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

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

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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 {gather*} -x-e^x+e^x (x+1)-\frac {30}{2 x+3}+\frac {15}{x} \end {gather*}

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 {gather*} \begin {aligned} \text {integral} &=\int \frac {-135-180 x-9 x^2-12 x^3-4 x^4+e^x \left (9 x^2+21 x^3+16 x^4+4 x^5\right )}{x^2 \left (9+12 x+4 x^2\right )} \, dx\\ &=\int \frac {-135-180 x-9 x^2-12 x^3-4 x^4+e^x \left (9 x^2+21 x^3+16 x^4+4 x^5\right )}{x^2 (3+2 x)^2} \, dx\\ &=\int \left (e^x (1+x)-\frac {9}{(3+2 x)^2}-\frac {135}{x^2 (3+2 x)^2}-\frac {180}{x (3+2 x)^2}-\frac {12 x}{(3+2 x)^2}-\frac {4 x^2}{(3+2 x)^2}\right ) \, dx\\ &=\frac {9}{2 (3+2 x)}-4 \int \frac {x^2}{(3+2 x)^2} \, dx-12 \int \frac {x}{(3+2 x)^2} \, dx-135 \int \frac {1}{x^2 (3+2 x)^2} \, dx-180 \int \frac {1}{x (3+2 x)^2} \, dx+\int e^x (1+x) \, dx\\ &=e^x (1+x)+\frac {9}{2 (3+2 x)}-4 \int \left (\frac {1}{4}+\frac {9}{4 (3+2 x)^2}-\frac {3}{2 (3+2 x)}\right ) \, dx-12 \int \left (-\frac {3}{2 (3+2 x)^2}+\frac {1}{2 (3+2 x)}\right ) \, dx-135 \int \left (\frac {1}{9 x^2}-\frac {4}{27 x}+\frac {4}{9 (3+2 x)^2}+\frac {8}{27 (3+2 x)}\right ) \, dx-180 \int \left (\frac {1}{9 x}-\frac {2}{3 (3+2 x)^2}-\frac {2}{9 (3+2 x)}\right ) \, dx-\int e^x \, dx\\ &=-e^x+\frac {15}{x}-x+e^x (1+x)-\frac {30}{3+2 x}\\ \end {aligned} \end {gather*}

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

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 {gather*} -\frac {2 \, x^{3} + 3 \, x^{2} - {\left (2 \, x^{3} + 3 \, x^{2}\right )} e^{x} - 45}{2 \, x^{2} + 3 \, x} \end {gather*}

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 {gather*} \frac {2 \, x^{3} e^{x} - 2 \, x^{3} + 3 \, x^{2} e^{x} - 3 \, x^{2} + 45}{2 \, x^{2} + 3 \, x} \end {gather*}

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	\(-x +\frac {45}{x \left (2 x +3\right )}+{\mathrm e}^{x} x\)	\(21\)
default	\(\frac {15}{x}-\frac {30}{2 x +3}-x +{\mathrm e}^{x} x\)	\(23\)
norman	\(\frac {45+\frac {9 x}{2}-2 x^{3}+3 \,{\mathrm e}^{x} x^{2}+2 \,{\mathrm e}^{x} x^{3}}{x \left (2 x +3\right )}\)	\(36\)

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 {gather*} x e^{x} - x - \frac {9 \, e^{\left (-\frac {3}{2}\right )} E_{2}\left (-x - \frac {3}{2}\right )}{2 \, {\left (2 \, x + 3\right )}} + \frac {15 \, {\left (4 \, x + 3\right )}}{2 \, x^{2} + 3 \, x} - \frac {60}{2 \, x + 3} - 9 \, \int \frac {e^{x}}{4 \, x^{2} + 12 \, x + 9}\,{d x} \end {gather*}

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 {gather*} x\,\left ({\mathrm {e}}^x-1\right )+\frac {45}{2\,x^2+3\,x} \end {gather*}

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 {gather*} x e^{x} - x + \frac {45}{2 x^{2} + 3 x} \end {gather*}

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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