3.27.31 \(\int \frac {-80 x+(240-48 x-3 x^2) \log (2)+(-20 x+(48-12 x-x^2) \log (2)) \log (\frac {-20 x+(48-12 x-x^2) \log (2)}{4 x \log (2)})}{20 x^3+(-48 x^2+12 x^3+x^4) \log (2)} \, dx\)

Optimal. Leaf size=33 \[ \frac {4+\log \left (-3+4 \left (\frac {3}{x}-\frac {x}{2}\right )+\frac {7 x}{4}-\frac {5}{\log (2)}\right )}{x} \]

________________________________________________________________________________________

Rubi [A]  time = 1.13, antiderivative size = 64, normalized size of antiderivative = 1.94, number of steps used = 17, number of rules used = 9, integrand size = 90, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6741, 6728, 1628, 634, 618, 206, 628, 2525, 12} \begin {gather*} \frac {\log \left (\frac {x^2 (-\log (2))-4 x (5+\log (8))+48 \log (2)}{x \log (16)}\right )}{x}+\frac {4}{x}-\frac {(5+\log (8)) \log (x)}{\log (4096)}+\frac {(5+\log (8)) \log (x)}{12 \log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-80*x + (240 - 48*x - 3*x^2)*Log[2] + (-20*x + (48 - 12*x - x^2)*Log[2])*Log[(-20*x + (48 - 12*x - x^2)*L
og[2])/(4*x*Log[2])])/(20*x^3 + (-48*x^2 + 12*x^3 + x^4)*Log[2]),x]

[Out]

4/x + ((5 + Log[8])*Log[x])/(12*Log[2]) - ((5 + Log[8])*Log[x])/Log[4096] + Log[(48*Log[2] - x^2*Log[2] - 4*x*
(5 + Log[8]))/(x*Log[16])]/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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
 /; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
 b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 2525

