3.9.98 \(\int \frac {(-3-8 x) \log (2)-5 \log (2) \log (x)}{-72 x+144 x^2+180 x \log (x)+(-24 x+48 x^2+60 x \log (x)) \log (50 x-100 x^2-125 x \log (x))+(-2 x+4 x^2+5 x \log (x)) \log ^2(50 x-100 x^2-125 x \log (x))} \, dx\)

Optimal. Leaf size=20 \[ \frac {\log (2)}{6+\log (25 x (2-4 x-5 \log (x)))} \]

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Rubi [A]  time = 0.25, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 94, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {6688, 12, 6686} \begin {gather*} \frac {\log (2)}{\log (25 x (-4 x-5 \log (x)+2))+6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-3 - 8*x)*Log[2] - 5*Log[2]*Log[x])/(-72*x + 144*x^2 + 180*x*Log[x] + (-24*x + 48*x^2 + 60*x*Log[x])*Log
[50*x - 100*x^2 - 125*x*Log[x]] + (-2*x + 4*x^2 + 5*x*Log[x])*Log[50*x - 100*x^2 - 125*x*Log[x]]^2),x]

[Out]

Log[2]/(6 + Log[25*x*(2 - 4*x - 5*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 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

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 {\log (2) (3+8 x+5 \log (x))}{x (2-4 x-5 \log (x)) (6+\log (-25 x (-2+4 x+5 \log (x))))^2} \, dx\\ &=\log (2) \int \frac {3+8 x+5 \log (x)}{x (2-4 x-5 \log (x)) (6+\log (-25 x (-2+4 x+5 \log (x))))^2} \, dx\\ &=\frac {\log (2)}{6+\log (25 x (2-4 x-5 \log (x)))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 20, normalized size = 1.00 \begin {gather*} \frac {\log (2)}{6+\log (-25 x (-2+4 x+5 \log (x)))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-3 - 8*x)*Log[2] - 5*Log[2]*Log[x])/(-72*x + 144*x^2 + 180*x*Log[x] + (-24*x + 48*x^2 + 60*x*Log[x
])*Log[50*x - 100*x^2 - 125*x*Log[x]] + (-2*x + 4*x^2 + 5*x*Log[x])*Log[50*x - 100*x^2 - 125*x*Log[x]]^2),x]

[Out]

Log[2]/(6 + Log[-25*x*(-2 + 4*x + 5*Log[x])])

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fricas [A]  time = 0.67, size = 22, normalized size = 1.10 \begin {gather*} \frac {\log \relax (2)}{\log \left (-100 \, x^{2} - 125 \, x \log \relax (x) + 50 \, x\right ) + 6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*log(2)*log(x)+(-8*x-3)*log(2))/((5*x*log(x)+4*x^2-2*x)*log(-125*x*log(x)-100*x^2+50*x)^2+(60*x*l
og(x)+48*x^2-24*x)*log(-125*x*log(x)-100*x^2+50*x)+180*x*log(x)+144*x^2-72*x),x, algorithm="fricas")

[Out]

log(2)/(log(-100*x^2 - 125*x*log(x) + 50*x) + 6)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*log(2)*log(x)+(-8*x-3)*log(2))/((5*x*log(x)+4*x^2-2*x)*log(-125*x*log(x)-100*x^2+50*x)^2+(60*x*l
og(x)+48*x^2-24*x)*log(-125*x*log(x)-100*x^2+50*x)+180*x*log(x)+144*x^2-72*x),x, algorithm="giac")

[Out]

Timed out

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maple [C]  time = 0.13, size = 149, normalized size = 7.45

