3.178 \(\int \frac{9}{5 x^2 (3-2 x^2)^3} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0208878, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {12, 290, 325, 206} \[ \frac{1}{8 x \left (3-2 x^2\right )}+\frac{3}{20 x \left (3-2 x^2\right )^2}-\frac{1}{8 x}+\frac{\tanh ^{-1}\left (\sqrt{\frac{2}{3}} x\right )}{4 \sqrt{6}} \]

Antiderivative was successfully verified.

[In]

Int[9/(5*x^2*(3 - 2*x^2)^3),x]

[Out]

-1/(8*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 290

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] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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 206

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

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*}

Mathematica [A]  time = 0.0642256, size = 65, normalized size = 1.1 \[ \frac{1}{240} \left (-\frac{12 \left (10 x^4-25 x^2+12\right )}{x \left (3-2 x^2\right )^2}-5 \sqrt{6} \log \left (\sqrt{6}-2 x\right )+5 \sqrt{6} \log \left (2 x+\sqrt{6}\right )\right ) \]

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 = 0.009, size = 39, normalized size = 0.7 \begin{align*} -{\frac{1}{15\,x}}-{\frac{8}{15\, \left ( 2\,{x}^{2}-3 \right ) ^{2}} \left ({\frac{7\,{x}^{3}}{16}}-{\frac{27\,x}{32}} \right ) }+{\frac{\sqrt{6}}{24}{\it Artanh} \left ({\frac{x\sqrt{6}}{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 1.41092, size = 76, normalized size = 1.29 \begin{align*} -\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{align*}

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 = 1.78168, size = 182, normalized size = 3.08 \begin{align*} -\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{align*}

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 = 0.152695, size = 58, normalized size = 0.98 \begin{align*} - \frac{9 \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{align*}

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 = 1.05502, size = 74, normalized size = 1.25 \begin{align*} -\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{align*}

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