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m
+ 1)*(a + b*Log[c*RFx^p])^n)/(e*(m + 1)), x] - Dist[(b*n*p)/(e*(m + 1)), Int[SimplifyIntegrand[((d + e*x)^(m +
 1)*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && RationalFunc
tionQ[RFx, x] && IGtQ[n, 0] && (EqQ[n, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 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} &=\int \frac {80 x-\left (240-48 x-3 x^2\right ) \log (2)-\left (-20 x+\left (48-12 x-x^2\right ) \log (2)\right ) \log \left (\frac {-20 x+\left (48-12 x-x^2\right ) \log (2)}{4 x \log (2)}\right )}{x^2 \left (48 \log (2)-x^2 \log (2)-4 x (5+\log (8))\right )} \, dx\\ &=\int \left (\frac {-240 \log (2)+x^2 \log (8)+16 x (5+\log (8))}{x^2 \left (48 \log (2)-x^2 \log (2)-4 x (5+\log (8))\right )}-\frac {\log \left (\frac {48 \log (2)-x^2 \log (2)-4 x (5+\log (8))}{x \log (16)}\right )}{x^2}\right ) \, dx\\ &=\int \frac {-240 \log (2)+x^2 \log (8)+16 x (5+\log (8))}{x^2 \left (48 \log (2)-x^2 \log (2)-4 x (5+\log (8))\right )} \, dx-\int \frac {\log \left (\frac {48 \log (2)-x^2 \log (2)-4 x (5+\log (8))}{x \log (16)}\right )}{x^2} \, dx\\ &=\frac {\log \left (\frac {48 \log (2)-x^2 \log (2)-4 x (5+\log (8))}{x \log (16)}\right )}{x}-\int \frac {\left (-48-x^2\right ) \log (2)}{x^2 \left (48 \log (2)-x^2 \log (2)-4 x (5+\log (8))\right )} \, dx+\int \left (-\frac {5}{x^2}+\frac {-x \log (2) (5+\log (8))-20 \left (5+3 \log ^2(2)+\log (64)\right )}{12 \log (2) \left (48 \log (2)-x^2 \log (2)-4 x (5+\log (8))\right )}+\frac {-5-\log (8)}{x \log (4096)}\right ) \, dx\\ &=\frac {5}{x}-\frac {(5+\log (8)) \log (x)}{\log (4096)}+\frac {\log \left (\frac {48 \log (2)-x^2 \log (2)-4 x (5+\log (8))}{x \log (16)}\right )}{x}+\frac {\int \frac {-x \log (2) (5+\log (8))-20 \left (5+3 \log ^2(2)+\log (64)\right )}{48 \log (2)-x^2 \log (2)-4 x (5+\log (8))} \, dx}{12 \log (2)}-\log (2) \int \frac {-48-x^2}{x^2 \left (48 \log (2)-x^2 \log (2)-4 x (5+\log (8))\right )} \, dx\\ &=\frac {5}{x}-\frac {(5+\log (8)) \log (x)}{\log (4096)}+\frac {\log \left (\frac {48 \log (2)-x^2 \log (2)-4 x (5+\log (8))}{x \log (16)}\right )}{x}-\log (2) \int \left (-\frac {1}{x^2 \log (2)}+\frac {-5-\log (8)}{12 x \log ^2(2)}+\frac {-x \log (2) (5+\log (8))-4 \left (6 \log ^2(2)+(5+\log (8))^2\right )}{12 \log ^2(2) \left (48 \log (2)-x^2 \log (2)-4 x (5+\log (8))\right )}\right ) \, dx+\frac {(5+\log (8)) \int \frac {-2 x \log (2)-4 (5+\log (8))}{48 \log (2)-x^2 \log (2)-4 x (5+\log (8))} \, dx}{24 \log (2)}+\frac {\left ((5+\log (8))^2-10 \left (5+3 \log ^2(2)+\log (64)\right )\right ) \int \frac {1}{48 \log (2)-x^2 \log (2)-4 x (5+\log (8))} \, dx}{6 \log (2)}\\ &=\frac {4}{x}+\frac {(5+\log (8)) \log (x)}{12 \log (2)}-\frac {(5+\log (8)) \log (x)}{\log (4096)}+\frac {(5+\log (8)) \log \left (48 \log (2)-x^2 \log (2)-4 x (5+\log (8))\right )}{24 \log (2)}+\frac {\log \left (\frac {48 \log (2)-x^2 \log (2)-4 x (5+\log (8))}{x \log (16)}\right )}{x}-\frac {\int \frac {-x \log (2) (5+\log (8))-4 \left (6 \log ^2(2)+(5+\log (8))^2\right )}{48 \log (2)-x^2 \log (2)-4 x (5+\log (8))} \, dx}{12 \log (2)}-\frac {\left ((5+\log (8))^2-10 \left (5+3 \log ^2(2)+\log (64)\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-x^2+16 \left (12 \log ^2(2)+(5+\log (8))^2\right )} \, dx,x,-2 x \log (2)-4 (5+\log (8))\right )}{3 \log (2)}\\ &=\frac {4}{x}-\frac {\tanh ^{-1}\left (\frac {10+x \log (2)+\log (64)}{2 \sqrt {12 \log ^2(2)+(5+\log (8))^2}}\right ) \sqrt {12 \log ^2(2)+(5+\log (8))^2}}{12 \log (2)}+\frac {(5+\log (8)) \log (x)}{12 \log (2)}-\frac {(5+\log (8)) \log (x)}{\log (4096)}+\frac {(5+\log (8)) \log \left (48 \log (2)-x^2 \log (2)-4 x (5+\log (8))\right )}{24 \log (2)}+\frac {\log \left (\frac {48 \log (2)-x^2 \log (2)-4 x (5+\log (8))}{x \log (16)}\right )}{x}-\frac {(5+\log (8)) \int \frac {-2 x \log (2)-4 (5+\log (8))}{48 \log (2)-x^2 \log (2)-4 x (5+\log (8))} \, dx}{24 \log (2)}+\frac {\left (12 \log ^2(2)+(5+\log (8))^2\right ) \int \frac {1}{48 \log (2)-x^2 \log (2)-4 x (5+\log (8))} \, dx}{6 \log (2)}\\ &=\frac {4}{x}-\frac {\tanh ^{-1}\left (\frac {10+x \log (2)+\log (64)}{2 \sqrt {12 \log ^2(2)+(5+\log (8))^2}}\right ) \sqrt {12 \log ^2(2)+(5+\log (8))^2}}{12 \log (2)}+\frac {(5+\log (8)) \log (x)}{12 \log (2)}-\frac {(5+\log (8)) \log (x)}{\log (4096)}+\frac {\log \left (\frac {48 \log (2)-x^2 \log (2)-4 x (5+\log (8))}{x \log (16)}\right )}{x}-\frac {\left (12 \log ^2(2)+(5+\log (8))^2\right ) \operatorname {Subst}\left (\int \frac {1}{-x^2+16 \left (12 \log ^2(2)+(5+\log (8))^2\right )} \, dx,x,-2 x \log (2)-4 (5+\log (8))\right )}{3 \log (2)}\\ &=\frac {4}{x}+\frac {(5+\log (8)) \log (x)}{12 \log (2)}-\frac {(5+\log (8)) \log (x)}{\log (4096)}+\frac {\log \left (\frac {48 \log (2)-x^2 \log (2)-4 x (5+\log (8))}{x \log (16)}\right )}{x}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [B]  time = 0.17, size = 96, normalized size = 2.91 \begin {gather*} \frac {4}{x}+\frac {\tanh ^{-1}\left (\frac {10+x \log (2)+\log (64)}{2 \sqrt {12 \log ^2(2)+(5+\log (8))^2}}\right ) (-3 \log (2)+\log (8))}{2 \sqrt {12 \log ^2(2)+(5+\log (8))^2}}+\frac {\log \left (\frac {48 \log (2)-x^2 \log (2)-4 x (5+\log (8))}{x \log (16)}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-80*x + (240 - 48*x - 3*x^2)*Log[2] + (-20*x + (48 - 12*x - x^2)*Log[2])*Log[(-20*x + (48 - 12*x -
x^2)*Log[2])/(4*x*Log[2])])/(20*x^3 + (-48*x^2 + 12*x^3 + x^4)*Log[2]),x]

