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

Optimal. Leaf size=66 \[ -\frac{8 x^2+7}{39 \sqrt{x^4+5 x^2+3}}-\frac{\tanh ^{-1}\left (\frac{5 x^2+6}{2 \sqrt{3} \sqrt{x^4+5 x^2+3}}\right )}{3 \sqrt{3}} \]

[Out]

-(7 + 8*x^2)/(39*Sqrt[3 + 5*x^2 + x^4]) - 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 = 0.0570446, antiderivative size = 66, 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, 12, 724, 206} \[ -\frac{8 x^2+7}{39 \sqrt{x^4+5 x^2+3}}-\frac{\tanh ^{-1}\left (\frac{5 x^2+6}{2 \sqrt{3} \sqrt{x^4+5 x^2+3}}\right )}{3 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(7 + 8*x^2)/(39*Sqrt[3 + 5*x^2 + x^4]) - 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 12

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

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 \left (3+5 x^2+x^4\right )^{3/2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{2+3 x}{x \left (3+5 x+x^2\right )^{3/2}} \, dx,x,x^2\right )\\ &=-\frac{7+8 x^2}{39 \sqrt{3+5 x^2+x^4}}-\frac{1}{39} \operatorname{Subst}\left (\int -\frac{13}{x \sqrt{3+5 x+x^2}} \, dx,x,x^2\right )\\ &=-\frac{7+8 x^2}{39 \sqrt{3+5 x^2+x^4}}+\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 \sqrt{3+5 x^2+x^4}}-\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 \sqrt{3+5 x^2+x^4}}-\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 = 0.0293901, size = 66, normalized size = 1. \[ -\frac{8 x^2+7}{39 \sqrt{x^4+5 x^2+3}}-\frac{\tanh ^{-1}\left (\frac{5 x^2+6}{2 \sqrt{3} \sqrt{x^4+5 x^2+3}}\right )}{3 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.018, size = 67, normalized size = 1. \begin{align*} -{\frac{8\,{x}^{2}+20}{39}{\frac{1}{\sqrt{{x}^{4}+5\,{x}^{2}+3}}}}+{\frac{1}{3}{\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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.43624, size = 88, normalized size = 1.33 \begin{align*} -\frac{8 \, 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{7}{39 \, \sqrt{x^{4} + 5 \, x^{2} + 3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/39*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) - 7/39/sqrt
(x^4 + 5*x^2 + 3)

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Fricas [B]  time = 1.37649, size = 281, normalized size = 4.26 \begin{align*} -\frac{24 \, x^{4} - 13 \, \sqrt{3}{\left (x^{4} + 5 \, x^{2} + 3\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 ) + 120 \, x^{2} + 3 \, \sqrt{x^{4} + 5 \, x^{2} + 3}{\left (8 \, x^{2} + 7\right )} + 72}{117 \,{\left (x^{4} + 5 \, x^{2} + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/117*(24*x^4 - 13*sqrt(3)*(x^4 + 5*x^2 + 3)*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) + 120*x^2 + 3*sqrt(x^4 + 5*x^2 + 3)*(8*x^2 + 7) + 72)/(x^4 + 5*x^2 + 3)

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

[Out]

Integral((3*x**2 + 2)/(x*(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}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2+2)/x/(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), x)