3.44.10 \(\int \frac {-4+38 x+(2-18 x) \log (\frac {25}{x^2 \log ^2(2)})+2 x \log ^2(\frac {25}{x^2 \log ^2(2)})}{16-8 \log (\frac {25}{x^2 \log ^2(2)})+\log ^2(\frac {25}{x^2 \log ^2(2)})} \, dx\)

Optimal. Leaf size=23 \[ x \left (x-\frac {-2+x}{-4+\log \left (\frac {25}{x^2 \log ^2(2)}\right )}\right ) \]

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Rubi [A]  time = 0.34, antiderivative size = 28, normalized size of antiderivative = 1.22, number of steps used = 18, number of rules used = 7, integrand size = 65, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {6741, 6742, 2320, 2330, 2300, 2178, 2310} \begin {gather*} x^2-\frac {(2-x) x}{4-\log \left (\frac {25}{x^2 \log ^2(2)}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4 + 38*x + (2 - 18*x)*Log[25/(x^2*Log[2]^2)] + 2*x*Log[25/(x^2*Log[2]^2)]^2)/(16 - 8*Log[25/(x^2*Log[2]^
2)] + Log[25/(x^2*Log[2]^2)]^2),x]

[Out]

x^2 - ((2 - x)*x)/(4 - Log[25/(x^2*Log[2]^2)])

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 2300

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

Rule 2310

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

Rule 2320

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

Rule 2330

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = Expand
Integrand[(a + b*Log[c*x^n])^p, (d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n, p, q, r}
, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[r]))

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} &=\int \frac {-4+38 x+(2-18 x) \log \left (\frac {25}{x^2 \log ^2(2)}\right )+2 x \log ^2\left (\frac {25}{x^2 \log ^2(2)}\right )}{\left (4-\log \left (\frac {25}{x^2 \log ^2(2)}\right )\right )^2} \, dx\\ &=\int \left (2 x-\frac {2 (-2+x)}{\left (-4+\log \left (\frac {25}{x^2 \log ^2(2)}\right )\right )^2}-\frac {2 (-1+x)}{-4+\log \left (\frac {25}{x^2 \log ^2(2)}\right )}\right ) \, dx\\ &=x^2-2 \int \frac {-2+x}{\left (-4+\log \left (\frac {25}{x^2 \log ^2(2)}\right )\right )^2} \, dx-2 \int \frac {-1+x}{-4+\log \left (\frac {25}{x^2 \log ^2(2)}\right )} \, dx\\ &=x^2-\frac {(2-x) x}{4-\log \left (\frac {25}{x^2 \log ^2(2)}\right )}-2 \int \left (-\frac {1}{-4+\log \left (\frac {25}{x^2 \log ^2(2)}\right )}+\frac {x}{-4+\log \left (\frac {25}{x^2 \log ^2(2)}\right )}\right ) \, dx+2 \int \frac {1}{-4+\log \left (\frac {25}{x^2 \log ^2(2)}\right )} \, dx+2 \int \frac {-2+x}{-4+\log \left (\frac {25}{x^2 \log ^2(2)}\right )} \, dx\\ &=x^2-\frac {(2-x) x}{4-\log \left (\frac {25}{x^2 \log ^2(2)}\right )}+2 \int \left (-\frac {2}{-4+\log \left (\frac {25}{x^2 \log ^2(2)}\right )}+\frac {x}{-4+\log \left (\frac {25}{x^2 \log ^2(2)}\right )}\right ) \, dx+2 \int \frac {1}{-4+\log \left (\frac {25}{x^2 \log ^2(2)}\right )} \, dx-2 \int \frac {x}{-4+\log \left (\frac {25}{x^2 \log ^2(2)}\right )} \, dx-\frac {\left (5 \sqrt {\frac {1}{x^2}} x\right ) \operatorname {Subst}\left (\int \frac {e^{-x/2}}{-4+x} \, dx,x,\log \left (\frac {25}{x^2 \log ^2(2)}\right )\right )}{\log (2)}\\ &=x^2-\frac {5 \sqrt {\frac {1}{x^2}} x \text {Ei}\left (\frac {1}{2} \left (4-\log \left (\frac {25}{x^2 \log ^2(2)}\right )\right )\right )}{e^2 \log (2)}-\frac {(2-x) x}{4-\log \left (\frac {25}{x^2 \log ^2(2)}\right )}+2 \int \frac {x}{-4+\log \left (\frac {25}{x^2 \log ^2(2)}\right )} \, dx-4 \int \frac {1}{-4+\log \left (\frac {25}{x^2 \log ^2(2)}\right )} \, dx+\frac {25 \operatorname {Subst}\left (\int \frac {e^{-x}}{-4+x} \, dx,x,\log \left (\frac {25}{x^2 \log ^2(2)}\right )\right )}{\log ^2(2)}-\frac {\left (5 \sqrt {\frac {1}{x^2}} x\right ) \operatorname {Subst}\left (\int \frac {e^{-x/2}}{-4+x} \, dx,x,\log \left (\frac {25}{x^2 \log ^2(2)}\right )\right )}{\log (2)}\\ &=x^2+\frac {25 \text {Ei}\left (4-\log \left (\frac {25}{x^2 \log ^2(2)}\right )\right )}{e^4 \log ^2(2)}-\frac {10 \sqrt {\frac {1}{x^2}} x \text {Ei}\left (\frac {1}{2} \left (4-\log \left (\frac {25}{x^2 \log ^2(2)}\right )\right )\right )}{e^2 \log (2)}-\frac {(2-x) x}{4-\log \left (\frac {25}{x^2 \log ^2(2)}\right )}-\frac {25 \operatorname {Subst}\left (\int \frac {e^{-x}}{-4+x} \, dx,x,\log \left (\frac {25}{x^2 \log ^2(2)}\right )\right )}{\log ^2(2)}+\frac {\left (10 \sqrt {\frac {1}{x^2}} x\right ) \operatorname {Subst}\left (\int \frac {e^{-x/2}}{-4+x} \, dx,x,\log \left (\frac {25}{x^2 \log ^2(2)}\right )\right )}{\log (2)}\\ &=x^2-\frac {(2-x) x}{4-\log \left (\frac {25}{x^2 \log ^2(2)}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 24, normalized size = 1.04 \begin {gather*} x \left (x+\frac {2-x}{-4+\log \left (\frac {25}{x^2 \log ^2(2)}\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 + 38*x + (2 - 18*x)*Log[25/(x^2*Log[2]^2)] + 2*x*Log[25/(x^2*Log[2]^2)]^2)/(16 - 8*Log[25/(x^2*L
og[2]^2)] + Log[25/(x^2*Log[2]^2)]^2),x]

