3.41.19 \(\int \frac {2+4 e^x+e^{1+x} (2+4 e^x)+(-4 e^x x+2 e^{1+x} x) \log (x)}{x+4 e^x x+4 e^{2 x} x} \, dx\)

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

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Rubi [F]  time = 0.85, 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 e^x+e^{1+x} \left (2+4 e^x\right )+\left (-4 e^x x+2 e^{1+x} x\right ) \log (x)}{x+4 e^x x+4 e^{2 x} x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.20, size = 20, normalized size = 0.83 \begin {gather*} \frac {2 \left (1+e^{1+x}\right ) \log (x)}{1+2 e^x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.51, size = 22, normalized size = 0.92 \begin {gather*} \frac {2 \, {\left (e + e^{\left (x + 2\right )}\right )} \log \relax (x)}{e + 2 \, e^{\left (x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(x+1)-4*exp(x)*x)*log(x)+(4*exp(x)+2)*exp(x+1)+4*exp(x)+2)/(4*x*exp(x)^2+4*exp(x)*x+x),x, a
lgorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 20, normalized size = 0.83 \begin {gather*} \frac {2 \, {\left (e^{\left (x + 1\right )} \log \relax (x) + \log \relax (x)\right )}}{2 \, e^{x} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(x+1)-4*exp(x)*x)*log(x)+(4*exp(x)+2)*exp(x+1)+4*exp(x)+2)/(4*x*exp(x)^2+4*exp(x)*x+x),x, a
lgorithm="giac")

[Out]

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

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maple [A]  time = 0.11, size = 23, normalized size = 0.96

method	result	size
norman	\(\frac {2 \ln \relax (x )+2 \,{\mathrm e} \,{\mathrm e}^{x} \ln \relax (x )}{2 \,{\mathrm e}^{x}+1}\)	\(23\)
risch	\(-\frac {\left ({\mathrm e}-2\right ) \ln \relax (x )}{2 \,{\mathrm e}^{x}+1}+{\mathrm e} \ln \relax (x )\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x*exp(x+1)-4*exp(x)*x)*ln(x)+(4*exp(x)+2)*exp(x+1)+4*exp(x)+2)/(4*x*exp(x)^2+4*exp(x)*x+x),x,method=_R
ETURNVERBOSE)

[Out]

(2*ln(x)+2*exp(1)*exp(x)*ln(x))/(2*exp(x)+1)

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maxima [A]  time = 0.42, size = 22, normalized size = 0.92 \begin {gather*} e \log \relax (x) - \frac {{\left (e - 2\right )} \log \relax (x)}{2 \, e^{x} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(x+1)-4*exp(x)*x)*log(x)+(4*exp(x)+2)*exp(x+1)+4*exp(x)+2)/(4*x*exp(x)^2+4*exp(x)*x+x),x, a
lgorithm="maxima")

[Out]

e*log(x) - (e - 2)*log(x)/(2*e^x + 1)

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mupad [B]  time = 3.19, size = 18, normalized size = 0.75 \begin {gather*} \frac {2\,\ln \relax (x)\,\left ({\mathrm {e}}^{x+1}+1\right )}{2\,{\mathrm {e}}^x+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*log(x)*(exp(x + 1) + 1))/(2*exp(x) + 1)

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sympy [A]  time = 0.25, size = 24, normalized size = 1.00 \begin {gather*} \frac {- e \log {\relax (x )} + 2 \log {\relax (x )}}{2 e^{x} + 1} + e \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-E*log(x) + 2*log(x))/(2*exp(x) + 1) + E*log(x)

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