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

Optimal. Leaf size=48 \[ \frac{1}{12} \log \left (9-x^4\right )-\frac{\log (x)}{3}+\frac{\tan ^{-1}\left (\frac{x}{\sqrt{3}}\right )}{\sqrt{3}}-\frac{\tanh ^{-1}\left (\frac{x}{\sqrt{3}}\right )}{\sqrt{3}} \]

[Out]

ArcTan[x/Sqrt[3]]/Sqrt[3] - ArcTanh[x/Sqrt[3]]/Sqrt[3] - Log[x]/3 + Log[9 - x^4]/12

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Rubi [A]  time = 0.0526631, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.529, Rules used = {1593, 1831, 266, 36, 31, 29, 298, 203, 206} \[ \frac{1}{12} \log \left (9-x^4\right )-\frac{\log (x)}{3}+\frac{\tan ^{-1}\left (\frac{x}{\sqrt{3}}\right )}{\sqrt{3}}-\frac{\tanh ^{-1}\left (\frac{x}{\sqrt{3}}\right )}{\sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[(3 + 2*x^3)/(-9*x + x^5),x]

[Out]

ArcTan[x/Sqrt[3]]/Sqrt[3] - ArcTanh[x/Sqrt[3]]/Sqrt[3] - Log[x]/3 + Log[9 - x^4]/12

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 1831

Int[((Pq_)*((c_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = Sum[((c*x)^(m + ii)*(Coeff[Pq,
x, ii] + Coeff[Pq, x, n/2 + ii]*x^(n/2)))/(c^ii*(a + b*x^n)), {ii, 0, n/2 - 1}]}, Int[v, x] /; SumQ[v]] /; Fr
eeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] && Expon[Pq, x] < n

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 31

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 203

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

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{3+2 x^3}{-9 x+x^5} \, dx &=\int \frac{3+2 x^3}{x \left (-9+x^4\right )} \, dx\\ &=\int \left (\frac{3}{x \left (-9+x^4\right )}+\frac{2 x^2}{-9+x^4}\right ) \, dx\\ &=2 \int \frac{x^2}{-9+x^4} \, dx+3 \int \frac{1}{x \left (-9+x^4\right )} \, dx\\ &=\frac{3}{4} \operatorname{Subst}\left (\int \frac{1}{(-9+x) x} \, dx,x,x^4\right )-\int \frac{1}{3-x^2} \, dx+\int \frac{1}{3+x^2} \, dx\\ &=\frac{\tan ^{-1}\left (\frac{x}{\sqrt{3}}\right )}{\sqrt{3}}-\frac{\tanh ^{-1}\left (\frac{x}{\sqrt{3}}\right )}{\sqrt{3}}+\frac{1}{12} \operatorname{Subst}\left (\int \frac{1}{-9+x} \, dx,x,x^4\right )-\frac{1}{12} \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^4\right )\\ &=\frac{\tan ^{-1}\left (\frac{x}{\sqrt{3}}\right )}{\sqrt{3}}-\frac{\tanh ^{-1}\left (\frac{x}{\sqrt{3}}\right )}{\sqrt{3}}-\frac{\log (x)}{3}+\frac{1}{12} \log \left (9-x^4\right )\\ \end{align*}

Mathematica [A]  time = 0.018192, size = 67, normalized size = 1.4 \[ \frac{1}{12} \left (\log \left (9-x^4\right )-4 \log (x)+2 \sqrt{3} \log \left (3-\sqrt{3} x\right )-2 \sqrt{3} \log \left (\sqrt{3} x+3\right )+4 \sqrt{3} \tan ^{-1}\left (\frac{x}{\sqrt{3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(3 + 2*x^3)/(-9*x + x^5),x]

[Out]

(4*Sqrt[3]*ArcTan[x/Sqrt[3]] - 4*Log[x] + 2*Sqrt[3]*Log[3 - Sqrt[3]*x] - 2*Sqrt[3]*Log[3 + Sqrt[3]*x] + Log[9
- x^4])/12

