\(\int \frac {e^6 (-2+3 x)}{-4 x+4 x^2} \, dx\) [6649]

Optimal result

Integrand size = 20, antiderivative size = 23 \[ \int \frac {e^6 (-2+3 x)}{-4 x+4 x^2} \, dx=\frac {1}{4} e^6 \log \left (\frac {3}{10} e^2 x \left (-x+x^2\right )\right ) \]

[Out]

1/4*exp(3)^2*ln(3/10*exp(2)*x*(x^2-x))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {12, 645} \[ \int \frac {e^6 (-2+3 x)}{-4 x+4 x^2} \, dx=\frac {1}{4} e^6 \log (1-x)+\frac {1}{2} e^6 \log (x) \]

[In]

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

[Out]

(E^6*Log[1 - x])/4 + (E^6*Log[x])/2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 645

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)
*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0]
|| EqQ[a, 0])

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {e^6 (-2+3 x)}{-4 x+4 x^2} \, dx=e^6 \left (\frac {1}{4} \log (1-x)+\frac {\log (x)}{2}\right ) \]

[In]

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

[Out]

E^6*(Log[1 - x]/4 + Log[x]/2)

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.70

method	result	size
default	\(\frac {{\mathrm e}^{6} \left (2 \ln \left (x \right )+\ln \left (-1+x \right )\right )}{4}\)	\(16\)
risch	\(\frac {{\mathrm e}^{6} \ln \left (x \right )}{2}+\frac {{\mathrm e}^{6} \ln \left (-1+x \right )}{4}\)	\(16\)
parallelrisch	\({\mathrm e}^{6} \left (\frac {\ln \left (x \right )}{2}+\frac {\ln \left (-1+x \right )}{4}\right )\)	\(17\)
norman	\(\frac {{\mathrm e}^{6} \ln \left (x \right )}{2}+\frac {{\mathrm e}^{6} \ln \left (-1+x \right )}{4}\)	\(20\)
meijerg	\(\frac {{\mathrm e}^{6} \left (\ln \left (x \right )+i \pi -\ln \left (1-x \right )\right )}{2}+\frac {3 \,{\mathrm e}^{6} \ln \left (1-x \right )}{4}\)	\(31\)

[In]

int((-2+3*x)*exp(3)^2/(4*x^2-4*x),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int \frac {e^6 (-2+3 x)}{-4 x+4 x^2} \, dx=\frac {1}{4} \, e^{6} \log \left (x - 1\right ) + \frac {1}{2} \, e^{6} \log \left (x\right ) \]

[In]

integrate((-2+3*x)*exp(3)^2/(4*x^2-4*x),x, algorithm="fricas")

[Out]

1/4*e^6*log(x - 1) + 1/2*e^6*log(x)

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {e^6 (-2+3 x)}{-4 x+4 x^2} \, dx=\frac {e^{6} \log {\left (x \right )}}{2} + \frac {e^{6} \log {\left (x - 1 \right )}}{4} \]

[In]

integrate((-2+3*x)*exp(3)**2/(4*x**2-4*x),x)

[Out]

exp(6)*log(x)/2 + exp(6)*log(x - 1)/4

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.57 \[ \int \frac {e^6 (-2+3 x)}{-4 x+4 x^2} \, dx=\frac {1}{4} \, {\left (\log \left (x - 1\right ) + 2 \, \log \left (x\right )\right )} e^{6} \]

[In]

integrate((-2+3*x)*exp(3)^2/(4*x^2-4*x),x, algorithm="maxima")

[Out]

1/4*(log(x - 1) + 2*log(x))*e^6

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int \frac {e^6 (-2+3 x)}{-4 x+4 x^2} \, dx=\frac {1}{4} \, {\left (\log \left ({\left | x - 1 \right |}\right ) + 2 \, \log \left ({\left | x \right |}\right )\right )} e^{6} \]

[In]

integrate((-2+3*x)*exp(3)^2/(4*x^2-4*x),x, algorithm="giac")

[Out]

1/4*(log(abs(x - 1)) + 2*log(abs(x)))*e^6

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int \frac {e^6 (-2+3 x)}{-4 x+4 x^2} \, dx=\frac {\ln \left (x-1\right )\,{\mathrm {e}}^6}{4}+\frac {{\mathrm {e}}^6\,\ln \left (x\right )}{2} \]

[In]

int(-(exp(6)*(3*x - 2))/(4*x - 4*x^2),x)

[Out]

(log(x - 1)*exp(6))/4 + (exp(6)*log(x))/2