3.43.82 \(\int \frac {5 e^x (1+x-2 x^2)+5 e^x (3 x-2 x^2) \log (\frac {5}{x})}{4 x-16 x^2+16 x^3+(8 x-32 x^2+32 x^3) \log (\frac {5}{x})+(4 x-16 x^2+16 x^3) \log ^2(\frac {5}{x})} \, dx\)

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

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Rubi [F]  time = 2.12, 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 {5 e^x \left (1+x-2 x^2\right )+5 e^x \left (3 x-2 x^2\right ) \log \left (\frac {5}{x}\right )}{4 x-16 x^2+16 x^3+\left (8 x-32 x^2+32 x^3\right ) \log \left (\frac {5}{x}\right )+\left (4 x-16 x^2+16 x^3\right ) \log ^2\left (\frac {5}{x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(5*E^x*(1 + x - 2*x^2) + 5*E^x*(3*x - 2*x^2)*Log[5/x])/(4*x - 16*x^2 + 16*x^3 + (8*x - 32*x^2 + 32*x^3)*Lo
g[5/x] + (4*x - 16*x^2 + 16*x^3)*Log[5/x]^2),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 e^x \left (1+x-2 x^2-x (-3+2 x) \log \left (\frac {5}{x}\right )\right )}{4 (1-2 x)^2 x \left (1+\log \left (\frac {5}{x}\right )\right )^2} \, dx\\ &=\frac {5}{4} \int \frac {e^x \left (1+x-2 x^2-x (-3+2 x) \log \left (\frac {5}{x}\right )\right )}{(1-2 x)^2 x \left (1+\log \left (\frac {5}{x}\right )\right )^2} \, dx\\ &=\frac {5}{4} \int \left (-\frac {e^x}{x (-1+2 x) \left (1+\log \left (\frac {5}{x}\right )\right )^2}+\frac {e^x (3-2 x)}{(-1+2 x)^2 \left (1+\log \left (\frac {5}{x}\right )\right )}\right ) \, dx\\ &=-\left (\frac {5}{4} \int \frac {e^x}{x (-1+2 x) \left (1+\log \left (\frac {5}{x}\right )\right )^2} \, dx\right )+\frac {5}{4} \int \frac {e^x (3-2 x)}{(-1+2 x)^2 \left (1+\log \left (\frac {5}{x}\right )\right )} \, dx\\ &=-\left (\frac {5}{4} \int \left (-\frac {e^x}{x \left (1+\log \left (\frac {5}{x}\right )\right )^2}+\frac {2 e^x}{(-1+2 x) \left (1+\log \left (\frac {5}{x}\right )\right )^2}\right ) \, dx\right )+\frac {5}{4} \int \left (\frac {2 e^x}{(-1+2 x)^2 \left (1+\log \left (\frac {5}{x}\right )\right )}-\frac {e^x}{(-1+2 x) \left (1+\log \left (\frac {5}{x}\right )\right )}\right ) \, dx\\ &=\frac {5}{4} \int \frac {e^x}{x \left (1+\log \left (\frac {5}{x}\right )\right )^2} \, dx-\frac {5}{4} \int \frac {e^x}{(-1+2 x) \left (1+\log \left (\frac {5}{x}\right )\right )} \, dx-\frac {5}{2} \int \frac {e^x}{(-1+2 x) \left (1+\log \left (\frac {5}{x}\right )\right )^2} \, dx+\frac {5}{2} \int \frac {e^x}{(-1+2 x)^2 \left (1+\log \left (\frac {5}{x}\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.30, size = 24, normalized size = 1.00 \begin {gather*} -\frac {5 e^x}{4 (-1+2 x) \left (1+\log \left (\frac {5}{x}\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5*E^x*(1 + x - 2*x^2) + 5*E^x*(3*x - 2*x^2)*Log[5/x])/(4*x - 16*x^2 + 16*x^3 + (8*x - 32*x^2 + 32*x
^3)*Log[5/x] + (4*x - 16*x^2 + 16*x^3)*Log[5/x]^2),x]

[Out]

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

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fricas [A]  time = 0.53, size = 26, normalized size = 1.08 \begin {gather*} -\frac {e^{\left (x + \log \relax (5)\right )}}{4 \, {\left ({\left (2 \, x - 1\right )} \log \left (\frac {5}{x}\right ) + 2 \, x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2+3*x)*exp(log(5)+x)*log(5/x)+(-2*x^2+x+1)*exp(log(5)+x))/((16*x^3-16*x^2+4*x)*log(5/x)^2+(32
*x^3-32*x^2+8*x)*log(5/x)+16*x^3-16*x^2+4*x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.24, size = 28, normalized size = 1.17 \begin {gather*} -\frac {5 \, e^{x}}{4 \, {\left (2 \, x \log \left (\frac {5}{x}\right ) + 2 \, x - \log \left (\frac {5}{x}\right ) - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2+3*x)*exp(log(5)+x)*log(5/x)+(-2*x^2+x+1)*exp(log(5)+x))/((16*x^3-16*x^2+4*x)*log(5/x)^2+(32
*x^3-32*x^2+8*x)*log(5/x)+16*x^3-16*x^2+4*x),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.11, size = 28, normalized size = 1.17

method	result	size
risch	\(-\frac {5 i {\mathrm e}^{x}}{2 \left (2 x -1\right ) \left (2 i \ln \relax (5)-2 i \ln \relax (x )+2 i\right )}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^2+3*x)*exp(ln(5)+x)*ln(5/x)+(-2*x^2+x+1)*exp(ln(5)+x))/((16*x^3-16*x^2+4*x)*ln(5/x)^2+(32*x^3-32*x^
2+8*x)*ln(5/x)+16*x^3-16*x^2+4*x),x,method=_RETURNVERBOSE)

[Out]

-5/2*I*exp(x)/(2*x-1)/(2*I*ln(5)-2*I*ln(x)+2*I)

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maxima [A]  time = 0.50, size = 28, normalized size = 1.17 \begin {gather*} -\frac {5 \, e^{x}}{4 \, {\left (2 \, x {\left (\log \relax (5) + 1\right )} - {\left (2 \, x - 1\right )} \log \relax (x) - \log \relax (5) - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2+3*x)*exp(log(5)+x)*log(5/x)+(-2*x^2+x+1)*exp(log(5)+x))/((16*x^3-16*x^2+4*x)*log(5/x)^2+(32
*x^3-32*x^2+8*x)*log(5/x)+16*x^3-16*x^2+4*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.71, size = 21, normalized size = 0.88 \begin {gather*} -\frac {5\,{\mathrm {e}}^x}{4\,\left (\ln \left (\frac {5}{x}\right )+1\right )\,\left (2\,x-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.37, size = 26, normalized size = 1.08 \begin {gather*} - \frac {5 e^{x}}{8 x \log {\left (\frac {5}{x} \right )} + 8 x - 4 \log {\left (\frac {5}{x} \right )} - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**2+3*x)*exp(ln(5)+x)*ln(5/x)+(-2*x**2+x+1)*exp(ln(5)+x))/((16*x**3-16*x**2+4*x)*ln(5/x)**2+(3
2*x**3-32*x**2+8*x)*ln(5/x)+16*x**3-16*x**2+4*x),x)

[Out]

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

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