3.85.67 $\int \frac{-3x^2+e^4{x}^2+e^x(-3+x+x^2)}{9x^2+6x^3+x^4}dx$

Optimal. Leaf size=21 $$\frac{e^x-x(e^4+x)}{x(3+x)}$$

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Rubi [A]  time = 0.45, antiderivative size = 36, normalized size of antiderivative = 1.71, number of steps used = 13, number of rules used = 6, integrand size = 40, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.150, Rules used = {6, 1594, 27, 6742, 2177, 2178} $$\begin{array}{c}-\frac{e^x}{3(x+3)}+\frac{3-e^4}{x+3}+\frac{e^x}{3x}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(-3*x^2 + E^4*x^2 + E^x*(-3 + x + x^2))/(9*x^2 + 6*x^3 + x^4),x]

[Out]

E^x/(3*x) - E^x/(3*(3 + x)) + (3 - E^4)/(3 + x)

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{(-3+e^4)x^2+e^x(-3+x+x^2)}{9x^2+6x^3+x^4}dx\\ & =\int \frac{(-3+e^4)x^2+e^x(-3+x+x^2)}{x^2(9+6x+x^2)}dx\\ & =\int \frac{(-3+e^4)x^2+e^x(-3+x+x^2)}{x^2(3+x{)}^2}dx\\ & =\int (\frac{-3+e^4}{(3+x{)}^2}+\frac{e^x(-3+x+x^2)}{x^2(3+x{)}^2})dx\\ & =\frac{3-e^4}{3+x}+\int \frac{e^x(-3+x+x^2)}{x^2(3+x{)}^2}dx\\ & =\frac{3-e^4}{3+x}+\int (-\frac{e^x}{3x^2}+\frac{e^x}{3x}+\frac{e^x}{3(3+x{)}^2}-\frac{e^x}{3(3+x)})dx\\ & =\frac{3-e^4}{3+x}-\frac{1}{3}\int \frac{e^x}{x^2}dx+\frac{1}{3}\int \frac{e^x}{x}dx+\frac{1}{3}\int \frac{e^x}{(3+x{)}^2}dx-\frac{1}{3}\int \frac{e^x}{3+x}dx\\ & =\frac{e^x}{3x}-\frac{e^x}{3(3+x)}+\frac{3-e^4}{3+x}+\frac{\text{Ei}(x)}{3}-\frac{\text{Ei}(3+x)}{3e^3}-\frac{1}{3}\int \frac{e^x}{x}dx+\frac{1}{3}\int \frac{e^x}{3+x}dx\\ & =\frac{e^x}{3x}-\frac{e^x}{3(3+x)}+\frac{3-e^4}{3+x}\end{array}\end{array}$$

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Mathematica [A]  time = 0.09, size = 23, normalized size = 1.10 $$\begin{array}{c}\frac{e^x+3x-e^{4}x}{3x+x^2}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-3*x^2 + E^4*x^2 + E^x*(-3 + x + x^2))/(9*x^2 + 6*x^3 + x^4),x]

[Out]

(E^x + 3*x - E^4*x)/(3*x + x^2)

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fricas [A]  time = 0.62, size = 23, normalized size = 1.10 $$\begin{array}{c}-\frac{x{e}^4-3x-e^x}{x^2+3x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+x-3)*exp(x)+x^2*exp(4)-3*x^2)/(x^4+6*x^3+9*x^2),x, algorithm="fricas")

[Out]

-(x*e^4 - 3*x - e^x)/(x^2 + 3*x)

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giac [A]  time = 0.12, size = 23, normalized size = 1.10 $$\begin{array}{c}-\frac{x{e}^4-3x-e^x}{x^2+3x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+x-3)*exp(x)+x^2*exp(4)-3*x^2)/(x^4+6*x^3+9*x^2),x, algorithm="giac")

[Out]

-(x*e^4 - 3*x - e^x)/(x^2 + 3*x)

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maple [A]  time = 0.06, size = 21, normalized size = 1.00

method	result	size
norman	$\frac{(3-{\mathrm{e}}^4)x+{\mathrm{e}}^x}{(3+x)x}$	$21$
risch	$-\frac{{\mathrm{e}}^4}{3+x}+\frac{3}{3+x}+\frac{{\mathrm{e}}^x}{(3+x)x}$	$29$
default	$-\frac{{\mathrm{e}}^4}{3+x}-\frac{{\mathrm{e}}^x}{3(3+x)}+\frac{3}{3+x}+\frac{{\mathrm{e}}^x}{3x}$	$34$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2+x-3)*exp(x)+x^2*exp(4)-3*x^2)/(x^4+6*x^3+9*x^2),x,method=_RETURNVERBOSE)

[Out]

((3-exp(4))*x+exp(x))/(3+x)/x

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maxima [A]  time = 0.43, size = 29, normalized size = 1.38 $$\begin{array}{c}-\frac{e^4}{x+3}+\frac{e^x}{x^2+3x}+\frac{3}{x+3}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+x-3)*exp(x)+x^2*exp(4)-3*x^2)/(x^4+6*x^3+9*x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.17, size = 20, normalized size = 0.95 $$\begin{array}{c}\frac{{\mathrm{e}}^x-x({\mathrm{e}}^4-3)}{x^2+3x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*exp(4) - 3*x^2 + exp(x)*(x + x^2 - 3))/(9*x^2 + 6*x^3 + x^4),x)

[Out]

(exp(x) - x*(exp(4) - 3))/(3*x + x^2)

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sympy [A]  time = 0.13, size = 17, normalized size = 0.81 $$\begin{array}{c}\frac{e^x}{x^2+3x}-\frac{-3+e^4}{x+3}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**2+x-3)*exp(x)+x**2*exp(4)-3*x**2)/(x**4+6*x**3+9*x**2),x)

[Out]

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

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