\(\int -\frac {e^3 (1-80 x)}{16 x (-x+80 x^2)} \, dx\) [2952]

Optimal result

Integrand size = 26, antiderivative size = 16 \[ \int -\frac {e^3 (1-80 x)}{16 x \left (-x+80 x^2\right )} \, dx=2-e^3 \left (-5+\frac {1}{16 x}\right ) \]

[Out]

2-exp(ln(1/16/x-5)+3)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {12, 777} \[ \int -\frac {e^3 (1-80 x)}{16 x \left (-x+80 x^2\right )} \, dx=-\frac {e^3}{16 x} \]

[In]

Int[-1/16*(E^3*(1 - 80*x))/(x*(-x + 80*x^2)),x]

[Out]

-1/16*E^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 777

Int[((e_.)*(x_))^(m_.)*((f_) + (g_.)*(x_))*((b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g*(e*x)^m*((b*
x + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), x] /; FreeQ[{b, c, e, f, g, m, p}, x] && EqQ[b*g*(m + p + 1) - c*f*(m +
2*p + 2), 0] && NeQ[m + 2*p + 2, 0]

Rubi steps \begin{align*} \text {integral}& = -\left (\frac {1}{16} e^3 \int \frac {1-80 x}{x \left (-x+80 x^2\right )} \, dx\right ) \\ & = -\frac {e^3}{16 x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int -\frac {e^3 (1-80 x)}{16 x \left (-x+80 x^2\right )} \, dx=-\frac {e^3}{16 x} \]

[In]

Integrate[-1/16*(E^3*(1 - 80*x))/(x*(-x + 80*x^2)),x]

[Out]

-1/16*E^3/x

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.50

method	result	size
norman	\(-\frac {{\mathrm e}^{3}}{16 x}\)	\(8\)
risch	\(-\frac {{\mathrm e}^{3}}{16 x}\)	\(8\)
gosper	\(\frac {{\mathrm e}^{\ln \left (-\frac {80 x -1}{16 x}\right )+3}}{80 x -1}\)	\(23\)
default	\(\frac {{\mathrm e}^{\ln \left (-\frac {80 x -1}{16 x}\right )+3}}{80 x -1}\)	\(23\)
parallelrisch	\(\frac {{\mathrm e}^{\ln \left (-\frac {80 x -1}{16 x}\right )+3}}{80 x -1}\)	\(23\)

[In]

int(-exp(ln(1/16*(-80*x+1)/x)+3)/(80*x^2-x),x,method=_RETURNVERBOSE)

[Out]

-1/16*exp(3)/x

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.44 \[ \int -\frac {e^3 (1-80 x)}{16 x \left (-x+80 x^2\right )} \, dx=-\frac {e^{3}}{16 \, x} \]

[In]

integrate(-exp(log(1/16*(-80*x+1)/x)+3)/(80*x^2-x),x, algorithm="fricas")

[Out]

-1/16*e^3/x

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.44 \[ \int -\frac {e^3 (1-80 x)}{16 x \left (-x+80 x^2\right )} \, dx=- \frac {e^{3}}{16 x} \]

[In]

integrate(-exp(ln(1/16*(-80*x+1)/x)+3)/(80*x**2-x),x)

[Out]

-exp(3)/(16*x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.44 \[ \int -\frac {e^3 (1-80 x)}{16 x \left (-x+80 x^2\right )} \, dx=-\frac {e^{3}}{16 \, x} \]

[In]

integrate(-exp(log(1/16*(-80*x+1)/x)+3)/(80*x^2-x),x, algorithm="maxima")

[Out]

-1/16*e^3/x

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.44 \[ \int -\frac {e^3 (1-80 x)}{16 x \left (-x+80 x^2\right )} \, dx=-\frac {e^{3}}{16 \, x} \]

[In]

integrate(-exp(log(1/16*(-80*x+1)/x)+3)/(80*x^2-x),x, algorithm="giac")

[Out]

-1/16*e^3/x

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.44 \[ \int -\frac {e^3 (1-80 x)}{16 x \left (-x+80 x^2\right )} \, dx=-\frac {{\mathrm {e}}^3}{16\,x} \]

[In]

int(exp(log(-(5*x - 1/16)/x) + 3)/(x - 80*x^2),x)

[Out]

-exp(3)/(16*x)