3.101.57 \(\int \frac {e^{\frac {e^{8 x} (-3+\log (2)+e^x \log (2))-\log (2) \log (5)}{-3+\log (2)+e^x \log (2)}} (e^{8 x} (72-48 \log (2)+8 \log ^2(2)+8 e^{2 x} \log ^2(2)+e^x (-48 \log (2)+16 \log ^2(2)))+e^x \log ^2(2) \log (5))}{9-6 \log (2)+\log ^2(2)+e^{2 x} \log ^2(2)+e^x (-6 \log (2)+2 \log ^2(2))} \, dx\)

Optimal. Leaf size=26 \[ e^{e^{8 x}+\frac {\log (5)}{-1-e^x+\frac {3}{\log (2)}}} \]

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Rubi [A]  time = 1.12, antiderivative size = 28, normalized size of antiderivative = 1.08, number of steps used = 2, number of rules used = 2, integrand size = 131, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.015, Rules used = {2282, 6706} \begin {gather*} e^{e^{8 x}} 2^{\frac {\log (5)}{-e^x \log (2)+3-\log (2)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((E^(8*x)*(-3 + Log[2] + E^x*Log[2]) - Log[2]*Log[5])/(-3 + Log[2] + E^x*Log[2]))*(E^(8*x)*(72 - 48*Log
[2] + 8*Log[2]^2 + 8*E^(2*x)*Log[2]^2 + E^x*(-48*Log[2] + 16*Log[2]^2)) + E^x*Log[2]^2*Log[5]))/(9 - 6*Log[2]
+ Log[2]^2 + E^(2*x)*Log[2]^2 + E^x*(-6*Log[2] + 2*Log[2]^2)),x]

[Out]

2^(Log[5]/(3 - Log[2] - E^x*Log[2]))*E^E^(8*x)

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\operatorname {Subst}\left (\int \frac {e^{x^8-\frac {\log (2) \log (5)}{-3+\log (2)+x \log (2)}} \left (8 x^7 (-3+\log (2))^2+16 x^8 (-3+\log (2)) \log (2)+8 x^9 \log ^2(2)+\log ^2(2) \log (5)\right )}{(3-\log (2)-x \log (2))^2} \, dx,x,e^x\right )\\ &=2^{\frac {\log (5)}{3-\log (2)-e^x \log (2)}} e^{e^{8 x}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 26, normalized size = 1.00 \begin {gather*} e^{e^{8 x}-\frac {\log (2) \log (5)}{-3+\log (2)+e^x \log (2)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((E^(8*x)*(-3 + Log[2] + E^x*Log[2]) - Log[2]*Log[5])/(-3 + Log[2] + E^x*Log[2]))*(E^(8*x)*(72 -
48*Log[2] + 8*Log[2]^2 + 8*E^(2*x)*Log[2]^2 + E^x*(-48*Log[2] + 16*Log[2]^2)) + E^x*Log[2]^2*Log[5]))/(9 - 6*L
og[2] + Log[2]^2 + E^(2*x)*Log[2]^2 + E^x*(-6*Log[2] + 2*Log[2]^2)),x]

[Out]

E^(E^(8*x) - (Log[2]*Log[5])/(-3 + Log[2] + E^x*Log[2]))

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fricas [A]  time = 0.90, size = 36, normalized size = 1.38 \begin {gather*} e^{\left (\frac {{\left (\log \relax (2) - 3\right )} e^{\left (8 \, x\right )} + e^{\left (9 \, x\right )} \log \relax (2) - \log \relax (5) \log \relax (2)}{e^{x} \log \relax (2) + \log \relax (2) - 3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*log(2)^2*exp(x)^2+(16*log(2)^2-48*log(2))*exp(x)+8*log(2)^2-48*log(2)+72)*exp(4*x)^2+log(2)^2*lo
g(5)*exp(x))*exp(((exp(x)*log(2)+log(2)-3)*exp(4*x)^2-log(2)*log(5))/(exp(x)*log(2)+log(2)-3))/(log(2)^2*exp(x
)^2+(2*log(2)^2-6*log(2))*exp(x)+log(2)^2-6*log(2)+9),x, algorithm="fricas")

[Out]

e^(((log(2) - 3)*e^(8*x) + e^(9*x)*log(2) - log(5)*log(2))/(e^x*log(2) + log(2) - 3))

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giac [B]  time = 0.73, size = 72, normalized size = 2.77 \begin {gather*} e^{\left (\frac {e^{\left (9 \, x\right )} \log \relax (2)}{e^{x} \log \relax (2) + \log \relax (2) - 3} + \frac {e^{\left (8 \, x\right )} \log \relax (2)}{e^{x} \log \relax (2) + \log \relax (2) - 3} - \frac {\log \relax (5) \log \relax (2)}{e^{x} \log \relax (2) + \log \relax (2) - 3} - \frac {3 \, e^{\left (8 \, x\right )}}{e^{x} \log \relax (2) + \log \relax (2) - 3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*log(2)^2*exp(x)^2+(16*log(2)^2-48*log(2))*exp(x)+8*log(2)^2-48*log(2)+72)*exp(4*x)^2+log(2)^2*lo
g(5)*exp(x))*exp(((exp(x)*log(2)+log(2)-3)*exp(4*x)^2-log(2)*log(5))/(exp(x)*log(2)+log(2)-3))/(log(2)^2*exp(x
)^2+(2*log(2)^2-6*log(2))*exp(x)+log(2)^2-6*log(2)+9),x, algorithm="giac")

