3.2.78 \(\int \frac {9}{5 x^2 (3-2 x^2)^3} \, dx\) [178]

Optimal. Leaf size=59 \[ -\frac {1}{8 x}+\frac {3}{20 x \left (3-2 x^2\right )^2}+\frac {1}{8 x \left (3-2 x^2\right )}+\frac {\tanh ^{-1}\left (\sqrt {\frac {2}{3}} x\right )}{4 \sqrt {6}} \]

[Out]

-1/8/x+3/20/x/(-2*x^2+3)^2+1/8/x/(-2*x^2+3)+1/24*arctanh(1/3*x*6^(1/2))*6^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {12, 296, 331, 212} \begin {gather*} \frac {1}{8 x \left (3-2 x^2\right )}+\frac {3}{20 x \left (3-2 x^2\right )^2}-\frac {1}{8 x}+\frac {\tanh ^{-1}\left (\sqrt {\frac {2}{3}} x\right )}{4 \sqrt {6}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[9/(5*x^2*(3 - 2*x^2)^3),x]

[Out]

-1/8*1/x + 3/(20*x*(3 - 2*x^2)^2) + 1/(8*x*(3 - 2*x^2)) + ArcTanh[Sqrt[2/3]*x]/(4*Sqrt[6])

Rule 12

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

Rule 212

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

Rule 296

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

Rule 331

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]

Rubi steps

\begin {align*} \int \frac {9}{5 x^2 \left (3-2 x^2\right )^3} \, dx &=\frac {9}{5} \int \frac {1}{x^2 \left (3-2 x^2\right )^3} \, dx\\ &=\frac {3}{20 x \left (3-2 x^2\right )^2}+\frac {3}{4} \int \frac {1}{x^2 \left (3-2 x^2\right )^2} \, dx\\ &=\frac {3}{20 x \left (3-2 x^2\right )^2}+\frac {1}{8 x \left (3-2 x^2\right )}+\frac {3}{8} \int \frac {1}{x^2 \left (3-2 x^2\right )} \, dx\\ &=-\frac {1}{8 x}+\frac {3}{20 x \left (3-2 x^2\right )^2}+\frac {1}{8 x \left (3-2 x^2\right )}+\frac {1}{4} \int \frac {1}{3-2 x^2} \, dx\\ &=-\frac {1}{8 x}+\frac {3}{20 x \left (3-2 x^2\right )^2}+\frac {1}{8 x \left (3-2 x^2\right )}+\frac {\tanh ^{-1}\left (\sqrt {\frac {2}{3}} x\right )}{4 \sqrt {6}}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 65, normalized size = 1.10 \begin {gather*} \frac {1}{240} \left (-\frac {12 \left (12-25 x^2+10 x^4\right )}{x \left (3-2 x^2\right )^2}-5 \sqrt {6} \log \left (\sqrt {6}-2 x\right )+5 \sqrt {6} \log \left (\sqrt {6}+2 x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[9/(5*x^2*(3 - 2*x^2)^3),x]

[Out]

((-12*(12 - 25*x^2 + 10*x^4))/(x*(3 - 2*x^2)^2) - 5*Sqrt[6]*Log[Sqrt[6] - 2*x] + 5*Sqrt[6]*Log[Sqrt[6] + 2*x])
/240

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Maple [A]
time = 0.06, size = 39, normalized size = 0.66

method result size
default \(-\frac {1}{15 x}-\frac {8 \left (\frac {7}{16} x^{3}-\frac {27}{32} x \right )}{15 \left (2 x^{2}-3\right )^{2}}+\frac {\arctanh \left (\frac {x \sqrt {6}}{3}\right ) \sqrt {6}}{24}\) \(39\)
meijerg \(\frac {i \sqrt {6}\, \left (\frac {i \sqrt {6}\, \left (\frac {20}{3} x^{4}-\frac {50}{3} x^{2}+8\right )}{4 x \left (-\frac {2 x^{2}}{3}+1\right )^{2}}-\frac {15 i \arctanh \left (\frac {x \sqrt {2}\, \sqrt {3}}{3}\right )}{2}\right )}{180}\) \(51\)
risch \(\frac {-\frac {1}{2} x^{4}+\frac {5}{4} x^{2}-\frac {3}{5}}{\left (2 x^{2}-3\right )^{2} x}+\frac {\sqrt {6}\, \ln \left (2 x +\sqrt {6}\right )}{48}-\frac {\sqrt {6}\, \ln \left (2 x -\sqrt {6}\right )}{48}\) \(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(9/5/x^2/(-2*x^2+3)^3,x,method=_RETURNVERBOSE)

