3.98.15 \(\int \frac {-400 \log (3)-4 \log (2) \log ^2(3)}{16 x^2 \log (2)-8 x \log (2) \log (3)+\log (2) \log ^2(3)} \, dx\)

Optimal. Leaf size=21 \[ \frac {x+\frac {25}{\log (2)}}{-\frac {1}{4}+\frac {x}{\log (3)}} \]

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Rubi [A]  time = 0.02, antiderivative size = 24, normalized size of antiderivative = 1.14, number of steps used = 4, number of rules used = 3, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {12, 27, 32} \begin {gather*} \frac {\log (3) (100+\log (2) \log (3))}{\log (2) (4 x-\log (3))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-400*Log[3] - 4*Log[2]*Log[3]^2)/(16*x^2*Log[2] - 8*x*Log[2]*Log[3] + Log[2]*Log[3]^2),x]

[Out]

(Log[3]*(100 + Log[2]*Log[3]))/(Log[2]*(4*x - Log[3]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left ((4 \log (3) (100+\log (2) \log (3))) \int \frac {1}{16 x^2 \log (2)-8 x \log (2) \log (3)+\log (2) \log ^2(3)} \, dx\right )\\ &=-\left ((4 \log (3) (100+\log (2) \log (3))) \int \frac {1}{\log (2) (4 x-\log (3))^2} \, dx\right )\\ &=-\frac {(4 \log (3) (100+\log (2) \log (3))) \int \frac {1}{(4 x-\log (3))^2} \, dx}{\log (2)}\\ &=\frac {\log (3) (100+\log (2) \log (3))}{\log (2) (4 x-\log (3))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 23, normalized size = 1.10 \begin {gather*} -\frac {\log (3) (100+\log (2) \log (3))}{\log (2) (-4 x+\log (3))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-400*Log[3] - 4*Log[2]*Log[3]^2)/(16*x^2*Log[2] - 8*x*Log[2]*Log[3] + Log[2]*Log[3]^2),x]

[Out]

-((Log[3]*(100 + Log[2]*Log[3]))/(Log[2]*(-4*x + Log[3])))

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fricas [A]  time = 0.80, size = 27, normalized size = 1.29 \begin {gather*} \frac {\log \relax (3)^{2} \log \relax (2) + 100 \, \log \relax (3)}{4 \, x \log \relax (2) - \log \relax (3) \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*log(2)*log(3)^2-400*log(3))/(log(2)*log(3)^2-8*x*log(2)*log(3)+16*x^2*log(2)),x, algorithm="fric
as")

[Out]

(log(3)^2*log(2) + 100*log(3))/(4*x*log(2) - log(3)*log(2))

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giac [A]  time = 0.31, size = 27, normalized size = 1.29 \begin {gather*} \frac {\log \relax (3)^{2} \log \relax (2) + 100 \, \log \relax (3)}{{\left (4 \, x - \log \relax (3)\right )} \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*log(2)*log(3)^2-400*log(3))/(log(2)*log(3)^2-8*x*log(2)*log(3)+16*x^2*log(2)),x, algorithm="giac
")

[Out]

(log(3)^2*log(2) + 100*log(3))/((4*x - log(3))*log(2))

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maple [A]  time = 0.14, size = 24, normalized size = 1.14

method	result	size
gosper	\(-\frac {\ln \relax (3) \left (\ln \relax (2) \ln \relax (3)+100\right )}{\ln \relax (2) \left (\ln \relax (3)-4 x \right )}\)	\(24\)
norman	\(-\frac {\ln \relax (3) \left (\ln \relax (2) \ln \relax (3)+100\right )}{\ln \relax (2) \left (\ln \relax (3)-4 x \right )}\)	\(24\)
default	\(-\frac {-4 \ln \relax (2) \ln \relax (3)^{2}-400 \ln \relax (3)}{4 \ln \relax (2) \left (-\ln \relax (3)+4 x \right )}\)	\(30\)
risch	\(-\frac {\ln \relax (3)^{2}}{\ln \relax (3)-4 x}-\frac {100 \ln \relax (3)}{\ln \relax (2) \left (\ln \relax (3)-4 x \right )}\)	\(32\)
meijerg	\(-\frac {4 x}{1-\frac {4 x}{\ln \relax (3)}}-\frac {400 x}{\ln \relax (3) \ln \relax (2) \left (1-\frac {4 x}{\ln \relax (3)}\right )}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*ln(2)*ln(3)^2-400*ln(3))/(ln(2)*ln(3)^2-8*x*ln(2)*ln(3)+16*x^2*ln(2)),x,method=_RETURNVERBOSE)

[Out]

-ln(3)*(ln(2)*ln(3)+100)/ln(2)/(ln(3)-4*x)

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maxima [A]  time = 0.38, size = 27, normalized size = 1.29 \begin {gather*} \frac {\log \relax (3)^{2} \log \relax (2) + 100 \, \log \relax (3)}{4 \, x \log \relax (2) - \log \relax (3) \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*log(2)*log(3)^2-400*log(3))/(log(2)*log(3)^2-8*x*log(2)*log(3)+16*x^2*log(2)),x, algorithm="maxi
ma")

[Out]

(log(3)^2*log(2) + 100*log(3))/(4*x*log(2) - log(3)*log(2))

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mupad [B]  time = 0.10, size = 24, normalized size = 1.14 \begin {gather*} \frac {\ln \relax (3)\,\left (\ln \relax (2)\,\ln \relax (3)+100\right )}{\ln \relax (2)\,\left (4\,x-\ln \relax (3)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(400*log(3) + 4*log(2)*log(3)^2)/(log(2)*log(3)^2 + 16*x^2*log(2) - 8*x*log(2)*log(3)),x)

[Out]

(log(3)*(log(2)*log(3) + 100))/(log(2)*(4*x - log(3)))

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sympy [B]  time = 0.12, size = 32, normalized size = 1.52 \begin {gather*} - \frac {- 400 \log {\relax (3 )} - 4 \log {\relax (2 )} \log {\relax (3 )}^{2}}{16 x \log {\relax (2 )} - 4 \log {\relax (2 )} \log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*ln(2)*ln(3)**2-400*ln(3))/(ln(2)*ln(3)**2-8*x*ln(2)*ln(3)+16*x**2*ln(2)),x)

[Out]

-(-400*log(3) - 4*log(2)*log(3)**2)/(16*x*log(2) - 4*log(2)*log(3))

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