}{(-5+4 x)^2} \, dx-\int e^{\frac {x}{5-4 x}} \, dx\\ &=2 x^2-x (1-\log (5))-\frac {5}{4} \int \frac {e^{-\frac {1}{4}+\frac {5}{4 (5-4 x)}}}{-5+4 x} \, dx-\frac {25}{4} \int \frac {e^{-\frac {1}{4}+\frac {5}{4 (5-4 x)}}}{(-5+4 x)^2} \, dx-\int e^{\frac {x}{5-4 x}} \, dx\\ &=-\frac {5}{4} e^{-\frac {1}{4}+\frac {5}{4 (5-4 x)}}+2 x^2+\frac {5 \text {Ei}\left (\frac {5}{4 (5-4 x)}\right )}{16 \sqrt [4]{e}}-x (1-\log (5))-\int e^{\frac {x}{5-4 x}} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 22, normalized size = 1.00 \begin {gather*} x \left (-1-e^{\frac {x}{5-4 x}}+2 x+\log (5)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-25 + 140*x - 176*x^2 + 64*x^3 + (-25 + 35*x - 16*x^2)/E^(x/(-5 + 4*x)) + (25 - 40*x + 16*x^2)*Log[
5])/(25 - 40*x + 16*x^2),x]

[Out]

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

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fricas [A]  time = 0.92, size = 27, normalized size = 1.23 \begin {gather*} 2 \, x^{2} - x e^{\left (-\frac {x}{4 \, x - 5}\right )} + x \log \relax (5) - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x^2+35*x-25)*exp(-x/(4*x-5))+(16*x^2-40*x+25)*log(5)+64*x^3-176*x^2+140*x-25)/(16*x^2-40*x+25)
,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.50, size = 100, normalized size = 4.55 \begin {gather*} -\frac {5 \, {\left (\frac {8 \, x e^{\left (-\frac {x}{4 \, x - 5}\right )}}{4 \, x - 5} - \frac {32 \, x^{2} e^{\left (-\frac {x}{4 \, x - 5}\right )}}{{\left (4 \, x - 5\right )}^{2}} + \frac {8 \, x \log \relax (5)}{4 \, x - 5} + \frac {32 \, x}{4 \, x - 5} - 2 \, \log \relax (5) - 3\right )}}{8 \, {\left (\frac {8 \, x}{4 \, x - 5} - \frac {16 \, x^{2}}{{\left (4 \, x - 5\right )}^{2}} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x^2+35*x-25)*exp(-x/(4*x-5))+(16*x^2-40*x+25)*log(5)+64*x^3-176*x^2+140*x-25)/(16*x^2-40*x+25)
,x, algorithm="giac")

[Out]

-5/8*(8*x*e^(-x/(4*x - 5))/(4*x - 5) - 32*x^2*e^(-x/(4*x - 5))/(4*x - 5)^2 + 8*x*log(5)/(4*x - 5) + 32*x/(4*x
- 5) - 2*log(5) - 3)/(8*x/(4*x - 5) - 16*x^2/(4*x - 5)^2 - 1)

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maple [A]  time = 0.20, size = 28, normalized size = 1.27

method	result	size
risch	\(x \ln \relax (5)+2 x^{2}-x \,{\mathrm e}^{-\frac {x}{4 x -5}}-x\)	\(28\)
derivativedivides	\(\frac {\ln \relax (5) \left (4 x -5\right )}{4}+\frac {\left (4 x -5\right )^{2}}{8}+4 x -5-\frac {{\mathrm e}^{-\frac {1}{4}-\frac {5}{4 \left (4 x -5\right )}} \left (4 x -5\right )}{4}-\frac {5 \,{\mathrm e}^{-\frac {1}{4}-\frac {5}{4 \left (4 x -5\right )}}}{4}\)	\(57\)
default	\(\frac {\ln \relax (5) \left (4 x -5\right )}{4}+\frac {\left (4 x -5\right )^{2}}{8}+4 x -5-\frac {{\mathrm e}^{-\frac {1}{4}-\frac {5}{4 \left (4 x -5\right )}} \left (4 x -5\right )}{4}-\frac {5 \,{\mathrm e}^{-\frac {1}{4}-\frac {5}{4 \left (4 x -5\right )}}}{4}\)	\(57\)
norman	\(\frac {\left (-14+4 \ln \relax (5)\right ) x^{2}+8 x^{3}+5 x \,{\mathrm e}^{-\frac {x}{4 x -5}}-4 x^{2} {\mathrm e}^{-\frac {x}{4 x -5}}+\frac {25}{4}-\frac {25 \ln \relax (5)}{4}}{4 x -5}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-16*x^2+35*x-25)*exp(-x/(4*x-5))+(16*x^2-40*x+25)*ln(5)+64*x^3-176*x^2+140*x-25)/(16*x^2-40*x+25),x,meth
od=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 2 \, x^{2} - x e^{\left (-\frac {5}{4 \, {\left (4 \, x - 5\right )}} - \frac {1}{4}\right )} + \frac {1}{4} \, {\left (4 \, x - \frac {25}{4 \, x - 5} + 10 \, \log \left (4 \, x - 5\right )\right )} \log \relax (5) + \frac {5}{2} \, {\left (\frac {5}{4 \, x - 5} - \log \left (4 \, x - 5\right )\right )} \log \relax (5) - x - \frac {25 \, \log \relax (5)}{4 \, {\left (4 \, x - 5\right )}} - 5 \, e^{\left (-\frac {5}{4 \, {\left (4 \, x - 5\right )}} - \frac {1}{4}\right )} + 25 \, \int \frac {e^{\left (-\frac {5}{4 \, {\left (4 \, x - 5\right )}}\right )}}{16 \, x^{2} e^{\frac {1}{4}} - 40 \, x e^{\frac {1}{4}} + 25 \, e^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x^2+35*x-25)*exp(-x/(4*x-5))+(16*x^2-40*x+25)*log(5)+64*x^3-176*x^2+140*x-25)/(16*x^2-40*x+25)
,x, algorithm="maxima")

[Out]

2*x^2 - x*e^(-5/4/(4*x - 5) - 1/4) + 1/4*(4*x - 25/(4*x - 5) + 10*log(4*x - 5))*log(5) + 5/2*(5/(4*x - 5) - lo
g(4*x - 5))*log(5) - x - 25/4*log(5)/(4*x - 5) - 5*e^(-5/4/(4*x - 5) - 1/4) + 25*integrate(e^(-5/4/(4*x - 5))/
(16*x^2*e^(1/4) - 40*x*e^(1/4) + 25*e^(1/4)), x)

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mupad [B]  time = 0.14, size = 26, normalized size = 1.18 \begin {gather*} 2\,x^2-x\,\left ({\mathrm {e}}^{-\frac {x}{4\,x-5}}-\ln \relax (5)+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((140*x + log(5)*(16*x^2 - 40*x + 25) - exp(-x/(4*x - 5))*(16*x^2 - 35*x + 25) - 176*x^2 + 64*x^3 - 25)/(16
*x^2 - 40*x + 25),x)

[Out]

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

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sympy [A]  time = 0.19, size = 20, normalized size = 0.91 \begin {gather*} 2 x^{2} + x \left (-1 + \log {\relax (5 )}\right ) - x e^{- \frac {x}{4 x - 5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x**2+35*x-25)*exp(-x/(4*x-5))+(16*x**2-40*x+25)*ln(5)+64*x**3-176*x**2+140*x-25)/(16*x**2-40*x
+25),x)

[Out]

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

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