[Out]

e^(e^(9*x)*log(2)/(e^x*log(2) + log(2) - 3) + e^(8*x)*log(2)/(e^x*log(2) + log(2) - 3) - log(5)*log(2)/(e^x*lo
g(2) + log(2) - 3) - 3*e^(8*x)/(e^x*log(2) + log(2) - 3))

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maple [A]  time = 0.21, size = 43, normalized size = 1.65

method	result	size
risch	\({\mathrm e}^{-\frac {\ln \relax (2) \ln \relax (5)-\ln \relax (2) {\mathrm e}^{8 x}-\ln \relax (2) {\mathrm e}^{9 x}+3 \,{\mathrm e}^{8 x}}{{\mathrm e}^{x} \ln \relax (2)+\ln \relax (2)-3}}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((8*ln(2)^2*exp(x)^2+(16*ln(2)^2-48*ln(2))*exp(x)+8*ln(2)^2-48*ln(2)+72)*exp(4*x)^2+ln(2)^2*ln(5)*exp(x))*
exp(((exp(x)*ln(2)+ln(2)-3)*exp(4*x)^2-ln(2)*ln(5))/(exp(x)*ln(2)+ln(2)-3))/(ln(2)^2*exp(x)^2+(2*ln(2)^2-6*ln(
2))*exp(x)+ln(2)^2-6*ln(2)+9),x,method=_RETURNVERBOSE)

[Out]

exp(-(ln(2)*ln(5)-ln(2)*exp(8*x)-ln(2)*exp(9*x)+3*exp(8*x))/(exp(x)*ln(2)+ln(2)-3))

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maxima [A]  time = 2.25, size = 23, normalized size = 0.88 \begin {gather*} e^{\left (-\frac {\log \relax (5) \log \relax (2)}{e^{x} \log \relax (2) + \log \relax (2) - 3} + e^{\left (8 \, x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*log(2)^2*exp(x)^2+(16*log(2)^2-48*log(2))*exp(x)+8*log(2)^2-48*log(2)+72)*exp(4*x)^2+log(2)^2*lo
g(5)*exp(x))*exp(((exp(x)*log(2)+log(2)-3)*exp(4*x)^2-log(2)*log(5))/(exp(x)*log(2)+log(2)-3))/(log(2)^2*exp(x
)^2+(2*log(2)^2-6*log(2))*exp(x)+log(2)^2-6*log(2)+9),x, algorithm="maxima")

[Out]

e^(-log(5)*log(2)/(e^x*log(2) + log(2) - 3) + e^(8*x))

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mupad [B]  time = 9.97, size = 47, normalized size = 1.81 \begin {gather*} {\mathrm {e}}^{\frac {\ln \relax (2)\,\left ({\mathrm {e}}^{8\,x}+{\mathrm {e}}^{9\,x}-\ln \relax (5)\right )}{\ln \relax (2)+{\mathrm {e}}^x\,\ln \relax (2)-3}}\,{\mathrm {e}}^{-\frac {3\,{\mathrm {e}}^{8\,x}}{\ln \relax (2)+{\mathrm {e}}^x\,\ln \relax (2)-3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((exp(8*x)*(log(2) + exp(x)*log(2) - 3) - log(2)*log(5))/(log(2) + exp(x)*log(2) - 3))*(exp(8*x)*(8*ex
p(2*x)*log(2)^2 - 48*log(2) - exp(x)*(48*log(2) - 16*log(2)^2) + 8*log(2)^2 + 72) + exp(x)*log(2)^2*log(5)))/(
exp(2*x)*log(2)^2 - 6*log(2) - exp(x)*(6*log(2) - 2*log(2)^2) + log(2)^2 + 9),x)

[Out]

exp((log(2)*(exp(8*x) + exp(9*x) - log(5)))/(log(2) + exp(x)*log(2) - 3))*exp(-(3*exp(8*x))/(log(2) + exp(x)*l
og(2) - 3))

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sympy [A]  time = 0.58, size = 36, normalized size = 1.38 \begin {gather*} e^{\frac {\left (e^{x} \log {\relax (2 )} - 3 + \log {\relax (2 )}\right ) e^{8 x} - \log {\relax (2 )} \log {\relax (5 )}}{e^{x} \log {\relax (2 )} - 3 + \log {\relax (2 )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*ln(2)**2*exp(x)**2+(16*ln(2)**2-48*ln(2))*exp(x)+8*ln(2)**2-48*ln(2)+72)*exp(4*x)**2+ln(2)**2*ln
(5)*exp(x))*exp(((exp(x)*ln(2)+ln(2)-3)*exp(4*x)**2-ln(2)*ln(5))/(exp(x)*ln(2)+ln(2)-3))/(ln(2)**2*exp(x)**2+(
2*ln(2)**2-6*ln(2))*exp(x)+ln(2)**2-6*ln(2)+9),x)

[Out]

exp(((exp(x)*log(2) - 3 + log(2))*exp(8*x) - log(2)*log(5))/(exp(x)*log(2) - 3 + log(2)))

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