[Out]

4/x + (ArcTanh[(10 + x*Log[2] + Log[64])/(2*Sqrt[12*Log[2]^2 + (5 + Log[8])^2])]*(-3*Log[2] + Log[8]))/(2*Sqrt
[12*Log[2]^2 + (5 + Log[8])^2]) + Log[(48*Log[2] - x^2*Log[2] - 4*x*(5 + Log[8]))/(x*Log[16])]/x

________________________________________________________________________________________

fricas [A]  time = 0.68, size = 31, normalized size = 0.94 \begin {gather*} \frac {\log \left (-\frac {{\left (x^{2} + 12 \, x - 48\right )} \log \relax (2) + 20 \, x}{4 \, x \log \relax (2)}\right ) + 4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^2-12*x+48)*log(2)-20*x)*log(1/4*((-x^2-12*x+48)*log(2)-20*x)/x/log(2))+(-3*x^2-48*x+240)*log(2
)-80*x)/((x^4+12*x^3-48*x^2)*log(2)+20*x^3),x, algorithm="fricas")

[Out]

(log(-1/4*((x^2 + 12*x - 48)*log(2) + 20*x)/(x*log(2))) + 4)/x

________________________________________________________________________________________

giac [A]  time = 0.39, size = 45, normalized size = 1.36 \begin {gather*} -\frac {2 \, {\left (\log \relax (2) - 2\right )}}{x} + \frac {\log \left (-x^{2} \log \relax (2) - 12 \, x \log \relax (2) - 20 \, x + 48 \, \log \relax (2)\right )}{x} - \frac {\log \left (x \log \relax (2)\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^2-12*x+48)*log(2)-20*x)*log(1/4*((-x^2-12*x+48)*log(2)-20*x)/x/log(2))+(-3*x^2-48*x+240)*log(2
)-80*x)/((x^4+12*x^3-48*x^2)*log(2)+20*x^3),x, algorithm="giac")

