\(\int \frac {e^{-5+(-2 x+x^2) \log (2)} (-6+6 x) \log (2)}{-20+3 e^{-5+(-2 x+x^2) \log (2)}} \, dx\) [9143]

Optimal result

Integrand size = 42, antiderivative size = 20 \[ \int \frac {e^{-5+\left (-2 x+x^2\right ) \log (2)} (-6+6 x) \log (2)}{-20+3 e^{-5+\left (-2 x+x^2\right ) \log (2)}} \, dx=3+\log \left (-5+\frac {3}{4} e^{-5+(-2+x) x \log (2)}\right ) \]

[Out]

3+ln(3/4*exp((-2+x)*x*ln(2)-5)-5)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {12, 6816} \[ \int \frac {e^{-5+\left (-2 x+x^2\right ) \log (2)} (-6+6 x) \log (2)}{-20+3 e^{-5+\left (-2 x+x^2\right ) \log (2)}} \, dx=\log \left (-2^{-2 x} \left (3\ 2^{x^2}-5 e^5 2^{2 x+2}\right )\right ) \]

[In]

Int[(E^(-5 + (-2*x + x^2)*Log[2])*(-6 + 6*x)*Log[2])/(-20 + 3*E^(-5 + (-2*x + x^2)*Log[2])),x]

[Out]

Log[-((3*2^x^2 - 5*2^(2 + 2*x)*E^5)/2^(2*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 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rubi steps \begin{align*} \text {integral}& = \log (2) \int \frac {e^{-5+\left (-2 x+x^2\right ) \log (2)} (-6+6 x)}{-20+3 e^{-5+\left (-2 x+x^2\right ) \log (2)}} \, dx \\ & = \log \left (-2^{-2 x} \left (3\ 2^{x^2}-5\ 2^{2+2 x} e^5\right )\right ) \\ \end{align*}

Mathematica [F]

\[ \int \frac {e^{-5+\left (-2 x+x^2\right ) \log (2)} (-6+6 x) \log (2)}{-20+3 e^{-5+\left (-2 x+x^2\right ) \log (2)}} \, dx=\int \frac {e^{-5+\left (-2 x+x^2\right ) \log (2)} (-6+6 x) \log (2)}{-20+3 e^{-5+\left (-2 x+x^2\right ) \log (2)}} \, dx \]

[In]

Integrate[(E^(-5 + (-2*x + x^2)*Log[2])*(-6 + 6*x)*Log[2])/(-20 + 3*E^(-5 + (-2*x + x^2)*Log[2])),x]

[Out]

Integrate[(E^(-5 + (-2*x + x^2)*Log[2])*(-6 + 6*x)*Log[2])/(-20 + 3*E^(-5 + (-2*x + x^2)*Log[2])), x]

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80

method	result	size
risch	\(\ln \left (2^{\left (-2+x \right ) x} {\mathrm e}^{-5}-\frac {20}{3}\right )+5\)	\(16\)
parallelrisch	\(\ln \left ({\mathrm e}^{\left (x^{2}-2 x \right ) \ln \left (2\right )-5}-\frac {20}{3}\right )\)	\(17\)
norman	\(\ln \left (3 \,{\mathrm e}^{\left (x^{2}-2 x \right ) \ln \left (2\right )-5}-20\right )\)	\(19\)

[In]

int((6*x-6)*ln(2)*exp((x^2-2*x)*ln(2)-5)/(3*exp((x^2-2*x)*ln(2)-5)-20),x,method=_RETURNVERBOSE)

[Out]

ln(2^((-2+x)*x)*exp(-5)-20/3)+5

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-5+\left (-2 x+x^2\right ) \log (2)} (-6+6 x) \log (2)}{-20+3 e^{-5+\left (-2 x+x^2\right ) \log (2)}} \, dx=\log \left (3 \, e^{\left ({\left (x^{2} - 2 \, x\right )} \log \left (2\right ) - 5\right )} - 20\right ) \]

[In]

integrate((6*x-6)*log(2)*exp((x^2-2*x)*log(2)-5)/(3*exp((x^2-2*x)*log(2)-5)-20),x, algorithm="fricas")

[Out]

log(3*e^((x^2 - 2*x)*log(2) - 5) - 20)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {e^{-5+\left (-2 x+x^2\right ) \log (2)} (-6+6 x) \log (2)}{-20+3 e^{-5+\left (-2 x+x^2\right ) \log (2)}} \, dx=\log {\left (e^{\left (x^{2} - 2 x\right ) \log {\left (2 \right )} - 5} - \frac {20}{3} \right )} \]

[In]

integrate((6*x-6)*ln(2)*exp((x**2-2*x)*ln(2)-5)/(3*exp((x**2-2*x)*ln(2)-5)-20),x)

[Out]

log(exp((x**2 - 2*x)*log(2) - 5) - 20/3)

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.55 \[ \int \frac {e^{-5+\left (-2 x+x^2\right ) \log (2)} (-6+6 x) \log (2)}{-20+3 e^{-5+\left (-2 x+x^2\right ) \log (2)}} \, dx=-{\left (2 \, x - \frac {\log \left (2^{\left (x^{2}\right )} - \frac {20}{3} \, e^{\left (2 \, x \log \left (2\right ) + 5\right )}\right )}{\log \left (2\right )}\right )} \log \left (2\right ) \]

[In]

integrate((6*x-6)*log(2)*exp((x^2-2*x)*log(2)-5)/(3*exp((x^2-2*x)*log(2)-5)-20),x, algorithm="maxima")

[Out]

-(2*x - log(2^(x^2) - 20/3*e^(2*x*log(2) + 5))/log(2))*log(2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-5+\left (-2 x+x^2\right ) \log (2)} (-6+6 x) \log (2)}{-20+3 e^{-5+\left (-2 x+x^2\right ) \log (2)}} \, dx=\log \left ({\left | 3 \, e^{\left (x^{2} \log \left (2\right ) - 2 \, x \log \left (2\right ) - 5\right )} - 20 \right |}\right ) \]

[In]

integrate((6*x-6)*log(2)*exp((x^2-2*x)*log(2)-5)/(3*exp((x^2-2*x)*log(2)-5)-20),x, algorithm="giac")

[Out]

log(abs(3*e^(x^2*log(2) - 2*x*log(2) - 5) - 20))

Mupad [B] (verification not implemented)

Time = 16.85 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {e^{-5+\left (-2 x+x^2\right ) \log (2)} (-6+6 x) \log (2)}{-20+3 e^{-5+\left (-2 x+x^2\right ) \log (2)}} \, dx=\ln \left (3\,2^{x^2}\,{\mathrm {e}}^{-5}-20\,2^{2\,x}\right )-2\,x\,\ln \left (2\right ) \]

[In]

int((exp(- log(2)*(2*x - x^2) - 5)*log(2)*(6*x - 6))/(3*exp(- log(2)*(2*x - x^2) - 5) - 20),x)

[Out]

log(3*2^(x^2)*exp(-5) - 20*2^(2*x)) - 2*x*log(2)