\(\int \frac {-12+4 e^2}{1444+4 e^4+e^2 (152-44 x)-836 x+121 x^2} \, dx\) [2800]

Optimal result

Integrand size = 34, antiderivative size = 16 \[ \int \frac {-12+4 e^2}{1444+4 e^4+e^2 (152-44 x)-836 x+121 x^2} \, dx=\frac {-4+x}{19+e^2-\frac {11 x}{2}} \]

[Out]

1/(19-11/2*x+exp(2))*(x-4)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.50, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 2006, 27, 32} \[ \int \frac {-12+4 e^2}{1444+4 e^4+e^2 (152-44 x)-836 x+121 x^2} \, dx=-\frac {4 \left (3-e^2\right )}{11 \left (2 \left (19+e^2\right )-11 x\right )} \]

[In]

Int[(-12 + 4*E^2)/(1444 + 4*E^4 + E^2*(152 - 44*x) - 836*x + 121*x^2),x]

[Out]

(-4*(3 - E^2))/(11*(2*(19 + E^2) - 11*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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2006

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && QuadraticQ[u, x] &&  !QuadraticMatch
Q[u, x]

Rubi steps \begin{align*} \text {integral}& = -\left (\left (4 \left (3-e^2\right )\right ) \int \frac {1}{1444+4 e^4+e^2 (152-44 x)-836 x+121 x^2} \, dx\right ) \\ & = -\left (\left (4 \left (3-e^2\right )\right ) \int \frac {1}{4 \left (19+e^2\right )^2-44 \left (19+e^2\right ) x+121 x^2} \, dx\right ) \\ & = -\left (\left (4 \left (3-e^2\right )\right ) \int \frac {1}{\left (38+2 e^2-11 x\right )^2} \, dx\right ) \\ & = -\frac {4 \left (3-e^2\right )}{11 \left (2 \left (19+e^2\right )-11 x\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.19 \[ \int \frac {-12+4 e^2}{1444+4 e^4+e^2 (152-44 x)-836 x+121 x^2} \, dx=\frac {4 \left (-3+e^2\right )}{418+22 e^2-121 x} \]

[In]

Integrate[(-12 + 4*E^2)/(1444 + 4*E^4 + E^2*(152 - 44*x) - 836*x + 121*x^2),x]

[Out]

(4*(-3 + E^2))/(418 + 22*E^2 - 121*x)

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.12

method	result	size
gosper	\(\frac {-\frac {12}{11}+\frac {4 \,{\mathrm e}^{2}}{11}}{2 \,{\mathrm e}^{2}-11 x +38}\)	\(18\)
norman	\(\frac {-\frac {12}{11}+\frac {4 \,{\mathrm e}^{2}}{11}}{2 \,{\mathrm e}^{2}-11 x +38}\)	\(19\)
parallelrisch	\(\frac {4 \,{\mathrm e}^{2}-12}{22 \,{\mathrm e}^{2}-121 x +418}\)	\(20\)
risch	\(\frac {2 \,{\mathrm e}^{2}}{11 \left (19-\frac {11 x}{2}+{\mathrm e}^{2}\right )}-\frac {6}{11 \left (19-\frac {11 x}{2}+{\mathrm e}^{2}\right )}\)	\(26\)
meijerg	\(\frac {6 x}{11 \left (-\frac {2 \,{\mathrm e}^{2}}{11}-\frac {38}{11}\right ) \left ({\mathrm e}^{2}+19\right ) \left (1-\frac {11 x}{2 \left ({\mathrm e}^{2}+19\right )}\right )}-\frac {2 \,{\mathrm e}^{2} x}{11 \left (-\frac {2 \,{\mathrm e}^{2}}{11}-\frac {38}{11}\right ) \left ({\mathrm e}^{2}+19\right ) \left (1-\frac {11 x}{2 \left ({\mathrm e}^{2}+19\right )}\right )}\)	\(64\)

[In]

int((4*exp(2)-12)/(4*exp(2)^2+(-44*x+152)*exp(2)+121*x^2-836*x+1444),x,method=_RETURNVERBOSE)

[Out]

4/11*(exp(2)-3)/(2*exp(2)-11*x+38)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \frac {-12+4 e^2}{1444+4 e^4+e^2 (152-44 x)-836 x+121 x^2} \, dx=-\frac {4 \, {\left (e^{2} - 3\right )}}{11 \, {\left (11 \, x - 2 \, e^{2} - 38\right )}} \]

[In]

integrate((4*exp(2)-12)/(4*exp(2)^2+(-44*x+152)*exp(2)+121*x^2-836*x+1444),x, algorithm="fricas")

[Out]

-4/11*(e^2 - 3)/(11*x - 2*e^2 - 38)

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \frac {-12+4 e^2}{1444+4 e^4+e^2 (152-44 x)-836 x+121 x^2} \, dx=- \frac {-12 + 4 e^{2}}{121 x - 418 - 22 e^{2}} \]

[In]

integrate((4*exp(2)-12)/(4*exp(2)**2+(-44*x+152)*exp(2)+121*x**2-836*x+1444),x)

[Out]

-(-12 + 4*exp(2))/(121*x - 418 - 22*exp(2))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \frac {-12+4 e^2}{1444+4 e^4+e^2 (152-44 x)-836 x+121 x^2} \, dx=-\frac {4 \, {\left (e^{2} - 3\right )}}{11 \, {\left (11 \, x - 2 \, e^{2} - 38\right )}} \]

[In]

integrate((4*exp(2)-12)/(4*exp(2)^2+(-44*x+152)*exp(2)+121*x^2-836*x+1444),x, algorithm="maxima")

[Out]

-4/11*(e^2 - 3)/(11*x - 2*e^2 - 38)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \frac {-12+4 e^2}{1444+4 e^4+e^2 (152-44 x)-836 x+121 x^2} \, dx=-\frac {4 \, {\left (e^{2} - 3\right )}}{11 \, {\left (11 \, x - 2 \, e^{2} - 38\right )}} \]

[In]

integrate((4*exp(2)-12)/(4*exp(2)^2+(-44*x+152)*exp(2)+121*x^2-836*x+1444),x, algorithm="giac")

[Out]

-4/11*(e^2 - 3)/(11*x - 2*e^2 - 38)

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.12 \[ \int \frac {-12+4 e^2}{1444+4 e^4+e^2 (152-44 x)-836 x+121 x^2} \, dx=\frac {\frac {4\,{\mathrm {e}}^2}{11}-\frac {12}{11}}{2\,{\mathrm {e}}^2-11\,x+38} \]

[In]

int((4*exp(2) - 12)/(4*exp(4) - 836*x + 121*x^2 - exp(2)*(44*x - 152) + 1444),x)

[Out]

((4*exp(2))/11 - 12/11)/(2*exp(2) - 11*x + 38)