3.92.89 \(\int -\frac {40 e^{e^{\frac {5}{4 \log ^2(x^2)}}+\frac {5}{4 \log ^2(x^2)}}}{x \log ^3(x^2)} \, dx\)

Optimal. Leaf size=16 \[ 8 e^{e^{\frac {5}{4 \log ^2\left (x^2\right )}}} \]

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Rubi [A]  time = 0.15, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {12, 6715, 2282, 2194} \begin {gather*} 8 e^{e^{\frac {5}{4 \log ^2\left (x^2\right )}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-40*E^(E^(5/(4*Log[x^2]^2)) + 5/(4*Log[x^2]^2)))/(x*Log[x^2]^3),x]

[Out]

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

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 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left (40 \int \frac {e^{e^{\frac {5}{4 \log ^2\left (x^2\right )}}+\frac {5}{4 \log ^2\left (x^2\right )}}}{x \log ^3\left (x^2\right )} \, dx\right )\\ &=-\left (20 \operatorname {Subst}\left (\int \frac {e^{e^{\frac {5}{4 x^2}}+\frac {5}{4 x^2}}}{x^3} \, dx,x,\log \left (x^2\right )\right )\right )\\ &=10 \operatorname {Subst}\left (\int e^{e^{5 x/4}+\frac {5 x}{4}} \, dx,x,\frac {1}{\log ^2\left (x^2\right )}\right )\\ &=8 \operatorname {Subst}\left (\int e^x \, dx,x,e^{\frac {5}{4 \log ^2\left (x^2\right )}}\right )\\ &=8 e^{e^{\frac {5}{4 \log ^2\left (x^2\right )}}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 16, normalized size = 1.00 \begin {gather*} 8 e^{e^{\frac {5}{4 \log ^2\left (x^2\right )}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-40*E^(E^(5/(4*Log[x^2]^2)) + 5/(4*Log[x^2]^2)))/(x*Log[x^2]^3),x]

[Out]

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

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fricas [B]  time = 1.35, size = 47, normalized size = 2.94 \begin {gather*} e^{\left (\frac {4 \, e^{\left (\frac {5}{4 \, \log \left (x^{2}\right )^{2}}\right )} \log \left (x^{2}\right )^{2} + 12 \, \log \relax (2) \log \left (x^{2}\right )^{2} + 5}{4 \, \log \left (x^{2}\right )^{2}} - \frac {5}{4 \, \log \left (x^{2}\right )^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5*exp(5/4/log(x^2)^2)*exp(exp(5/4/log(x^2)^2)+3*log(2))/x/log(x^2)^3,x, algorithm="fricas")

[Out]

e^(1/4*(4*e^(5/4/log(x^2)^2)*log(x^2)^2 + 12*log(2)*log(x^2)^2 + 5)/log(x^2)^2 - 5/4/log(x^2)^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5*exp(5/4/log(x^2)^2)*exp(exp(5/4/log(x^2)^2)+3*log(2))/x/log(x^2)^3,x, algorithm="giac")

[Out]

undef

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maple [A]  time = 0.02, size = 13, normalized size = 0.81

method	result	size
risch	\(8 \,{\mathrm e}^{{\mathrm e}^{\frac {5}{4 \ln \left (x^{2}\right )^{2}}}}\)	\(13\)
derivativedivides	\({\mathrm e}^{{\mathrm e}^{\frac {5}{4 \ln \left (x^{2}\right )^{2}}}+3 \ln \relax (2)}\)	\(16\)
default	\({\mathrm e}^{{\mathrm e}^{\frac {5}{4 \ln \left (x^{2}\right )^{2}}}+3 \ln \relax (2)}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-5*exp(5/4/ln(x^2)^2)*exp(exp(5/4/ln(x^2)^2)+3*ln(2))/x/ln(x^2)^3,x,method=_RETURNVERBOSE)

[Out]

8*exp(exp(5/4/ln(x^2)^2))

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maxima [A]  time = 0.37, size = 10, normalized size = 0.62 \begin {gather*} 8 \, e^{\left (e^{\left (\frac {5}{16 \, \log \relax (x)^{2}}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5*exp(5/4/log(x^2)^2)*exp(exp(5/4/log(x^2)^2)+3*log(2))/x/log(x^2)^3,x, algorithm="maxima")

[Out]

8*e^(e^(5/16/log(x)^2))

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mupad [B]  time = 7.60, size = 12, normalized size = 0.75 \begin {gather*} 8\,{\mathrm {e}}^{{\mathrm {e}}^{\frac {5}{4\,{\ln \left (x^2\right )}^2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*exp(5/(4*log(x^2)^2))*exp(exp(5/(4*log(x^2)^2)) + 3*log(2)))/(x*log(x^2)^3),x)

[Out]

8*exp(exp(5/(4*log(x^2)^2)))

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sympy [A]  time = 0.41, size = 14, normalized size = 0.88 \begin {gather*} 8 e^{e^{\frac {5}{4 \log {\left (x^{2} \right )}^{2}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5*exp(5/4/ln(x**2)**2)*exp(exp(5/4/ln(x**2)**2)+3*ln(2))/x/ln(x**2)**3,x)

[Out]

8*exp(exp(5/(4*log(x**2)**2)))

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