3.89.89 $\int \frac{-12-4x-3x^2+e^{5x}(2+13x)}{-4+e^{5x}x-x^2}dx$

Optimal. Leaf size=23 $$-4+x+\log (3)+2(x+\log (4+x(-e^{5x}+x)))$$

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Rubi [F]  time = 0.57, 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 = {} $$\begin{array}{c}\int \frac{-12-4x-3x^2+e^{5x}(2+13x)}{-4+e^{5x}x-x^2}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(-12 - 4*x - 3*x^2 + E^(5*x)*(2 + 13*x))/(-4 + E^(5*x)*x - x^2),x]

[Out]

13*x + 2*Log[x] + 40*Defer[Int][(-4 + E^(5*x)*x - x^2)^(-1), x] - 8*Defer[Int][1/(x*(4 - E^(5*x)*x + x^2)), x]
+ 2*Defer[Int][x/(4 - E^(5*x)*x + x^2), x] - 10*Defer[Int][x^2/(4 - E^(5*x)*x + x^2), x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (\frac{2+13x}{x}-\frac{2(4+20x-x^2+5x^3)}{x(4-e^{5x}x+x^2)})dx\\ & =-(2\int \frac{4+20x-x^2+5x^3}{x(4-e^{5x}x+x^2)}dx)+\int \frac{2+13x}{x}dx\\ & =-(2\int (-\frac{20}{-4+e^{5x}x-x^2}+\frac{4}{x(4-e^{5x}x+x^2)}-\frac{x}{4-e^{5x}x+x^2}+\frac{5x^2}{4-e^{5x}x+x^2})dx)+\int (13+\frac{2}{x})dx\\ & =13x+2\log (x)+2\int \frac{x}{4-e^{5x}x+x^2}dx-8\int \frac{1}{x(4-e^{5x}x+x^2)}dx-10\int \frac{x^2}{4-e^{5x}x+x^2}dx+40\int \frac{1}{-4+e^{5x}x-x^2}dx\end{array}\end{array}$$

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Mathematica [A]  time = 0.15, size = 20, normalized size = 0.87 $$\begin{array}{c}3x+2\log (4-e^{5x}x+x^2)\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-12 - 4*x - 3*x^2 + E^(5*x)*(2 + 13*x))/(-4 + E^(5*x)*x - x^2),x]

[Out]

3*x + 2*Log[4 - E^(5*x)*x + x^2]

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fricas [A]  time = 0.52, size = 28, normalized size = 1.22 $$\begin{array}{c}3x+2\log (x)+2\log (-\frac{x^2-x{e}^{\left(5x\right)}+4}{x})\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((13*x+2)*exp(5*x)-3*x^2-4*x-12)/(x*exp(5*x)-x^2-4),x, algorithm="fricas")

[Out]

3*x + 2*log(x) + 2*log(-(x^2 - x*e^(5*x) + 4)/x)

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giac [A]  time = 0.21, size = 20, normalized size = 0.87 $$\begin{array}{c}3x+2\log (-x^2+x{e}^{\left(5x\right)}-4)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((13*x+2)*exp(5*x)-3*x^2-4*x-12)/(x*exp(5*x)-x^2-4),x, algorithm="giac")

[Out]

3*x + 2*log(-x^2 + x*e^(5*x) - 4)

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maple [A]  time = 0.07, size = 20, normalized size = 0.87

method	result	size
norman	$3x+2\ln (x^2-x{\mathrm{e}}^{5x}+4)$	$20$
risch	$3x+2\ln (x)+2\ln ({\mathrm{e}}^{5x}-\frac{x^2+4}{x})$	$27$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((13*x+2)*exp(5*x)-3*x^2-4*x-12)/(x*exp(5*x)-x^2-4),x,method=_RETURNVERBOSE)

[Out]

3*x+2*ln(x^2-x*exp(5*x)+4)

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maxima [A]  time = 0.40, size = 28, normalized size = 1.22 $$\begin{array}{c}3x+2\log (x)+2\log (-\frac{x^2-x{e}^{\left(5x\right)}+4}{x})\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((13*x+2)*exp(5*x)-3*x^2-4*x-12)/(x*exp(5*x)-x^2-4),x, algorithm="maxima")

[Out]

3*x + 2*log(x) + 2*log(-(x^2 - x*e^(5*x) + 4)/x)

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mupad [B]  time = 5.59, size = 19, normalized size = 0.83 $$\begin{array}{c}3x+2\ln (x^2-x{\mathrm{e}}^{5x}+4)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x - exp(5*x)*(13*x + 2) + 3*x^2 + 12)/(x^2 - x*exp(5*x) + 4),x)

[Out]

3*x + 2*log(x^2 - x*exp(5*x) + 4)

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sympy [A]  time = 0.20, size = 24, normalized size = 1.04 $$\begin{array}{c}3x+2\log (x)+2\log (e^{5x}+\frac{-x^2-4}{x})\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((13*x+2)*exp(5*x)-3*x**2-4*x-12)/(x*exp(5*x)-x**2-4),x)

[Out]

3*x + 2*log(x) + 2*log(exp(5*x) + (-x**2 - 4)/x)

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