3.85.17 $\int \frac{-135-180x-9x^2-12x^3-4x^4+e^x(9x^2+21x^3+16x^4+4x^5)}{9x^2+12x^3+4x^4}dx$

Optimal. Leaf size=34 $$5(2+\frac{4}{2+\frac{3}{x}}+\frac{3}{x})+(-1+e^x-x)x+x^2$$

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Rubi [A]  time = 0.44, antiderivative size = 30, normalized size of antiderivative = 0.88, number of steps used = 14, number of rules used = 7, integrand size = 64, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.109, Rules used = {1594, 27, 6742, 2176, 2194, 44, 43} $$\begin{array}{c}-x-e^x+e^x(x+1)-\frac{30}{2x+3}+\frac{15}{x}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(-135 - 180*x - 9*x^2 - 12*x^3 - 4*x^4 + E^x*(9*x^2 + 21*x^3 + 16*x^4 + 4*x^5))/(9*x^2 + 12*x^3 + 4*x^4),x
]

[Out]

-E^x + 15/x - x + E^x*(1 + x) - 30/(3 + 2*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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

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 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), 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] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 6742

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{-135-180x-9x^2-12x^3-4x^4+e^x(9x^2+21x^3+16x^4+4x^5)}{x^2(9+12x+4x^2)}dx\\ & =\int \frac{-135-180x-9x^2-12x^3-4x^4+e^x(9x^2+21x^3+16x^4+4x^5)}{x^2(3+2x{)}^2}dx\\ & =\int (e^x(1+x)-\frac{9}{(3+2x{)}^2}-\frac{135}{x^2(3+2x{)}^2}-\frac{180}{x(3+2x{)}^2}-\frac{12x}{(3+2x{)}^2}-\frac{4x^2}{(3+2x{)}^2})dx\\ & =\frac{9}{2(3+2x)}-4\int \frac{x^2}{(3+2x{)}^2}dx-12\int \frac{x}{(3+2x{)}^2}dx-135\int \frac{1}{x^2(3+2x{)}^2}dx-180\int \frac{1}{x(3+2x{)}^2}dx+\int e^x(1+x)dx\\ & =e^x(1+x)+\frac{9}{2(3+2x)}-4\int (\frac{1}{4}+\frac{9}{4(3+2x{)}^2}-\frac{3}{2(3+2x)})dx-12\int (-\frac{3}{2(3+2x{)}^2}+\frac{1}{2(3+2x)})dx-135\int (\frac{1}{9x^2}-\frac{4}{27x}+\frac{4}{9(3+2x{)}^2}+\frac{8}{27(3+2x)})dx-180\int (\frac{1}{9x}-\frac{2}{3(3+2x{)}^2}-\frac{2}{9(3+2x)})dx-\int e^{x}dx\\ & =-e^x+\frac{15}{x}-x+e^x(1+x)-\frac{30}{3+2x}\end{array}\end{array}$$

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Mathematica [A]  time = 0.16, size = 23, normalized size = 0.68 $$\begin{array}{c}\frac{15}{x}-x+e^{x}x-\frac{30}{3+2x}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-135 - 180*x - 9*x^2 - 12*x^3 - 4*x^4 + E^x*(9*x^2 + 21*x^3 + 16*x^4 + 4*x^5))/(9*x^2 + 12*x^3 + 4*
x^4),x]

[Out]

15/x - x + E^x*x - 30/(3 + 2*x)

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fricas [A]  time = 0.95, size = 40, normalized size = 1.18 $$\begin{array}{c}-\frac{2x^3+3x^2-(2x^3+3x^2)e^x-45}{2x^2+3x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^5+16*x^4+21*x^3+9*x^2)*exp(x)-4*x^4-12*x^3-9*x^2-180*x-135)/(4*x^4+12*x^3+9*x^2),x, algorithm=
"fricas")

[Out]

-(2*x^3 + 3*x^2 - (2*x^3 + 3*x^2)*e^x - 45)/(2*x^2 + 3*x)

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giac [A]  time = 0.19, size = 38, normalized size = 1.12 $$\begin{array}{c}\frac{2x^3{e}^x-2x^3+3x^2{e}^x-3x^2+45}{2x^2+3x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^5+16*x^4+21*x^3+9*x^2)*exp(x)-4*x^4-12*x^3-9*x^2-180*x-135)/(4*x^4+12*x^3+9*x^2),x, algorithm=
"giac")

[Out]

(2*x^3*e^x - 2*x^3 + 3*x^2*e^x - 3*x^2 + 45)/(2*x^2 + 3*x)

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maple [A]  time = 0.06, size = 21, normalized size = 0.62

method	result	size
risch	$-x+\frac{45}{x(2x+3)}+{\mathrm{e}}^{x}x$	$21$
default	$\frac{15}{x}-\frac{30}{2x+3}-x+{\mathrm{e}}^{x}x$	$23$
norman	$\frac{45+\frac{9x}{2}-2x^3+3{\mathrm{e}}^x{x}^2+2{\mathrm{e}}^x{x}^3}{x(2x+3)}$	$36$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^5+16*x^4+21*x^3+9*x^2)*exp(x)-4*x^4-12*x^3-9*x^2-180*x-135)/(4*x^4+12*x^3+9*x^2),x,method=_RETURNVER
BOSE)

[Out]

-x+45/x/(2*x+3)+exp(x)*x

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}x{e}^x-x-\frac{9e^{(-\frac{3}{2})}{E}_2(-x-\frac{3}{2})}{2(2x+3)}+\frac{15(4x+3)}{2x^2+3x}-\frac{60}{2x+3}-9\int \frac{e^x}{4x^2+12x+9}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^5+16*x^4+21*x^3+9*x^2)*exp(x)-4*x^4-12*x^3-9*x^2-180*x-135)/(4*x^4+12*x^3+9*x^2),x, algorithm=
"maxima")

[Out]

x*e^x - x - 9/2*e^(-3/2)*exp_integral_e(2, -x - 3/2)/(2*x + 3) + 15*(4*x + 3)/(2*x^2 + 3*x) - 60/(2*x + 3) - 9
*integrate(e^x/(4*x^2 + 12*x + 9), x)

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mupad [B]  time = 0.13, size = 20, normalized size = 0.59 $$\begin{array}{c}x({\mathrm{e}}^x-1)+\frac{45}{2x^2+3x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(180*x - exp(x)*(9*x^2 + 21*x^3 + 16*x^4 + 4*x^5) + 9*x^2 + 12*x^3 + 4*x^4 + 135)/(9*x^2 + 12*x^3 + 4*x^4
),x)

[Out]

x*(exp(x) - 1) + 45/(3*x + 2*x^2)

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sympy [A]  time = 0.16, size = 15, normalized size = 0.44 $$\begin{array}{c}x{e}^x-x+\frac{45}{2x^2+3x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**5+16*x**4+21*x**3+9*x**2)*exp(x)-4*x**4-12*x**3-9*x**2-180*x-135)/(4*x**4+12*x**3+9*x**2),x)

[Out]

x*exp(x) - x + 45/(2*x**2 + 3*x)

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