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

Optimal. Leaf size=24 \[ \frac {3 \sqrt {x^4+5}}{2}+\sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1248, 641, 215} \[ \frac {3 \sqrt {x^4+5}}{2}+\sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(x*(2 + 3*x^2))/Sqrt[5 + x^4],x]

[Out]

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

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 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 1248

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

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 24, normalized size = 1.00 \[ \frac {3 \sqrt {x^4+5}}{2}+\sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.66, size = 26, normalized size = 1.08 \[ \frac {3}{2} \, \sqrt {x^{4} + 5} - \log \left (-x^{2} + \sqrt {x^{4} + 5}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.20, size = 26, normalized size = 1.08 \[ \frac {3}{2} \, \sqrt {x^{4} + 5} - \log \left (-x^{2} + \sqrt {x^{4} + 5}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 20, normalized size = 0.83 \[ \arcsinh \left (\frac {\sqrt {5}\, x^{2}}{5}\right )+\frac {3 \sqrt {x^{4}+5}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.12, size = 42, normalized size = 1.75 \[ \frac {3}{2} \, \sqrt {x^{4} + 5} + \frac {1}{2} \, \log \left (\frac {\sqrt {x^{4} + 5}}{x^{2}} + 1\right ) - \frac {1}{2} \, \log \left (\frac {\sqrt {x^{4} + 5}}{x^{2}} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.29, size = 19, normalized size = 0.79 \[ \mathrm {asinh}\left (\frac {\sqrt {5}\,x^2}{5}\right )+\frac {3\,\sqrt {x^4+5}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.13, size = 22, normalized size = 0.92 \[ \frac {3 \sqrt {x^{4} + 5}}{2} + \operatorname {asinh}{\left (\frac {\sqrt {5} x^{2}}{5} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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