3.89.45 $\int \frac{1764x+e^x(3+588x-3x^2)+(-9x-3e^{x}x)\log (x)}{e^{2x}x+6e^x{x}^2+9x^3}dx$

Optimal. Leaf size=17 $$\frac{-195+x+\log (x)}{\frac{e^x}{3}+x}$$

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Rubi [F]  time = 1.46, 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 = {} $$\begin{array}{c}\int \frac{1764x+e^x(3+588x-3x^2)+(-9x-3e^{x}x)\log (x)}{e^{2x}x+6e^x{x}^2+9x^3}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(1764*x + E^x*(3 + 588*x - 3*x^2) + (-9*x - 3*E^x*x)*Log[x])/(E^(2*x)*x + 6*E^x*x^2 + 9*x^3),x]

[Out]

1755*Defer[Int][(E^x + 3*x)^(-2), x] - 9*Log[x]*Defer[Int][(E^x + 3*x)^(-2), x] - 1764*Defer[Int][x/(E^x + 3*x
)^2, x] + 9*Log[x]*Defer[Int][x/(E^x + 3*x)^2, x] + 9*Defer[Int][x^2/(E^x + 3*x)^2, x] + 588*Defer[Int][(E^x +
3*x)^(-1), x] - 3*Log[x]*Defer[Int][(E^x + 3*x)^(-1), x] + 3*Defer[Int][1/(x*(E^x + 3*x)), x] - 3*Defer[Int][
x/(E^x + 3*x), x] + 9*Defer[Int][Defer[Int][(E^x + 3*x)^(-2), x]/x, x] - 9*Defer[Int][Defer[Int][x/(E^x + 3*x)
^2, x]/x, x] + 3*Defer[Int][Defer[Int][(E^x + 3*x)^(-1), x]/x, x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{3(588x-e^x(-1-196x+x^2)-(3+e^x)x\log (x))}{x{(e^x+3x)}^2}dx\\ & =3\int \frac{588x-e^x(-1-196x+x^2)-(3+e^x)x\log (x)}{x{(e^x+3x)}^2}dx\\ & =3\int (\frac{3(-1+x)(-195+x+\log (x))}{{(e^x+3x)}^2}-\frac{-1-196x+x^2+x\log (x)}{x(e^x+3x)})dx\\ & =-(3\int \frac{-1-196x+x^2+x\log (x)}{x(e^x+3x)}dx)+9\int \frac{(-1+x)(-195+x+\log (x))}{{(e^x+3x)}^2}dx\\ & =-(3\int (-\frac{196}{e^x+3x}-\frac{1}{x(e^x+3x)}+\frac{x}{e^x+3x}+\frac{\log (x)}{e^x+3x})dx)+9\int (-\frac{-195+x+\log (x)}{{(e^x+3x)}^2}+\frac{x(-195+x+\log (x))}{{(e^x+3x)}^2})dx\\ & =3\int \frac{1}{x(e^x+3x)}dx-3\int \frac{x}{e^x+3x}dx-3\int \frac{\log (x)}{e^x+3x}dx-9\int \frac{-195+x+\log (x)}{{(e^x+3x)}^2}dx+9\int \frac{x(-195+x+\log (x))}{{(e^x+3x)}^2}dx+588\int \frac{1}{e^x+3x}dx\\ & =3\int \frac{1}{x(e^x+3x)}dx-3\int \frac{x}{e^x+3x}dx+3\int \frac{\int \frac{1}{e^x+3x}dx}{x}dx-9\int (-\frac{195}{{(e^x+3x)}^2}+\frac{x}{{(e^x+3x)}^2}+\frac{\log (x)}{{(e^x+3x)}^2})dx+9\int (-\frac{195x}{{(e^x+3x)}^2}+\frac{x^2}{{(e^x+3x)}^2}+\frac{x\log (x)}{{(e^x+3x)}^2})dx+588\int \frac{1}{e^x+3x}dx-(3\log (x))\int \frac{1}{e^x+3x}dx\\ & =3\int \frac{1}{x(e^x+3x)}dx-3\int \frac{x}{e^x+3x}dx+3\int \frac{\int \frac{1}{e^x+3x}dx}{x}dx-9\int \frac{x}{{(e^x+3x)}^2}dx+9\int \frac{x^2}{{(e^x+3x)}^2}dx-9\int \frac{\log (x)}{{(e^x+3x)}^2}dx+9\int \frac{x\log (x)}{{(e^x+3x)}^2}dx+588\int \frac{1}{e^x+3x}dx+1755\int \frac{1}{{(e^x+3x)}^2}dx-1755\int \frac{x}{{(e^x+3x)}^2}dx-(3\log (x))\int \frac{1}{e^x+3x}dx\\ & =3\int \frac{1}{x(e^x+3x)}dx-3\int \frac{x}{e^x+3x}dx+3\int \frac{\int \frac{1}{e^x+3x}dx}{x}dx-9\int \frac{x}{{(e^x+3x)}^2}dx+9\int \frac{x^2}{{(e^x+3x)}^2}dx+9\int \frac{\int \frac{1}{{(e^x+3x)}^2}dx}{x}dx-9\int \frac{\int \frac{x}{{(e^x+3x)}^2}dx}{x}dx+588\int \frac{1}{e^x+3x}dx+1755\int \frac{1}{{(e^x+3x)}^2}dx-1755\int \frac{x}{{(e^x+3x)}^2}dx-(3\log (x))\int \frac{1}{e^x+3x}dx-(9\log (x))\int \frac{1}{{(e^x+3x)}^2}dx+(9\log (x))\int \frac{x}{{(e^x+3x)}^2}dx\end{array}\end{array}$$

