3.17.98 $\int \frac{10-10x-5x^2+e^x(5+5x)+e^x(5+5x+5x^2)\log (x)}{4x^2-4x^3+x^4+e^x(4x^2-2x^3)\log (x)+e^{2x}{x}^2{\log }^2(x)}dx$

Optimal. Leaf size=20 $$\frac{5(1+x)}{x(-2+x-e^x\log (x))}$$

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Rubi [F]  time = 4.28, 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 = {} $$\begin{array}{c}\int \frac{10-10x-5x^2+e^x(5+5x)+e^x(5+5x+5x^2)\log (x)}{4x^2-4x^3+x^4+e^x(4x^2-2x^3)\log (x)+e^{2x}{x}^2{\log }^2(x)}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(10 - 10*x - 5*x^2 + E^x*(5 + 5*x) + E^x*(5 + 5*x + 5*x^2)*Log[x])/(4*x^2 - 4*x^3 + x^4 + E^x*(4*x^2 - 2*x
^3)*Log[x] + E^(2*x)*x^2*Log[x]^2),x]

[Out]

-10*Defer[Int][(-2 + x - E^x*Log[x])^(-2), x] - 15*Defer[Int][1/(x*(-2 + x - E^x*Log[x])^2), x] + 5*Defer[Int]
[x/(-2 + x - E^x*Log[x])^2, x] - 10*Defer[Int][1/(x^2*Log[x]*(-2 + x - E^x*Log[x])^2), x] - 5*Defer[Int][1/(x*
Log[x]*(-2 + x - E^x*Log[x])^2), x] - 5*Defer[Int][(-2 + x - E^x*Log[x])^(-1), x] - 5*Defer[Int][1/(x^2*(-2 +
x - E^x*Log[x])), x] - 5*Defer[Int][1/(x*(-2 + x - E^x*Log[x])), x] - 5*Defer[Int][1/(x^2*Log[x]*(-2 + x - E^x
*Log[x])), x] - 5*Defer[Int][1/(x*Log[x]*(-2 + x - E^x*Log[x])), x] + 5*Defer[Int][1/(Log[x]*(2 - x + E^x*Log[
x])^2), x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{5(2-2x-x^2+e^x(1+x)+e^x(1+x+x^2)\log (x))}{x^2{(2-x+e^x\log (x))}^2}dx\\ & =5\int \frac{2-2x-x^2+e^x(1+x)+e^x(1+x+x^2)\log (x)}{x^2{(2-x+e^x\log (x))}^2}dx\\ & =5\int (\frac{(1+x)(-2+x-3x\log (x)+x^2\log (x))}{x^2\log (x){(-2+x-e^x\log (x))}^2}-\frac{1+x+\log (x)+x\log (x)+x^2\log (x)}{x^2\log (x)(-2+x-e^x\log (x))})dx\\ & =5\int \frac{(1+x)(-2+x-3x\log (x)+x^2\log (x))}{x^2\log (x){(-2+x-e^x\log (x))}^2}dx-5\int \frac{1+x+\log (x)+x\log (x)+x^2\log (x)}{x^2\log (x)(-2+x-e^x\log (x))}dx\\ & =-(5\int (\frac{1}{-2+x-e^x\log (x)}+\frac{1}{x^2(-2+x-e^x\log (x))}+\frac{1}{x(-2+x-e^x\log (x))}+\frac{1}{x^2\log (x)(-2+x-e^x\log (x))}+\frac{1}{x\log (x)(-2+x-e^x\log (x))})dx)+5\int (\frac{-2+x-3x\log (x)+x^2\log (x)}{x^2\log (x){(-2+x-e^x\log (x))}^2}+\frac{-2+x-3x\log (x)+x^2\log (x)}{x\log (x){(-2+x-e^x\log (x))}^2})dx\\ & =-(5\int \frac{1}{-2+x-e^x\log (x)}dx)-5\int \frac{1}{x^2(-2+x-e^x\log (x))}dx-5\int \frac{1}{x(-2+x-e^x\log (x))}dx-5\int \frac{1}{x^2\log (x)(-2+x-e^x\log (x))}dx-5\int \frac{1}{x\log (x)(-2+x-e^x\log (x))}dx+5\int \frac{-2+x-3x\log (x)+x^2\log (x)}{x^2\log (x){(-2+x-e^x\log (x))}^2}dx+5\int \frac{-2+x-3x\log (x)+x^2\log (x)}{x\log (x){(-2+x-e^x\log (x))}^2}dx\\ & =-(5\int \frac{1}{-2+x-e^x\log (x)}dx)-5\int \frac{1}{x^2(-2+x-e^x\log (x))}dx-5\int \frac{1}{x(-2+x-e^x\log (x))}dx-5\int \frac{1}{x^2\log (x)(-2+x-e^x\log (x))}dx-5\int \frac{1}{x\log (x)(-2+x-e^x\log (x))}dx+5\int (\frac{1}{{(-2+x-e^x\log (x))}^2}-\frac{3}{x{(-2+x-e^x\log (x))}^2}-\frac{2}{x^2\log (x){(-2+x-e^x\log (x))}^2}+\frac{1}{x\log (x){(-2+x-e^x\log (x))}^2})dx+5\int (-\frac{3}{{(-2+x-e^x\log (x))}^2}+\frac{x}{{(-2+x-e^x\log (x))}^2}-\frac{2}{x\log (x){(-2+x-e^x\log (x))}^2}+\frac{1}{\log (x){(2-x+e^x\log (x))}^2})dx\\ & =5\int \frac{1}{{(-2+x-e^x\log (x))}^2}dx+5\int \frac{x}{{(-2+x-e^x\log (x))}^2}dx+5\int \frac{1}{x\log (x){(-2+x-e^x\log (x))}^2}dx-5\int \frac{1}{-2+x-e^x\log (x)}dx-5\int \frac{1}{x^2(-2+x-e^x\log (x))}dx-5\int \frac{1}{x(-2+x-e^x\log (x))}dx-5\int \frac{1}{x^2\log (x)(-2+x-e^x\log (x))}dx-5\int \frac{1}{x\log (x)(-2+x-e^x\log (x))}dx+5\int \frac{1}{\log (x){(2-x+e^x\log (x))}^2}dx-10\int \frac{1}{x^2\log (x){(-2+x-e^x\log (x))}^2}dx-10\int \frac{1}{x\log (x){(-2+x-e^x\log (x))}^2}dx-15\int \frac{1}{{(-2+x-e^x\log (x))}^2}dx-15\int \frac{1}{x{(-2+x-e^x\log (x))}^2}dx\end{array}\end{array}$$

