3.16.92 \(\int \frac {48 x+4 x^3+4 x^2 \log (\frac {5}{2})+(-48 x+4 x^3+(-24+2 x^2) \log (\frac {5}{2})) \log (\frac {-12+x^2}{2 e^2 x+e^2 \log (\frac {5}{2})})}{-24 x+2 x^3+(-12+x^2) \log (\frac {5}{2})} \, dx\)

Optimal. Leaf size=28 \[ 2 x \log \left (\frac {4 \left (-3+\frac {x^2}{4}\right )}{e^2 \left (2 x+\log \left (\frac {5}{2}\right )\right )}\right ) \]

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Rubi [A]  time = 0.70, antiderivative size = 27, normalized size of antiderivative = 0.96, number of steps used = 13, number of rules used = 7, integrand size = 86, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {6741, 6688, 12, 6725, 1629, 207, 2523} \begin {gather*} 2 x \log \left (-\frac {12-x^2}{2 x+\log \left (\frac {5}{2}\right )}\right )-4 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(48*x + 4*x^3 + 4*x^2*Log[5/2] + (-48*x + 4*x^3 + (-24 + 2*x^2)*Log[5/2])*Log[(-12 + x^2)/(2*E^2*x + E^2*L
og[5/2])])/(-24*x + 2*x^3 + (-12 + x^2)*Log[5/2]),x]

[Out]

-4*x + 2*x*Log[-((12 - x^2)/(2*x + Log[5/2]))]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[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 1629

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 2523

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*Log[c*RFx^p])^n, x] - Dist[b*n*p
, Int[SimplifyIntegrand[(x*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x], x] /; FreeQ[{a, b, c, p}, x] &
& RationalFunctionQ[RFx, x] && IGtQ[n, 0]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

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]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-48 x-4 x^3-4 x^2 \log \left (\frac {5}{2}\right )-\left (-48 x+4 x^3+\left (-24+2 x^2\right ) \log \left (\frac {5}{2}\right )\right ) \log \left (\frac {-12+x^2}{2 e^2 x+e^2 \log \left (\frac {5}{2}\right )}\right )}{24 x-2 x^3+12 \log \left (\frac {5}{2}\right )-x^2 \log \left (\frac {5}{2}\right )} \, dx\\ &=\int \frac {2 \left (-72 x+2 x^3-24 \log \left (\frac {5}{2}\right )-\left (-12+x^2\right ) \left (2 x+\log \left (\frac {5}{2}\right )\right ) \log \left (\frac {-12+x^2}{2 x+\log \left (\frac {5}{2}\right )}\right )\right )}{\left (12-x^2\right ) \left (2 x+\log \left (\frac {5}{2}\right )\right )} \, dx\\ &=2 \int \frac {-72 x+2 x^3-24 \log \left (\frac {5}{2}\right )-\left (-12+x^2\right ) \left (2 x+\log \left (\frac {5}{2}\right )\right ) \log \left (\frac {-12+x^2}{2 x+\log \left (\frac {5}{2}\right )}\right )}{\left (12-x^2\right ) \left (2 x+\log \left (\frac {5}{2}\right )\right )} \, dx\\ &=2 \int \left (-\frac {2 \left (-36 x+x^3-12 \log \left (\frac {5}{2}\right )\right )}{\left (-12+x^2\right ) \left (2 x+\log \left (\frac {5}{2}\right )\right )}+\log \left (\frac {-12+x^2}{2 x+\log \left (\frac {5}{2}\right )}\right )\right ) \, dx\\ &=2 \int \log \left (\frac {-12+x^2}{2 x+\log \left (\frac {5}{2}\right )}\right ) \, dx-4 \int \frac {-36 x+x^3-12 \log \left (\frac {5}{2}\right )}{\left (-12+x^2\right ) \left (2 x+\log \left (\frac {5}{2}\right )\right )} \, dx\\ &=2 x \log \left (-\frac {12-x^2}{2 x+\log \left (\frac {5}{2}\right )}\right )-2 \int \frac {2 x \left (-12-x^2-x \log \left (\frac {5}{2}\right )\right )}{\left (12-x^2\right ) \left (2 x+\log \left (\frac {5}{2}\right )\right )} \, dx-4 \int \left (\frac {1}{2}-\frac {12}{-12+x^2}-\frac {\log \left (\frac {5}{2}\right )}{4 x+\log \left (\frac {25}{4}\right )}\right ) \, dx\\ &=-2 x+2 x \log \left (-\frac {12-x^2}{2 x+\log \left (\frac {5}{2}\right )}\right )+\log \left (\frac {5}{2}\right ) \log \left (4 x+\log \left (\frac {25}{4}\right )\right )-4 \int \frac {x \left (-12-x^2-x \log \left (\frac {5}{2}\right )\right )}{\left (12-x^2\right ) \left (2 x+\log \left (\frac {5}{2}\right )\right )} \, dx+48 \int \frac {1}{-12+x^2} \, dx\\ &=-2 x-8 \sqrt {3} \tanh ^{-1}\left (\frac {x}{2 \sqrt {3}}\right )+2 x \log \left (-\frac {12-x^2}{2 x+\log \left (\frac {5}{2}\right )}\right )+\log \left (\frac {5}{2}\right ) \log \left (4 x+\log \left (\frac {25}{4}\right )\right )-4 \int \left (\frac {1}{2}+\frac {12}{-12+x^2}+\frac {\log \left (\frac {5}{2}\right )}{4 x+\log \left (\frac {25}{4}\right )}\right ) \, dx\\ &=-4 x-8 \sqrt {3} \tanh ^{-1}\left (\frac {x}{2 \sqrt {3}}\right )+2 x \log \left (-\frac {12-x^2}{2 x+\log \left (\frac {5}{2}\right )}\right )-48 \int \frac {1}{-12+x^2} \, dx\\ &=-4 x+2 x \log \left (-\frac {12-x^2}{2 x+\log \left (\frac {5}{2}\right )}\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.09, size = 59, normalized size = 2.11 \begin {gather*} 2 \left (-2 x+x \log \left (\frac {-12+x^2}{2 x+\log \left (\frac {5}{2}\right )}\right )+\frac {1}{2} \log \left (\frac {5}{2}\right ) \log \left (2 x+\log \left (\frac {5}{2}\right )\right )-\frac {1}{2} \log \left (\frac {5}{2}\right ) \log \left (4 x+\log \left (\frac {25}{4}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(48*x + 4*x^3 + 4*x^2*Log[5/2] + (-48*x + 4*x^3 + (-24 + 2*x^2)*Log[5/2])*Log[(-12 + x^2)/(2*E^2*x +
E^2*Log[5/2])])/(-24*x + 2*x^3 + (-12 + x^2)*Log[5/2]),x]

