\(\int \frac {e^{\frac {8-4 x+64 e^{2 x} x+(128-64 x) \log (2)}{x+16 x \log (2)}} (-8+128 e^{2 x} x^2-128 \log (2))}{x^2+16 x^2 \log (2)} \, dx\) [4356]

Optimal result

Integrand size = 63, antiderivative size = 25 \[ \int \frac {e^{\frac {8-4 x+64 e^{2 x} x+(128-64 x) \log (2)}{x+16 x \log (2)}} \left (-8+128 e^{2 x} x^2-128 \log (2)\right )}{x^2+16 x^2 \log (2)} \, dx=e^{4 \left (-1+\frac {2}{x}+\frac {e^{2 x}}{\frac {1}{16}+\log (2)}\right )} \]

[Out]

exp(8/x+4*exp(x)^2/(1/16+ln(2))-4)

Rubi [A] (verified)

Time = 2.36 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.96, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.063, Rules used = {6, 12, 6873, 6838} \[ \int \frac {e^{\frac {8-4 x+64 e^{2 x} x+(128-64 x) \log (2)}{x+16 x \log (2)}} \left (-8+128 e^{2 x} x^2-128 \log (2)\right )}{x^2+16 x^2 \log (2)} \, dx=2^{\frac {64 (2-x)}{x (1+16 \log (2))}} e^{\frac {4 \left (16 e^{2 x} x-x+2\right )}{x (1+16 \log (2))}} \]

[In]

Int[(E^((8 - 4*x + 64*E^(2*x)*x + (128 - 64*x)*Log[2])/(x + 16*x*Log[2]))*(-8 + 128*E^(2*x)*x^2 - 128*Log[2]))
/(x^2 + 16*x^2*Log[2]),x]

[Out]

2^((64*(2 - x))/(x*(1 + 16*Log[2])))*E^((4*(2 - x + 16*E^(2*x)*x))/(x*(1 + 16*Log[2])))

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rule 6873

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (\frac {8-4 x+64 e^{2 x} x+(128-64 x) \log (2)}{x+16 x \log (2)}\right ) \left (-8+128 e^{2 x} x^2-128 \log (2)\right )}{x^2 (1+16 \log (2))} \, dx \\ & = \frac {\int \frac {\exp \left (\frac {8-4 x+64 e^{2 x} x+(128-64 x) \log (2)}{x+16 x \log (2)}\right ) \left (-8+128 e^{2 x} x^2-128 \log (2)\right )}{x^2} \, dx}{1+16 \log (2)} \\ & = \frac {\int \frac {8 \exp \left (\frac {8-4 x+64 e^{2 x} x+(128-64 x) \log (2)}{x (1+16 \log (2))}\right ) \left (-1+16 e^{2 x} x^2-16 \log (2)\right )}{x^2} \, dx}{1+16 \log (2)} \\ & = \frac {8 \int \frac {\exp \left (\frac {8-4 x+64 e^{2 x} x+(128-64 x) \log (2)}{x (1+16 \log (2))}\right ) \left (-1+16 e^{2 x} x^2-16 \log (2)\right )}{x^2} \, dx}{1+16 \log (2)} \\ & = 2^{\frac {64 (2-x)}{x (1+16 \log (2))}} e^{\frac {4 \left (2-x+16 e^{2 x} x\right )}{x (1+16 \log (2))}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(51\) vs. \(2(25)=50\).

Time = 1.02 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.04 \[ \int \frac {e^{\frac {8-4 x+64 e^{2 x} x+(128-64 x) \log (2)}{x+16 x \log (2)}} \left (-8+128 e^{2 x} x^2-128 \log (2)\right )}{x^2+16 x^2 \log (2)} \, dx=2^{3+\frac {128-x (67+48 \log (2))}{x+16 x \log (2)}} e^{\frac {8+\left (-4+64 e^{2 x}\right ) x}{x+16 x \log (2)}} \]

[In]

