3.41.49 $\int \frac{8+16x+e^3(6x-3x^2-6x^3)}{-8+3e^3{x}^2}dx$

Optimal. Leaf size=22 $$-x-x^2+\log (\frac{8}{3}-e^3{x}^2)$$

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Rubi [A]  time = 0.02, antiderivative size = 20, normalized size of antiderivative = 0.91, number of steps used = 3, number of rules used = 2, integrand size = 36, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.056, Rules used = {1810, 260} $$\begin{array}{c}-x^2+\log (8-3e^3{x}^2)-x\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(8 + 16*x + E^3*(6*x - 3*x^2 - 6*x^3))/(-8 + 3*E^3*x^2),x]

[Out]

-x - x^2 + Log[8 - 3*E^3*x^2]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 1810

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a,
b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (-1-2x+\frac{6e^{3}x}{-8+3e^3{x}^2})dx\\ & =-x-x^2+\left(6e^3\right)\int \frac{x}{-8+3e^3{x}^2}dx\\ & =-x-x^2+\log (8-3e^3{x}^2)\end{array}\end{array}$$

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Mathematica [A]  time = 0.01, size = 20, normalized size = 0.91 $$\begin{array}{c}-x-x^2+\log (8-3e^3{x}^2)\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(8 + 16*x + E^3*(6*x - 3*x^2 - 6*x^3))/(-8 + 3*E^3*x^2),x]

[Out]

-x - x^2 + Log[8 - 3*E^3*x^2]

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fricas [A]  time = 0.50, size = 19, normalized size = 0.86 $$\begin{array}{c}-x^2-x+\log (3x^2{e}^3-8)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x^3-3*x^2+6*x)*exp(3)+16*x+8)/(3*x^2*exp(3)-8),x, algorithm="fricas")

[Out]

-x^2 - x + log(3*x^2*e^3 - 8)

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giac [A]  time = 0.22, size = 27, normalized size = 1.23 $$\begin{array}{c}-(x^2{e}^6+x{e}^6)e^{(-6)}+\log \left(|3x^2{e}^3-8|\right)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x^3-3*x^2+6*x)*exp(3)+16*x+8)/(3*x^2*exp(3)-8),x, algorithm="giac")

[Out]

-(x^2*e^6 + x*e^6)*e^(-6) + log(abs(3*x^2*e^3 - 8))

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maple [A]  time = 0.09, size = 20, normalized size = 0.91

method	result	size
default	$-x^2-x+\ln (3x^2{\mathrm{e}}^3-8)$	$20$
norman	$-x^2-x+\ln (3x^2{\mathrm{e}}^3-8)$	$20$
risch	$-x^2-x+\ln (3x^2{\mathrm{e}}^3-8)$	$20$
meijerg	$-\frac{2\sqrt{3}\sqrt{2}{\mathrm{e}}^{-\frac{3}{2}}\mathrm{arctanh}\left(\frac{x\sqrt{3}\sqrt{2}{\mathrm{e}}^{\frac{3}{2}}}{4}\right)}{3}+\frac{8{\mathrm{e}}^{-3}(-\frac{3x^2{\mathrm{e}}^3}{8}-\ln (1-\frac{3x^2{\mathrm{e}}^3}{8}))}{3}+\frac{i\sqrt{6}{\mathrm{e}}^{-\frac{3}{2}}(\frac{i\sqrt{6}x{\mathrm{e}}^{\frac{3}{2}}}{2}-2i\mathrm{arctanh}\left(\frac{x\sqrt{3}\sqrt{2}{\mathrm{e}}^{\frac{3}{2}}}{4}\right))}{3}+\frac{(6{\mathrm{e}}^3+16){\mathrm{e}}^{-3}\ln (1-\frac{3x^2{\mathrm{e}}^3}{8})}{6}$	$101$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-6*x^3-3*x^2+6*x)*exp(3)+16*x+8)/(3*x^2*exp(3)-8),x,method=_RETURNVERBOSE)

[Out]

-x^2-x+ln(3*x^2*exp(3)-8)

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maxima [A]  time = 0.36, size = 19, normalized size = 0.86 $$\begin{array}{c}-x^2-x+\log (3x^2{e}^3-8)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x^3-3*x^2+6*x)*exp(3)+16*x+8)/(3*x^2*exp(3)-8),x, algorithm="maxima")

[Out]

-x^2 - x + log(3*x^2*e^3 - 8)

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mupad [B]  time = 0.11, size = 18, normalized size = 0.82 $$\begin{array}{c}\ln (x^2-\frac{8{\mathrm{e}}^{-3}}{3})-x-x^2\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*x - exp(3)*(3*x^2 - 6*x + 6*x^3) + 8)/(3*x^2*exp(3) - 8),x)

[Out]

log(x^2 - (8*exp(-3))/3) - x - x^2

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sympy [A]  time = 0.14, size = 15, normalized size = 0.68 $$\begin{array}{c}-x^2-x+\log (3x^2{e}^3-8)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x**3-3*x**2+6*x)*exp(3)+16*x+8)/(3*x**2*exp(3)-8),x)

[Out]

-x**2 - x + log(3*x**2*exp(3) - 8)

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