[Out]

2*(-2*x + x*Log[(-12 + x^2)/(2*x + Log[5/2])] + (Log[5/2]*Log[2*x + Log[5/2]])/2 - (Log[5/2]*Log[4*x + Log[25/
4]])/2)

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fricas [A]  time = 0.82, size = 23, normalized size = 0.82 \begin {gather*} 2 \, x \log \left (\frac {x^{2} - 12}{2 \, x e^{2} + e^{2} \log \left (\frac {5}{2}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2-24)*log(5/2)+4*x^3-48*x)*log((x^2-12)/(exp(2)*log(5/2)+2*exp(2)*x))+4*x^2*log(5/2)+4*x^3+48
*x)/((x^2-12)*log(5/2)+2*x^3-24*x),x, algorithm="fricas")

[Out]

2*x*log((x^2 - 12)/(2*x*e^2 + e^2*log(5/2)))

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giac [B]  time = 0.31, size = 53, normalized size = 1.89 \begin {gather*} 2 \, x \log \left (x^{2} - 12\right ) + {\left (\log \relax (5) - \log \relax (2)\right )} \log \left (2 \, x + \log \relax (5) - \log \relax (2)\right ) - 2 \, x \log \left (2 \, x + \log \left (\frac {5}{2}\right )\right ) - \log \left (\frac {5}{2}\right ) \log \left (2 \, x + \log \left (\frac {5}{2}\right )\right ) - 4 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2-24)*log(5/2)+4*x^3-48*x)*log((x^2-12)/(exp(2)*log(5/2)+2*exp(2)*x))+4*x^2*log(5/2)+4*x^3+48
*x)/((x^2-12)*log(5/2)+2*x^3-24*x),x, algorithm="giac")

[Out]

2*x*log(x^2 - 12) + (log(5) - log(2))*log(2*x + log(5) - log(2)) - 2*x*log(2*x + log(5/2)) - log(5/2)*log(2*x
+ log(5/2)) - 4*x

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maple [A]  time = 0.18, size = 24, normalized size = 0.86

method	result	size
norman	\(2 x \ln \left (\frac {x^{2}-12}{{\mathrm e}^{2} \ln \left (\frac {5}{2}\right )+2 \,{\mathrm e}^{2} x}\right )\)	\(24\)
risch	\(2 x \ln \left (\frac {x^{2}-12}{{\mathrm e}^{2} \left (\ln \relax (5)-\ln \relax (2)\right )+2 \,{\mathrm e}^{2} x}\right )\)	\(29\)
default	\(-4 x +\ln \left (-\ln \relax (2)+\ln \relax (5)+2 x \right ) \ln \relax (5)-\ln \left (-\ln \relax (2)+\ln \relax (5)+2 x \right ) \ln \relax (2)+2 x \ln \left (\frac {x^{2}-12}{\ln \left (\frac {5}{2}\right )+2 x}\right )-\ln \left (\frac {5}{2}\right ) \ln \left (\ln \left (\frac {5}{2}\right )+2 x \right )\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x^2-24)*ln(5/2)+4*x^3-48*x)*ln((x^2-12)/(exp(2)*ln(5/2)+2*exp(2)*x))+4*x^2*ln(5/2)+4*x^3+48*x)/((x^2-
12)*ln(5/2)+2*x^3-24*x),x,method=_RETURNVERBOSE)

