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

Optimal. Leaf size=42 \[ -\frac{\sqrt{x^4+5}}{5 x^2}-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{x^4+5}}{\sqrt{5}}\right )}{2 \sqrt{5}} \]

[Out]

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

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Rubi [A]  time = 0.0372282, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1252, 807, 266, 63, 207} \[ -\frac{\sqrt{x^4+5}}{5 x^2}-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{x^4+5}}{\sqrt{5}}\right )}{2 \sqrt{5}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[5 + x^4]/(5*x^2) - (3*ArcTanh[Sqrt[5 + x^4]/Sqrt[5]])/(2*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 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(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]
&& EqQ[Simplify[m + 2*p + 3], 0]

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

Mathematica [A]  time = 0.0248819, size = 42, normalized size = 1. \[ -\frac{\sqrt{x^4+5}}{5 x^2}-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{x^4+5}}{\sqrt{5}}\right )}{2 \sqrt{5}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.01, size = 31, normalized size = 0.7 \begin{align*} -{\frac{3\,\sqrt{5}}{10}{\it Artanh} \left ({\sqrt{5}{\frac{1}{\sqrt{{x}^{4}+5}}}} \right ) }-{\frac{1}{5\,{x}^{2}}\sqrt{{x}^{4}+5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.42862, size = 63, normalized size = 1.5 \begin{align*} \frac{3}{20} \, \sqrt{5} \log \left (-\frac{\sqrt{5} - \sqrt{x^{4} + 5}}{\sqrt{5} + \sqrt{x^{4} + 5}}\right ) - \frac{\sqrt{x^{4} + 5}}{5 \, 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)^(1/2),x, algorithm="maxima")

[Out]

3/20*sqrt(5)*log(-(sqrt(5) - sqrt(x^4 + 5))/(sqrt(5) + sqrt(x^4 + 5))) - 1/5*sqrt(x^4 + 5)/x^2

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Fricas [A]  time = 1.55037, size = 119, normalized size = 2.83 \begin{align*} \frac{3 \, \sqrt{5} x^{2} \log \left (-\frac{\sqrt{5} - \sqrt{x^{4} + 5}}{x^{2}}\right ) - 2 \, x^{2} - 2 \, \sqrt{x^{4} + 5}}{10 \, 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)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 3.03978, size = 31, normalized size = 0.74 \begin{align*} - \frac{\sqrt{1 + \frac{5}{x^{4}}}}{5} - \frac{3 \sqrt{5} \operatorname{asinh}{\left (\frac{\sqrt{5}}{x^{2}} \right )}}{10} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.18384, size = 65, normalized size = 1.55 \begin{align*} -\frac{3}{20} \, \sqrt{5} \log \left (\sqrt{5} + \sqrt{x^{4} + 5}\right ) + \frac{3}{20} \, \sqrt{5} \log \left (-\sqrt{5} + \sqrt{x^{4} + 5}\right ) - \frac{1}{5} \, \sqrt{\frac{5}{x^{4}} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/20*sqrt(5)*log(sqrt(5) + sqrt(x^4 + 5)) + 3/20*sqrt(5)*log(-sqrt(5) + sqrt(x^4 + 5)) - 1/5*sqrt(5/x^4 + 1)