Integrate[(E^((8 - 4*x + 64*E^(2*x)*x + (128 - 64*x)*Log[2])/(x + 16*x*Log[2]))*(-8 + 128*E^(2*x)*x^2 - 128*Lo
g[2]))/(x^2 + 16*x^2*Log[2]),x]

[Out]

2^(3 + (128 - x*(67 + 48*Log[2]))/(x + 16*x*Log[2]))*E^((8 + (-4 + 64*E^(2*x))*x)/(x + 16*x*Log[2]))

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28

method	result	size
norman	\({\mathrm e}^{\frac {64 x \,{\mathrm e}^{2 x}+\left (-64 x +128\right ) \ln \left (2\right )-4 x +8}{16 x \ln \left (2\right )+x}}\)	\(32\)
risch	\({\mathrm e}^{-\frac {4 \left (-16 x \,{\mathrm e}^{2 x}+16 x \ln \left (2\right )-32 \ln \left (2\right )+x -2\right )}{x \left (16 \ln \left (2\right )+1\right )}}\)	\(34\)
parallelrisch	\(\frac {16 \,{\mathrm e}^{\frac {64 x \,{\mathrm e}^{2 x}+\left (-64 x +128\right ) \ln \left (2\right )-4 x +8}{x \left (16 \ln \left (2\right )+1\right )}} \ln \left (2\right )+{\mathrm e}^{\frac {64 x \,{\mathrm e}^{2 x}+\left (-64 x +128\right ) \ln \left (2\right )-4 x +8}{x \left (16 \ln \left (2\right )+1\right )}}}{16 \ln \left (2\right )+1}\)	\(81\)

[In]

int((128*exp(x)^2*x^2-128*ln(2)-8)*exp((64*x*exp(x)^2+(-64*x+128)*ln(2)-4*x+8)/(16*x*ln(2)+x))/(16*x^2*ln(2)+x
^2),x,method=_RETURNVERBOSE)

[Out]

exp((64*x*exp(x)^2+(-64*x+128)*ln(2)-4*x+8)/(16*x*ln(2)+x))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {e^{\frac {8-4 x+64 e^{2 x} x+(128-64 x) \log (2)}{x+16 x \log (2)}} \left (-8+128 e^{2 x} x^2-128 \log (2)\right )}{x^2+16 x^2 \log (2)} \, dx=e^{\left (\frac {4 \, {\left (16 \, x e^{\left (2 \, x\right )} - 16 \, {\left (x - 2\right )} \log \left (2\right ) - x + 2\right )}}{16 \, x \log \left (2\right ) + x}\right )} \]

[In]

integrate((128*exp(x)^2*x^2-128*log(2)-8)*exp((64*x*exp(x)^2+(-64*x+128)*log(2)-4*x+8)/(16*x*log(2)+x))/(16*x^
2*log(2)+x^2),x, algorithm="fricas")

[Out]

e^(4*(16*x*e^(2*x) - 16*(x - 2)*log(2) - x + 2)/(16*x*log(2) + x))

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {e^{\frac {8-4 x+64 e^{2 x} x+(128-64 x) \log (2)}{x+16 x \log (2)}} \left (-8+128 e^{2 x} x^2-128 \log (2)\right )}{x^2+16 x^2 \log (2)} \, dx=e^{\frac {64 x e^{2 x} - 4 x + \left (128 - 64 x\right ) \log {\left (2 \right )} + 8}{x + 16 x \log {\left (2 \right )}}} \]

[In]

integrate((128*exp(x)**2*x**2-128*ln(2)-8)*exp((64*x*exp(x)**2+(-64*x+128)*ln(2)-4*x+8)/(16*x*ln(2)+x))/(16*x*
*2*ln(2)+x**2),x)

[Out]

exp((64*x*exp(2*x) - 4*x + (128 - 64*x)*log(2) + 8)/(x + 16*x*log(2)))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (22) = 44\).

