3.89.35 \(\int \frac {-2+2 e^{25 e^2}+2 \log (x)}{e^{50 e^2}+2 e^{25 e^2} \log (x)+\log ^2(x)} \, dx\)

Optimal. Leaf size=17 \[ -1+\frac {2 x}{e^{25 e^2}+\log (x)} \]

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Rubi [A]  time = 0.18, antiderivative size = 15, normalized size of antiderivative = 0.88, number of steps used = 9, number of rules used = 6, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {6688, 12, 2360, 2297, 2299, 2178} \begin {gather*} \frac {2 x}{\log (x)+e^{25 e^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2 + 2*E^(25*E^2) + 2*Log[x])/(E^(50*E^2) + 2*E^(25*E^2)*Log[x] + Log[x]^2),x]

[Out]

(2*x)/(E^(25*E^2) + 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 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2299

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

Rule 2360

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(Log[(c_.)*(x_)^(n_.)]*(e_.) + (d_))^(q_.), x_Symbol] :> Int[E
xpandIntegrand[(a + b*Log[c*x^n])^p*(d + e*Log[c*x^n])^q, x], x] /; FreeQ[{a, b, c, d, e, n}, x] && IntegerQ[p
] && IntegerQ[q]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \left (-1+e^{25 e^2}+\log (x)\right )}{\left (e^{25 e^2}+\log (x)\right )^2} \, dx\\ &=2 \int \frac {-1+e^{25 e^2}+\log (x)}{\left (e^{25 e^2}+\log (x)\right )^2} \, dx\\ &=2 \int \left (-\frac {1}{\left (e^{25 e^2}+\log (x)\right )^2}+\frac {1}{e^{25 e^2}+\log (x)}\right ) \, dx\\ &=-\left (2 \int \frac {1}{\left (e^{25 e^2}+\log (x)\right )^2} \, dx\right )+2 \int \frac {1}{e^{25 e^2}+\log (x)} \, dx\\ &=\frac {2 x}{e^{25 e^2}+\log (x)}-2 \int \frac {1}{e^{25 e^2}+\log (x)} \, dx+2 \operatorname {Subst}\left (\int \frac {e^x}{e^{25 e^2}+x} \, dx,x,\log (x)\right )\\ &=2 e^{-e^{25 e^2}} \text {Ei}\left (e^{25 e^2}+\log (x)\right )+\frac {2 x}{e^{25 e^2}+\log (x)}-2 \operatorname {Subst}\left (\int \frac {e^x}{e^{25 e^2}+x} \, dx,x,\log (x)\right )\\ &=\frac {2 x}{e^{25 e^2}+\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 15, normalized size = 0.88 \begin {gather*} \frac {2 x}{e^{25 e^2}+\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 + 2*E^(25*E^2) + 2*Log[x])/(E^(50*E^2) + 2*E^(25*E^2)*Log[x] + Log[x]^2),x]

[Out]

(2*x)/(E^(25*E^2) + Log[x])

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fricas [A]  time = 0.58, size = 13, normalized size = 0.76 \begin {gather*} \frac {2 \, x}{e^{\left (25 \, e^{2}\right )} + \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(x)+2*exp(25/4*exp(1)^2)^4-2)/(log(x)^2+2*exp(25/4*exp(1)^2)^4*log(x)+exp(25/4*exp(1)^2)^8),x,
algorithm="fricas")

[Out]

2*x/(e^(25*e^2) + log(x))

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giac [A]  time = 0.18, size = 13, normalized size = 0.76 \begin {gather*} \frac {2 \, x}{e^{\left (25 \, e^{2}\right )} + \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(x)+2*exp(25/4*exp(1)^2)^4-2)/(log(x)^2+2*exp(25/4*exp(1)^2)^4*log(x)+exp(25/4*exp(1)^2)^8),x,
algorithm="giac")

[Out]

2*x/(e^(25*e^2) + log(x))

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maple [A]  time = 0.52, size = 14, normalized size = 0.82

method	result	size
risch	\(\frac {2 x}{{\mathrm e}^{25 \,{\mathrm e}^{2}}+\ln \relax (x )}\)	\(14\)
default	\(\frac {2 x}{{\mathrm e}^{25 \,{\mathrm e}^{2}}+\ln \relax (x )}\)	\(16\)
norman	\(\frac {2 x}{{\mathrm e}^{25 \,{\mathrm e}^{2}}+\ln \relax (x )}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*ln(x)+2*exp(25/4*exp(1)^2)^4-2)/(ln(x)^2+2*exp(25/4*exp(1)^2)^4*ln(x)+exp(25/4*exp(1)^2)^8),x,method=_R
ETURNVERBOSE)

[Out]

2*x/(exp(25*exp(2))+ln(x))

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maxima [A]  time = 0.41, size = 13, normalized size = 0.76 \begin {gather*} \frac {2 \, x}{e^{\left (25 \, e^{2}\right )} + \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(x)+2*exp(25/4*exp(1)^2)^4-2)/(log(x)^2+2*exp(25/4*exp(1)^2)^4*log(x)+exp(25/4*exp(1)^2)^8),x,
algorithm="maxima")

[Out]

2*x/(e^(25*e^2) + log(x))

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mupad [B]  time = 5.31, size = 13, normalized size = 0.76 \begin {gather*} \frac {2\,x}{{\mathrm {e}}^{25\,{\mathrm {e}}^2}+\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*exp(25*exp(2)) + 2*log(x) - 2)/(exp(50*exp(2)) + log(x)^2 + 2*exp(25*exp(2))*log(x)),x)

[Out]

(2*x)/(exp(25*exp(2)) + log(x))

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sympy [A]  time = 0.09, size = 12, normalized size = 0.71 \begin {gather*} \frac {2 x}{\log {\relax (x )} + e^{25 e^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*ln(x)+2*exp(25/4*exp(1)**2)**4-2)/(ln(x)**2+2*exp(25/4*exp(1)**2)**4*ln(x)+exp(25/4*exp(1)**2)**8
),x)

[Out]

2*x/(log(x) + exp(25*exp(2)))

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