3.98.98 \(\int \frac {6-6 x+(9-6 x-18 x^2+12 x^3) \log (\frac {-1+2 x^2}{x^2})}{-2 x^4+4 x^6} \, dx\)

Optimal. Leaf size=20 \[ \frac {3 (1-x) \log \left (2-\frac {1}{x^2}\right )}{2 x^3} \]

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Rubi [B]  time = 0.32, antiderivative size = 95, normalized size of antiderivative = 4.75, number of steps used = 18, number of rules used = 13, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.277, Rules used = {1593, 6725, 801, 633, 31, 2466, 2455, 263, 325, 207, 2454, 2389, 2295} \begin {gather*} \frac {3}{2} \left (2-\frac {1}{x^2}\right ) \log \left (2-\frac {1}{x^2}\right )+\frac {3 \log \left (2-\frac {1}{x^2}\right )}{2 x^3}+6 \log (x)-3 \left (1-\sqrt {2}\right ) \log \left (1-\sqrt {2} x\right )-3 \left (1+\sqrt {2}\right ) \log \left (\sqrt {2} x+1\right )+6 \sqrt {2} \tanh ^{-1}\left (\sqrt {2} x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(6 - 6*x + (9 - 6*x - 18*x^2 + 12*x^3)*Log[(-1 + 2*x^2)/x^2])/(-2*x^4 + 4*x^6),x]

[Out]

6*Sqrt[2]*ArcTanh[Sqrt[2]*x] + (3*(2 - x^(-2))*Log[2 - x^(-2)])/2 + (3*Log[2 - x^(-2)])/(2*x^3) + 6*Log[x] - 3
*(1 - Sqrt[2])*Log[1 - Sqrt[2]*x] - 3*(1 + Sqrt[2])*Log[1 + Sqrt[2]*x]

Rule 31

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

Rule 207

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

Rule 263

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

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 1593

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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2389

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

Rule 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rule 2455

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 2466

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_.) + (g_.)*(x_))^(r_.), x_S
ymbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x^n)^p])^q, x^m*(f + g*x)^r, x], x] /; FreeQ[{a, b, c, d, e,
f, g, n, p, q}, x] && IntegerQ[m] && IntegerQ[r]

