3.27.94 \(\int \frac {-80 \log (\frac {1}{2 x})+(80-80 \log (\frac {1}{2 x})) \log (x)+(20-20 \log (\frac {1}{2 x})) \log ^2(x)}{16+8 \log (x)+\log ^2(x)} \, dx\)

Optimal. Leaf size=21 \[ 24-\frac {20 x \log \left (\frac {1}{2 x}\right ) \log (x)}{4+\log (x)} \]

________________________________________________________________________________________

Rubi [C]  time = 0.24, antiderivative size = 79, normalized size of antiderivative = 3.76, number of steps used = 15, number of rules used = 9, integrand size = 56, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {6688, 6742, 2295, 2297, 2299, 2178, 2361, 6482, 12} \begin {gather*} -\frac {80 \left (1-\log \left (\frac {1}{2 x}\right )\right ) \text {Ei}(\log (x)+4)}{e^4}-\frac {80 \log \left (\frac {1}{2 x}\right ) \text {Ei}(\log (x)+4)}{e^4}+\frac {80 \text {Ei}(\log (x)+4)}{e^4}-20 x \log \left (\frac {1}{2 x}\right )+\frac {80 x \log \left (\frac {1}{2 x}\right )}{\log (x)+4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-80*Log[1/(2*x)] + (80 - 80*Log[1/(2*x)])*Log[x] + (20 - 20*Log[1/(2*x)])*Log[x]^2)/(16 + 8*Log[x] + Log[
x]^2),x]

[Out]

(80*ExpIntegralEi[4 + Log[x]])/E^4 - (80*ExpIntegralEi[4 + Log[x]]*(1 - Log[1/(2*x)]))/E^4 - 20*x*Log[1/(2*x)]
 - (80*ExpIntegralEi[4 + Log[x]]*Log[1/(2*x)])/E^4 + (80*x*Log[1/(2*x)])/(4 + 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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

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 2361

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.)), x_Symbol] :> With[{u =
IntHide[(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[SimplifyIntegrand[u/x, x], x],
 x]] /; FreeQ[{a, b, c, d, e, f, n, p, r}, x]

Rule 6482

Int[ExpIntegralEi[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[((a + b*x)*ExpIntegralEi[a + b*x])/b, x] - Simp[E^(a
+ b*x)/b, x] /; FreeQ[{a, b}, x]

Rule 6688

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

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 {-20 \log \left (\frac {1}{2 x}\right ) (2+\log (x))^2+20 \log (x) (4+\log (x))}{(4+\log (x))^2} \, dx\\ &=\int \left (-20 \left (-1+\log \left (\frac {1}{2 x}\right )\right )-\frac {80 \log \left (\frac {1}{2 x}\right )}{(4+\log (x))^2}+\frac {80 \left (-1+\log \left (\frac {1}{2 x}\right )\right )}{4+\log (x)}\right ) \, dx\\ &=-\left (20 \int \left (-1+\log \left (\frac {1}{2 x}\right )\right ) \, dx\right )-80 \int \frac {\log \left (\frac {1}{2 x}\right )}{(4+\log (x))^2} \, dx+80 \int \frac {-1+\log \left (\frac {1}{2 x}\right )}{4+\log (x)} \, dx\\ &=20 x-\frac {80 \text {Ei}(4+\log (x)) \left (1-\log \left (\frac {1}{2 x}\right )\right )}{e^4}-\frac {80 \text {Ei}(4+\log (x)) \log \left (\frac {1}{2 x}\right )}{e^4}+\frac {80 x \log \left (\frac {1}{2 x}\right )}{4+\log (x)}-20 \int \log \left (\frac {1}{2 x}\right ) \, dx+80 \int \frac {\text {Ei}(4+\log (x))}{e^4 x} \, dx-80 \int \left (\frac {\text {Ei}(4+\log (x))}{e^4 x}-\frac {1}{4+\log (x)}\right ) \, dx\\ &=-\frac {80 \text {Ei}(4+\log (x)) \left (1-\log \left (\frac {1}{2 x}\right )\right )}{e^4}-20 x \log \left (\frac {1}{2 x}\right )-\frac {80 \text {Ei}(4+\log (x)) \log \left (\frac {1}{2 x}\right )}{e^4}+\frac {80 x \log \left (\frac {1}{2 x}\right )}{4+\log (x)}+80 \int \frac {1}{4+\log (x)} \, dx\\ &=-\frac {80 \text {Ei}(4+\log (x)) \left (1-\log \left (\frac {1}{2 x}\right )\right )}{e^4}-20 x \log \left (\frac {1}{2 x}\right )-\frac {80 \text {Ei}(4+\log (x)) \log \left (\frac {1}{2 x}\right )}{e^4}+\frac {80 x \log \left (\frac {1}{2 x}\right )}{4+\log (x)}+80 \operatorname {Subst}\left (\int \frac {e^x}{4+x} \, dx,x,\log (x)\right )\\ &=\frac {80 \text {Ei}(4+\log (x))}{e^4}-\frac {80 \text {Ei}(4+\log (x)) \left (1-\log \left (\frac {1}{2 x}\right )\right )}{e^4}-20 x \log \left (\frac {1}{2 x}\right )-\frac {80 \text {Ei}(4+\log (x)) \log \left (\frac {1}{2 x}\right )}{e^4}+\frac {80 x \log \left (\frac {1}{2 x}\right )}{4+\log (x)}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.08, size = 19, normalized size = 0.90 \begin {gather*} -\frac {20 x \log \left (\frac {1}{2 x}\right ) \log (x)}{4+\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-80*Log[1/(2*x)] + (80 - 80*Log[1/(2*x)])*Log[x] + (20 - 20*Log[1/(2*x)])*Log[x]^2)/(16 + 8*Log[x]
+ Log[x]^2),x]

