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

Optimal. Leaf size=44 \[ \frac{1}{2} \sqrt{x^4+5} x^2+\frac{1}{2} \left (x^4+5\right )^{3/2}+\frac{5}{2} \sinh ^{-1}\left (\frac{x^2}{\sqrt{5}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0214319, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {1248, 641, 195, 215} \[ \frac{1}{2} \sqrt{x^4+5} x^2+\frac{1}{2} \left (x^4+5\right )^{3/2}+\frac{5}{2} \sinh ^{-1}\left (\frac{x^2}{\sqrt{5}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x*(2 + 3*x^2)*Sqrt[5 + x^4],x]

[Out]

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

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 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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

Mathematica [A]  time = 0.032063, size = 36, normalized size = 0.82 \[ \frac{1}{2} \sqrt{x^4+5} \left (x^4+x^2+5\right )+\frac{5}{2} \sinh ^{-1}\left (\frac{x^2}{\sqrt{5}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.01, size = 34, normalized size = 0.8 \begin{align*}{\frac{1}{2} \left ({x}^{4}+5 \right ) ^{{\frac{3}{2}}}}+{\frac{5}{2}{\it Arcsinh} \left ({\frac{{x}^{2}\sqrt{5}}{5}} \right ) }+{\frac{{x}^{2}}{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]

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

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Maxima [A]  time = 1.43529, size = 90, normalized size = 2.05 \begin{align*} \frac{1}{2} \,{\left (x^{4} + 5\right )}^{\frac{3}{2}} + \frac{5 \, \sqrt{x^{4} + 5}}{2 \, x^{2}{\left (\frac{x^{4} + 5}{x^{4}} - 1\right )}} + \frac{5}{4} \, \log \left (\frac{\sqrt{x^{4} + 5}}{x^{2}} + 1\right ) - \frac{5}{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(x*(3*x^2+2)*(x^4+5)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.51203, size = 90, normalized size = 2.05 \begin{align*} \frac{1}{2} \,{\left (x^{4} + x^{2} + 5\right )} \sqrt{x^{4} + 5} - \frac{5}{2} \, \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]

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

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Sympy [A]  time = 2.77543, size = 53, normalized size = 1.2 \begin{align*} \frac{x^{6}}{2 \sqrt{x^{4} + 5}} + \frac{5 x^{2}}{2 \sqrt{x^{4} + 5}} + \frac{\left (x^{4} + 5\right )^{\frac{3}{2}}}{2} + \frac{5 \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(x*(3*x**2+2)*(x**4+5)**(1/2),x)

[Out]

x**6/(2*sqrt(x**4 + 5)) + 5*x**2/(2*sqrt(x**4 + 5)) + (x**4 + 5)**(3/2)/2 + 5*asinh(sqrt(5)*x**2/5)/2

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Giac [A]  time = 1.12636, size = 50, normalized size = 1.14 \begin{align*} \frac{1}{2} \, \sqrt{x^{4} + 5}{\left ({\left (x^{2} + 1\right )} x^{2} + 5\right )} - \frac{5}{2} \, \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]

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