[Out]

2*x*ln((x^2-12)/(exp(2)*ln(5/2)+2*exp(2)*x))

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maxima [B]  time = 1.05, size = 233, normalized size = 8.32 \begin {gather*} -\frac {\log \left (\frac {5}{2}\right )^{3} \log \left (2 \, x + \log \left (\frac {5}{2}\right )\right )}{\log \left (\frac {5}{2}\right )^{2} - 48} + 2 \, {\left (\frac {\log \left (\frac {5}{2}\right )^{2} \log \left (2 \, x + \log \left (\frac {5}{2}\right )\right )}{\log \left (\frac {5}{2}\right )^{2} - 48} + \frac {2 \, \sqrt {3} \log \left (\frac {5}{2}\right ) \log \left (\frac {x - 2 \, \sqrt {3}}{x + 2 \, \sqrt {3}}\right )}{\log \left (\frac {5}{2}\right )^{2} - 48} - \frac {24 \, \log \left (x^{2} - 12\right )}{\log \left (\frac {5}{2}\right )^{2} - 48}\right )} \log \left (\frac {5}{2}\right ) + 2 \, x \log \left (x^{2} - 12\right ) - 2 \, x \log \left (2 \, x + \log \relax (5) - \log \relax (2)\right ) - {\left (\log \relax (5) - \log \relax (2)\right )} \log \left (2 \, x + \log \relax (5) - \log \relax (2)\right ) - 4 \, \sqrt {3} \log \left (\frac {x - 2 \, \sqrt {3}}{x + 2 \, \sqrt {3}}\right ) - 4 \, x + \frac {48 \, \log \left (\frac {5}{2}\right ) \log \left (x^{2} - 12\right )}{\log \left (\frac {5}{2}\right )^{2} - 48} - \frac {48 \, \log \left (\frac {5}{2}\right ) \log \left (2 \, x + \log \left (\frac {5}{2}\right )\right )}{\log \left (\frac {5}{2}\right )^{2} - 48} - \frac {192 \, \sqrt {3} \log \left (\frac {x - 2 \, \sqrt {3}}{x + 2 \, \sqrt {3}}\right )}{\log \left (\frac {5}{2}\right )^{2} - 48} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2-24)*log(5/2)+4*x^3-48*x)*log((x^2-12)/(exp(2)*log(5/2)+2*exp(2)*x))+4*x^2*log(5/2)+4*x^3+48
*x)/((x^2-12)*log(5/2)+2*x^3-24*x),x, algorithm="maxima")

[Out]

-log(5/2)^3*log(2*x + log(5/2))/(log(5/2)^2 - 48) + 2*(log(5/2)^2*log(2*x + log(5/2))/(log(5/2)^2 - 48) + 2*sq
rt(3)*log(5/2)*log((x - 2*sqrt(3))/(x + 2*sqrt(3)))/(log(5/2)^2 - 48) - 24*log(x^2 - 12)/(log(5/2)^2 - 48))*lo
g(5/2) + 2*x*log(x^2 - 12) - 2*x*log(2*x + log(5) - log(2)) - (log(5) - log(2))*log(2*x + log(5) - log(2)) - 4
*sqrt(3)*log((x - 2*sqrt(3))/(x + 2*sqrt(3))) - 4*x + 48*log(5/2)*log(x^2 - 12)/(log(5/2)^2 - 48) - 48*log(5/2
)*log(2*x + log(5/2))/(log(5/2)^2 - 48) - 192*sqrt(3)*log((x - 2*sqrt(3))/(x + 2*sqrt(3)))/(log(5/2)^2 - 48)

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((48*x + log((x^2 - 12)/(exp(2)*log(5/2) + 2*x*exp(2)))*(log(5/2)*(2*x^2 - 24) - 48*x + 4*x^3) + 4*x^2*log(
5/2) + 4*x^3)/(log(5/2)*(x^2 - 12) - 24*x + 2*x^3),x)

[Out]

\text{Hanged}

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sympy [A]  time = 0.30, size = 24, normalized size = 0.86 \begin {gather*} 2 x \log {\left (\frac {x^{2} - 12}{2 x e^{2} + e^{2} \log {\left (\frac {5}{2} \right )}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x**2-24)*ln(5/2)+4*x**3-48*x)*ln((x**2-12)/(exp(2)*ln(5/2)+2*exp(2)*x))+4*x**2*ln(5/2)+4*x**3+4
8*x)/((x**2-12)*ln(5/2)+2*x**3-24*x),x)

[Out]

2*x*log((x**2 - 12)/(2*x*exp(2) + exp(2)*log(5/2)))

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