∫ −15−258x−79x2+40x3+12x4+ex(−45−30x−5x2)___________________________________

Optimal. Leaf size=32 \[ -2-e^x+x+4 \left (-\frac {1}{5} (4-x) x^2+\frac {3}{3+x}\right )+\log (6) \]

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Rubi [A] time = 0.18, antiderivative size = 28, normalized size of antiderivative = 0.88, number of steps used = 13, number of rules used = 5, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.106, Rules used = {27, 12, 6742, 2194, 43} \begin {gather*} \frac {4 x^3}{5}-\frac {16 x^2}{5}+x-e^x+\frac {12}{x+3} \end {gather*}

Int[(-15 - 258*x - 79*x^2 + 40*x^3 + 12*x^4 + E^x*(-45 - 30*x - 5*x^2))/(45 + 30*x + 5*x^2),x]

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr

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])

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

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

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-15-258 x-79 x^2+40 x^3+12 x^4+e^x \left (-45-30 x-5 x^2\right )}{5 (3+x)^2} \, dx\\ &=\frac {1}{5} \int \frac {-15-258 x-79 x^2+40 x^3+12 x^4+e^x \left (-45-30 x-5 x^2\right )}{(3+x)^2} \, dx\\ &=\frac {1}{5} \int \left (-5 e^x-\frac {15}{(3+x)^2}-\frac {258 x}{(3+x)^2}-\frac {79 x^2}{(3+x)^2}+\frac {40 x^3}{(3+x)^2}+\frac {12 x^4}{(3+x)^2}\right ) \, dx\\ &=\frac {3}{3+x}+\frac {12}{5} \int \frac {x^4}{(3+x)^2} \, dx+8 \int \frac {x^3}{(3+x)^2} \, dx-\frac {79}{5} \int \frac {x^2}{(3+x)^2} \, dx-\frac {258}{5} \int \frac {x}{(3+x)^2} \, dx-\int e^x \, dx\\ &=-e^x+\frac {3}{3+x}+\frac {12}{5} \int \left (27-6 x+x^2+\frac {81}{(3+x)^2}-\frac {108}{3+x}\right ) \, dx+8 \int \left (-6+x-\frac {27}{(3+x)^2}+\frac {27}{3+x}\right ) \, dx-\frac {79}{5} \int \left (1+\frac {9}{(3+x)^2}-\frac {6}{3+x}\right ) \, dx-\frac {258}{5} \int \left (-\frac {3}{(3+x)^2}+\frac {1}{3+x}\right ) \, dx\\ &=-e^x+x-\frac {16 x^2}{5}+\frac {4 x^3}{5}+\frac {12}{3+x}\\ \end {aligned} \end {gather*}

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Mathematica [A] time = 0.09, size = 30, normalized size = 0.94 \begin {gather*} \frac {1}{5} \left (-5 e^x+5 x-16 x^2+4 x^3+\frac {60}{3+x}\right ) \end {gather*}

Integrate[(-15 - 258*x - 79*x^2 + 40*x^3 + 12*x^4 + E^x*(-45 - 30*x - 5*x^2))/(45 + 30*x + 5*x^2),x]

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fricas [A] time = 0.78, size = 34, normalized size = 1.06 \begin {gather*} \frac {4 \, x^{4} - 4 \, x^{3} - 43 \, x^{2} - 5 \, {\left (x + 3\right )} e^{x} + 15 \, x + 60}{5 \, {\left (x + 3\right )}} \end {gather*}

integrate(((-5*x^2-30*x-45)*exp(x)+12*x^4+40*x^3-79*x^2-258*x-15)/(5*x^2+30*x+45),x, algorithm="fricas")

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giac [A] time = 0.28, size = 36, normalized size = 1.12 \begin {gather*} \frac {4 \, x^{4} - 4 \, x^{3} - 43 \, x^{2} - 5 \, x e^{x} + 15 \, x - 15 \, e^{x} + 60}{5 \, {\left (x + 3\right )}} \end {gather*}

integrate(((-5*x^2-30*x-45)*exp(x)+12*x^4+40*x^3-79*x^2-258*x-15)/(5*x^2+30*x+45),x, algorithm="giac")

1/5*(4*x^4 - 4*x^3 - 43*x^2 - 5*x*e^x + 15*x - 15*e^x + 60)/(x + 3)

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method	result	size

default	\(\frac {12}{3+x}+x -\frac {16 x^{2}}{5}+\frac {4 x^{3}}{5}-{\mathrm e}^{x}\)	\(24\)
risch	\(\frac {12}{3+x}+x -\frac {16 x^{2}}{5}+\frac {4 x^{3}}{5}-{\mathrm e}^{x}\)	\(24\)
norman	\(\frac {-\frac {43 x^{2}}{5}-\frac {4 x^{3}}{5}+\frac {4 x^{4}}{5}-{\mathrm e}^{x} x -3 \,{\mathrm e}^{x}+3}{3+x}\)	\(33\)

int(((-5*x^2-30*x-45)*exp(x)+12*x^4+40*x^3-79*x^2-258*x-15)/(5*x^2+30*x+45),x,method=_RETURNVERBOSE)

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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {4}{5} \, x^{3} - \frac {16}{5} \, x^{2} + x - \frac {{\left (x^{2} + 6 \, x\right )} e^{x}}{x^{2} + 6 \, x + 9} + \frac {9 \, e^{\left (-3\right )} E_{2}\left (-x - 3\right )}{x + 3} + \frac {12}{x + 3} + 18 \, \int \frac {e^{x}}{x^{3} + 9 \, x^{2} + 27 \, x + 27}\,{d x} \end {gather*}

integrate(((-5*x^2-30*x-45)*exp(x)+12*x^4+40*x^3-79*x^2-258*x-15)/(5*x^2+30*x+45),x, algorithm="maxima")

4/5*x^3 - 16/5*x^2 + x - (x^2 + 6*x)*e^x/(x^2 + 6*x + 9) + 9*e^(-3)*exp_integral_e(2, -x - 3)/(x + 3) + 12/(x

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mupad [B] time = 0.09, size = 23, normalized size = 0.72 \begin {gather*} x-{\mathrm {e}}^x+\frac {12}{x+3}-\frac {16\,x^2}{5}+\frac {4\,x^3}{5} \end {gather*}

int(-(258*x + exp(x)*(30*x + 5*x^2 + 45) + 79*x^2 - 40*x^3 - 12*x^4 + 15)/(30*x + 5*x^2 + 45),x)

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sympy [A] time = 0.14, size = 22, normalized size = 0.69 \begin {gather*} \frac {4 x^{3}}{5} - \frac {16 x^{2}}{5} + x - e^{x} + \frac {12}{x + 3} \end {gather*}

integrate(((-5*x**2-30*x-45)*exp(x)+12*x**4+40*x**3-79*x**2-258*x-15)/(5*x**2+30*x+45),x)

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3.2.66 \(\int \frac {-15-258 x-79 x^2+40 x^3+12 x^4+e^x (-45-30 x-5 x^2)}{45+30 x+5 x^2} \, dx\)