Rule 6725

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 {6-6 x+\left (9-6 x-18 x^2+12 x^3\right ) \log \left (\frac {-1+2 x^2}{x^2}\right )}{x^4 \left (-2+4 x^2\right )} \, dx\\ &=\int \left (-\frac {3 (-1+x)}{x^4 \left (-1+2 x^2\right )}+\frac {3 (-3+2 x) \log \left (2-\frac {1}{x^2}\right )}{2 x^4}\right ) \, dx\\ &=\frac {3}{2} \int \frac {(-3+2 x) \log \left (2-\frac {1}{x^2}\right )}{x^4} \, dx-3 \int \frac {-1+x}{x^4 \left (-1+2 x^2\right )} \, dx\\ &=\frac {3}{2} \int \left (-\frac {3 \log \left (2-\frac {1}{x^2}\right )}{x^4}+\frac {2 \log \left (2-\frac {1}{x^2}\right )}{x^3}\right ) \, dx-3 \int \left (\frac {1}{x^4}-\frac {1}{x^3}+\frac {2}{x^2}-\frac {2}{x}+\frac {4 (-1+x)}{-1+2 x^2}\right ) \, dx\\ &=\frac {1}{x^3}-\frac {3}{2 x^2}+\frac {6}{x}+6 \log (x)+3 \int \frac {\log \left (2-\frac {1}{x^2}\right )}{x^3} \, dx-\frac {9}{2} \int \frac {\log \left (2-\frac {1}{x^2}\right )}{x^4} \, dx-12 \int \frac {-1+x}{-1+2 x^2} \, dx\\ &=\frac {1}{x^3}-\frac {3}{2 x^2}+\frac {6}{x}+\frac {3 \log \left (2-\frac {1}{x^2}\right )}{2 x^3}+6 \log (x)-\frac {3}{2} \operatorname {Subst}\left (\int \log (2-x) \, dx,x,\frac {1}{x^2}\right )-3 \int \frac {1}{\left (2-\frac {1}{x^2}\right ) x^6} \, dx-\left (6 \left (1-\sqrt {2}\right )\right ) \int \frac {1}{-\sqrt {2}+2 x} \, dx-\left (6 \left (1+\sqrt {2}\right )\right ) \int \frac {1}{\sqrt {2}+2 x} \, dx\\ &=\frac {1}{x^3}-\frac {3}{2 x^2}+\frac {6}{x}+\frac {3 \log \left (2-\frac {1}{x^2}\right )}{2 x^3}+6 \log (x)-3 \left (1-\sqrt {2}\right ) \log \left (1-\sqrt {2} x\right )-3 \left (1+\sqrt {2}\right ) \log \left (1+\sqrt {2} x\right )+\frac {3}{2} \operatorname {Subst}\left (\int \log (x) \, dx,x,2-\frac {1}{x^2}\right )-3 \int \frac {1}{x^4 \left (-1+2 x^2\right )} \, dx\\ &=\frac {6}{x}+\frac {3}{2} \left (2-\frac {1}{x^2}\right ) \log \left (2-\frac {1}{x^2}\right )+\frac {3 \log \left (2-\frac {1}{x^2}\right )}{2 x^3}+6 \log (x)-3 \left (1-\sqrt {2}\right ) \log \left (1-\sqrt {2} x\right )-3 \left (1+\sqrt {2}\right ) \log \left (1+\sqrt {2} x\right )-6 \int \frac {1}{x^2 \left (-1+2 x^2\right )} \, dx\\ &=\frac {3}{2} \left (2-\frac {1}{x^2}\right ) \log \left (2-\frac {1}{x^2}\right )+\frac {3 \log \left (2-\frac {1}{x^2}\right )}{2 x^3}+6 \log (x)-3 \left (1-\sqrt {2}\right ) \log \left (1-\sqrt {2} x\right )-3 \left (1+\sqrt {2}\right ) \log \left (1+\sqrt {2} x\right )-12 \int \frac {1}{-1+2 x^2} \, dx\\ &=6 \sqrt {2} \tanh ^{-1}\left (\sqrt {2} x\right )+\frac {3}{2} \left (2-\frac {1}{x^2}\right ) \log \left (2-\frac {1}{x^2}\right )+\frac {3 \log \left (2-\frac {1}{x^2}\right )}{2 x^3}+6 \log (x)-3 \left (1-\sqrt {2}\right ) \log \left (1-\sqrt {2} x\right )-3 \left (1+\sqrt {2}\right ) \log \left (1+\sqrt {2} x\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.15, size = 86, normalized size = 4.30 \begin {gather*} 6 \sqrt {2} \tanh ^{-1}\left (\frac {1}{\sqrt {2} x}\right )+\frac {3 \left (1-x+2 x^3\right ) \log \left (2-\frac {1}{x^2}\right )}{2 x^3}+6 \log (x)+3 \left (-1+\sqrt {2}\right ) \log \left (1-\sqrt {2} x\right )-3 \left (1+\sqrt {2}\right ) \log \left (1+\sqrt {2} x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(6 - 6*x + (9 - 6*x - 18*x^2 + 12*x^3)*Log[(-1 + 2*x^2)/x^2])/(-2*x^4 + 4*x^6),x]

[Out]

6*Sqrt[2]*ArcTanh[1/(Sqrt[2]*x)] + (3*(1 - x + 2*x^3)*Log[2 - x^(-2)])/(2*x^3) + 6*Log[x] + 3*(-1 + Sqrt[2])*L
og[1 - Sqrt[2]*x] - 3*(1 + Sqrt[2])*Log[1 + Sqrt[2]*x]

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fricas [A]  time = 0.95, size = 20, normalized size = 1.00 \begin {gather*} -\frac {3 \, {\left (x - 1\right )} \log \left (\frac {2 \, x^{2} - 1}{x^{2}}\right )}{2 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x^3-18*x^2-6*x+9)*log((2*x^2-1)/x^2)+6-6*x)/(4*x^6-2*x^4),x, algorithm="fricas")

