3.70.82 \(\int \frac {e^8+4 e^{12}+(e^8+4 e^{12}) \log (x)-16 e^4 \log ^2(x)}{4 e^4 x^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=22 \[ \frac {4-\frac {e^4 \left (\frac {1}{4}+e^4\right )}{\log (x)}}{x} \]

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Rubi [A]  time = 0.37, antiderivative size = 27, normalized size of antiderivative = 1.23, number of steps used = 10, number of rules used = 6, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {12, 6741, 6742, 2306, 2309, 2178} \begin {gather*} \frac {4}{x}-\frac {e^4 \left (1+4 e^4\right )}{4 x \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^8 + 4*E^12 + (E^8 + 4*E^12)*Log[x] - 16*E^4*Log[x]^2)/(4*E^4*x^2*Log[x]^2),x]

[Out]

4/x - (E^4*(1 + 4*E^4))/(4*x*Log[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 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2306

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log
[c*x^n])^(p + 1))/(b*d*n*(p + 1)), x] - Dist[(m + 1)/(b*n*(p + 1)), Int[(d*x)^m*(a + b*Log[c*x^n])^(p + 1), x]
, x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1] && LtQ[p, -1]

Rule 2309

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

Rule 6741

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

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 \frac {e^8+4 e^{12}+\left (e^8+4 e^{12}\right ) \log (x)-16 e^4 \log ^2(x)}{x^2 \log ^2(x)} \, dx}{4 e^4}\\ &=\frac {\int \frac {e^4 \left (e^4 \left (1+4 e^4\right )+e^4 \left (1+4 e^4\right ) \log (x)-16 \log ^2(x)\right )}{x^2 \log ^2(x)} \, dx}{4 e^4}\\ &=\frac {1}{4} \int \frac {e^4 \left (1+4 e^4\right )+e^4 \left (1+4 e^4\right ) \log (x)-16 \log ^2(x)}{x^2 \log ^2(x)} \, dx\\ &=\frac {1}{4} \int \left (-\frac {16}{x^2}+\frac {e^4+4 e^8}{x^2 \log ^2(x)}+\frac {e^4+4 e^8}{x^2 \log (x)}\right ) \, dx\\ &=\frac {4}{x}+\frac {1}{4} \left (e^4 \left (1+4 e^4\right )\right ) \int \frac {1}{x^2 \log ^2(x)} \, dx+\frac {1}{4} \left (e^4 \left (1+4 e^4\right )\right ) \int \frac {1}{x^2 \log (x)} \, dx\\ &=\frac {4}{x}-\frac {e^4 \left (1+4 e^4\right )}{4 x \log (x)}-\frac {1}{4} \left (e^4 \left (1+4 e^4\right )\right ) \int \frac {1}{x^2 \log (x)} \, dx+\frac {1}{4} \left (e^4 \left (1+4 e^4\right )\right ) \operatorname {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log (x)\right )\\ &=\frac {4}{x}+\frac {1}{4} e^4 \left (1+4 e^4\right ) \text {Ei}(-\log (x))-\frac {e^4 \left (1+4 e^4\right )}{4 x \log (x)}-\frac {1}{4} \left (e^4 \left (1+4 e^4\right )\right ) \operatorname {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log (x)\right )\\ &=\frac {4}{x}-\frac {e^4 \left (1+4 e^4\right )}{4 x \log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 32, normalized size = 1.45 \begin {gather*} \frac {4}{x}-\frac {e^4}{4 x \log (x)}-\frac {e^8}{x \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^8 + 4*E^12 + (E^8 + 4*E^12)*Log[x] - 16*E^4*Log[x]^2)/(4*E^4*x^2*Log[x]^2),x]

[Out]

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

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fricas [A]  time = 1.25, size = 20, normalized size = 0.91 \begin {gather*} -\frac {4 \, e^{8} + e^{4} - 16 \, \log \relax (x)}{4 \, x \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-16*exp(2)^2*log(x)^2+(4*exp(4)^3+exp(4)^2)*log(x)+4*exp(4)^3+exp(4)^2)/x^2/exp(2)^2/log(x)^2,x
, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 26, normalized size = 1.18 \begin {gather*} \frac {{\left (16 \, e^{4} \log \relax (x) - 4 \, e^{12} - e^{8}\right )} e^{\left (-4\right )}}{4 \, x \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-16*exp(2)^2*log(x)^2+(4*exp(4)^3+exp(4)^2)*log(x)+4*exp(4)^3+exp(4)^2)/x^2/exp(2)^2/log(x)^2,x
, algorithm="giac")

