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]

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

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Rubi [A]  time = 0.04, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, 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 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])

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 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 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]

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*}

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Mathematica [A]  time = 0.02, size = 38, normalized size = 1.00 \[ \frac {3}{2} \sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right )-\frac {\tanh ^{-1}\left (\frac {\sqrt {x^4+5}}{\sqrt {5}}\right )}{\sqrt {5}} \]

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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fricas [A]  time = 0.70, size = 41, normalized size = 1.08 \[ \frac {1}{5} \, \sqrt {5} \log \left (-\frac {\sqrt {5} - \sqrt {x^{4} + 5}}{x^{2}}\right ) - \frac {3}{2} \, \log \left (-x^{2} + \sqrt {x^{4} + 5}\right ) \]

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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giac [B]  time = 0.20, size = 61, normalized size = 1.61 \[ \frac {1}{5} \, \sqrt {5} \log \left (-\frac {x^{2} + \sqrt {5} - \sqrt {x^{4} + 5}}{x^{2} - \sqrt {5} - \sqrt {x^{4} + 5}}\right ) - \frac {3}{2} \, \log \left (-x^{2} + \sqrt {x^{4} + 5}\right ) \]

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]

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

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maple [A]  time = 0.01, size = 30, normalized size = 0.79 \[ \frac {3 \arcsinh \left (\frac {\sqrt {5}\, x^{2}}{5}\right )}{2}-\frac {\sqrt {5}\, \arctanh \left (\frac {\sqrt {5}}{\sqrt {x^{4}+5}}\right )}{5} \]

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

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maxima [B]  time = 1.17, size = 67, normalized size = 1.76 \[ \frac {1}{10} \, \sqrt {5} \log \left (-\frac {\sqrt {5} - \sqrt {x^{4} + 5}}{\sqrt {5} + \sqrt {x^{4} + 5}}\right ) + \frac {3}{4} \, \log \left (\frac {\sqrt {x^{4} + 5}}{x^{2}} + 1\right ) - \frac {3}{4} \, \log \left (\frac {\sqrt {x^{4} + 5}}{x^{2}} - 1\right ) \]

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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mupad [B]  time = 0.61, size = 30, normalized size = 0.79 \[ \frac {3\,\mathrm {asinh}\left (\frac {\sqrt {5}\,x^2}{5}\right )}{2}-\frac {\sqrt {5}\,\mathrm {atanh}\left (\frac {\sqrt {5}\,\sqrt {x^4+5}}{5}\right )}{5} \]

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*asinh((5^(1/2)*x^2)/5))/2 - (5^(1/2)*atanh((5^(1/2)*(x^4 + 5)^(1/2))/5))/5

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sympy [A]  time = 6.01, size = 31, normalized size = 0.82 \[ - \frac {\sqrt {5} \operatorname {asinh}{\left (\frac {\sqrt {5}}{x^{2}} \right )}}{5} + \frac {3 \operatorname {asinh}{\left (\frac {\sqrt {5} x^{2}}{5} \right )}}{2} \]

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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