[Out]

-2*(log(2) - 2)/x + log(-x^2*log(2) - 12*x*log(2) - 20*x + 48*log(2))/x - log(x*log(2))/x

________________________________________________________________________________________

maple [A]  time = 0.15, size = 34, normalized size = 1.03




method result size



norman \(\frac {4+\ln \left (\frac {\left (-x^{2}-12 x +48\right ) \ln \relax (2)-20 x}{4 x \ln \relax (2)}\right )}{x}\) \(34\)
risch \(\frac {\ln \left (\frac {\left (-x^{2}-12 x +48\right ) \ln \relax (2)-20 x}{4 x \ln \relax (2)}\right )}{x}+\frac {4}{x}\) \(38\)
default \(-\frac {\ln \left (\ln \relax (2)\right )}{x}-\frac {2 \ln \relax (2)}{x}+\frac {4}{x}+\frac {\ln \left (\frac {-x^{2} \ln \relax (2)-12 x \ln \relax (2)+48 \ln \relax (2)-20 x}{x}\right )}{x}\) \(51\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-x^2-12*x+48)*ln(2)-20*x)*ln(1/4*((-x^2-12*x+48)*ln(2)-20*x)/x/ln(2))+(-3*x^2-48*x+240)*ln(2)-80*x)/((x
^4+12*x^3-48*x^2)*ln(2)+20*x^3),x,method=_RETURNVERBOSE)

[Out]

(4+ln(1/4*((-x^2-12*x+48)*ln(2)-20*x)/x/ln(2)))/x

________________________________________________________________________________________

maxima [B]  time = 0.93, size = 317, normalized size = 9.61 \begin {gather*} \frac {5}{24} \, {\left (\frac {{\left (3 \, \log \relax (2) + 5\right )} \log \left (x^{2} \log \relax (2) + 4 \, x {\left (3 \, \log \relax (2) + 5\right )} - 48 \, \log \relax (2)\right )}{\log \relax (2)^{2}} - \frac {2 \, {\left (3 \, \log \relax (2) + 5\right )} \log \relax (x)}{\log \relax (2)^{2}} + \frac {5 \, {\left (3 \, \log \relax (2)^{2} + 6 \, \log \relax (2) + 5\right )} \log \left (\frac {x \log \relax (2) - 2 \, \sqrt {21 \, \log \relax (2)^{2} + 30 \, \log \relax (2) + 25} + 6 \, \log \relax (2) + 10}{x \log \relax (2) + 2 \, \sqrt {21 \, \log \relax (2)^{2} + 30 \, \log \relax (2) + 25} + 6 \, \log \relax (2) + 10}\right )}{\sqrt {21 \, \log \relax (2)^{2} + 30 \, \log \relax (2) + 25} \log \relax (2)^{2}} + \frac {24}{x \log \relax (2)}\right )} \log \relax (2) - \frac {25 \, {\left (3 \, \log \relax (2)^{2} + 6 \, \log \relax (2) + 5\right )} \log \left (\frac {x \log \relax (2) - 2 \, \sqrt {21 \, \log \relax (2)^{2} + 30 \, \log \relax (2) + 25} + 6 \, \log \relax (2) + 10}{x \log \relax (2) + 2 \, \sqrt {21 \, \log \relax (2)^{2} + 30 \, \log \relax (2) + 25} + 6 \, \log \relax (2) + 10}\right )}{24 \, \sqrt {21 \, \log \relax (2)^{2} + 30 \, \log \relax (2) + 25} \log \relax (2)} - \frac {24 \, {\left (\log \left (\log \relax (2)\right ) + 1\right )} \log \relax (2) + 48 \, \log \relax (2)^{2} + {\left (5 \, x {\left (3 \, \log \relax (2) + 5\right )} - 24 \, \log \relax (2)\right )} \log \left (-x^{2} \log \relax (2) - 4 \, x {\left (3 \, \log \relax (2) + 5\right )} + 48 \, \log \relax (2)\right ) - 2 \, {\left (5 \, x {\left (3 \, \log \relax (2) + 5\right )} - 12 \, \log \relax (2)\right )} \log \relax (x)}{24 \, x \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^2-12*x+48)*log(2)-20*x)*log(1/4*((-x^2-12*x+48)*log(2)-20*x)/x/log(2))+(-3*x^2-48*x+240)*log(2
)-80*x)/((x^4+12*x^3-48*x^2)*log(2)+20*x^3),x, algorithm="maxima")