[Out]

(-20*x*Log[1/(2*x)]*Log[x])/(4 + Log[x])

________________________________________________________________________________________

fricas [A]  time = 0.49, size = 35, normalized size = 1.67 \begin {gather*} -\frac {20 \, {\left (x \log \relax (2) \log \left (\frac {1}{2 \, x}\right ) + x \log \left (\frac {1}{2 \, x}\right )^{2}\right )}}{\log \relax (2) + \log \left (\frac {1}{2 \, x}\right ) - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*log(1/2/x)+20)*log(x)^2+(-80*log(1/2/x)+80)*log(x)-80*log(1/2/x))/(log(x)^2+8*log(x)+16),x, al
gorithm="fricas")

[Out]

-20*(x*log(2)*log(1/2/x) + x*log(1/2/x)^2)/(log(2) + log(1/2/x) - 4)

________________________________________________________________________________________

giac [A]  time = 0.29, size = 27, normalized size = 1.29 \begin {gather*} \frac {20 \, x \log \relax (2) \log \relax (x)}{\log \relax (x) + 4} + \frac {20 \, x \log \relax (x)^{2}}{\log \relax (x) + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*log(1/2/x)+20)*log(x)^2+(-80*log(1/2/x)+80)*log(x)-80*log(1/2/x))/(log(x)^2+8*log(x)+16),x, al
gorithm="giac")

[Out]

20*x*log(2)*log(x)/(log(x) + 4) + 20*x*log(x)^2/(log(x) + 4)

________________________________________________________________________________________

maple [A]  time = 0.78, size = 18, normalized size = 0.86




method result size



norman \(-\frac {20 \ln \relax (x ) \ln \left (\frac {1}{2 x}\right ) x}{\ln \relax (x )+4}\) \(18\)
risch \(20 x \ln \relax (x )+20 x \ln \relax (2)-80 x -\frac {40 x \left (2 \ln \relax (2)-8\right )}{\ln \relax (x )+4}\) \(30\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-20*ln(1/2/x)+20)*ln(x)^2+(-80*ln(1/2/x)+80)*ln(x)-80*ln(1/2/x))/(ln(x)^2+8*ln(x)+16),x,method=_RETURNVE
RBOSE)

[Out]

-20/(ln(x)+4)*ln(x)*ln(1/2/x)*x

________________________________________________________________________________________

maxima [A]  time = 0.62, size = 21, normalized size = 1.00 \begin {gather*} \frac {20 \, {\left (x \log \relax (2) \log \relax (x) + x \log \relax (x)^{2}\right )}}{\log \relax (x) + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*log(1/2/x)+20)*log(x)^2+(-80*log(1/2/x)+80)*log(x)-80*log(1/2/x))/(log(x)^2+8*log(x)+16),x, al
gorithm="maxima")

[Out]

20*(x*log(2)*log(x) + x*log(x)^2)/(log(x) + 4)

________________________________________________________________________________________

mupad [B]  time = 1.61, size = 17, normalized size = 0.81 \begin {gather*} -\frac {20\,x\,\ln \left (\frac {1}{2\,x}\right )\,\ln \relax (x)}{\ln \relax (x)+4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(80*log(1/(2*x)) + log(x)^2*(20*log(1/(2*x)) - 20) + log(x)*(80*log(1/(2*x)) - 80))/(8*log(x) + log(x)^2
+ 16),x)

[Out]

-(20*x*log(1/(2*x))*log(x))/(log(x) + 4)

________________________________________________________________________________________

sympy [A]  time = 0.25, size = 29, normalized size = 1.38 \begin {gather*} 20 x \log {\relax (x )} + x \left (-80 + 20 \log {\relax (2 )}\right ) + \frac {- 80 x \log {\relax (2 )} + 320 x}{\log {\relax (x )} + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*ln(1/2/x)+20)*ln(x)**2+(-80*ln(1/2/x)+80)*ln(x)-80*ln(1/2/x))/(ln(x)**2+8*ln(x)+16),x)

[Out]

20*x*log(x) + x*(-80 + 20*log(2)) + (-80*x*log(2) + 320*x)/(log(x) + 4)

________________________________________________________________________________________