3.41.30 $\int \frac{2-2x+e^x(-2x^2-x^3)}{-2x^2-e^x{x}^3+2x\log (\frac{x}{3e^5})}dx$

Optimal. Leaf size=27 $$\log \left(x(e^{x}x-\frac{2(-x+\log \left(\frac{x}{3e^5}\right))}{x})\right)$$

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Rubi [F]  time = 1.57, 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 = {} $$\begin{array}{c}\int \frac{2-2x+e^x(-2x^2-x^3)}{-2x^2-e^x{x}^3+2x\log \left(\frac{x}{3e^5}\right)}dx\end{array}$$

Verification is not applicable to the result.

[In]

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

[Out]

x + 2*Log[x] - 2*(11 + Log[9])*Defer[Int][1/(x*(2*x + E^x*x^2 + 10*(1 + Log[3]/5) - 2*Log[x])), x] + 2*Defer[I
nt][Log[x/3]/(2*x + E^x*x^2 + 10*(1 + Log[3]/5) - 2*Log[x]), x] + 4*Defer[Int][Log[x]/(x*(2*x + E^x*x^2 + 10*(
1 + Log[3]/5) - 2*Log[x])), x] + 12*Defer[Int][(-2*x - E^x*x^2 - 10*(1 + Log[3]/5) + 2*Log[x])^(-1), x] + 2*De
fer[Int][x/(-2*x - E^x*x^2 - 10*(1 + Log[3]/5) + 2*Log[x]), x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (\frac{2+x}{x}+\frac{2(-6x-x^2-11(1+\frac{2\log (3)}{11})+x\log \left(\frac{x}{3}\right)+2\log (x))}{x(2x+e^x{x}^2+10(1+\frac{\log (3)}{5})-2\log (x))})dx\\ & =2\int \frac{-6x-x^2-11(1+\frac{2\log (3)}{11})+x\log \left(\frac{x}{3}\right)+2\log (x)}{x(2x+e^x{x}^2+10(1+\frac{\log (3)}{5})-2\log (x))}dx+\int \frac{2+x}{x}dx\\ & =2\int (\frac{-11-\log (9)}{x(2x+e^x{x}^2+10(1+\frac{\log (3)}{5})-2\log (x))}+\frac{\log \left(\frac{x}{3}\right)}{2x+e^x{x}^2+10(1+\frac{\log (3)}{5})-2\log (x)}+\frac{2\log (x)}{x(2x+e^x{x}^2+10(1+\frac{\log (3)}{5})-2\log (x))}+\frac{6}{-2x-e^x{x}^2-10(1+\frac{\log (3)}{5})+2\log (x)}+\frac{x}{-2x-e^x{x}^2-10(1+\frac{\log (3)}{5})+2\log (x)})dx+\int (1+\frac{2}{x})dx\\ & =x+2\log (x)+2\int \frac{\log \left(\frac{x}{3}\right)}{2x+e^x{x}^2+10(1+\frac{\log (3)}{5})-2\log (x)}dx+2\int \frac{x}{-2x-e^x{x}^2-10(1+\frac{\log (3)}{5})+2\log (x)}dx+4\int \frac{\log (x)}{x(2x+e^x{x}^2+10(1+\frac{\log (3)}{5})-2\log (x))}dx+12\int \frac{1}{-2x-e^x{x}^2-10(1+\frac{\log (3)}{5})+2\log (x)}dx-(2(11+\log (9)))\int \frac{1}{x(2x+e^x{x}^2+10(1+\frac{\log (3)}{5})-2\log (x))}dx\end{array}\end{array}$$

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Mathematica [A]  time = 0.43, size = 19, normalized size = 0.70 $$\begin{array}{c}\log (10+2x+e^x{x}^2+\log (9)-2\log (x))\end{array}$$

Antiderivative was successfully verified.

[In]

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

[Out]

Log[10 + 2*x + E^x*x^2 + Log[9] - 2*Log[x]]

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fricas [A]  time = 0.57, size = 20, normalized size = 0.74 $$\begin{array}{c}\log (-x^2{e}^x-2x+2\log \left(\frac{1}{3}x{e}^{(-5)}\right))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3-2*x^2)*exp(x)-2*x+2)/(2*x*log(1/3*x/exp(5))-exp(x)*x^3-2*x^2),x, algorithm="fricas")

[Out]

log(-x^2*e^x - 2*x + 2*log(1/3*x*e^(-5)))

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giac [A]  time = 0.17, size = 19, normalized size = 0.70 $$\begin{array}{c}\log (-x^2{e}^x-2x+2\log \left(\frac{1}{3}x\right)-10)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3-2*x^2)*exp(x)-2*x+2)/(2*x*log(1/3*x/exp(5))-exp(x)*x^3-2*x^2),x, algorithm="giac")

[Out]

log(-x^2*e^x - 2*x + 2*log(1/3*x) - 10)

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maple [A]  time = 0.04, size = 19, normalized size = 0.70

method	result	size
risch	$\ln (-\frac{{\mathrm{e}}^x{x}^2}{2}-x+\ln \left(\frac{x{\mathrm{e}}^{-5}}{3}\right))$	$19$
norman	$\ln ({\mathrm{e}}^x{x}^2-2\ln \left(\frac{x{\mathrm{e}}^{-5}}{3}\right)+2x)$	$22$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^3-2*x^2)*exp(x)-2*x+2)/(2*x*ln(1/3*x/exp(5))-exp(x)*x^3-2*x^2),x,method=_RETURNVERBOSE)

[Out]

ln(-1/2*exp(x)*x^2-x+ln(1/3*x*exp(-5)))

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maxima [A]  time = 0.48, size = 29, normalized size = 1.07 $$\begin{array}{c}2\log (x)+\log \left(\frac{x^2{e}^x+2x+2\log (3)-2\log (x)+10}{x^2}\right)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3-2*x^2)*exp(x)-2*x+2)/(2*x*log(1/3*x/exp(5))-exp(x)*x^3-2*x^2),x, algorithm="maxima")

[Out]

2*log(x) + log((x^2*e^x + 2*x + 2*log(3) - 2*log(x) + 10)/x^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 $$\begin{array}{c}\int \frac{2x+{\mathrm{e}}^x(x^3+2x^2)-2}{x^3{\mathrm{e}}^x-2x\ln \left(\frac{x{\mathrm{e}}^{-5}}{3}\right)+2x^2}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.36, size = 26, normalized size = 0.96 $$\begin{array}{c}2\log (x)+\log (e^x+\frac{2x-2\log \left(\frac{x}{3e^5}\right)}{x^2})\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**3-2*x**2)*exp(x)-2*x+2)/(2*x*ln(1/3*x/exp(5))-exp(x)*x**3-2*x**2),x)

[Out]

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

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