3.88.8 \(\int \frac {e^{-2 x} (-8 e^{\frac {4 e^{-2 x}}{\log (5)}}+2 e^{2 x} x \log (5) \log (\log (2)))}{\log (5)} \, dx\)

Optimal. Leaf size=21 \[ e^{\frac {4 e^{-2 x}}{\log (5)}}+x^2 \log (\log (2)) \]

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Rubi [A]  time = 0.18, antiderivative size = 30, normalized size of antiderivative = 1.43, number of steps used = 5, number of rules used = 4, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {12, 6742, 2282, 2194} \begin {gather*} \frac {x^2 \log (25) \log (\log (2))}{2 \log (5)}+e^{\frac {4 e^{-2 x}}{\log (5)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-8*E^(4/(E^(2*x)*Log[5])) + 2*E^(2*x)*x*Log[5]*Log[Log[2]])/(E^(2*x)*Log[5]),x]

[Out]

E^(4/(E^(2*x)*Log[5])) + (x^2*Log[25]*Log[Log[2]])/(2*Log[5])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, 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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int e^{-2 x} \left (-8 e^{\frac {4 e^{-2 x}}{\log (5)}}+2 e^{2 x} x \log (5) \log (\log (2))\right ) \, dx}{\log (5)}\\ &=\frac {\int \left (-8 e^{-2 x+\frac {4 e^{-2 x}}{\log (5)}}+x \log (25) \log (\log (2))\right ) \, dx}{\log (5)}\\ &=\frac {x^2 \log (25) \log (\log (2))}{2 \log (5)}-\frac {8 \int e^{-2 x+\frac {4 e^{-2 x}}{\log (5)}} \, dx}{\log (5)}\\ &=\frac {x^2 \log (25) \log (\log (2))}{2 \log (5)}+\frac {4 \operatorname {Subst}\left (\int e^{\frac {4 x}{\log (5)}} \, dx,x,e^{-2 x}\right )}{\log (5)}\\ &=e^{\frac {4 e^{-2 x}}{\log (5)}}+\frac {x^2 \log (25) \log (\log (2))}{2 \log (5)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 21, normalized size = 1.00 \begin {gather*} e^{\frac {4 e^{-2 x}}{\log (5)}}+x^2 \log (\log (2)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8*E^(4/(E^(2*x)*Log[5])) + 2*E^(2*x)*x*Log[5]*Log[Log[2]])/(E^(2*x)*Log[5]),x]

[Out]

E^(4/(E^(2*x)*Log[5])) + x^2*Log[Log[2]]

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fricas [A]  time = 0.49, size = 19, normalized size = 0.90 \begin {gather*} x^{2} \log \left (\log \relax (2)\right ) + e^{\left (\frac {4 \, e^{\left (-2 \, x\right )}}{\log \relax (5)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*exp(4/log(5)/exp(x)^2)+2*x*log(5)*exp(x)^2*log(log(2)))/log(5)/exp(x)^2,x, algorithm="fricas")

[Out]

x^2*log(log(2)) + e^(4*e^(-2*x)/log(5))

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giac [B]  time = 0.12, size = 44, normalized size = 2.10 \begin {gather*} \frac {{\left (x^{2} e^{\left (-2 \, x\right )} \log \relax (5) \log \left (\log \relax (2)\right ) + e^{\left (-\frac {2 \, {\left (x \log \relax (5) - 2 \, e^{\left (-2 \, x\right )}\right )}}{\log \relax (5)}\right )} \log \relax (5)\right )} e^{\left (2 \, x\right )}}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*exp(4/log(5)/exp(x)^2)+2*x*log(5)*exp(x)^2*log(log(2)))/log(5)/exp(x)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 20, normalized size = 0.95

method	result	size
risch	\({\mathrm e}^{\frac {4 \,{\mathrm e}^{-2 x}}{\ln \relax (5)}}+x^{2} \ln \left (\ln \relax (2)\right )\)	\(20\)
default	\(\frac {x^{2} \ln \relax (5) \ln \left (\ln \relax (2)\right )+{\mathrm e}^{\frac {4 \,{\mathrm e}^{-2 x}}{\ln \relax (5)}} \ln \relax (5)}{\ln \relax (5)}\)	\(34\)
norman	\(\left ({\mathrm e}^{2 x} {\mathrm e}^{\frac {4 \,{\mathrm e}^{-2 x}}{\ln \relax (5)}}+x^{2} \ln \left (\ln \relax (2)\right ) {\mathrm e}^{2 x}\right ) {\mathrm e}^{-2 x}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-8*exp(4/ln(5)/exp(x)^2)+2*x*ln(5)*exp(x)^2*ln(ln(2)))/ln(5)/exp(x)^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.44, size = 29, normalized size = 1.38 \begin {gather*} \frac {x^{2} \log \relax (5) \log \left (\log \relax (2)\right ) + e^{\left (\frac {4 \, e^{\left (-2 \, x\right )}}{\log \relax (5)}\right )} \log \relax (5)}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*exp(4/log(5)/exp(x)^2)+2*x*log(5)*exp(x)^2*log(log(2)))/log(5)/exp(x)^2,x, algorithm="maxima")

[Out]

(x^2*log(5)*log(log(2)) + e^(4*e^(-2*x)/log(5))*log(5))/log(5)

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mupad [B]  time = 5.47, size = 19, normalized size = 0.90 \begin {gather*} {\mathrm {e}}^{\frac {4\,{\mathrm {e}}^{-2\,x}}{\ln \relax (5)}}+x^2\,\ln \left (\ln \relax (2)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-2*x)*(8*exp((4*exp(-2*x))/log(5)) - 2*x*exp(2*x)*log(5)*log(log(2))))/log(5),x)

[Out]

exp((4*exp(-2*x))/log(5)) + x^2*log(log(2))

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sympy [A]  time = 0.16, size = 19, normalized size = 0.90 \begin {gather*} x^{2} \log {\left (\log {\relax (2 )} \right )} + e^{\frac {4 e^{- 2 x}}{\log {\relax (5 )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*exp(4/ln(5)/exp(x)**2)+2*x*ln(5)*exp(x)**2*ln(ln(2)))/ln(5)/exp(x)**2,x)

[Out]

x**2*log(log(2)) + exp(4*exp(-2*x)/log(5))

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