\(\int \frac {752+256 e^3}{2209+256 e^6+e^3 (1504-2560 x)-7520 x+6400 x^2} \, dx\) [6625]

Optimal result

Integrand size = 34, antiderivative size = 16 \[ \int \frac {752+256 e^3}{2209+256 e^6+e^3 (1504-2560 x)-7520 x+6400 x^2} \, dx=-1+\frac {x}{\frac {47}{16}+e^3-5 x} \]

[Out]

x/(47/16-5*x+exp(3))-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 = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 2006, 27, 32} \[ \int \frac {752+256 e^3}{2209+256 e^6+e^3 (1504-2560 x)-7520 x+6400 x^2} \, dx=\frac {47+16 e^3}{5 \left (-80 x+16 e^3+47\right )} \]

[In]

Int[(752 + 256*E^3)/(2209 + 256*E^6 + E^3*(1504 - 2560*x) - 7520*x + 6400*x^2),x]

[Out]

(47 + 16*E^3)/(5*(47 + 16*E^3 - 80*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 (16 \left (47+16 e^3\right )\right ) \int \frac {1}{2209+256 e^6+e^3 (1504-2560 x)-7520 x+6400 x^2} \, dx \\ & = \left (16 \left (47+16 e^3\right )\right ) \int \frac {1}{\left (47+16 e^3\right )^2-160 \left (47+16 e^3\right ) x+6400 x^2} \, dx \\ & = \left (16 \left (47+16 e^3\right )\right ) \int \frac {1}{\left (47+16 e^3-80 x\right )^2} \, dx \\ & = \frac {47+16 e^3}{5 \left (47+16 e^3-80 x\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.44 \[ \int \frac {752+256 e^3}{2209+256 e^6+e^3 (1504-2560 x)-7520 x+6400 x^2} \, dx=\frac {47+16 e^3}{5 \left (47+16 e^3-80 x\right )} \]

[In]

Integrate[(752 + 256*E^3)/(2209 + 256*E^6 + E^3*(1504 - 2560*x) - 7520*x + 6400*x^2),x]

[Out]

(47 + 16*E^3)/(5*(47 + 16*E^3 - 80*x))

Maple [A] (verified)

Time = 0.63 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.19

method	result	size
norman	\(\frac {\frac {47}{5}+\frac {16 \,{\mathrm e}^{3}}{5}}{16 \,{\mathrm e}^{3}-80 x +47}\)	\(19\)
gosper	\(\frac {16 \,{\mathrm e}^{3}+47}{80 \,{\mathrm e}^{3}-400 x +235}\)	\(20\)
parallelrisch	\(\frac {256 \,{\mathrm e}^{3}+752}{1280 \,{\mathrm e}^{3}-6400 x +3760}\)	\(20\)
risch	\(\frac {{\mathrm e}^{3}}{\frac {235}{16}-25 x +5 \,{\mathrm e}^{3}}+\frac {47}{80 \left (\frac {47}{16}-5 x +{\mathrm e}^{3}\right )}\)	\(26\)
meijerg	\(-\frac {47 x}{5 \left (-\frac {{\mathrm e}^{3}}{5}-\frac {47}{80}\right ) \left (16 \,{\mathrm e}^{3}+47\right ) \left (1-\frac {80 x}{16 \,{\mathrm e}^{3}+47}\right )}-\frac {16 \,{\mathrm e}^{3} x}{5 \left (-\frac {{\mathrm e}^{3}}{5}-\frac {47}{80}\right ) \left (16 \,{\mathrm e}^{3}+47\right ) \left (1-\frac {80 x}{16 \,{\mathrm e}^{3}+47}\right )}\)	\(72\)

[In]

int((256*exp(3)+752)/(256*exp(3)^2+(-2560*x+1504)*exp(3)+6400*x^2-7520*x+2209),x,method=_RETURNVERBOSE)

[Out]

(47/5+16/5*exp(3))/(16*exp(3)-80*x+47)

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.19 \[ \int \frac {752+256 e^3}{2209+256 e^6+e^3 (1504-2560 x)-7520 x+6400 x^2} \, dx=-\frac {16 \, e^{3} + 47}{5 \, {\left (80 \, x - 16 \, e^{3} - 47\right )}} \]

[In]

integrate((256*exp(3)+752)/(256*exp(3)^2+(-2560*x+1504)*exp(3)+6400*x^2-7520*x+2209),x, algorithm="fricas")

[Out]

-1/5*(16*e^3 + 47)/(80*x - 16*e^3 - 47)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \frac {752+256 e^3}{2209+256 e^6+e^3 (1504-2560 x)-7520 x+6400 x^2} \, dx=- \frac {752 + 256 e^{3}}{6400 x - 1280 e^{3} - 3760} \]

[In]

integrate((256*exp(3)+752)/(256*exp(3)**2+(-2560*x+1504)*exp(3)+6400*x**2-7520*x+2209),x)

[Out]

-(752 + 256*exp(3))/(6400*x - 1280*exp(3) - 3760)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.19 \[ \int \frac {752+256 e^3}{2209+256 e^6+e^3 (1504-2560 x)-7520 x+6400 x^2} \, dx=-\frac {16 \, e^{3} + 47}{5 \, {\left (80 \, x - 16 \, e^{3} - 47\right )}} \]

[In]

integrate((256*exp(3)+752)/(256*exp(3)^2+(-2560*x+1504)*exp(3)+6400*x^2-7520*x+2209),x, algorithm="maxima")

[Out]

-1/5*(16*e^3 + 47)/(80*x - 16*e^3 - 47)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.19 \[ \int \frac {752+256 e^3}{2209+256 e^6+e^3 (1504-2560 x)-7520 x+6400 x^2} \, dx=-\frac {16 \, e^{3} + 47}{5 \, {\left (80 \, x - 16 \, e^{3} - 47\right )}} \]

[In]

integrate((256*exp(3)+752)/(256*exp(3)^2+(-2560*x+1504)*exp(3)+6400*x^2-7520*x+2209),x, algorithm="giac")

[Out]

-1/5*(16*e^3 + 47)/(80*x - 16*e^3 - 47)

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.12 \[ \int \frac {752+256 e^3}{2209+256 e^6+e^3 (1504-2560 x)-7520 x+6400 x^2} \, dx=\frac {\frac {16\,{\mathrm {e}}^3}{5}+\frac {47}{5}}{16\,{\mathrm {e}}^3-80\,x+47} \]

[In]

int((256*exp(3) + 752)/(256*exp(6) - 7520*x + 6400*x^2 - exp(3)*(2560*x - 1504) + 2209),x)

[Out]

((16*exp(3))/5 + 47/5)/(16*exp(3) - 80*x + 47)