3.43.45 \(\int \frac {156-30 x^2+(-52+12 x^2) \log (x)+(13-3 x^2) \log (\frac {1}{3} (-52+12 x^2))}{-52 x^2+12 x^4} \, dx\) [4245]

Optimal. Leaf size=27 \[ \frac {2-x-\log (x)+\frac {1}{4} \log \left (4 \left (-\frac {13}{3}+x^2\right )\right )}{x} \]

[Out]

(2+1/4*ln(4*x^2-52/3)-x-ln(x))/x

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Rubi [A]
time = 0.46, antiderivative size = 30, normalized size of antiderivative = 1.11, number of steps used = 10, number of rules used = 6, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {1607, 6857, 464, 213, 2341, 2505} \begin {gather*} \frac {\log \left (4 x^2-\frac {52}{3}\right )}{4 x}+\frac {2}{x}-\frac {\log (x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(156 - 30*x^2 + (-52 + 12*x^2)*Log[x] + (13 - 3*x^2)*Log[(-52 + 12*x^2)/3])/(-52*x^2 + 12*x^4),x]

[Out]

2/x - Log[x]/x + Log[-52/3 + 4*x^2]/(4*x)

Rule 213

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

Rule 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2341

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^
n])/(d*(m + 1))), x] - Simp[b*n*((d*x)^(m + 1)/(d*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2505

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.)*(x_))^(m_.), x_Symbol] :> Simp[(f*x)^(m +
1)*((a + b*Log[c*(d + e*x^n)^p])/(f*(m + 1))), x] - Dist[b*e*n*(p/(f*(m + 1))), Int[x^(n - 1)*((f*x)^(m + 1)/
(d + e*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && NeQ[m, -1]

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {156-30 x^2+\left (-52+12 x^2\right ) \log (x)+\left (13-3 x^2\right ) \log \left (\frac {1}{3} \left (-52+12 x^2\right )\right )}{x^2 \left (-52+12 x^2\right )} \, dx\\ &=\int \left (\frac {78-15 x^2-26 \log (x)+6 x^2 \log (x)}{2 x^2 \left (-13+3 x^2\right )}-\frac {\log \left (-\frac {52}{3}+4 x^2\right )}{4 x^2}\right ) \, dx\\ &=-\left (\frac {1}{4} \int \frac {\log \left (-\frac {52}{3}+4 x^2\right )}{x^2} \, dx\right )+\frac {1}{2} \int \frac {78-15 x^2-26 \log (x)+6 x^2 \log (x)}{x^2 \left (-13+3 x^2\right )} \, dx\\ &=\frac {\log \left (-\frac {52}{3}+4 x^2\right )}{4 x}+\frac {1}{2} \int \left (-\frac {3 \left (-26+5 x^2\right )}{x^2 \left (-13+3 x^2\right )}+\frac {2 \log (x)}{x^2}\right ) \, dx-2 \int \frac {1}{-\frac {52}{3}+4 x^2} \, dx\\ &=\frac {1}{2} \sqrt {\frac {3}{13}} \tanh ^{-1}\left (\sqrt {\frac {3}{13}} x\right )+\frac {\log \left (-\frac {52}{3}+4 x^2\right )}{4 x}-\frac {3}{2} \int \frac {-26+5 x^2}{x^2 \left (-13+3 x^2\right )} \, dx+\int \frac {\log (x)}{x^2} \, dx\\ &=\frac {2}{x}+\frac {1}{2} \sqrt {\frac {3}{13}} \tanh ^{-1}\left (\sqrt {\frac {3}{13}} x\right )-\frac {\log (x)}{x}+\frac {\log \left (-\frac {52}{3}+4 x^2\right )}{4 x}+\frac {3}{2} \int \frac {1}{-13+3 x^2} \, dx\\ &=\frac {2}{x}-\frac {\log (x)}{x}+\frac {\log \left (-\frac {52}{3}+4 x^2\right )}{4 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.04, size = 31, normalized size = 1.15 \begin {gather*} \frac {1}{4} \left (\frac {8}{x}-\frac {4 \log (x)}{x}+\frac {\log \left (-\frac {52}{3}+4 x^2\right )}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(156 - 30*x^2 + (-52 + 12*x^2)*Log[x] + (13 - 3*x^2)*Log[(-52 + 12*x^2)/3])/(-52*x^2 + 12*x^4),x]

