3.63.4 \(\int \frac {2+4 x+e^{-2 x} (-x+2 x^2)}{10 x+4 x^2-e^{-2 x} x^2+2 x \log (x)} \, dx\)

Optimal. Leaf size=25 \[ \log \left (-5-2 x+\frac {1}{2} e^{3+x-3 (1+x)} x-\log (x)\right ) \]

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Rubi [F]  time = 2.46, 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 {2+4 x+e^{-2 x} \left (-x+2 x^2\right )}{10 x+4 x^2-e^{-2 x} x^2+2 x \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(2 + 4*x + (-x + 2*x^2)/E^(2*x))/(10*x + 4*x^2 - x^2/E^(2*x) + 2*x*Log[x]),x]

[Out]

Log[5 + 2*x + Log[x]] - 4*Defer[Int][1/((5 + 2*x + Log[x])*(10*E^(2*x) - x + 4*E^(2*x)*x + 2*E^(2*x)*Log[x])),
x] + 10*Defer[Int][x/((5 + 2*x + Log[x])*(10*E^(2*x) - x + 4*E^(2*x)*x + 2*E^(2*x)*Log[x])), x] + 4*Defer[Int
][x^2/((5 + 2*x + Log[x])*(10*E^(2*x) - x + 4*E^(2*x)*x + 2*E^(2*x)*Log[x])), x] - Defer[Int][Log[x]/((5 + 2*x
+ Log[x])*(10*E^(2*x) - x + 4*E^(2*x)*x + 2*E^(2*x)*Log[x])), x] + 2*Defer[Int][(x*Log[x])/((5 + 2*x + Log[x]
)*(10*E^(2*x) - x + 4*E^(2*x)*x + 2*E^(2*x)*Log[x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {1+2 x}{x (5+2 x+\log (x))}+\frac {-4+10 x+4 x^2-\log (x)+2 x \log (x)}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )}\right ) \, dx\\ &=\int \frac {1+2 x}{x (5+2 x+\log (x))} \, dx+\int \frac {-4+10 x+4 x^2-\log (x)+2 x \log (x)}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )} \, dx\\ &=\log (5+2 x+\log (x))+\int \left (-\frac {4}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )}+\frac {10 x}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )}+\frac {4 x^2}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )}-\frac {\log (x)}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )}+\frac {2 x \log (x)}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )}\right ) \, dx\\ &=\log (5+2 x+\log (x))+2 \int \frac {x \log (x)}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )} \, dx-4 \int \frac {1}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )} \, dx+4 \int \frac {x^2}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )} \, dx+10 \int \frac {x}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )} \, dx-\int \frac {\log (x)}{(5+2 x+\log (x)) \left (10 e^{2 x}-x+4 e^{2 x} x+2 e^{2 x} \log (x)\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 2.02, size = 18, normalized size = 0.72 \begin {gather*} \log \left (10+\left (4-e^{-2 x}\right ) x+2 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + 4*x + (-x + 2*x^2)/E^(2*x))/(10*x + 4*x^2 - x^2/E^(2*x) + 2*x*Log[x]),x]

[Out]

Log[10 + (4 - E^(-2*x))*x + 2*Log[x]]

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fricas [A]  time = 1.11, size = 17, normalized size = 0.68 \begin {gather*} \log \left (-x e^{\left (-2 \, x\right )} + 4 \, x + 2 \, \log \relax (x) + 10\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2-x)*exp(-3*x)*exp(x)+4*x+2)/(2*x*log(x)-x^2*exp(-3*x)*exp(x)+4*x^2+10*x),x, algorithm="fricas
")

[Out]

log(-x*e^(-2*x) + 4*x + 2*log(x) + 10)

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giac [A]  time = 0.15, size = 16, normalized size = 0.64 \begin {gather*} \log \left (x e^{\left (-2 \, x\right )} - 4 \, x - 2 \, \log \relax (x) - 10\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2-x)*exp(-3*x)*exp(x)+4*x+2)/(2*x*log(x)-x^2*exp(-3*x)*exp(x)+4*x^2+10*x),x, algorithm="giac")

[Out]

log(x*e^(-2*x) - 4*x - 2*log(x) - 10)

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maple [A]  time = 0.03, size = 28, normalized size = 1.12

method	result	size
risch	\(\ln \left (\ln \relax (x )+\frac {\left (4 x \,{\mathrm e}^{2 x}+10 \,{\mathrm e}^{2 x}-x \right ) {\mathrm e}^{-2 x}}{2}\right )\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^2-x)*exp(-3*x)*exp(x)+4*x+2)/(2*x*ln(x)-x^2*exp(-3*x)*exp(x)+4*x^2+10*x),x,method=_RETURNVERBOSE)

[Out]

ln(ln(x)+1/2*(4*x*exp(2*x)+10*exp(2*x)-x)*exp(-2*x))

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maxima [B]  time = 0.39, size = 41, normalized size = 1.64 \begin {gather*} -2 \, x + \log \left (2 \, x + \log \relax (x) + 5\right ) + \log \left (\frac {2 \, {\left (2 \, x + \log \relax (x) + 5\right )} e^{\left (2 \, x\right )} - x}{2 \, {\left (2 \, x + \log \relax (x) + 5\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2-x)*exp(-3*x)*exp(x)+4*x+2)/(2*x*log(x)-x^2*exp(-3*x)*exp(x)+4*x^2+10*x),x, algorithm="maxima
")

[Out]

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

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mupad [B]  time = 4.18, size = 15, normalized size = 0.60 \begin {gather*} \ln \left (2\,x+\ln \relax (x)-\frac {x\,{\mathrm {e}}^{-2\,x}}{2}+5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x - exp(-2*x)*(x - 2*x^2) + 2)/(10*x - x^2*exp(-2*x) + 2*x*log(x) + 4*x^2),x)

[Out]

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

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sympy [A]  time = 0.38, size = 24, normalized size = 0.96 \begin {gather*} \log {\relax (x )} + \log {\left (e^{- 2 x} + \frac {- 4 x - 2 \log {\relax (x )} - 10}{x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**2-x)*exp(-3*x)*exp(x)+4*x+2)/(2*x*ln(x)-x**2*exp(-3*x)*exp(x)+4*x**2+10*x),x)

[Out]

log(x) + log(exp(-2*x) + (-4*x - 2*log(x) - 10)/x)

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