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Mathematica [A]  time = 0.60, size = 20, normalized size = 1.00 $$\begin{array}{c}\frac{5(1+x)}{x(-2+x-e^x\log (x))}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(10 - 10*x - 5*x^2 + E^x*(5 + 5*x) + E^x*(5 + 5*x + 5*x^2)*Log[x])/(4*x^2 - 4*x^3 + x^4 + E^x*(4*x^2
- 2*x^3)*Log[x] + E^(2*x)*x^2*Log[x]^2),x]

[Out]

(5*(1 + x))/(x*(-2 + x - E^x*Log[x]))

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fricas [A]  time = 0.62, size = 22, normalized size = 1.10 $$\begin{array}{c}-\frac{5(x+1)}{x{e}^x\log (x)-x^2+2x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^2+5*x+5)*exp(x)*log(x)+(5*x+5)*exp(x)-5*x^2-10*x+10)/(x^2*exp(x)^2*log(x)^2+(-2*x^3+4*x^2)*exp
(x)*log(x)+x^4-4*x^3+4*x^2),x, algorithm="fricas")

[Out]

-5*(x + 1)/(x*e^x*log(x) - x^2 + 2*x)

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giac [A]  time = 0.30, size = 22, normalized size = 1.10 $$\begin{array}{c}-\frac{5(x+1)}{x{e}^x\log (x)-x^2+2x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^2+5*x+5)*exp(x)*log(x)+(5*x+5)*exp(x)-5*x^2-10*x+10)/(x^2*exp(x)^2*log(x)^2+(-2*x^3+4*x^2)*exp
(x)*log(x)+x^4-4*x^3+4*x^2),x, algorithm="giac")

[Out]

-5*(x + 1)/(x*e^x*log(x) - x^2 + 2*x)

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maple [A]  time = 0.03, size = 20, normalized size = 1.00

method	result	size
risch	$\frac{5x+5}{(x-2-{\mathrm{e}}^x\ln (x))x}$	$20$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*x^2+5*x+5)*exp(x)*ln(x)+(5*x+5)*exp(x)-5*x^2-10*x+10)/(x^2*exp(x)^2*ln(x)^2+(-2*x^3+4*x^2)*exp(x)*ln(x
)+x^4-4*x^3+4*x^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.54, size = 22, normalized size = 1.10 $$\begin{array}{c}-\frac{5(x+1)}{x{e}^x\log (x)-x^2+2x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^2+5*x+5)*exp(x)*log(x)+(5*x+5)*exp(x)-5*x^2-10*x+10)/(x^2*exp(x)^2*log(x)^2+(-2*x^3+4*x^2)*exp
(x)*log(x)+x^4-4*x^3+4*x^2),x, algorithm="maxima")

[Out]

-5*(x + 1)/(x*e^x*log(x) - x^2 + 2*x)

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mupad [B]  time = 1.16, size = 68, normalized size = 3.40 $$\begin{array}{c}-\frac{5{\mathrm{e}}^{2x}+5x{\mathrm{e}}^{2x}-{\mathrm{e}}^x(-5x^3+10x^2+15x)}{({\mathrm{e}}^x\ln (x)-x+2)(x{\mathrm{e}}^{2x}-3x^2{\mathrm{e}}^x+x^3{\mathrm{e}}^x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(5*x + 5) - 10*x - 5*x^2 + exp(x)*log(x)*(5*x + 5*x^2 + 5) + 10)/(4*x^2 - 4*x^3 + x^4 + x^2*exp(2*
x)*log(x)^2 + exp(x)*log(x)*(4*x^2 - 2*x^3)),x)

[Out]

-(5*exp(2*x) + 5*x*exp(2*x) - exp(x)*(15*x + 10*x^2 - 5*x^3))/((exp(x)*log(x) - x + 2)*(x*exp(2*x) - 3*x^2*exp
(x) + x^3*exp(x)))

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sympy [A]  time = 0.39, size = 20, normalized size = 1.00 $$\begin{array}{c}\frac{-5x-5}{-x^2+x{e}^x\log (x)+2x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x**2+5*x+5)*exp(x)*ln(x)+(5*x+5)*exp(x)-5*x**2-10*x+10)/(x**2*exp(x)**2*ln(x)**2+(-2*x**3+4*x**2
)*exp(x)*ln(x)+x**4-4*x**3+4*x**2),x)

[Out]

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

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