Time = 0.38 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.76 \[ \int \frac {e^{\frac {8-4 x+64 e^{2 x} x+(128-64 x) \log (2)}{x+16 x \log (2)}} \left (-8+128 e^{2 x} x^2-128 \log (2)\right )}{x^2+16 x^2 \log (2)} \, dx=\frac {e^{\left (\frac {64 \, e^{\left (2 \, x\right )}}{16 \, \log \left (2\right ) + 1} - \frac {4}{16 \, \log \left (2\right ) + 1} + \frac {128 \, \log \left (2\right )}{x {\left (16 \, \log \left (2\right ) + 1\right )}} + \frac {8}{x {\left (16 \, \log \left (2\right ) + 1\right )}}\right )}}{2^{\frac {64}{16 \, \log \left (2\right ) + 1}}} \]

[In]

integrate((128*exp(x)^2*x^2-128*log(2)-8)*exp((64*x*exp(x)^2+(-64*x+128)*log(2)-4*x+8)/(16*x*log(2)+x))/(16*x^
2*log(2)+x^2),x, algorithm="maxima")

[Out]

e^(64*e^(2*x)/(16*log(2) + 1) - 4/(16*log(2) + 1) + 128*log(2)/(x*(16*log(2) + 1)) + 8/(x*(16*log(2) + 1)))/2^
(64/(16*log(2) + 1))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (22) = 44\).

Time = 0.29 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.72 \[ \int \frac {e^{\frac {8-4 x+64 e^{2 x} x+(128-64 x) \log (2)}{x+16 x \log (2)}} \left (-8+128 e^{2 x} x^2-128 \log (2)\right )}{x^2+16 x^2 \log (2)} \, dx=e^{\left (\frac {64 \, x e^{\left (2 \, x\right )}}{16 \, x \log \left (2\right ) + x} - \frac {64 \, x \log \left (2\right )}{16 \, x \log \left (2\right ) + x} - \frac {4 \, x}{16 \, x \log \left (2\right ) + x} + \frac {128 \, \log \left (2\right )}{16 \, x \log \left (2\right ) + x} + \frac {8}{16 \, x \log \left (2\right ) + x}\right )} \]

[In]

integrate((128*exp(x)^2*x^2-128*log(2)-8)*exp((64*x*exp(x)^2+(-64*x+128)*log(2)-4*x+8)/(16*x*log(2)+x))/(16*x^
2*log(2)+x^2),x, algorithm="giac")

[Out]

e^(64*x*e^(2*x)/(16*x*log(2) + x) - 64*x*log(2)/(16*x*log(2) + x) - 4*x/(16*x*log(2) + x) + 128*log(2)/(16*x*l
og(2) + x) + 8/(16*x*log(2) + x))

Mupad [B] (verification not implemented)

Time = 11.22 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.32 \[ \int \frac {e^{\frac {8-4 x+64 e^{2 x} x+(128-64 x) \log (2)}{x+16 x \log (2)}} \left (-8+128 e^{2 x} x^2-128 \log (2)\right )}{x^2+16 x^2 \log (2)} \, dx={\left (\frac {1}{18446744073709551616}\right )}^{\frac {x-2}{x+16\,x\,\ln \left (2\right )}}\,{\mathrm {e}}^{\frac {8}{x+16\,x\,\ln \left (2\right )}}\,{\mathrm {e}}^{-\frac {4\,x}{x+16\,x\,\ln \left (2\right )}}\,{\mathrm {e}}^{\frac {64\,x\,{\mathrm {e}}^{2\,x}}{x+16\,x\,\ln \left (2\right )}} \]

[In]

int(-(exp(-(4*x + log(2)*(64*x - 128) - 64*x*exp(2*x) - 8)/(x + 16*x*log(2)))*(128*log(2) - 128*x^2*exp(2*x) +
8))/(16*x^2*log(2) + x^2),x)

[Out]

(1/18446744073709551616)^((x - 2)/(x + 16*x*log(2)))*exp(8/(x + 16*x*log(2)))*exp(-(4*x)/(x + 16*x*log(2)))*ex
p((64*x*exp(2*x))/(x + 16*x*log(2)))