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

Optimal. Leaf size=51 \[ \frac {1}{2} \sqrt {x^4+5} x^4-\frac {5}{2} \sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right )-\frac {1}{2} \left (10-x^2\right ) \sqrt {x^4+5} \]

[Out]

-5/2*arcsinh(1/5*x^2*5^(1/2))+1/2*x^4*(x^4+5)^(1/2)-1/2*(-x^2+10)*(x^4+5)^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1252, 833, 780, 215} \[ \frac {1}{2} \sqrt {x^4+5} x^4-\frac {1}{2} \left (10-x^2\right ) \sqrt {x^4+5}-\frac {5}{2} \sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^4*Sqrt[5 + x^4])/2 - ((10 - x^2)*Sqrt[5 + x^4])/2 - (5*ArcSinh[x^2/Sqrt[5]])/2

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 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 833

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(g*(d + e*x)
^m*(a + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p*
Simp[c*d*f*(m + 2*p + 2) - a*e*g*m + c*(e*f*(m + 2*p + 2) + d*g*m)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, p
}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p]) &&  !(IGtQ[m, 0] && EqQ[f, 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 {x^5 \left (2+3 x^2\right )}{\sqrt {5+x^4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^2 (2+3 x)}{\sqrt {5+x^2}} \, dx,x,x^2\right )\\ &=\frac {1}{2} x^4 \sqrt {5+x^4}+\frac {1}{6} \operatorname {Subst}\left (\int \frac {x (-30+6 x)}{\sqrt {5+x^2}} \, dx,x,x^2\right )\\ &=\frac {1}{2} x^4 \sqrt {5+x^4}-\frac {1}{2} \left (10-x^2\right ) \sqrt {5+x^4}-\frac {5}{2} \operatorname {Subst}\left (\int \frac {1}{\sqrt {5+x^2}} \, dx,x,x^2\right )\\ &=\frac {1}{2} x^4 \sqrt {5+x^4}-\frac {1}{2} \left (10-x^2\right ) \sqrt {5+x^4}-\frac {5}{2} \sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 35, normalized size = 0.69 \[ \frac {1}{2} \left (\sqrt {x^4+5} \left (x^4+x^2-10\right )-5 \sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[5 + x^4]*(-10 + x^2 + x^4) - 5*ArcSinh[x^2/Sqrt[5]])/2

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fricas [A]  time = 0.70, size = 34, normalized size = 0.67 \[ \frac {1}{2} \, {\left (x^{4} + x^{2} - 10\right )} \sqrt {x^{4} + 5} + \frac {5}{2} \, \log \left (-x^{2} + \sqrt {x^{4} + 5}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(x^4 + x^2 - 10)*sqrt(x^4 + 5) + 5/2*log(-x^2 + sqrt(x^4 + 5))

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giac [A]  time = 0.22, size = 37, normalized size = 0.73 \[ \frac {1}{2} \, \sqrt {x^{4} + 5} {\left ({\left (x^{2} + 1\right )} x^{2} - 10\right )} + \frac {5}{2} \, \log \left (-x^{2} + \sqrt {x^{4} + 5}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(x^4 + 5)*((x^2 + 1)*x^2 - 10) + 5/2*log(-x^2 + sqrt(x^4 + 5))

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maple [A]  time = 0.01, size = 39, normalized size = 0.76 \[ \frac {\sqrt {x^{4}+5}\, x^{2}}{2}-\frac {5 \arcsinh \left (\frac {\sqrt {5}\, x^{2}}{5}\right )}{2}+\frac {\sqrt {x^{4}+5}\, \left (x^{4}-10\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(x^4+5)^(1/2)*(x^4-10)+1/2*(x^4+5)^(1/2)*x^2-5/2*arcsinh(1/5*5^(1/2)*x^2)

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maxima [A]  time = 1.22, size = 76, normalized size = 1.49 \[ \frac {1}{2} \, {\left (x^{4} + 5\right )}^{\frac {3}{2}} - \frac {15}{2} \, \sqrt {x^{4} + 5} + \frac {5 \, \sqrt {x^{4} + 5}}{2 \, x^{2} {\left (\frac {x^{4} + 5}{x^{4}} - 1\right )}} - \frac {5}{4} \, \log \left (\frac {\sqrt {x^{4} + 5}}{x^{2}} + 1\right ) + \frac {5}{4} \, \log \left (\frac {\sqrt {x^{4} + 5}}{x^{2}} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(x^4 + 5)^(3/2) - 15/2*sqrt(x^4 + 5) + 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(sqrt(x^4 + 5)/x^2 - 1)

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mupad [B]  time = 0.31, size = 32, normalized size = 0.63 \[ \sqrt {x^4+5}\,\left (\frac {x^4}{2}+\frac {x^2}{2}-5\right )-\frac {5\,\mathrm {asinh}\left (\frac {\sqrt {5}\,x^2}{5}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 5.52, size = 66, normalized size = 1.29 \[ \frac {x^{6}}{2 \sqrt {x^{4} + 5}} + \frac {x^{4} \sqrt {x^{4} + 5}}{2} + \frac {5 x^{2}}{2 \sqrt {x^{4} + 5}} - 5 \sqrt {x^{4} + 5} - \frac {5 \operatorname {asinh}{\left (\frac {\sqrt {5} x^{2}}{5} \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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