[Out]

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

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fricas [A]  time = 0.58, size = 38, normalized size = 1.65 \begin {gather*} \frac {x^{2} \log \left (\frac {25}{x^{2} \log \relax (2)^{2}}\right ) - 5 \, x^{2} + 2 \, x}{\log \left (\frac {25}{x^{2} \log \relax (2)^{2}}\right ) - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(25/x^2/log(2)^2)^2+(-18*x+2)*log(25/x^2/log(2)^2)+38*x-4)/(log(25/x^2/log(2)^2)^2-8*log(25/
x^2/log(2)^2)+16),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.18, size = 32, normalized size = 1.39 \begin {gather*} x^{2} - \frac {x^{2} - 2 \, x}{2 \, \log \relax (5) - \log \left (x^{2} \log \relax (2)^{2}\right ) - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(25/x^2/log(2)^2)^2+(-18*x+2)*log(25/x^2/log(2)^2)+38*x-4)/(log(25/x^2/log(2)^2)^2-8*log(25/
x^2/log(2)^2)+16),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.09, size = 25, normalized size = 1.09

method	result	size
risch	\(x^{2}-\frac {\left (x -2\right ) x}{\ln \left (\frac {25}{x^{2} \ln \relax (2)^{2}}\right )-4}\)	\(25\)
norman	\(\frac {x^{2} \ln \left (\frac {25}{x^{2} \ln \relax (2)^{2}}\right )+2 x -5 x^{2}}{\ln \left (\frac {25}{x^{2} \ln \relax (2)^{2}}\right )-4}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x*ln(25/x^2/ln(2)^2)^2+(-18*x+2)*ln(25/x^2/ln(2)^2)+38*x-4)/(ln(25/x^2/ln(2)^2)^2-8*ln(25/x^2/ln(2)^2)+
16),x,method=_RETURNVERBOSE)

[Out]

x^2-(x-2)*x/(ln(25/x^2/ln(2)^2)-4)

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maxima [A]  time = 0.47, size = 43, normalized size = 1.87 \begin {gather*} \frac {x^{2} {\left (2 \, \log \relax (5) - 2 \, \log \left (\log \relax (2)\right ) - 5\right )} - 2 \, x^{2} \log \relax (x) + 2 \, x}{2 \, {\left (\log \relax (5) - \log \relax (x) - \log \left (\log \relax (2)\right ) - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(25/x^2/log(2)^2)^2+(-18*x+2)*log(25/x^2/log(2)^2)+38*x-4)/(log(25/x^2/log(2)^2)^2-8*log(25/
x^2/log(2)^2)+16),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.43, size = 24, normalized size = 1.04 \begin {gather*} x^2-\frac {x\,\left (x-2\right )}{\ln \left (\frac {25}{x^2\,{\ln \relax (2)}^2}\right )-4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((38*x - log(25/(x^2*log(2)^2))*(18*x - 2) + 2*x*log(25/(x^2*log(2)^2))^2 - 4)/(log(25/(x^2*log(2)^2))^2 -
8*log(25/(x^2*log(2)^2)) + 16),x)

[Out]

x^2 - (x*(x - 2))/(log(25/(x^2*log(2)^2)) - 4)

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sympy [A]  time = 0.15, size = 22, normalized size = 0.96 \begin {gather*} x^{2} + \frac {- x^{2} + 2 x}{\log {\left (\frac {25}{x^{2} \log {\relax (2 )}^{2}} \right )} - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*ln(25/x**2/ln(2)**2)**2+(-18*x+2)*ln(25/x**2/ln(2)**2)+38*x-4)/(ln(25/x**2/ln(2)**2)**2-8*ln(25
/x**2/ln(2)**2)+16),x)

[Out]

x**2 + (-x**2 + 2*x)/(log(25/(x**2*log(2)**2)) - 4)

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