3.21.83 \(\int \frac {240 x-24 x \log (4)}{16 x^4-24 x^2 \log (5)+9 \log ^2(5)} \, dx\)

Optimal. Leaf size=19 \[ \frac {-10+\log (4)}{\frac {4 x^2}{3}-\log (5)} \]

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Rubi [A]  time = 0.01, antiderivative size = 20, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {6, 12, 28, 261} \begin {gather*} -\frac {3 (10-\log (4))}{4 x^2-3 \log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(240*x - 24*x*Log[4])/(16*x^4 - 24*x^2*Log[5] + 9*Log[5]^2),x]

[Out]

(-3*(10 - Log[4]))/(4*x^2 - 3*Log[5])

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x (240-24 \log (4))}{16 x^4-24 x^2 \log (5)+9 \log ^2(5)} \, dx\\ &=(24 (10-\log (4))) \int \frac {x}{16 x^4-24 x^2 \log (5)+9 \log ^2(5)} \, dx\\ &=(384 (10-\log (4))) \int \frac {x}{\left (16 x^2-12 \log (5)\right )^2} \, dx\\ &=-\frac {3 (10-\log (4))}{4 x^2-3 \log (5)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 18, normalized size = 0.95 \begin {gather*} \frac {3 (-10+\log (4))}{4 x^2-3 \log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(240*x - 24*x*Log[4])/(16*x^4 - 24*x^2*Log[5] + 9*Log[5]^2),x]

[Out]

(3*(-10 + Log[4]))/(4*x^2 - 3*Log[5])

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fricas [A]  time = 1.11, size = 18, normalized size = 0.95 \begin {gather*} \frac {6 \, {\left (\log \relax (2) - 5\right )}}{4 \, x^{2} - 3 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-48*x*log(2)+240*x)/(9*log(5)^2-24*x^2*log(5)+16*x^4),x, algorithm="fricas")

[Out]

6*(log(2) - 5)/(4*x^2 - 3*log(5))

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giac [B]  time = 2.99, size = 37, normalized size = 1.95 \begin {gather*} \frac {6 \, {\left (\log \relax (2)^{2} - 10 \, \log \relax (2) + 25\right )}}{4 \, x^{2} \log \relax (2) - 20 \, x^{2} - 3 \, \log \relax (5) \log \relax (2) + 15 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-48*x*log(2)+240*x)/(9*log(5)^2-24*x^2*log(5)+16*x^4),x, algorithm="giac")

[Out]

6*(log(2)^2 - 10*log(2) + 25)/(4*x^2*log(2) - 20*x^2 - 3*log(5)*log(2) + 15*log(5))

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maple [A]  time = 0.05, size = 19, normalized size = 1.00

method	result	size
gosper	\(-\frac {6 \left (\ln \relax (2)-5\right )}{-4 x^{2}+3 \ln \relax (5)}\)	\(19\)
default	\(\frac {-30+6 \ln \relax (2)}{-3 \ln \relax (5)+4 x^{2}}\)	\(19\)
norman	\(\frac {30-6 \ln \relax (2)}{-4 x^{2}+3 \ln \relax (5)}\)	\(20\)
risch	\(-\frac {2 \ln \relax (2)}{-\frac {4 x^{2}}{3}+\ln \relax (5)}+\frac {10}{-\frac {4 x^{2}}{3}+\ln \relax (5)}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-48*x*ln(2)+240*x)/(9*ln(5)^2-24*x^2*ln(5)+16*x^4),x,method=_RETURNVERBOSE)

[Out]

-6*(ln(2)-5)/(-4*x^2+3*ln(5))

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maxima [A]  time = 0.51, size = 18, normalized size = 0.95 \begin {gather*} \frac {6 \, {\left (\log \relax (2) - 5\right )}}{4 \, x^{2} - 3 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-48*x*log(2)+240*x)/(9*log(5)^2-24*x^2*log(5)+16*x^4),x, algorithm="maxima")

[Out]

6*(log(2) - 5)/(4*x^2 - 3*log(5))

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mupad [B]  time = 1.42, size = 113, normalized size = 5.95 \begin {gather*} -\frac {30\,\mathrm {atanh}\left (\frac {4\,x^2\,\sqrt {3\,\ln \relax (5)+\ln \left (125\right )}\,\sqrt {3\,\ln \relax (5)-\ln \left (125\right )}}{12\,x^2\,\ln \relax (5)-{\ln \left (125\right )}^2}\right )-2\,\ln \relax (8)\,\mathrm {atanh}\left (\frac {4\,x^2\,\sqrt {3\,\ln \relax (5)+\ln \left (125\right )}\,\sqrt {3\,\ln \relax (5)-\ln \left (125\right )}}{12\,x^2\,\ln \relax (5)-{\ln \left (125\right )}^2}\right )}{\sqrt {3\,\ln \relax (5)+\ln \left (125\right )}\,\sqrt {3\,\ln \relax (5)-\ln \left (125\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((240*x - 48*x*log(2))/(9*log(5)^2 - 24*x^2*log(5) + 16*x^4),x)

[Out]

-(30*atanh((4*x^2*(3*log(5) + log(125))^(1/2)*(3*log(5) - log(125))^(1/2))/(12*x^2*log(5) - log(125)^2)) - 2*l
og(8)*atanh((4*x^2*(3*log(5) + log(125))^(1/2)*(3*log(5) - log(125))^(1/2))/(12*x^2*log(5) - log(125)^2)))/((3
*log(5) + log(125))^(1/2)*(3*log(5) - log(125))^(1/2))

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sympy [A]  time = 0.13, size = 17, normalized size = 0.89 \begin {gather*} - \frac {240 - 48 \log {\relax (2 )}}{32 x^{2} - 24 \log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-48*x*ln(2)+240*x)/(9*ln(5)**2-24*x**2*ln(5)+16*x**4),x)

[Out]

-(240 - 48*log(2))/(32*x**2 - 24*log(5))

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