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

Optimal. Leaf size=69 \[ \frac{3}{2} \tanh ^{-1}\left (\frac{2 x^2+5}{2 \sqrt{x^4+5 x^2+3}}\right )-\frac{\tanh ^{-1}\left (\frac{5 x^2+6}{2 \sqrt{3} \sqrt{x^4+5 x^2+3}}\right )}{\sqrt{3}} \]

[Out]

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

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Rubi [A]  time = 0.0624769, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {1251, 843, 621, 206, 724} \[ \frac{3}{2} \tanh ^{-1}\left (\frac{2 x^2+5}{2 \sqrt{x^4+5 x^2+3}}\right )-\frac{\tanh ^{-1}\left (\frac{5 x^2+6}{2 \sqrt{3} \sqrt{x^4+5 x^2+3}}\right )}{\sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*ArcTanh[(5 + 2*x^2)/(2*Sqrt[3 + 5*x^2 + x^4])])/2 - ArcTanh[(6 + 5*x^2)/(2*Sqrt[3]*Sqrt[3 + 5*x^2 + x^4])]/
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 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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]
&&  !IGtQ[m, 0]

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 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])

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]

Rubi steps

\begin{align*} \int \frac{2+3 x^2}{x \sqrt{3+5 x^2+x^4}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{2+3 x}{x \sqrt{3+5 x+x^2}} \, dx,x,x^2\right )\\ &=\frac{3}{2} \operatorname{Subst}\left (\int \frac{1}{\sqrt{3+5 x+x^2}} \, dx,x,x^2\right )+\operatorname{Subst}\left (\int \frac{1}{x \sqrt{3+5 x+x^2}} \, dx,x,x^2\right )\\ &=-\left (2 \operatorname{Subst}\left (\int \frac{1}{12-x^2} \, dx,x,\frac{6+5 x^2}{\sqrt{3+5 x^2+x^4}}\right )\right )+3 \operatorname{Subst}\left (\int \frac{1}{4-x^2} \, dx,x,\frac{5+2 x^2}{\sqrt{3+5 x^2+x^4}}\right )\\ &=\frac{3}{2} \tanh ^{-1}\left (\frac{5+2 x^2}{2 \sqrt{3+5 x^2+x^4}}\right )-\frac{\tanh ^{-1}\left (\frac{6+5 x^2}{2 \sqrt{3} \sqrt{3+5 x^2+x^4}}\right )}{\sqrt{3}}\\ \end{align*}

Mathematica [A]  time = 0.0156079, size = 69, normalized size = 1. \[ \frac{3}{2} \tanh ^{-1}\left (\frac{2 x^2+5}{2 \sqrt{x^4+5 x^2+3}}\right )-\frac{\tanh ^{-1}\left (\frac{5 x^2+6}{2 \sqrt{3} \sqrt{x^4+5 x^2+3}}\right )}{\sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.01, size = 52, normalized size = 0.8 \begin{align*}{\frac{3}{2}\ln \left ({\frac{5}{2}}+{x}^{2}+\sqrt{{x}^{4}+5\,{x}^{2}+3} \right ) }-{\frac{\sqrt{3}}{3}{\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)^(1/2),x)

[Out]

3/2*ln(5/2+x^2+(x^4+5*x^2+3)^(1/2))-1/3*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 = 1.43019, size = 78, normalized size = 1.13 \begin{align*} -\frac{1}{3} \, \sqrt{3} \log \left (\frac{2 \, \sqrt{3} \sqrt{x^{4} + 5 \, x^{2} + 3}}{x^{2}} + \frac{6}{x^{2}} + 5\right ) + \frac{3}{2} \, \log \left (2 \, x^{2} + 2 \, \sqrt{x^{4} + 5 \, x^{2} + 3} + 5\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)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.46738, size = 203, normalized size = 2.94 \begin{align*} \frac{1}{3} \, \sqrt{3} \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 ) - \frac{3}{2} \, \log \left (-2 \, x^{2} + 2 \, \sqrt{x^{4} + 5 \, x^{2} + 3} - 5\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)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{3 x^{2} + 2}{x \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/(x**4+5*x**2+3)**(1/2),x)

[Out]

Integral((3*x**2 + 2)/(x*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}\,{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)^(1/2),x, algorithm="giac")

[Out]

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