[Out]

1/4*(16*e^4*log(x) - 4*e^12 - e^8)*e^(-4)/(x*log(x))

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maple [A]  time = 0.04, size = 24, normalized size = 1.09

method	result	size
risch	\(\frac {4}{x}-\frac {{\mathrm e}^{4} \left (4 \,{\mathrm e}^{4}+1\right )}{4 x \ln \relax (x )}\)	\(24\)
norman	\(\frac {\left (4 \,{\mathrm e}^{2} \ln \relax (x )-\frac {{\mathrm e}^{8} \left (4 \,{\mathrm e}^{4}+1\right ) {\mathrm e}^{-2}}{4}\right ) {\mathrm e}^{-2}}{x \ln \relax (x )}\)	\(36\)
default	\(\frac {{\mathrm e}^{-4} \left (\frac {16 \,{\mathrm e}^{4}}{x}-4 \,{\mathrm e}^{12} \expIntegralEi \left (1, \ln \relax (x )\right )-{\mathrm e}^{8} \expIntegralEi \left (1, \ln \relax (x )\right )+4 \,{\mathrm e}^{12} \left (-\frac {1}{x \ln \relax (x )}+\expIntegralEi \left (1, \ln \relax (x )\right )\right )+{\mathrm e}^{8} \left (-\frac {1}{x \ln \relax (x )}+\expIntegralEi \left (1, \ln \relax (x )\right )\right )\right )}{4}\)	\(76\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(-16*exp(2)^2*ln(x)^2+(4*exp(4)^3+exp(4)^2)*ln(x)+4*exp(4)^3+exp(4)^2)/x^2/exp(2)^2/ln(x)^2,x,method=_
RETURNVERBOSE)

[Out]

4/x-1/4*exp(4)/x*(4*exp(4)+1)/ln(x)

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maxima [C]  time = 0.43, size = 45, normalized size = 2.05 \begin {gather*} \frac {1}{4} \, {\left (4 \, {\rm Ei}\left (-\log \relax (x)\right ) e^{12} + {\rm Ei}\left (-\log \relax (x)\right ) e^{8} - 4 \, e^{12} \Gamma \left (-1, \log \relax (x)\right ) - e^{8} \Gamma \left (-1, \log \relax (x)\right ) + \frac {16 \, e^{4}}{x}\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-16*exp(2)^2*log(x)^2+(4*exp(4)^3+exp(4)^2)*log(x)+4*exp(4)^3+exp(4)^2)/x^2/exp(2)^2/log(x)^2,x
, algorithm="maxima")

[Out]

1/4*(4*Ei(-log(x))*e^12 + Ei(-log(x))*e^8 - 4*e^12*gamma(-1, log(x)) - e^8*gamma(-1, log(x)) + 16*e^4/x)*e^(-4
)

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mupad [B]  time = 4.08, size = 22, normalized size = 1.00 \begin {gather*} \frac {4}{x}-\frac {{\mathrm {e}}^4+4\,{\mathrm {e}}^8}{4\,x\,\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-4)*(exp(8)/4 + exp(12) - 4*exp(4)*log(x)^2 + (log(x)*(exp(8) + 4*exp(12)))/4))/(x^2*log(x)^2),x)

[Out]

4/x - (exp(4) + 4*exp(8))/(4*x*log(x))

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sympy [A]  time = 0.10, size = 19, normalized size = 0.86 \begin {gather*} \frac {4}{x} + \frac {- 4 e^{8} - e^{4}}{4 x \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-16*exp(2)**2*ln(x)**2+(4*exp(4)**3+exp(4)**2)*ln(x)+4*exp(4)**3+exp(4)**2)/x**2/exp(2)**2/ln(x
)**2,x)

[Out]

4/x + (-4*exp(8) - exp(4))/(4*x*log(x))

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