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Mathematica [A]  time = 0.26, size = 16, normalized size = 0.94 $$\begin{array}{c}\frac{3(-195+x+\log (x))}{e^x+3x}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(1764*x + E^x*(3 + 588*x - 3*x^2) + (-9*x - 3*E^x*x)*Log[x])/(E^(2*x)*x + 6*E^x*x^2 + 9*x^3),x]

[Out]

(3*(-195 + x + Log[x]))/(E^x + 3*x)

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fricas [A]  time = 0.49, size = 15, normalized size = 0.88 $$\begin{array}{c}\frac{3(x+\log (x)-195)}{3x+e^x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*exp(x)*x-9*x)*log(x)+(-3*x^2+588*x+3)*exp(x)+1764*x)/(x*exp(x)^2+6*exp(x)*x^2+9*x^3),x, algorit
hm="fricas")

[Out]

3*(x + log(x) - 195)/(3*x + e^x)

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giac [A]  time = 0.15, size = 15, normalized size = 0.88 $$\begin{array}{c}\frac{3(x+\log (x)-195)}{3x+e^x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*exp(x)*x-9*x)*log(x)+(-3*x^2+588*x+3)*exp(x)+1764*x)/(x*exp(x)^2+6*exp(x)*x^2+9*x^3),x, algorit
hm="giac")

[Out]

3*(x + log(x) - 195)/(3*x + e^x)

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maple [A]  time = 0.04, size = 27, normalized size = 1.59

method	result	size
risch	$\frac{3\ln (x)}{3x+{\mathrm{e}}^x}+\frac{3x-585}{3x+{\mathrm{e}}^x}$	$27$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-3*exp(x)*x-9*x)*ln(x)+(-3*x^2+588*x+3)*exp(x)+1764*x)/(x*exp(x)^2+6*exp(x)*x^2+9*x^3),x,method=_RETURNV
ERBOSE)

[Out]

3/(3*x+exp(x))*ln(x)+3*(x-195)/(3*x+exp(x))

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maxima [A]  time = 0.41, size = 15, normalized size = 0.88 $$\begin{array}{c}\frac{3(x+\log (x)-195)}{3x+e^x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*exp(x)*x-9*x)*log(x)+(-3*x^2+588*x+3)*exp(x)+1764*x)/(x*exp(x)^2+6*exp(x)*x^2+9*x^3),x, algorit
hm="maxima")

[Out]

3*(x + log(x) - 195)/(3*x + e^x)

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mupad [B]  time = 5.26, size = 15, normalized size = 0.88 $$\begin{array}{c}\frac{3(x+\ln (x)-195)}{3x+{\mathrm{e}}^x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1764*x - log(x)*(9*x + 3*x*exp(x)) + exp(x)*(588*x - 3*x^2 + 3))/(x*exp(2*x) + 6*x^2*exp(x) + 9*x^3),x)

[Out]

(3*(x + log(x) - 195))/(3*x + exp(x))

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sympy [A]  time = 0.25, size = 15, normalized size = 0.88 $$\begin{array}{c}\frac{3x+3\log (x)-585}{3x+e^x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*exp(x)*x-9*x)*ln(x)+(-3*x**2+588*x+3)*exp(x)+1764*x)/(x*exp(x)**2+6*exp(x)*x**2+9*x**3),x)

[Out]

(3*x + 3*log(x) - 585)/(3*x + exp(x))

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