3.40.65 \(\int \frac {-18 x+2 x \log (25)}{-8-9 x^2+x^2 \log (25)} \, dx\)

Optimal. Leaf size=23 \[ \log \left (\frac {1}{4} \left (-4-5 x^2+\frac {1}{2} x (x+x \log (25))\right )\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 13, normalized size of antiderivative = 0.57, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {6, 12, 260} \begin {gather*} \log \left (x^2 (9-\log (25))+8\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-18*x + 2*x*Log[25])/(-8 - 9*x^2 + x^2*Log[25]),x]

[Out]

Log[8 + x^2*(9 - Log[25])]

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 260

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x (-18+2 \log (25))}{-8-9 x^2+x^2 \log (25)} \, dx\\ &=\int \frac {x (-18+2 \log (25))}{-8+x^2 (-9+\log (25))} \, dx\\ &=(-18+2 \log (25)) \int \frac {x}{-8+x^2 (-9+\log (25))} \, dx\\ &=\log \left (8+x^2 (9-\log (25))\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 23, normalized size = 1.00 \begin {gather*} \frac {2 (-9+\log (25)) \log \left (-8+x^2 (-9+\log (25))\right )}{-18+\log (625)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-18*x + 2*x*Log[25])/(-8 - 9*x^2 + x^2*Log[25]),x]

[Out]

(2*(-9 + Log[25])*Log[-8 + x^2*(-9 + Log[25])])/(-18 + Log[625])

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fricas [A]  time = 0.61, size = 15, normalized size = 0.65 \begin {gather*} \log \left (2 \, x^{2} \log \relax (5) - 9 \, x^{2} - 8\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*log(5)-18*x)/(2*x^2*log(5)-9*x^2-8),x, algorithm="fricas")

[Out]

log(2*x^2*log(5) - 9*x^2 - 8)

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giac [A]  time = 0.23, size = 15, normalized size = 0.65 \begin {gather*} \log \left (-2 \, x^{2} \log \relax (5) + 9 \, x^{2} + 8\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*log(5)-18*x)/(2*x^2*log(5)-9*x^2-8),x, algorithm="giac")

[Out]

log(-2*x^2*log(5) + 9*x^2 + 8)

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maple [A]  time = 0.05, size = 14, normalized size = 0.61

method	result	size
risch	\(\ln \left (-8+\left (2 \ln \relax (5)-9\right ) x^{2}\right )\)	\(14\)
derivativedivides	\(\ln \left (2 x^{2} \ln \relax (5)-9 x^{2}-8\right )\)	\(16\)
default	\(\ln \left (2 x^{2} \ln \relax (5)-9 x^{2}-8\right )\)	\(16\)
norman	\(\ln \left (2 x^{2} \ln \relax (5)-9 x^{2}-8\right )\)	\(16\)
meijerg	\(-\frac {\left (4 \ln \relax (5)-18\right ) \ln \left (1+\frac {x^{2} \left (-2 \ln \relax (5)+9\right )}{8}\right )}{2 \left (-2 \ln \relax (5)+9\right )}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x*ln(5)-18*x)/(2*x^2*ln(5)-9*x^2-8),x,method=_RETURNVERBOSE)

[Out]

ln(-8+(2*ln(5)-9)*x^2)

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maxima [A]  time = 0.40, size = 15, normalized size = 0.65 \begin {gather*} \log \left (2 \, x^{2} \log \relax (5) - 9 \, x^{2} - 8\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*log(5)-18*x)/(2*x^2*log(5)-9*x^2-8),x, algorithm="maxima")

[Out]

log(2*x^2*log(5) - 9*x^2 - 8)

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mupad [B]  time = 0.10, size = 11, normalized size = 0.48 \begin {gather*} \ln \left (x^2\,\left (\ln \left (25\right )-9\right )-8\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((18*x - 4*x*log(5))/(9*x^2 - 2*x^2*log(5) + 8),x)

[Out]

log(x^2*(log(25) - 9) - 8)

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sympy [A]  time = 0.20, size = 26, normalized size = 1.13 \begin {gather*} \frac {\left (-18 + 4 \log {\relax (5 )}\right ) \log {\left (x^{2} \left (-9 + 2 \log {\relax (5 )}\right ) - 8 \right )}}{2 \left (-9 + 2 \log {\relax (5 )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*ln(5)-18*x)/(2*x**2*ln(5)-9*x**2-8),x)

[Out]

(-18 + 4*log(5))*log(x**2*(-9 + 2*log(5)) - 8)/(2*(-9 + 2*log(5)))

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