3.198 \(\int \frac{2+3 x^2}{x^3 (3+5 x^2+x^4)^{3/2}} \, dx\)

Optimal. Leaf size=90 \[ -\frac{8 x^2+7}{39 x^2 \sqrt{x^4+5 x^2+3}}-\frac{2 \sqrt{x^4+5 x^2+3}}{39 x^2}+\frac{\tanh ^{-1}\left (\frac{5 x^2+6}{2 \sqrt{3} \sqrt{x^4+5 x^2+3}}\right )}{3 \sqrt{3}} \]

[Out]

-(7 + 8*x^2)/(39*x^2*Sqrt[3 + 5*x^2 + x^4]) - (2*Sqrt[3 + 5*x^2 + x^4])/(39*x^2) + ArcTanh[(6 + 5*x^2)/(2*Sqrt
[3]*Sqrt[3 + 5*x^2 + x^4])]/(3*Sqrt[3])

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Rubi [A]  time = 0.0707172, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {1251, 822, 806, 724, 206} \[ -\frac{8 x^2+7}{39 x^2 \sqrt{x^4+5 x^2+3}}-\frac{2 \sqrt{x^4+5 x^2+3}}{39 x^2}+\frac{\tanh ^{-1}\left (\frac{5 x^2+6}{2 \sqrt{3} \sqrt{x^4+5 x^2+3}}\right )}{3 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[(2 + 3*x^2)/(x^3*(3 + 5*x^2 + x^4)^(3/2)),x]

[Out]

-(7 + 8*x^2)/(39*x^2*Sqrt[3 + 5*x^2 + x^4]) - (2*Sqrt[3 + 5*x^2 + x^4])/(39*x^2) + ArcTanh[(6 + 5*x^2)/(2*Sqrt
[3]*Sqrt[3 + 5*x^2 + x^4])]/(3*Sqrt[3])

Rule 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

Rule 822

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2*a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*e))*x
)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*
c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2
*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d*m + b*e*m) - b*d*(3*c*d -
b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, b,
c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] ||
IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2)), x] - Dist[(b
*(e*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x],
x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[Sim
plify[m + 2*p + 3], 0]

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 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{2+3 x^2}{x^3 \left (3+5 x^2+x^4\right )^{3/2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{2+3 x}{x^2 \left (3+5 x+x^2\right )^{3/2}} \, dx,x,x^2\right )\\ &=-\frac{7+8 x^2}{39 x^2 \sqrt{3+5 x^2+x^4}}-\frac{1}{39} \operatorname{Subst}\left (\int \frac{-6+8 x}{x^2 \sqrt{3+5 x+x^2}} \, dx,x,x^2\right )\\ &=-\frac{7+8 x^2}{39 x^2 \sqrt{3+5 x^2+x^4}}-\frac{2 \sqrt{3+5 x^2+x^4}}{39 x^2}-\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{x \sqrt{3+5 x+x^2}} \, dx,x,x^2\right )\\ &=-\frac{7+8 x^2}{39 x^2 \sqrt{3+5 x^2+x^4}}-\frac{2 \sqrt{3+5 x^2+x^4}}{39 x^2}+\frac{2}{3} \operatorname{Subst}\left (\int \frac{1}{12-x^2} \, dx,x,\frac{6+5 x^2}{\sqrt{3+5 x^2+x^4}}\right )\\ &=-\frac{7+8 x^2}{39 x^2 \sqrt{3+5 x^2+x^4}}-\frac{2 \sqrt{3+5 x^2+x^4}}{39 x^2}+\frac{\tanh ^{-1}\left (\frac{6+5 x^2}{2 \sqrt{3} \sqrt{3+5 x^2+x^4}}\right )}{3 \sqrt{3}}\\ \end{align*}

Mathematica [A]  time = 0.0183794, size = 88, normalized size = 0.98 \[ \frac{-6 x^4-54 x^2+13 \sqrt{3} \sqrt{x^4+5 x^2+3} x^2 \tanh ^{-1}\left (\frac{5 x^2+6}{2 \sqrt{3} \sqrt{x^4+5 x^2+3}}\right )-39}{117 x^2 \sqrt{x^4+5 x^2+3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(2 + 3*x^2)/(x^3*(3 + 5*x^2 + x^4)^(3/2)),x]

[Out]

(-39 - 54*x^2 - 6*x^4 + 13*Sqrt[3]*x^2*Sqrt[3 + 5*x^2 + x^4]*ArcTanh[(6 + 5*x^2)/(2*Sqrt[3]*Sqrt[3 + 5*x^2 + x
^4])])/(117*x^2*Sqrt[3 + 5*x^2 + x^4])

