\(\int \frac {e^{\frac {46+2 x+(8+x) \log (2)}{23+x+4 \log (2)}} (-23 \log (2)-4 \log ^2(2))}{529+46 x+x^2+(184+8 x) \log (2)+16 \log ^2(2)} \, dx\) [5924]

Optimal result

Integrand size = 59, antiderivative size = 29 \[ \int \frac {e^{\frac {46+2 x+(8+x) \log (2)}{23+x+4 \log (2)}} \left (-23 \log (2)-4 \log ^2(2)\right )}{529+46 x+x^2+(184+8 x) \log (2)+16 \log ^2(2)} \, dx=-1+e^{e^{e^5}}-e^{2+\frac {x}{4+\frac {23+x}{\log (2)}}} \]

[Out]

exp(exp(exp(5)))-1-exp(x/(4+(x+23)/ln(2))+2)

Rubi [F]

\[ \int \frac {e^{\frac {46+2 x+(8+x) \log (2)}{23+x+4 \log (2)}} \left (-23 \log (2)-4 \log ^2(2)\right )}{529+46 x+x^2+(184+8 x) \log (2)+16 \log ^2(2)} \, dx=\int \frac {e^{\frac {46+2 x+(8+x) \log (2)}{23+x+4 \log (2)}} \left (-23 \log (2)-4 \log ^2(2)\right )}{529+46 x+x^2+(184+8 x) \log (2)+16 \log ^2(2)} \, dx \]

[In]

Int[(E^((46 + 2*x + (8 + x)*Log[2])/(23 + x + 4*Log[2]))*(-23*Log[2] - 4*Log[2]^2))/(529 + 46*x + x^2 + (184 +
8*x)*Log[2] + 16*Log[2]^2),x]

[Out]

-(Log[2]*(23 + Log[16])*Defer[Int][(2^((8 + x)/(23 + x + Log[16]))*E^((2*(23 + x))/(23 + x + Log[16])))/(23 +
x + Log[16])^2, x])

Rubi steps \begin{align*} \text {integral}& = -\left ((\log (2) (23+\log (16))) \int \frac {e^{\frac {46+2 x+(8+x) \log (2)}{23+x+4 \log (2)}}}{529+46 x+x^2+(184+8 x) \log (2)+16 \log ^2(2)} \, dx\right ) \\ & = -\left ((\log (2) (23+\log (16))) \int \frac {2^{\frac {8+x}{23+x+\log (16)}} e^{\frac {2 (23+x)}{23+x+\log (16)}}}{(23+x+\log (16))^2} \, dx\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {e^{\frac {46+2 x+(8+x) \log (2)}{23+x+4 \log (2)}} \left (-23 \log (2)-4 \log ^2(2)\right )}{529+46 x+x^2+(184+8 x) \log (2)+16 \log ^2(2)} \, dx=-\frac {2^{\frac {x}{23+x+\log (16)}} e^2 \log (2) (23+\log (16))}{\log (2) \log (16)+\log (8388608)} \]

[In]

Integrate[(E^((46 + 2*x + (8 + x)*Log[2])/(23 + x + 4*Log[2]))*(-23*Log[2] - 4*Log[2]^2))/(529 + 46*x + x^2 +
(184 + 8*x)*Log[2] + 16*Log[2]^2),x]

[Out]

-((2^(x/(23 + x + Log[16]))*E^2*Log[2]*(23 + Log[16]))/(Log[2]*Log[16] + Log[8388608]))

Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93

method	result	size
gosper	\(-{\mathrm e}^{\frac {x \ln \left (2\right )+8 \ln \left (2\right )+2 x +46}{4 \ln \left (2\right )+x +23}}\)	\(27\)
parallelrisch	\(\frac {\left (-4 \ln \left (2\right )^{2}-23 \ln \left (2\right )\right ) {\mathrm e}^{\frac {\left (x +8\right ) \ln \left (2\right )+2 x +46}{4 \ln \left (2\right )+x +23}}}{\ln \left (2\right ) \left (4 \ln \left (2\right )+23\right )}\)	\(47\)
default	\(-\frac {\left (-4 \ln \left (2\right )^{2}-23 \ln \left (2\right )\right )^{2} {\mathrm e}^{\frac {-4 \ln \left (2\right )^{2}-23 \ln \left (2\right )}{4 \ln \left (2\right )+x +23}+\ln \left (2\right )+2}}{\ln \left (2\right )^{2} \left (16 \ln \left (2\right )^{2}+184 \ln \left (2\right )+529\right )}\)	\(60\)
norman	\(\frac {\left (-23-4 \ln \left (2\right )\right ) {\mathrm e}^{\frac {\left (x +8\right ) \ln \left (2\right )+2 x +46}{4 \ln \left (2\right )+x +23}}-x \,{\mathrm e}^{\frac {\left (x +8\right ) \ln \left (2\right )+2 x +46}{4 \ln \left (2\right )+x +23}}}{4 \ln \left (2\right )+x +23}\)	\(66\)
derivativedivides	\(\frac {\left (4 \ln \left (2\right )^{2}+23 \ln \left (2\right )\right ) \left (-4 \ln \left (2\right )^{2}-23 \ln \left (2\right )\right ) {\mathrm e}^{\frac {-4 \ln \left (2\right )^{2}-23 \ln \left (2\right )}{4 \ln \left (2\right )+x +23}+\ln \left (2\right )+2}}{\ln \left (2\right )^{2} \left (16 \ln \left (2\right )^{2}+184 \ln \left (2\right )+529\right )}\)	\(68\)
risch	\(-\frac {4 \ln \left (2\right ) {\mathrm e}^{\frac {x \ln \left (2\right )+8 \ln \left (2\right )+2 x +46}{4 \ln \left (2\right )+x +23}}}{4 \ln \left (2\right )+23}-\frac {23 \,{\mathrm e}^{\frac {x \ln \left (2\right )+8 \ln \left (2\right )+2 x +46}{4 \ln \left (2\right )+x +23}}}{4 \ln \left (2\right )+23}\)	\(72\)