[Out]

-3/2*(x - 1)*log((2*x^2 - 1)/x^2)/x^3

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giac [A]  time = 0.15, size = 20, normalized size = 1.00 \begin {gather*} -\frac {3 \, {\left (x - 1\right )} \log \left (\frac {2 \, x^{2} - 1}{x^{2}}\right )}{2 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x^3-18*x^2-6*x+9)*log((2*x^2-1)/x^2)+6-6*x)/(4*x^6-2*x^4),x, algorithm="giac")

[Out]

-3/2*(x - 1)*log((2*x^2 - 1)/x^2)/x^3

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maple [A]  time = 0.11, size = 21, normalized size = 1.05

method	result	size
risch	\(-\frac {3 \left (x -1\right ) \ln \left (\frac {2 x^{2}-1}{x^{2}}\right )}{2 x^{3}}\)	\(21\)
norman	\(\frac {-\frac {3 \ln \left (\frac {2 x^{2}-1}{x^{2}}\right ) x}{2}+\frac {3 \ln \left (\frac {2 x^{2}-1}{x^{2}}\right )}{2}}{x^{3}}\)	\(35\)
derivativedivides	\(-3 \ln \left (\frac {1}{x^{2}}-2\right )+\frac {3 \ln \left (2-\frac {1}{x^{2}}\right ) \left (2-\frac {1}{x^{2}}\right )}{2}-3+\frac {3 \ln \left (2-\frac {1}{x^{2}}\right )}{2 x^{3}}\)	\(41\)
default	\(-3 \ln \left (\frac {1}{x^{2}}-2\right )+\frac {3 \ln \left (2-\frac {1}{x^{2}}\right ) \left (2-\frac {1}{x^{2}}\right )}{2}-3+\frac {3 \ln \left (2-\frac {1}{x^{2}}\right )}{2 x^{3}}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((12*x^3-18*x^2-6*x+9)*ln((2*x^2-1)/x^2)+6-6*x)/(4*x^6-2*x^4),x,method=_RETURNVERBOSE)

[Out]

-3/2*(x-1)/x^3*ln((2*x^2-1)/x^2)

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maxima [B]  time = 0.51, size = 44, normalized size = 2.20 \begin {gather*} -\frac {12 \, x^{2} + 3 \, {\left (x - 1\right )} \log \left (2 \, x^{2} - 1\right ) - 6 \, {\left (x - 1\right )} \log \relax (x) + 2}{2 \, x^{3}} + \frac {6 \, x^{2} + 1}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x^3-18*x^2-6*x+9)*log((2*x^2-1)/x^2)+6-6*x)/(4*x^6-2*x^4),x, algorithm="maxima")

[Out]

-1/2*(12*x^2 + 3*(x - 1)*log(2*x^2 - 1) - 6*(x - 1)*log(x) + 2)/x^3 + (6*x^2 + 1)/x^3

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mupad [B]  time = 5.96, size = 20, normalized size = 1.00 \begin {gather*} -\frac {3\,\ln \left (\frac {2\,x^2-1}{x^2}\right )\,\left (x-1\right )}{2\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*x + log((2*x^2 - 1)/x^2)*(6*x + 18*x^2 - 12*x^3 - 9) - 6)/(2*x^4 - 4*x^6),x)

[Out]

-(3*log((2*x^2 - 1)/x^2)*(x - 1))/(2*x^3)

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sympy [A]  time = 0.16, size = 20, normalized size = 1.00 \begin {gather*} \frac {\left (3 - 3 x\right ) \log {\left (\frac {2 x^{2} - 1}{x^{2}} \right )}}{2 x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x**3-18*x**2-6*x+9)*ln((2*x**2-1)/x**2)+6-6*x)/(4*x**6-2*x**4),x)

[Out]

(3 - 3*x)*log((2*x**2 - 1)/x**2)/(2*x**3)

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