\(\int \frac {-15-258 x-79 x^2+40 x^3+12 x^4+e^x (-45-30 x-5 x^2)}{45+30 x+5 x^2} \, dx\) [166]

Optimal result

Integrand size = 47, antiderivative size = 32 \[ \int \frac {-15-258 x-79 x^2+40 x^3+12 x^4+e^x \left (-45-30 x-5 x^2\right )}{45+30 x+5 x^2} \, dx=-2-e^x+x+4 \left (-\frac {1}{5} (4-x) x^2+\frac {3}{3+x}\right )+\log (6) \]

[Out]

12/(3+x)-4/5*(-x+4)*x^2-2-exp(x)+ln(6)+x

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88, number of steps used = 13, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.106, Rules used = {27, 12, 6874, 2225, 45} \[ \int \frac {-15-258 x-79 x^2+40 x^3+12 x^4+e^x \left (-45-30 x-5 x^2\right )}{45+30 x+5 x^2} \, dx=\frac {4 x^3}{5}-\frac {16 x^2}{5}+x-e^x+\frac {12}{x+3} \]

[In]

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]

[Out]

-E^x + x - (16*x^2)/5 + (4*x^3)/5 + 12/(3 + x)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 45

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 2225

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 6874

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

Rubi steps \begin{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94 \[ \int \frac {-15-258 x-79 x^2+40 x^3+12 x^4+e^x \left (-45-30 x-5 x^2\right )}{45+30 x+5 x^2} \, dx=\frac {1}{5} \left (-5 e^x+5 x-16 x^2+4 x^3+\frac {60}{3+x}\right ) \]

[In]

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]

[Out]

(-5*E^x + 5*x - 16*x^2 + 4*x^3 + 60/(3 + x))/5

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75

method	result	size
default	\(\frac {12}{3+x}+x -\frac {16 x^{2}}{5}+\frac {4 x^{3}}{5}-{\mathrm e}^{x}\)	\(24\)
risch	\(\frac {12}{3+x}+x -\frac {16 x^{2}}{5}+\frac {4 x^{3}}{5}-{\mathrm e}^{x}\)	\(24\)
parts	\(\frac {12}{3+x}+x -\frac {16 x^{2}}{5}+\frac {4 x^{3}}{5}-{\mathrm e}^{x}\)	\(24\)
norman	\(\frac {-\frac {43 x^{2}}{5}-\frac {4 x^{3}}{5}+\frac {4 x^{4}}{5}-{\mathrm e}^{x} x -3 \,{\mathrm e}^{x}+3}{3+x}\)	\(33\)
parallelrisch	\(\frac {4 x^{4}-4 x^{3}-43 x^{2}-5 \,{\mathrm e}^{x} x +15-15 \,{\mathrm e}^{x}}{5 x +15}\)	\(34\)

[In]

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)

[Out]

12/(3+x)+x-16/5*x^2+4/5*x^3-exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {-15-258 x-79 x^2+40 x^3+12 x^4+e^x \left (-45-30 x-5 x^2\right )}{45+30 x+5 x^2} \, dx=\frac {4 \, x^{4} - 4 \, x^{3} - 43 \, x^{2} - 5 \, {\left (x + 3\right )} e^{x} + 15 \, x + 60}{5 \, {\left (x + 3\right )}} \]

[In]

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")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.69 \[ \int \frac {-15-258 x-79 x^2+40 x^3+12 x^4+e^x \left (-45-30 x-5 x^2\right )}{45+30 x+5 x^2} \, dx=\frac {4 x^{3}}{5} - \frac {16 x^{2}}{5} + x - e^{x} + \frac {12}{x + 3} \]

[In]

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)

[Out]

4*x**3/5 - 16*x**2/5 + x - exp(x) + 12/(x + 3)

Maxima [F]

\[ \int \frac {-15-258 x-79 x^2+40 x^3+12 x^4+e^x \left (-45-30 x-5 x^2\right )}{45+30 x+5 x^2} \, dx=\int { \frac {12 \, x^{4} + 40 \, x^{3} - 79 \, x^{2} - 5 \, {\left (x^{2} + 6 \, x + 9\right )} e^{x} - 258 \, x - 15}{5 \, {\left (x^{2} + 6 \, x + 9\right )}} \,d x } \]

[In]

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")

[Out]

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
+ 3) + 18*integrate(e^x/(x^3 + 9*x^2 + 27*x + 27), x)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int \frac {-15-258 x-79 x^2+40 x^3+12 x^4+e^x \left (-45-30 x-5 x^2\right )}{45+30 x+5 x^2} \, dx=\frac {4 \, x^{4} - 4 \, x^{3} - 43 \, x^{2} - 5 \, x e^{x} + 15 \, x - 15 \, e^{x} + 60}{5 \, {\left (x + 3\right )}} \]

[In]

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")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.72 \[ \int \frac {-15-258 x-79 x^2+40 x^3+12 x^4+e^x \left (-45-30 x-5 x^2\right )}{45+30 x+5 x^2} \, dx=x-{\mathrm {e}}^x+\frac {12}{x+3}-\frac {16\,x^2}{5}+\frac {4\,x^3}{5} \]

[In]

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)

[Out]

x - exp(x) + 12/(x + 3) - (16*x^2)/5 + (4*x^3)/5