[Out]

(8/x - (4*Log[x])/x + Log[-52/3 + 4*x^2]/x)/4

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Maple [A]
time = 0.44, size = 41, normalized size = 1.52

method	result	size
risch	\(\frac {\ln \left (4 x^{2}-\frac {52}{3}\right )}{4 x}-\frac {\ln \left (x \right )-2}{x}\)	\(24\)
default	\(-\frac {\ln \left (3\right )}{4 x}+\frac {\ln \left (3 x^{2}-13\right )}{4 x}+\frac {\ln \left (2\right )}{2 x}-\frac {\ln \left (x \right )}{x}+\frac {2}{x}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-3*x^2+13)*ln(4*x^2-52/3)+(12*x^2-52)*ln(x)-30*x^2+156)/(12*x^4-52*x^2),x,method=_RETURNVERBOSE)

[Out]

-1/4*ln(3)/x+1/4/x*ln(3*x^2-13)+1/2*ln(2)/x-ln(x)/x+2/x

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Maxima [A]
time = 0.54, size = 33, normalized size = 1.22 \begin {gather*} -\frac {\log \left (3\right ) - 2 \, \log \left (2\right ) - \log \left (3 \, x^{2} - 13\right ) + 4 \, \log \left (x\right ) + 4}{4 \, x} + \frac {3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^2+13)*log(4*x^2-52/3)+(12*x^2-52)*log(x)-30*x^2+156)/(12*x^4-52*x^2),x, algorithm="maxima")

[Out]

-1/4*(log(3) - 2*log(2) - log(3*x^2 - 13) + 4*log(x) + 4)/x + 3/x

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Fricas [A]
time = 0.41, size = 19, normalized size = 0.70 \begin {gather*} \frac {\log \left (4 \, x^{2} - \frac {52}{3}\right ) - 4 \, \log \left (x\right ) + 8}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^2+13)*log(4*x^2-52/3)+(12*x^2-52)*log(x)-30*x^2+156)/(12*x^4-52*x^2),x, algorithm="fricas")

[Out]

1/4*(log(4*x^2 - 52/3) - 4*log(x) + 8)/x

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Sympy [A]
time = 0.14, size = 20, normalized size = 0.74 \begin {gather*} - \frac {\log {\left (x \right )}}{x} + \frac {\log {\left (4 x^{2} - \frac {52}{3} \right )}}{4 x} + \frac {2}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x**2+13)*ln(4*x**2-52/3)+(12*x**2-52)*ln(x)-30*x**2+156)/(12*x**4-52*x**2),x)

[Out]

-log(x)/x + log(4*x**2 - 52/3)/(4*x) + 2/x

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Giac [A]
time = 0.43, size = 30, normalized size = 1.11 \begin {gather*} -\frac {\log \left (3\right ) - 8}{4 \, x} + \frac {\log \left (12 \, x^{2} - 52\right )}{4 \, x} - \frac {\log \left (x\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^2+13)*log(4*x^2-52/3)+(12*x^2-52)*log(x)-30*x^2+156)/(12*x^4-52*x^2),x, algorithm="giac")

[Out]

-1/4*(log(3) - 8)/x + 1/4*log(12*x^2 - 52)/x - log(x)/x

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Mupad [B]
time = 3.37, size = 19, normalized size = 0.70 \begin {gather*} \frac {\ln \left (4\,x^2-\frac {52}{3}\right )-4\,\ln \left (x\right )+8}{4\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((30*x^2 - log(x)*(12*x^2 - 52) + log(4*x^2 - 52/3)*(3*x^2 - 13) - 156)/(52*x^2 - 12*x^4),x)

[Out]

(log(4*x^2 - 52/3) - 4*log(x) + 8)/(4*x)

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