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

Optimal. Leaf size=67 \[ \frac {3}{10} \left (x^4+5\right )^{3/2} x^4-\frac {25}{8} \sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right )-\frac {5}{8} \sqrt {x^4+5} x^2-\frac {1}{4} \left (4-x^2\right ) \left (x^4+5\right )^{3/2} \]

[Out]

3/10*x^4*(x^4+5)^(3/2)-1/4*(-x^2+4)*(x^4+5)^(3/2)-25/8*arcsinh(1/5*x^2*5^(1/2))-5/8*x^2*(x^4+5)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1252, 833, 780, 195, 215} \[ \frac {3}{10} \left (x^4+5\right )^{3/2} x^4-\frac {5}{8} \sqrt {x^4+5} x^2-\frac {1}{4} \left (4-x^2\right ) \left (x^4+5\right )^{3/2}-\frac {25}{8} \sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^5*(2 + 3*x^2)*Sqrt[5 + x^4],x]

[Out]

(-5*x^2*Sqrt[5 + x^4])/8 + (3*x^4*(5 + x^4)^(3/2))/10 - ((4 - x^2)*(5 + x^4)^(3/2))/4 - (25*ArcSinh[x^2/Sqrt[5
]])/8

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]

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 x^5 \left (2+3 x^2\right ) \sqrt {5+x^4} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int x^2 (2+3 x) \sqrt {5+x^2} \, dx,x,x^2\right )\\ &=\frac {3}{10} x^4 \left (5+x^4\right )^{3/2}+\frac {1}{10} \operatorname {Subst}\left (\int x (-30+10 x) \sqrt {5+x^2} \, dx,x,x^2\right )\\ &=\frac {3}{10} x^4 \left (5+x^4\right )^{3/2}-\frac {1}{4} \left (4-x^2\right ) \left (5+x^4\right )^{3/2}-\frac {5}{4} \operatorname {Subst}\left (\int \sqrt {5+x^2} \, dx,x,x^2\right )\\ &=-\frac {5}{8} x^2 \sqrt {5+x^4}+\frac {3}{10} x^4 \left (5+x^4\right )^{3/2}-\frac {1}{4} \left (4-x^2\right ) \left (5+x^4\right )^{3/2}-\frac {25}{8} \operatorname {Subst}\left (\int \frac {1}{\sqrt {5+x^2}} \, dx,x,x^2\right )\\ &=-\frac {5}{8} x^2 \sqrt {5+x^4}+\frac {3}{10} x^4 \left (5+x^4\right )^{3/2}-\frac {1}{4} \left (4-x^2\right ) \left (5+x^4\right )^{3/2}-\frac {25}{8} \sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 50, normalized size = 0.75 \[ \frac {1}{40} \sqrt {x^4+5} \left (12 x^8+10 x^6+20 x^4+25 x^2-200\right )-\frac {25}{8} \sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[5 + x^4]*(-200 + 25*x^2 + 20*x^4 + 10*x^6 + 12*x^8))/40 - (25*ArcSinh[x^2/Sqrt[5]])/8

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fricas [A]  time = 0.49, size = 48, normalized size = 0.72 \[ \frac {1}{40} \, {\left (12 \, x^{8} + 10 \, x^{6} + 20 \, x^{4} + 25 \, x^{2} - 200\right )} \sqrt {x^{4} + 5} + \frac {25}{8} \, \log \left (-x^{2} + \sqrt {x^{4} + 5}\right ) \]

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]

1/40*(12*x^8 + 10*x^6 + 20*x^4 + 25*x^2 - 200)*sqrt(x^4 + 5) + 25/8*log(-x^2 + sqrt(x^4 + 5))

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giac [A]  time = 0.23, size = 54, normalized size = 0.81 \[ \frac {1}{8} \, {\left (2 \, x^{4} + 5\right )} \sqrt {x^{4} + 5} x^{2} + \frac {3}{10} \, {\left (x^{4} + 5\right )}^{\frac {5}{2}} - \frac {5}{2} \, {\left (x^{4} + 5\right )}^{\frac {3}{2}} + \frac {25}{8} \, \log \left (-x^{2} + \sqrt {x^{4} + 5}\right ) \]

