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]

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

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Rubi [A]  time = 0.0168086, antiderivative size = 24, normalized size of antiderivative = 1., 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 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]

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 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]

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*}

Mathematica [A]  time = 0.0252598, size = 24, normalized size = 1. \[ \frac{3 \sqrt{x^4+5}}{2}+\sinh ^{-1}\left (\frac{x^2}{\sqrt{5}}\right ) \]

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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Maple [A]  time = 0.009, size = 20, normalized size = 0.8 \begin{align*}{\it Arcsinh} \left ({\frac{{x}^{2}\sqrt{5}}{5}} \right ) +{\frac{3}{2}\sqrt{{x}^{4}+5}} \end{align*}

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

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Maxima [B]  time = 1.44408, size = 57, normalized size = 2.38 \begin{align*} \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 ) \end{align*}

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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Fricas [A]  time = 1.5046, size = 63, normalized size = 2.62 \begin{align*} \frac{3}{2} \, \sqrt{x^{4} + 5} - \log \left (-x^{2} + \sqrt{x^{4} + 5}\right ) \end{align*}

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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Sympy [A]  time = 1.71826, size = 22, normalized size = 0.92 \begin{align*} \frac{3 \sqrt{x^{4} + 5}}{2} + \operatorname{asinh}{\left (\frac{\sqrt{5} x^{2}}{5} \right )} \end{align*}

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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Giac [A]  time = 1.12754, size = 35, normalized size = 1.46 \begin{align*} \frac{3}{2} \, \sqrt{x^{4} + 5} - \log \left (-x^{2} + \sqrt{x^{4} + 5}\right ) \end{align*}

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))