[Out]

-1/15/x-8/15*(7/16*x^3-27/32*x)/(2*x^2-3)^2+1/24*arctanh(1/3*x*6^(1/2))*6^(1/2)

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Maxima [A]
time = 3.24, size = 56, normalized size = 0.95 \begin {gather*} -\frac {1}{48} \, \sqrt {6} \log \left (\frac {2 \, x - \sqrt {6}}{2 \, x + \sqrt {6}}\right ) - \frac {10 \, x^{4} - 25 \, x^{2} + 12}{20 \, {\left (4 \, x^{5} - 12 \, x^{3} + 9 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(9/5/x^2/(-2*x^2+3)^3,x, algorithm="maxima")

[Out]

-1/48*sqrt(6)*log((2*x - sqrt(6))/(2*x + sqrt(6))) - 1/20*(10*x^4 - 25*x^2 + 12)/(4*x^5 - 12*x^3 + 9*x)

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Fricas [A]
time = 0.47, size = 73, normalized size = 1.24 \begin {gather*} -\frac {120 \, x^{4} - 5 \, \sqrt {6} {\left (4 \, x^{5} - 12 \, x^{3} + 9 \, x\right )} \log \left (\frac {2 \, x^{2} + 2 \, \sqrt {6} x + 3}{2 \, x^{2} - 3}\right ) - 300 \, x^{2} + 144}{240 \, {\left (4 \, x^{5} - 12 \, x^{3} + 9 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(9/5/x^2/(-2*x^2+3)^3,x, algorithm="fricas")

[Out]

-1/240*(120*x^4 - 5*sqrt(6)*(4*x^5 - 12*x^3 + 9*x)*log((2*x^2 + 2*sqrt(6)*x + 3)/(2*x^2 - 3)) - 300*x^2 + 144)
/(4*x^5 - 12*x^3 + 9*x)

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Sympy [A]
time = 0.06, size = 58, normalized size = 0.98 \begin {gather*} - \frac {9 \cdot \left (10 x^{4} - 25 x^{2} + 12\right )}{720 x^{5} - 2160 x^{3} + 1620 x} - \frac {\sqrt {6} \log {\left (x - \frac {\sqrt {6}}{2} \right )}}{48} + \frac {\sqrt {6} \log {\left (x + \frac {\sqrt {6}}{2} \right )}}{48} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(9/5/x**2/(-2*x**2+3)**3,x)

[Out]

-9*(10*x**4 - 25*x**2 + 12)/(720*x**5 - 2160*x**3 + 1620*x) - sqrt(6)*log(x - sqrt(6)/2)/48 + sqrt(6)*log(x +
sqrt(6)/2)/48

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Giac [A]
time = 0.52, size = 55, normalized size = 0.93 \begin {gather*} -\frac {1}{48} \, \sqrt {6} \log \left (\frac {{\left | 4 \, x - 2 \, \sqrt {6} \right |}}{{\left | 4 \, x + 2 \, \sqrt {6} \right |}}\right ) - \frac {14 \, x^{3} - 27 \, x}{60 \, {\left (2 \, x^{2} - 3\right )}^{2}} - \frac {1}{15 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(9/5/x^2/(-2*x^2+3)^3,x, algorithm="giac")

[Out]

-1/48*sqrt(6)*log(abs(4*x - 2*sqrt(6))/abs(4*x + 2*sqrt(6))) - 1/60*(14*x^3 - 27*x)/(2*x^2 - 3)^2 - 1/15/x

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Mupad [B]
time = 0.24, size = 41, normalized size = 0.69 \begin {gather*} \frac {\sqrt {6}\,\mathrm {atanh}\left (\frac {\sqrt {6}\,x}{3}\right )}{24}-\frac {\frac {x^4}{8}-\frac {5\,x^2}{16}+\frac {3}{20}}{x^5-3\,x^3+\frac {9\,x}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-9/(5*x^2*(2*x^2 - 3)^3),x)

[Out]

(6^(1/2)*atanh((6^(1/2)*x)/3))/24 - (x^4/8 - (5*x^2)/16 + 3/20)/((9*x)/4 - 3*x^3 + x^5)

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