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Maple [A]  time = 0.007, size = 46, normalized size = 1. \begin{align*} -{\frac{\ln \left ( x \right ) }{3}}+{\frac{\ln \left ({x}^{2}+3 \right ) }{12}}+{\frac{\sqrt{3}}{3}\arctan \left ({\frac{x\sqrt{3}}{3}} \right ) }+{\frac{\ln \left ({x}^{2}-3 \right ) }{12}}-{\frac{\sqrt{3}}{3}{\it Artanh} \left ({\frac{x\sqrt{3}}{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*ln(x)+1/12*ln(x^2+3)+1/3*arctan(1/3*x*3^(1/2))*3^(1/2)+1/12*ln(x^2-3)-1/3*arctanh(1/3*x*3^(1/2))*3^(1/2)

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Maxima [A]  time = 1.41503, size = 73, normalized size = 1.52 \begin{align*} \frac{1}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3} x\right ) + \frac{1}{6} \, \sqrt{3} \log \left (\frac{x - \sqrt{3}}{x + \sqrt{3}}\right ) + \frac{1}{12} \, \log \left (x^{2} + 3\right ) + \frac{1}{12} \, \log \left (x^{2} - 3\right ) - \frac{1}{3} \, \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*sqrt(3)*arctan(1/3*sqrt(3)*x) + 1/6*sqrt(3)*log((x - sqrt(3))/(x + sqrt(3))) + 1/12*log(x^2 + 3) + 1/12*lo
g(x^2 - 3) - 1/3*log(x)

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Fricas [A]  time = 2.21509, size = 190, normalized size = 3.96 \begin{align*} \frac{1}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3} x\right ) + \frac{1}{6} \, \sqrt{3} \log \left (\frac{x^{2} - 2 \, \sqrt{3} x + 3}{x^{2} - 3}\right ) + \frac{1}{12} \, \log \left (x^{2} + 3\right ) + \frac{1}{12} \, \log \left (x^{2} - 3\right ) - \frac{1}{3} \, \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 1.32339, size = 172, normalized size = 3.58 \begin{align*} - \frac{\log{\left (x \right )}}{3} + \frac{\log{\left (x^{2} + 3 \right )}}{12} + \left (\frac{1}{12} - \frac{\sqrt{3}}{6}\right ) \log{\left (x - \frac{108000 \left (\frac{1}{12} - \frac{\sqrt{3}}{6}\right )^{4}}{481} + \frac{1368 \left (\frac{1}{12} - \frac{\sqrt{3}}{6}\right )^{3}}{481} + \frac{943 \sqrt{3}}{5772} + \frac{4158 \left (\frac{1}{12} - \frac{\sqrt{3}}{6}\right )^{2}}{481} + \frac{17413}{11544} \right )} + \left (\frac{1}{12} + \frac{\sqrt{3}}{6}\right ) \log{\left (x - \frac{108000 \left (\frac{1}{12} + \frac{\sqrt{3}}{6}\right )^{4}}{481} - \frac{943 \sqrt{3}}{5772} + \frac{1368 \left (\frac{1}{12} + \frac{\sqrt{3}}{6}\right )^{3}}{481} + \frac{4158 \left (\frac{1}{12} + \frac{\sqrt{3}}{6}\right )^{2}}{481} + \frac{17413}{11544} \right )} + \frac{\sqrt{3} \operatorname{atan}{\left (\frac{\sqrt{3} x}{3} \right )}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(x)/3 + log(x**2 + 3)/12 + (1/12 - sqrt(3)/6)*log(x - 108000*(1/12 - sqrt(3)/6)**4/481 + 1368*(1/12 - sqrt
(3)/6)**3/481 + 943*sqrt(3)/5772 + 4158*(1/12 - sqrt(3)/6)**2/481 + 17413/11544) + (1/12 + sqrt(3)/6)*log(x -
108000*(1/12 + sqrt(3)/6)**4/481 - 943*sqrt(3)/5772 + 1368*(1/12 + sqrt(3)/6)**3/481 + 4158*(1/12 + sqrt(3)/6)
**2/481 + 17413/11544) + sqrt(3)*atan(sqrt(3)*x/3)/3

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Giac [A]  time = 1.0724, size = 86, normalized size = 1.79 \begin{align*} \frac{1}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3} x\right ) + \frac{1}{6} \, \sqrt{3} \log \left (\frac{{\left | 2 \, x - 2 \, \sqrt{3} \right |}}{{\left | 2 \, x + 2 \, \sqrt{3} \right |}}\right ) + \frac{1}{12} \, \log \left (x^{2} + 3\right ) + \frac{1}{12} \, \log \left ({\left | x^{2} - 3 \right |}\right ) - \frac{1}{3} \, \log \left ({\left | x \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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