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

Optimal. Leaf size=24 $$\frac{(x+e^{1+x}x)\log (x)}{(\frac{1}{2}+e^x)x}$$

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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 = {} $$\begin{array}{c}\int \frac{2+4e^x+e^{1+x}(2+4e^x)+(-4e^{x}x+2e^{1+x}x)\log (x)}{x+4e^{x}x+4e^{2x}x}dx\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{2((1+2e^x)(1+e^{1+x})+(-2+e)e^{x}x\log (x))}{{(1+2e^x)}^{2}x}dx\\ & =2\int \frac{(1+2e^x)(1+e^{1+x})+(-2+e)e^{x}x\log (x)}{{(1+2e^x)}^{2}x}dx\\ & =2\int (\frac{e}{2x}-\frac{(-2+e)\log (x)}{2{(1+2e^x)}^2}+\frac{(-2+e)(-1+x\log (x))}{2(1+2e^x)x})dx\\ & =e\log (x)+(2-e)\int \frac{\log (x)}{{(1+2e^x)}^2}dx+(-2+e)\int \frac{-1+x\log (x)}{(1+2e^x)x}dx\\ & =e\log (x)+(2-e)\int \frac{\log (x)}{{(1+2e^x)}^2}dx+(-2+e)\int (-\frac{1}{(1+2e^x)x}+\frac{\log (x)}{1+2e^x})dx\\ & =e\log (x)+(2-e)\int \frac{1}{(1+2e^x)x}dx+(2-e)\int \frac{\log (x)}{{(1+2e^x)}^2}dx+(-2+e)\int \frac{\log (x)}{1+2e^x}dx\end{array}\end{array}$$

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Mathematica [A]  time = 0.20, size = 20, normalized size = 0.83 $$\begin{array}{c}\frac{2(1+e^{1+x})\log (x)}{1+2e^x}\end{array}$$

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 $$\begin{array}{c}\frac{2(e+e^{(x+2)})\log (x)}{e+2e^{(x+1)}}\end{array}$$

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 $$\begin{array}{c}\frac{2(e^{(x+1)}\log (x)+\log (x))}{2e^x+1}\end{array}$$

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	$\frac{2\ln (x)+2\mathrm{e}{\mathrm{e}}^x\ln (x)}{2{\mathrm{e}}^x+1}$	$23$
risch	$-\frac{(\mathrm{e}-2)\ln (x)}{2{\mathrm{e}}^x+1}+\mathrm{e}\ln (x)$	$23$

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 $$\begin{array}{c}e\log (x)-\frac{(e-2)\log (x)}{2e^x+1}\end{array}$$

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 $$\begin{array}{c}\frac{2\ln (x)({\mathrm{e}}^{x+1}+1)}{2{\mathrm{e}}^x+1}\end{array}$$

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 $$\begin{array}{c}\frac{-e\log (x)+2\log (x)}{2e^x+1}+e\log (x)\end{array}$$

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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