\(\int \frac {e^4 (2 x+3 x^2)}{1+6 x+9 x^2} \, dx\) [2302]

Optimal result

Integrand size = 25, antiderivative size = 16 \[ \int \frac {e^4 \left (2 x+3 x^2\right )}{1+6 x+9 x^2} \, dx=e \left (3+\frac {e^3 x}{3+\frac {1}{x}}\right ) \]

[Out]

(3+x*exp(3)/(1/x+3))*exp(1)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.44, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {12, 27, 697} \[ \int \frac {e^4 \left (2 x+3 x^2\right )}{1+6 x+9 x^2} \, dx=\frac {e^4 x}{3}+\frac {e^4}{9 (3 x+1)} \]

[In]

Int[(E^4*(2*x + 3*x^2))/(1 + 6*x + 9*x^2),x]

[Out]

(E^4*x)/3 + E^4/(9*(1 + 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 697

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && IGtQ[p, 0] &&  !(EqQ[m, 3] && NeQ[p, 1])

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.31 \[ \int \frac {e^4 \left (2 x+3 x^2\right )}{1+6 x+9 x^2} \, dx=\frac {e^4 \left (2+6 x+9 x^2\right )}{9+27 x} \]

[In]

Integrate[(E^4*(2*x + 3*x^2))/(1 + 6*x + 9*x^2),x]

[Out]

(E^4*(2 + 6*x + 9*x^2))/(9 + 27*x)

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88

method	result	size
gosper	\(\frac {x^{2} {\mathrm e}^{4}}{1+3 x}\)	\(14\)
norman	\(\frac {x^{2} {\mathrm e}^{3} {\mathrm e}}{1+3 x}\)	\(16\)
risch	\(\frac {x \,{\mathrm e}^{4}}{3}+\frac {{\mathrm e}^{4}}{27 x +9}\)	\(16\)
parallelrisch	\(\frac {x^{2} {\mathrm e}^{3} {\mathrm e}}{1+3 x}\)	\(16\)
default	\({\mathrm e} \,{\mathrm e}^{3} \left (\frac {x}{3}+\frac {1}{27 x +9}\right )\)	\(19\)
meijerg	\(\frac {{\mathrm e}^{4} \left (\frac {x \left (9 x +6\right )}{1+3 x}-2 \ln \left (1+3 x \right )\right )}{9}+\frac {2 \,{\mathrm e}^{4} \left (-\frac {3 x}{1+3 x}+\ln \left (1+3 x \right )\right )}{9}\)	\(50\)

[In]

int((3*x^2+2*x)*exp(1)*exp(3)/(9*x^2+6*x+1),x,method=_RETURNVERBOSE)

[Out]

x^2*exp(4)/(1+3*x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.31 \[ \int \frac {e^4 \left (2 x+3 x^2\right )}{1+6 x+9 x^2} \, dx=\frac {{\left (9 \, x^{2} + 3 \, x + 1\right )} e^{4}}{9 \, {\left (3 \, x + 1\right )}} \]

[In]

integrate((3*x^2+2*x)*exp(1)*exp(3)/(9*x^2+6*x+1),x, algorithm="fricas")

[Out]

1/9*(9*x^2 + 3*x + 1)*e^4/(3*x + 1)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {e^4 \left (2 x+3 x^2\right )}{1+6 x+9 x^2} \, dx=\frac {x e^{4}}{3} + \frac {e^{4}}{27 x + 9} \]

[In]

integrate((3*x**2+2*x)*exp(1)*exp(3)/(9*x**2+6*x+1),x)

[Out]

x*exp(4)/3 + exp(4)/(27*x + 9)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {e^4 \left (2 x+3 x^2\right )}{1+6 x+9 x^2} \, dx=\frac {1}{9} \, {\left (3 \, x + \frac {1}{3 \, x + 1}\right )} e^{4} \]

[In]

integrate((3*x^2+2*x)*exp(1)*exp(3)/(9*x^2+6*x+1),x, algorithm="maxima")

[Out]

1/9*(3*x + 1/(3*x + 1))*e^4

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {e^4 \left (2 x+3 x^2\right )}{1+6 x+9 x^2} \, dx=\frac {1}{9} \, {\left (3 \, x + \frac {1}{3 \, x + 1}\right )} e^{4} \]

[In]

integrate((3*x^2+2*x)*exp(1)*exp(3)/(9*x^2+6*x+1),x, algorithm="giac")

[Out]

1/9*(3*x + 1/(3*x + 1))*e^4

Mupad [B] (verification not implemented)

Time = 7.89 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {e^4 \left (2 x+3 x^2\right )}{1+6 x+9 x^2} \, dx=\frac {x\,{\mathrm {e}}^4}{3}+\frac {{\mathrm {e}}^4}{9\,\left (3\,x+1\right )} \]

[In]

int((exp(4)*(2*x + 3*x^2))/(6*x + 9*x^2 + 1),x)

[Out]

(x*exp(4))/3 + exp(4)/(9*(3*x + 1))