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Maple [A]  time = 0.013, size = 84, normalized size = 0.9 \begin{align*} -{\frac{1}{3}{\frac{1}{\sqrt{{x}^{4}+5\,{x}^{2}+3}}}}-{\frac{2\,{x}^{2}+5}{39}{\frac{1}{\sqrt{{x}^{4}+5\,{x}^{2}+3}}}}+{\frac{\sqrt{3}}{9}{\it Artanh} \left ({\frac{ \left ( 5\,{x}^{2}+6 \right ) \sqrt{3}}{6}{\frac{1}{\sqrt{{x}^{4}+5\,{x}^{2}+3}}}} \right ) }-{\frac{1}{3\,{x}^{2}}{\frac{1}{\sqrt{{x}^{4}+5\,{x}^{2}+3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^2+2)/x^3/(x^4+5*x^2+3)^(3/2),x)

[Out]

-1/3/(x^4+5*x^2+3)^(1/2)-1/39*(2*x^2+5)/(x^4+5*x^2+3)^(1/2)+1/9*arctanh(1/6*(5*x^2+6)*3^(1/2)/(x^4+5*x^2+3)^(1
/2))*3^(1/2)-1/3/x^2/(x^4+5*x^2+3)^(1/2)

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Maxima [A]  time = 1.66629, size = 111, normalized size = 1.23 \begin{align*} -\frac{2 \, x^{2}}{39 \, \sqrt{x^{4} + 5 \, x^{2} + 3}} + \frac{1}{9} \, \sqrt{3} \log \left (\frac{2 \, \sqrt{3} \sqrt{x^{4} + 5 \, x^{2} + 3}}{x^{2}} + \frac{6}{x^{2}} + 5\right ) - \frac{6}{13 \, \sqrt{x^{4} + 5 \, x^{2} + 3}} - \frac{1}{3 \, \sqrt{x^{4} + 5 \, x^{2} + 3} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2+2)/x^3/(x^4+5*x^2+3)^(3/2),x, algorithm="maxima")

[Out]

-2/39*x^2/sqrt(x^4 + 5*x^2 + 3) + 1/9*sqrt(3)*log(2*sqrt(3)*sqrt(x^4 + 5*x^2 + 3)/x^2 + 6/x^2 + 5) - 6/13/sqrt
(x^4 + 5*x^2 + 3) - 1/3/(sqrt(x^4 + 5*x^2 + 3)*x^2)

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Fricas [A]  time = 1.349, size = 308, normalized size = 3.42 \begin{align*} -\frac{6 \, x^{6} + 30 \, x^{4} - 13 \, \sqrt{3}{\left (x^{6} + 5 \, x^{4} + 3 \, x^{2}\right )} \log \left (\frac{25 \, x^{2} + 2 \, \sqrt{3}{\left (5 \, x^{2} + 6\right )} + 2 \, \sqrt{x^{4} + 5 \, x^{2} + 3}{\left (5 \, \sqrt{3} + 6\right )} + 30}{x^{2}}\right ) + 18 \, x^{2} + 3 \,{\left (2 \, x^{4} + 18 \, x^{2} + 13\right )} \sqrt{x^{4} + 5 \, x^{2} + 3}}{117 \,{\left (x^{6} + 5 \, x^{4} + 3 \, x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2+2)/x^3/(x^4+5*x^2+3)^(3/2),x, algorithm="fricas")

[Out]

-1/117*(6*x^6 + 30*x^4 - 13*sqrt(3)*(x^6 + 5*x^4 + 3*x^2)*log((25*x^2 + 2*sqrt(3)*(5*x^2 + 6) + 2*sqrt(x^4 + 5
*x^2 + 3)*(5*sqrt(3) + 6) + 30)/x^2) + 18*x^2 + 3*(2*x^4 + 18*x^2 + 13)*sqrt(x^4 + 5*x^2 + 3))/(x^6 + 5*x^4 +
3*x^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{3 x^{2} + 2}{x^{3} \left (x^{4} + 5 x^{2} + 3\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x**2+2)/x**3/(x**4+5*x**2+3)**(3/2),x)

[Out]

Integral((3*x**2 + 2)/(x**3*(x**4 + 5*x**2 + 3)**(3/2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{3 \, x^{2} + 2}{{\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac{3}{2}} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2+2)/x^3/(x^4+5*x^2+3)^(3/2),x, algorithm="giac")

[Out]

integrate((3*x^2 + 2)/((x^4 + 5*x^2 + 3)^(3/2)*x^3), x)