[Out]

5/24*((3*log(2) + 5)*log(x^2*log(2) + 4*x*(3*log(2) + 5) - 48*log(2))/log(2)^2 - 2*(3*log(2) + 5)*log(x)/log(2
)^2 + 5*(3*log(2)^2 + 6*log(2) + 5)*log((x*log(2) - 2*sqrt(21*log(2)^2 + 30*log(2) + 25) + 6*log(2) + 10)/(x*l
og(2) + 2*sqrt(21*log(2)^2 + 30*log(2) + 25) + 6*log(2) + 10))/(sqrt(21*log(2)^2 + 30*log(2) + 25)*log(2)^2) +
 24/(x*log(2)))*log(2) - 25/24*(3*log(2)^2 + 6*log(2) + 5)*log((x*log(2) - 2*sqrt(21*log(2)^2 + 30*log(2) + 25
) + 6*log(2) + 10)/(x*log(2) + 2*sqrt(21*log(2)^2 + 30*log(2) + 25) + 6*log(2) + 10))/(sqrt(21*log(2)^2 + 30*l
og(2) + 25)*log(2)) - 1/24*(24*(log(log(2)) + 1)*log(2) + 48*log(2)^2 + (5*x*(3*log(2) + 5) - 24*log(2))*log(-
x^2*log(2) - 4*x*(3*log(2) + 5) + 48*log(2)) - 2*(5*x*(3*log(2) + 5) - 12*log(2))*log(x))/(x*log(2))

________________________________________________________________________________________

mupad [B]  time = 2.06, size = 32, normalized size = 0.97 \begin {gather*} \frac {\ln \left (-\frac {5\,x+\frac {\ln \relax (2)\,\left (x^2+12\,x-48\right )}{4}}{x\,\ln \relax (2)}\right )+4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(80*x + log(2)*(48*x + 3*x^2 - 240) + log(-(5*x + (log(2)*(12*x + x^2 - 48))/4)/(x*log(2)))*(20*x + log(2
)*(12*x + x^2 - 48)))/(log(2)*(12*x^3 - 48*x^2 + x^4) + 20*x^3),x)

[Out]

(log(-(5*x + (log(2)*(12*x + x^2 - 48))/4)/(x*log(2))) + 4)/x

________________________________________________________________________________________

sympy [A]  time = 0.25, size = 27, normalized size = 0.82 \begin {gather*} \frac {\log {\left (\frac {- 5 x + \frac {\left (- x^{2} - 12 x + 48\right ) \log {\relax (2 )}}{4}}{x \log {\relax (2 )}} \right )}}{x} + \frac {4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x**2-12*x+48)*ln(2)-20*x)*ln(1/4*((-x**2-12*x+48)*ln(2)-20*x)/x/ln(2))+(-3*x**2-48*x+240)*ln(2)-
80*x)/((x**4+12*x**3-48*x**2)*ln(2)+20*x**3),x)

[Out]

log((-5*x + (-x**2 - 12*x + 48)*log(2)/4)/(x*log(2)))/x + 4/x

________________________________________________________________________________________