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

Optimal. Leaf size=25 \[ e^{4 \left (-1+\frac {2}{x}+\frac {e^{2 x}}{\frac {1}{16}+\log (2)}\right )} \]

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Rubi [A]  time = 3.44, antiderivative size = 49, normalized size of antiderivative = 1.96, number of steps used = 5, number of rules used = 4, integrand size = 63, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.063, Rules used = {6, 12, 6741, 6706} \begin {gather*} 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))}} \end {gather*}

Antiderivative was successfully verified.

[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 6706

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 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 1.09, size = 24, normalized size = 0.96 \begin {gather*} e^{-4+\frac {8}{x}+\frac {64 e^{2 x}}{1+16 \log (2)}} \end {gather*}

Antiderivative was successfully verified.

[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]

E^(-4 + 8/x + (64*E^(2*x))/(1 + 16*Log[2]))

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fricas [A]  time = 0.56, size = 31, normalized size = 1.24 \begin {gather*} e^{\left (\frac {4 \, {\left (16 \, x e^{\left (2 \, x\right )} - 16 \, {\left (x - 2\right )} \log \relax (2) - x + 2\right )}}{16 \, x \log \relax (2) + x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [B]  time = 0.27, size = 68, normalized size = 2.72 \begin {gather*} e^{\left (\frac {64 \, x e^{\left (2 \, x\right )}}{16 \, x \log \relax (2) + x} - \frac {64 \, x \log \relax (2)}{16 \, x \log \relax (2) + x} - \frac {4 \, x}{16 \, x \log \relax (2) + x} + \frac {128 \, \log \relax (2)}{16 \, x \log \relax (2) + x} + \frac {8}{16 \, x \log \relax (2) + x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [A]  time = 0.19, size = 32, normalized size = 1.28

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

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maxima [B]  time = 0.59, size = 69, normalized size = 2.76 \begin {gather*} \frac {e^{\left (\frac {64 \, e^{\left (2 \, x\right )}}{16 \, \log \relax (2) + 1} - \frac {4}{16 \, \log \relax (2) + 1} + \frac {128 \, \log \relax (2)}{x {\left (16 \, \log \relax (2) + 1\right )}} + \frac {8}{x {\left (16 \, \log \relax (2) + 1\right )}}\right )}}{2^{\frac {64}{16 \, \log \relax (2) + 1}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [B]  time = 3.31, size = 58, normalized size = 2.32 \begin {gather*} {\left (\frac {1}{18446744073709551616}\right )}^{\frac {x-2}{x+16\,x\,\ln \relax (2)}}\,{\mathrm {e}}^{\frac {8}{x+16\,x\,\ln \relax (2)}}\,{\mathrm {e}}^{-\frac {4\,x}{x+16\,x\,\ln \relax (2)}}\,{\mathrm {e}}^{\frac {64\,x\,{\mathrm {e}}^{2\,x}}{x+16\,x\,\ln \relax (2)}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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sympy [A]  time = 0.30, size = 31, normalized size = 1.24 \begin {gather*} e^{\frac {64 x e^{2 x} - 4 x + \left (128 - 64 x\right ) \log {\relax (2 )} + 8}{x + 16 x \log {\relax (2 )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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