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

Optimal. Leaf size=38 \[ \frac{3}{2} \sinh ^{-1}\left (\frac{x^2}{\sqrt{5}}\right )-\frac{\tanh ^{-1}\left (\frac{\sqrt{x^4+5}}{\sqrt{5}}\right )}{\sqrt{5}} \]

[Out]

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

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Rubi [A]  time = 0.0382031, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {1252, 844, 215, 266, 63, 207} \[ \frac{3}{2} \sinh ^{-1}\left (\frac{x^2}{\sqrt{5}}\right )-\frac{\tanh ^{-1}\left (\frac{\sqrt{x^4+5}}{\sqrt{5}}\right )}{\sqrt{5}} \]

Antiderivative was successfully verified.

[In]

Int[(2 + 3*x^2)/(x*Sqrt[5 + x^4]),x]

[Out]

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

Rule 1252

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

Rule 844

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

Rule 215

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

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 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 207

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

Rubi steps

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

Mathematica [A]  time = 0.0219835, size = 38, normalized size = 1. \[ \frac{3}{2} \sinh ^{-1}\left (\frac{x^2}{\sqrt{5}}\right )-\frac{\tanh ^{-1}\left (\frac{\sqrt{x^4+5}}{\sqrt{5}}\right )}{\sqrt{5}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.009, size = 30, normalized size = 0.8 \begin{align*}{\frac{3}{2}{\it Arcsinh} \left ({\frac{{x}^{2}\sqrt{5}}{5}} \right ) }-{\frac{\sqrt{5}}{5}{\it Artanh} \left ({\sqrt{5}{\frac{1}{\sqrt{{x}^{4}+5}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.43429, size = 90, normalized size = 2.37 \begin{align*} \frac{1}{10} \, \sqrt{5} \log \left (-\frac{\sqrt{5} - \sqrt{x^{4} + 5}}{\sqrt{5} + \sqrt{x^{4} + 5}}\right ) + \frac{3}{4} \, \log \left (\frac{\sqrt{x^{4} + 5}}{x^{2}} + 1\right ) - \frac{3}{4} \, \log \left (\frac{\sqrt{x^{4} + 5}}{x^{2}} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*sqrt(5)*log(-(sqrt(5) - sqrt(x^4 + 5))/(sqrt(5) + sqrt(x^4 + 5))) + 3/4*log(sqrt(x^4 + 5)/x^2 + 1) - 3/4*
log(sqrt(x^4 + 5)/x^2 - 1)

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Fricas [A]  time = 1.52611, size = 109, normalized size = 2.87 \begin{align*} \frac{1}{5} \, \sqrt{5} \log \left (-\frac{\sqrt{5} - \sqrt{x^{4} + 5}}{x^{2}}\right ) - \frac{3}{2} \, \log \left (-x^{2} + \sqrt{x^{4} + 5}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 4.8927, size = 31, normalized size = 0.82 \begin{align*} - \frac{\sqrt{5} \operatorname{asinh}{\left (\frac{\sqrt{5}}{x^{2}} \right )}}{5} + \frac{3 \operatorname{asinh}{\left (\frac{\sqrt{5} x^{2}}{5} \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(5)*asinh(sqrt(5)/x**2)/5 + 3*asinh(sqrt(5)*x**2/5)/2

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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}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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