[In]

int((-4*ln(2)^2-23*ln(2))*exp(((x+8)*ln(2)+2*x+46)/(4*ln(2)+x+23))/(16*ln(2)^2+(8*x+184)*ln(2)+x^2+46*x+529),x
,method=_RETURNVERBOSE)

[Out]

-exp((x*ln(2)+8*ln(2)+2*x+46)/(4*ln(2)+x+23))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {e^{\frac {46+2 x+(8+x) \log (2)}{23+x+4 \log (2)}} \left (-23 \log (2)-4 \log ^2(2)\right )}{529+46 x+x^2+(184+8 x) \log (2)+16 \log ^2(2)} \, dx=-e^{\left (\frac {{\left (x + 8\right )} \log \left (2\right ) + 2 \, x + 46}{x + 4 \, \log \left (2\right ) + 23}\right )} \]

[In]

integrate((-4*log(2)^2-23*log(2))*exp(((x+8)*log(2)+2*x+46)/(4*log(2)+x+23))/(16*log(2)^2+(8*x+184)*log(2)+x^2
+46*x+529),x, algorithm="fricas")

[Out]

-e^(((x + 8)*log(2) + 2*x + 46)/(x + 4*log(2) + 23))

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {e^{\frac {46+2 x+(8+x) \log (2)}{23+x+4 \log (2)}} \left (-23 \log (2)-4 \log ^2(2)\right )}{529+46 x+x^2+(184+8 x) \log (2)+16 \log ^2(2)} \, dx=- e^{\frac {2 x + \left (x + 8\right ) \log {\left (2 \right )} + 46}{x + 4 \log {\left (2 \right )} + 23}} \]

[In]

integrate((-4*ln(2)**2-23*ln(2))*exp(((x+8)*ln(2)+2*x+46)/(4*ln(2)+x+23))/(16*ln(2)**2+(8*x+184)*ln(2)+x**2+46
*x+529),x)

[Out]

-exp((2*x + (x + 8)*log(2) + 46)/(x + 4*log(2) + 23))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {e^{\frac {46+2 x+(8+x) \log (2)}{23+x+4 \log (2)}} \left (-23 \log (2)-4 \log ^2(2)\right )}{529+46 x+x^2+(184+8 x) \log (2)+16 \log ^2(2)} \, dx=-2 \, e^{\left (-\frac {4 \, \log \left (2\right )^{2}}{x + 4 \, \log \left (2\right ) + 23} - \frac {23 \, \log \left (2\right )}{x + 4 \, \log \left (2\right ) + 23} + 2\right )} \]

[In]

integrate((-4*log(2)^2-23*log(2))*exp(((x+8)*log(2)+2*x+46)/(4*log(2)+x+23))/(16*log(2)^2+(8*x+184)*log(2)+x^2
+46*x+529),x, algorithm="maxima")

[Out]

-2*e^(-4*log(2)^2/(x + 4*log(2) + 23) - 23*log(2)/(x + 4*log(2) + 23) + 2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (25) = 50\).

Time = 0.29 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.83 \[ \int \frac {e^{\frac {46+2 x+(8+x) \log (2)}{23+x+4 \log (2)}} \left (-23 \log (2)-4 \log ^2(2)\right )}{529+46 x+x^2+(184+8 x) \log (2)+16 \log ^2(2)} \, dx=-e^{\left (\frac {x \log \left (2\right )}{x + 4 \, \log \left (2\right ) + 23} + \frac {2 \, x}{x + 4 \, \log \left (2\right ) + 23} + \frac {8 \, \log \left (2\right )}{x + 4 \, \log \left (2\right ) + 23} + \frac {46}{x + 4 \, \log \left (2\right ) + 23}\right )} \]

[In]

integrate((-4*log(2)^2-23*log(2))*exp(((x+8)*log(2)+2*x+46)/(4*log(2)+x+23))/(16*log(2)^2+(8*x+184)*log(2)+x^2
+46*x+529),x, algorithm="giac")

[Out]

-e^(x*log(2)/(x + 4*log(2) + 23) + 2*x/(x + 4*log(2) + 23) + 8*log(2)/(x + 4*log(2) + 23) + 46/(x + 4*log(2) +
23))

Mupad [B] (verification not implemented)

Time = 12.57 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {e^{\frac {46+2 x+(8+x) \log (2)}{23+x+4 \log (2)}} \left (-23 \log (2)-4 \log ^2(2)\right )}{529+46 x+x^2+(184+8 x) \log (2)+16 \log ^2(2)} \, dx=-2^{\frac {x+8}{x+\ln \left (16\right )+23}}\,{\mathrm {e}}^{\frac {2\,x}{x+\ln \left (16\right )+23}+\frac {46}{x+\ln \left (16\right )+23}} \]

[In]

int(-(exp((2*x + log(2)*(x + 8) + 46)/(x + 4*log(2) + 23))*(23*log(2) + 4*log(2)^2))/(46*x + log(2)*(8*x + 184
) + 16*log(2)^2 + x^2 + 529),x)

[Out]

-2^((x + 8)/(x + log(16) + 23))*exp((2*x)/(x + log(16) + 23) + 46/(x + log(16) + 23))