3.25.12 \(\int \frac {2}{-10+5 x+(-2+x) \log (\frac {(100-100 x+25 x^2) \log (5)}{16 e^5})} \, dx\)

Optimal. Leaf size=20 \[ \log \left (2 \left (5+\log \left (\frac {25 (-2+x)^2 \log (5)}{16 e^5}\right )\right )\right ) \]

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Rubi [A]  time = 0.07, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {12, 6741, 2390, 2302, 29} \begin {gather*} \log \left (\log \left (\frac {25 (2-x)^2 \log (5)}{16 e^5}\right )+5\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[2/(-10 + 5*x + (-2 + x)*Log[((100 - 100*x + 25*x^2)*Log[5])/(16*E^5)]),x]

[Out]

Log[5 + Log[(25*(2 - x)^2*Log[5])/(16*E^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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2302

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

Rule 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[((f*x)/d)^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rule 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=2 \int \frac {1}{-10+5 x+(-2+x) \log \left (\frac {\left (100-100 x+25 x^2\right ) \log (5)}{16 e^5}\right )} \, dx\\ &=2 \int \frac {1}{(-2+x) \left (5+\log \left (\frac {25 (-2+x)^2 \log (5)}{16 e^5}\right )\right )} \, dx\\ &=2 \operatorname {Subst}\left (\int \frac {1}{x \left (5+\log \left (\frac {25 x^2 \log (5)}{16 e^5}\right )\right )} \, dx,x,-2+x\right )\\ &=\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,5+\log \left (\frac {25 (-2+x)^2 \log (5)}{16 e^5}\right )\right )\\ &=\log \left (5+\log \left (\frac {25 (-2+x)^2 \log (5)}{16 e^5}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 18, normalized size = 0.90 \begin {gather*} \log \left (5+\log \left (\frac {25 (-2+x)^2 \log (5)}{16 e^5}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[2/(-10 + 5*x + (-2 + x)*Log[((100 - 100*x + 25*x^2)*Log[5])/(16*E^5)]),x]

[Out]

Log[5 + Log[(25*(-2 + x)^2*Log[5])/(16*E^5)]]

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fricas [A]  time = 0.51, size = 18, normalized size = 0.90 \begin {gather*} \log \left (\log \left (\frac {25}{16} \, {\left (x^{2} - 4 \, x + 4\right )} e^{\left (-5\right )} \log \relax (5)\right ) + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/((x-2)*log(1/16*(25*x^2-100*x+100)*log(5)/exp(5))+5*x-10),x, algorithm="fricas")

[Out]

log(log(25/16*(x^2 - 4*x + 4)*e^(-5)*log(5)) + 5)

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giac [A]  time = 0.32, size = 27, normalized size = 1.35 \begin {gather*} \log \left (2 \, \log \relax (5) - 4 \, \log \relax (2) + \log \left (x^{2} \log \relax (5) - 4 \, x \log \relax (5) + 4 \, \log \relax (5)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/((x-2)*log(1/16*(25*x^2-100*x+100)*log(5)/exp(5))+5*x-10),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 21, normalized size = 1.05

method	result	size
risch	\(\ln \left (\ln \left (\frac {\left (25 x^{2}-100 x +100\right ) \ln \relax (5) {\mathrm e}^{-5}}{16}\right )+5\right )\)	\(21\)
norman	\(\ln \left (\ln \left (\frac {\left (25 x^{2}-100 x +100\right ) \ln \relax (5) {\mathrm e}^{-5}}{16}\right )+5\right )\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2/((x-2)*ln(1/16*(25*x^2-100*x+100)*ln(5)/exp(5))+5*x-10),x,method=_RETURNVERBOSE)

[Out]

ln(ln(1/16*(25*x^2-100*x+100)*ln(5)*exp(-5))+5)

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maxima [A]  time = 0.58, size = 17, normalized size = 0.85 \begin {gather*} \log \left (\log \relax (5) - 2 \, \log \relax (2) + \log \left (x - 2\right ) + \frac {1}{2} \, \log \left (\log \relax (5)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/((x-2)*log(1/16*(25*x^2-100*x+100)*log(5)/exp(5))+5*x-10),x, algorithm="maxima")

[Out]

log(log(5) - 2*log(2) + log(x - 2) + 1/2*log(log(5)))

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mupad [B]  time = 1.02, size = 20, normalized size = 1.00 \begin {gather*} \ln \left (\ln \left (\frac {{\mathrm {e}}^{-5}\,\ln \relax (5)\,\left (25\,x^2-100\,x+100\right )}{16}\right )+5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2/(5*x + log((exp(-5)*log(5)*(25*x^2 - 100*x + 100))/16)*(x - 2) - 10),x)

[Out]

log(log((exp(-5)*log(5)*(25*x^2 - 100*x + 100))/16) + 5)

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sympy [A]  time = 0.15, size = 26, normalized size = 1.30 \begin {gather*} \log {\left (\log {\left (\frac {\left (\frac {25 x^{2}}{16} - \frac {25 x}{4} + \frac {25}{4}\right ) \log {\relax (5 )}}{e^{5}} \right )} + 5 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/((x-2)*ln(1/16*(25*x**2-100*x+100)*ln(5)/exp(5))+5*x-10),x)

[Out]

log(log((25*x**2/16 - 25*x/4 + 25/4)*exp(-5)*log(5)) + 5)

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