\(\int \frac {-108-72 x-48 x^2-24 x^3-4 x^4+e^5 (11 x^2+6 x^3+x^4)}{9 x^2+6 x^3+x^4} \, dx\) [9955]

Optimal result

Integrand size = 55, antiderivative size = 29 \[ \int \frac {-108-72 x-48 x^2-24 x^3-4 x^4+e^5 \left (11 x^2+6 x^3+x^4\right )}{9 x^2+6 x^3+x^4} \, dx=-3+\frac {12}{x}-e^5 \left (-x+\frac {2}{3+x}\right )-4 (x+\log (3)) \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.055, Rules used = {1608, 27, 1634} \[ \int \frac {-108-72 x-48 x^2-24 x^3-4 x^4+e^5 \left (11 x^2+6 x^3+x^4\right )}{9 x^2+6 x^3+x^4} \, dx=-\left (\left (4-e^5\right ) x\right )-\frac {2 e^5}{x+3}+\frac {12}{x} \]

[In]

Int[(-108 - 72*x - 48*x^2 - 24*x^3 - 4*x^4 + E^5*(11*x^2 + 6*x^3 + x^4))/(9*x^2 + 6*x^3 + x^4),x]

[Out]

12/x - (4 - E^5)*x - (2*E^5)/(3 + 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 1608

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 1634

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-108-72 x-48 x^2-24 x^3-4 x^4+e^5 \left (11 x^2+6 x^3+x^4\right )}{x^2 \left (9+6 x+x^2\right )} \, dx \\ & = \int \frac {-108-72 x-48 x^2-24 x^3-4 x^4+e^5 \left (11 x^2+6 x^3+x^4\right )}{x^2 (3+x)^2} \, dx \\ & = \int \left (-4 \left (1-\frac {e^5}{4}\right )-\frac {12}{x^2}+\frac {2 e^5}{(3+x)^2}\right ) \, dx \\ & = \frac {12}{x}-\left (4-e^5\right ) x-\frac {2 e^5}{3+x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {-108-72 x-48 x^2-24 x^3-4 x^4+e^5 \left (11 x^2+6 x^3+x^4\right )}{9 x^2+6 x^3+x^4} \, dx=\frac {12}{x}+\left (-4+e^5\right ) x-\frac {2 e^5}{3+x} \]

[In]

Integrate[(-108 - 72*x - 48*x^2 - 24*x^3 - 4*x^4 + E^5*(11*x^2 + 6*x^3 + x^4))/(9*x^2 + 6*x^3 + x^4),x]

[Out]

12/x + (-4 + E^5)*x - (2*E^5)/(3 + x)

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79

method	result	size
default	\(x \,{\mathrm e}^{5}-4 x +\frac {12}{x}-\frac {2 \,{\mathrm e}^{5}}{3+x}\)	\(23\)
norman	\(\frac {\left ({\mathrm e}^{5}-4\right ) x^{3}+36+\left (-11 \,{\mathrm e}^{5}+48\right ) x}{\left (3+x \right ) x}\)	\(28\)
risch	\(x \,{\mathrm e}^{5}-4 x +\frac {\left (-2 \,{\mathrm e}^{5}+12\right ) x +36}{\left (3+x \right ) x}\)	\(28\)
gosper	\(\frac {x^{3} {\mathrm e}^{5}-4 x^{3}-11 x \,{\mathrm e}^{5}+48 x +36}{x \left (3+x \right )}\)	\(31\)
parallelrisch	\(\frac {x^{3} {\mathrm e}^{5}-4 x^{3}-11 x \,{\mathrm e}^{5}+48 x +36}{x \left (3+x \right )}\)	\(31\)

[In]

int(((x^4+6*x^3+11*x^2)*exp(5)-4*x^4-24*x^3-48*x^2-72*x-108)/(x^4+6*x^3+9*x^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45 \[ \int \frac {-108-72 x-48 x^2-24 x^3-4 x^4+e^5 \left (11 x^2+6 x^3+x^4\right )}{9 x^2+6 x^3+x^4} \, dx=-\frac {4 \, x^{3} + 12 \, x^{2} - {\left (x^{3} + 3 \, x^{2} - 2 \, x\right )} e^{5} - 12 \, x - 36}{x^{2} + 3 \, x} \]

[In]

integrate(((x^4+6*x^3+11*x^2)*exp(5)-4*x^4-24*x^3-48*x^2-72*x-108)/(x^4+6*x^3+9*x^2),x, algorithm="fricas")

[Out]

-(4*x^3 + 12*x^2 - (x^3 + 3*x^2 - 2*x)*e^5 - 12*x - 36)/(x^2 + 3*x)

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {-108-72 x-48 x^2-24 x^3-4 x^4+e^5 \left (11 x^2+6 x^3+x^4\right )}{9 x^2+6 x^3+x^4} \, dx=- x \left (4 - e^{5}\right ) - \frac {x \left (-12 + 2 e^{5}\right ) - 36}{x^{2} + 3 x} \]

[In]

integrate(((x**4+6*x**3+11*x**2)*exp(5)-4*x**4-24*x**3-48*x**2-72*x-108)/(x**4+6*x**3+9*x**2),x)

[Out]

-x*(4 - exp(5)) - (x*(-12 + 2*exp(5)) - 36)/(x**2 + 3*x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {-108-72 x-48 x^2-24 x^3-4 x^4+e^5 \left (11 x^2+6 x^3+x^4\right )}{9 x^2+6 x^3+x^4} \, dx=x {\left (e^{5} - 4\right )} - \frac {2 \, {\left (x {\left (e^{5} - 6\right )} - 18\right )}}{x^{2} + 3 \, x} \]

[In]

integrate(((x^4+6*x^3+11*x^2)*exp(5)-4*x^4-24*x^3-48*x^2-72*x-108)/(x^4+6*x^3+9*x^2),x, algorithm="maxima")

[Out]

x*(e^5 - 4) - 2*(x*(e^5 - 6) - 18)/(x^2 + 3*x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {-108-72 x-48 x^2-24 x^3-4 x^4+e^5 \left (11 x^2+6 x^3+x^4\right )}{9 x^2+6 x^3+x^4} \, dx=x e^{5} - 4 \, x - \frac {2 \, {\left (x e^{5} - 6 \, x - 18\right )}}{x^{2} + 3 \, x} \]

[In]

integrate(((x^4+6*x^3+11*x^2)*exp(5)-4*x^4-24*x^3-48*x^2-72*x-108)/(x^4+6*x^3+9*x^2),x, algorithm="giac")

[Out]

x*e^5 - 4*x - 2*(x*e^5 - 6*x - 18)/(x^2 + 3*x)

Mupad [B] (verification not implemented)

Time = 14.31 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {-108-72 x-48 x^2-24 x^3-4 x^4+e^5 \left (11 x^2+6 x^3+x^4\right )}{9 x^2+6 x^3+x^4} \, dx=x\,\left ({\mathrm {e}}^5-4\right )-\frac {x\,\left (2\,{\mathrm {e}}^5-12\right )-36}{x\,\left (x+3\right )} \]

[In]

int(-(72*x - exp(5)*(11*x^2 + 6*x^3 + x^4) + 48*x^2 + 24*x^3 + 4*x^4 + 108)/(9*x^2 + 6*x^3 + x^4),x)

[Out]

x*(exp(5) - 4) - (x*(2*exp(5) - 12) - 36)/(x*(x + 3))