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

Optimal. Leaf size=83 \[ -\frac{\sqrt{x^4+5 x^2+3}}{12 x^2}-\frac{\sqrt{x^4+5 x^2+3}}{6 x^4}+\frac{1}{8} \sqrt{3} \tanh ^{-1}\left (\frac{5 x^2+6}{2 \sqrt{3} \sqrt{x^4+5 x^2+3}}\right ) \]

[Out]

-Sqrt[3 + 5*x^2 + x^4]/(6*x^4) - Sqrt[3 + 5*x^2 + x^4]/(12*x^2) + (Sqrt[3]*ArcTanh[(6 + 5*x^2)/(2*Sqrt[3]*Sqrt
[3 + 5*x^2 + x^4])])/8

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Rubi [A]  time = 0.0699484, antiderivative size = 83, 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, 834, 806, 724, 206} \[ -\frac{\sqrt{x^4+5 x^2+3}}{12 x^2}-\frac{\sqrt{x^4+5 x^2+3}}{6 x^4}+\frac{1}{8} \sqrt{3} \tanh ^{-1}\left (\frac{5 x^2+6}{2 \sqrt{3} \sqrt{x^4+5 x^2+3}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[3 + 5*x^2 + x^4]/(6*x^4) - Sqrt[3 + 5*x^2 + x^4]/(12*x^2) + (Sqrt[3]*ArcTanh[(6 + 5*x^2)/(2*Sqrt[3]*Sqrt
[3 + 5*x^2 + x^4])])/8

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 834

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

Mathematica [A]  time = 0.0227131, size = 67, normalized size = 0.81 \[ \frac{1}{8} \sqrt{3} \tanh ^{-1}\left (\frac{5 x^2+6}{2 \sqrt{3} \sqrt{x^4+5 x^2+3}}\right )-\frac{\left (x^2+2\right ) \sqrt{x^4+5 x^2+3}}{12 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2 + x^2)*Sqrt[3 + 5*x^2 + x^4])/(12*x^4) + (Sqrt[3]*ArcTanh[(6 + 5*x^2)/(2*Sqrt[3]*Sqrt[3 + 5*x^2 + x^4])])
/8

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Maple [A]  time = 0.013, size = 66, normalized size = 0.8 \begin{align*}{\frac{\sqrt{3}}{8}{\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}{6\,{x}^{4}}\sqrt{{x}^{4}+5\,{x}^{2}+3}}-{\frac{1}{12\,{x}^{2}}\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^5/(x^4+5*x^2+3)^(1/2),x)

[Out]

1/8*arctanh(1/6*(5*x^2+6)*3^(1/2)/(x^4+5*x^2+3)^(1/2))*3^(1/2)-1/6*(x^4+5*x^2+3)^(1/2)/x^4-1/12*(x^4+5*x^2+3)^
(1/2)/x^2

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Maxima [A]  time = 1.44558, size = 92, normalized size = 1.11 \begin{align*} \frac{1}{8} \, \sqrt{3} \log \left (\frac{2 \, \sqrt{3} \sqrt{x^{4} + 5 \, x^{2} + 3}}{x^{2}} + \frac{6}{x^{2}} + 5\right ) - \frac{\sqrt{x^{4} + 5 \, x^{2} + 3}}{12 \, x^{2}} - \frac{\sqrt{x^{4} + 5 \, x^{2} + 3}}{6 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2+2)/x^5/(x^4+5*x^2+3)^(1/2),x, algorithm="maxima")

[Out]

1/8*sqrt(3)*log(2*sqrt(3)*sqrt(x^4 + 5*x^2 + 3)/x^2 + 6/x^2 + 5) - 1/12*sqrt(x^4 + 5*x^2 + 3)/x^2 - 1/6*sqrt(x
^4 + 5*x^2 + 3)/x^4

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Fricas [A]  time = 1.39538, size = 215, normalized size = 2.59 \begin{align*} \frac{3 \, \sqrt{3} x^{4} \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 ) - 2 \, x^{4} - 2 \, \sqrt{x^{4} + 5 \, x^{2} + 3}{\left (x^{2} + 2\right )}}{24 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2+2)/x^5/(x^4+5*x^2+3)^(1/2),x, algorithm="fricas")

[Out]

1/24*(3*sqrt(3)*x^4*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) -
2*x^4 - 2*sqrt(x^4 + 5*x^2 + 3)*(x^2 + 2))/x^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x**2+2)/x**5/(x**4+5*x**2+3)**(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2+2)/x^5/(x^4+5*x^2+3)^(1/2),x, algorithm="giac")

[Out]

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