3.1.10 \(\int x (2+3 x^2) \sqrt {5+x^4} \, dx\) [10]

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

[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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Rubi [A]
time = 0.01, antiderivative size = 44, normalized size of antiderivative = 1.00, 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 = {1262, 655, 201, 221} \begin {gather*} \frac {1}{2} \left (x^4+5\right )^{3/2}+\frac {5}{2} \sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right )+\frac {1}{2} \sqrt {x^4+5} x^2 \end {gather*}

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 201

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 221

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 655

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 1262

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 x \left (2+3 x^2\right ) \sqrt {5+x^4} \, dx &=\frac {1}{2} \text {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}+\text {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} \text {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*}

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Mathematica [A]
time = 0.06, size = 40, normalized size = 0.91 \begin {gather*} \frac {1}{2} \sqrt {5+x^4} \left (5+x^2+x^4\right )+\frac {5}{2} \tanh ^{-1}\left (\frac {x^2}{\sqrt {5+x^4}}\right ) \end {gather*}

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*ArcTanh[x^2/Sqrt[5 + x^4]])/2

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Maple [A]
time = 0.12, size = 34, normalized size = 0.77

method	result	size
risch	\(\frac {\left (x^{4}+x^{2}+5\right ) \sqrt {x^{4}+5}}{2}+\frac {5 \arcsinh \left (\frac {x^{2} \sqrt {5}}{5}\right )}{2}\)	\(30\)
default	\(\frac {\left (x^{4}+5\right )^{\frac {3}{2}}}{2}+\frac {5 \arcsinh \left (\frac {x^{2} \sqrt {5}}{5}\right )}{2}+\frac {x^{2} \sqrt {x^{4}+5}}{2}\)	\(34\)
trager	\(\left (\frac {1}{2} x^{4}+\frac {1}{2} x^{2}+\frac {5}{2}\right ) \sqrt {x^{4}+5}-\frac {5 \ln \left (x^{2}-\sqrt {x^{4}+5}\right )}{2}\)	\(38\)
elliptic	\(\frac {x^{4} \sqrt {x^{4}+5}}{2}+\frac {5 \sqrt {x^{4}+5}}{2}+\frac {x^{2} \sqrt {x^{4}+5}}{2}+\frac {5 \arcsinh \left (\frac {x^{2} \sqrt {5}}{5}\right )}{2}\)	\(46\)
meijerg	\(-\frac {15 \sqrt {5}\, \left (\frac {4 \sqrt {\pi }}{3}-\frac {2 \sqrt {\pi }\, \left (2+\frac {2 x^{4}}{5}\right ) \sqrt {1+\frac {x^{4}}{5}}}{3}\right )}{8 \sqrt {\pi }}-\frac {5 \left (-\frac {2 \sqrt {\pi }\, x^{2} \sqrt {5}\, \sqrt {1+\frac {x^{4}}{5}}}{5}-2 \sqrt {\pi }\, \arcsinh \left (\frac {x^{2} \sqrt {5}}{5}\right )\right )}{4 \sqrt {\pi }}\)	\(77\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 [B] Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (33) = 66\).
time = 0.49, size = 67, normalized size = 1.52 \begin {gather*} \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 {gather*}

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 = 0.36, size = 34, normalized size = 0.77 \begin {gather*} \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 {gather*}

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 = 1.60, size = 53, normalized size = 1.20 \begin {gather*} \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 {gather*}

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 = 3.28, size = 38, normalized size = 0.86 \begin {gather*} \frac {1}{2} \, \sqrt {x^{4} + 5} x^{2} + \frac {1}{2} \, {\left (x^{4} + 5\right )}^{\frac {3}{2}} - \frac {5}{2} \, \log \left (-x^{2} + \sqrt {x^{4} + 5}\right ) \end {gather*}

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

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Mupad [B]
time = 0.14, size = 32, normalized size = 0.73 \begin {gather*} \frac {5\,\mathrm {asinh}\left (\frac {\sqrt {5}\,x^2}{5}\right )}{2}+\sqrt {x^4+5}\,\left (\frac {x^4}{2}+\frac {x^2}{2}+\frac {5}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(x^4 + 5)^(1/2)*(3*x^2 + 2),x)

[Out]

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

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