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]

1/8*(2*x^4 + 5)*sqrt(x^4 + 5)*x^2 + 3/10*(x^4 + 5)^(5/2) - 5/2*(x^4 + 5)^(3/2) + 25/8*log(-x^2 + sqrt(x^4 + 5)
)

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maple [A]  time = 0.05, size = 53, normalized size = 0.79 \[ \frac {\left (x^{4}+5\right )^{\frac {3}{2}} x^{2}}{4}-\frac {5 \sqrt {x^{4}+5}\, x^{2}}{8}-\frac {25 \arcsinh \left (\frac {\sqrt {5}\, x^{2}}{5}\right )}{8}+\frac {\left (x^{4}+5\right )^{\frac {3}{2}} \left (3 x^{4}-10\right )}{10} \]

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]

1/10*(x^4+5)^(3/2)*(3*x^4-10)+1/4*x^2*(x^4+5)^(3/2)-5/8*x^2*(x^4+5)^(1/2)-25/8*arcsinh(1/5*x^2*5^(1/2))

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maxima [B]  time = 1.01, size = 102, normalized size = 1.52 \[ \frac {3}{10} \, {\left (x^{4} + 5\right )}^{\frac {5}{2}} - \frac {5}{2} \, {\left (x^{4} + 5\right )}^{\frac {3}{2}} - \frac {25 \, {\left (\frac {\sqrt {x^{4} + 5}}{x^{2}} + \frac {{\left (x^{4} + 5\right )}^{\frac {3}{2}}}{x^{6}}\right )}}{8 \, {\left (\frac {2 \, {\left (x^{4} + 5\right )}}{x^{4}} - \frac {{\left (x^{4} + 5\right )}^{2}}{x^{8}} - 1\right )}} - \frac {25}{16} \, \log \left (\frac {\sqrt {x^{4} + 5}}{x^{2}} + 1\right ) + \frac {25}{16} \, \log \left (\frac {\sqrt {x^{4} + 5}}{x^{2}} - 1\right ) \]

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]

3/10*(x^4 + 5)^(5/2) - 5/2*(x^4 + 5)^(3/2) - 25/8*(sqrt(x^4 + 5)/x^2 + (x^4 + 5)^(3/2)/x^6)/(2*(x^4 + 5)/x^4 -
(x^4 + 5)^2/x^8 - 1) - 25/16*log(sqrt(x^4 + 5)/x^2 + 1) + 25/16*log(sqrt(x^4 + 5)/x^2 - 1)

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mupad [B]  time = 0.37, size = 42, normalized size = 0.63 \[ \sqrt {x^4+5}\,\left (\frac {3\,x^8}{10}+\frac {x^6}{4}+\frac {x^4}{2}+\frac {5\,x^2}{8}-5\right )-\frac {25\,\mathrm {asinh}\left (\frac {\sqrt {5}\,x^2}{5}\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^4 + 5)^(1/2)*((5*x^2)/8 + x^4/2 + x^6/4 + (3*x^8)/10 - 5) - (25*asinh((5^(1/2)*x^2)/5))/8

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sympy [A]  time = 6.01, size = 97, normalized size = 1.45 \[ \frac {x^{10}}{4 \sqrt {x^{4} + 5}} + \frac {3 x^{8} \sqrt {x^{4} + 5}}{10} + \frac {15 x^{6}}{8 \sqrt {x^{4} + 5}} + \frac {x^{4} \sqrt {x^{4} + 5}}{2} + \frac {25 x^{2}}{8 \sqrt {x^{4} + 5}} - 5 \sqrt {x^{4} + 5} - \frac {25 \operatorname {asinh}{\left (\frac {\sqrt {5} x^{2}}{5} \right )}}{8} \]

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**10/(4*sqrt(x**4 + 5)) + 3*x**8*sqrt(x**4 + 5)/10 + 15*x**6/(8*sqrt(x**4 + 5)) + x**4*sqrt(x**4 + 5)/2 + 25*
x**2/(8*sqrt(x**4 + 5)) - 5*sqrt(x**4 + 5) - 25*asinh(sqrt(5)*x**2/5)/8

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