method	result	size
risch	\(\frac {2 i \ln \relax (2)}{\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (x +\frac {5 \ln \relax (x )}{4}-\frac {1}{2}\right )\right ) \mathrm {csgn}\left (i x \left (x +\frac {5 \ln \relax (x )}{4}-\frac {1}{2}\right )\right )-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left (x +\frac {5 \ln \relax (x )}{4}-\frac {1}{2}\right )\right )^{2}+2 \pi \mathrm {csgn}\left (i x \left (x +\frac {5 \ln \relax (x )}{4}-\frac {1}{2}\right )\right )^{2}-\pi \,\mathrm {csgn}\left (i \left (x +\frac {5 \ln \relax (x )}{4}-\frac {1}{2}\right )\right ) \mathrm {csgn}\left (i x \left (x +\frac {5 \ln \relax (x )}{4}-\frac {1}{2}\right )\right )^{2}-\pi \mathrm {csgn}\left (i x \left (x +\frac {5 \ln \relax (x )}{4}-\frac {1}{2}\right )\right )^{3}-2 \pi +4 i \ln \relax (5)+2 i \ln \relax (x )+2 i \ln \left (x +\frac {5 \ln \relax (x )}{4}-\frac {1}{2}\right )+12 i}\)	\(149\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-5*ln(2)*ln(x)+(-8*x-3)*ln(2))/((5*x*ln(x)+4*x^2-2*x)*ln(-125*x*ln(x)-100*x^2+50*x)^2+(60*x*ln(x)+48*x^2-
24*x)*ln(-125*x*ln(x)-100*x^2+50*x)+180*x*ln(x)+144*x^2-72*x),x,method=_RETURNVERBOSE)

[Out]

2*I*ln(2)/(Pi*csgn(I*x)*csgn(I*(x+5/4*ln(x)-1/2))*csgn(I*x*(x+5/4*ln(x)-1/2))-Pi*csgn(I*x)*csgn(I*x*(x+5/4*ln(
x)-1/2))^2+2*Pi*csgn(I*x*(x+5/4*ln(x)-1/2))^2-Pi*csgn(I*(x+5/4*ln(x)-1/2))*csgn(I*x*(x+5/4*ln(x)-1/2))^2-Pi*cs
gn(I*x*(x+5/4*ln(x)-1/2))^3-2*Pi+4*I*ln(5)+2*I*ln(x)+2*I*ln(x+5/4*ln(x)-1/2)+12*I)

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maxima [C]  time = 0.73, size = 26, normalized size = 1.30 \begin {gather*} \frac {\log \relax (2)}{i \, \pi + 2 \, \log \relax (5) + \log \left (4 \, x + 5 \, \log \relax (x) - 2\right ) + \log \relax (x) + 6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*log(2)*log(x)+(-8*x-3)*log(2))/((5*x*log(x)+4*x^2-2*x)*log(-125*x*log(x)-100*x^2+50*x)^2+(60*x*l
og(x)+48*x^2-24*x)*log(-125*x*log(x)-100*x^2+50*x)+180*x*log(x)+144*x^2-72*x),x, algorithm="maxima")

[Out]

log(2)/(I*pi + 2*log(5) + log(4*x + 5*log(x) - 2) + log(x) + 6)

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mupad [B]  time = 1.63, size = 22, normalized size = 1.10 \begin {gather*} \frac {\ln \relax (2)}{\ln \left (50\,x-125\,x\,\ln \relax (x)-100\,x^2\right )+6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(2)*(8*x + 3) + 5*log(2)*log(x))/(log(50*x - 125*x*log(x) - 100*x^2)*(60*x*log(x) - 24*x + 48*x^2) -
72*x + log(50*x - 125*x*log(x) - 100*x^2)^2*(5*x*log(x) - 2*x + 4*x^2) + 180*x*log(x) + 144*x^2),x)

[Out]

log(2)/(log(50*x - 125*x*log(x) - 100*x^2) + 6)

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sympy [A]  time = 0.49, size = 20, normalized size = 1.00 \begin {gather*} \frac {\log {\relax (2 )}}{\log {\left (- 100 x^{2} - 125 x \log {\relax (x )} + 50 x \right )} + 6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*ln(2)*ln(x)+(-8*x-3)*ln(2))/((5*x*ln(x)+4*x**2-2*x)*ln(-125*x*ln(x)-100*x**2+50*x)**2+(60*x*ln(x
)+48*x**2-24*x)*ln(-125*x*ln(x)-100*x**2+50*x)+180*x*ln(x)+144*x**2-72*x),x)

[Out]

log(2)/(log(-100*x**2 - 125*x*log(x) + 50*x) + 6)

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