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

Optimal. Leaf size=74 \[ \frac {1}{2} \left (x^4+5 x^2+3\right )^{3/2}-\frac {11}{16} \left (2 x^2+5\right ) \sqrt {x^4+5 x^2+3}+\frac {143}{32} \tanh ^{-1}\left (\frac {2 x^2+5}{2 \sqrt {x^4+5 x^2+3}}\right ) \]

[Out]

1/2*(x^4+5*x^2+3)^(3/2)+143/32*arctanh(1/2*(2*x^2+5)/(x^4+5*x^2+3)^(1/2))-11/16*(2*x^2+5)*(x^4+5*x^2+3)^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {1247, 640, 612, 621, 206} \[ \frac {1}{2} \left (x^4+5 x^2+3\right )^{3/2}-\frac {11}{16} \left (2 x^2+5\right ) \sqrt {x^4+5 x^2+3}+\frac {143}{32} \tanh ^{-1}\left (\frac {2 x^2+5}{2 \sqrt {x^4+5 x^2+3}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-11*(5 + 2*x^2)*Sqrt[3 + 5*x^2 + x^4])/16 + (3 + 5*x^2 + x^4)^(3/2)/2 + (143*ArcTanh[(5 + 2*x^2)/(2*Sqrt[3 +
5*x^2 + x^4])])/32

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 1247

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]

Rubi steps

\begin {align*} \int x \left (2+3 x^2\right ) \sqrt {3+5 x^2+x^4} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int (2+3 x) \sqrt {3+5 x+x^2} \, dx,x,x^2\right )\\ &=\frac {1}{2} \left (3+5 x^2+x^4\right )^{3/2}-\frac {11}{4} \operatorname {Subst}\left (\int \sqrt {3+5 x+x^2} \, dx,x,x^2\right )\\ &=-\frac {11}{16} \left (5+2 x^2\right ) \sqrt {3+5 x^2+x^4}+\frac {1}{2} \left (3+5 x^2+x^4\right )^{3/2}+\frac {143}{32} \operatorname {Subst}\left (\int \frac {1}{\sqrt {3+5 x+x^2}} \, dx,x,x^2\right )\\ &=-\frac {11}{16} \left (5+2 x^2\right ) \sqrt {3+5 x^2+x^4}+\frac {1}{2} \left (3+5 x^2+x^4\right )^{3/2}+\frac {143}{16} \operatorname {Subst}\left (\int \frac {1}{4-x^2} \, dx,x,\frac {5+2 x^2}{\sqrt {3+5 x^2+x^4}}\right )\\ &=-\frac {11}{16} \left (5+2 x^2\right ) \sqrt {3+5 x^2+x^4}+\frac {1}{2} \left (3+5 x^2+x^4\right )^{3/2}+\frac {143}{32} \tanh ^{-1}\left (\frac {5+2 x^2}{2 \sqrt {3+5 x^2+x^4}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 61, normalized size = 0.82 \[ \frac {1}{32} \left (2 \sqrt {x^4+5 x^2+3} \left (8 x^4+18 x^2-31\right )+143 \tanh ^{-1}\left (\frac {2 x^2+5}{2 \sqrt {x^4+5 x^2+3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[3 + 5*x^2 + x^4]*(-31 + 18*x^2 + 8*x^4) + 143*ArcTanh[(5 + 2*x^2)/(2*Sqrt[3 + 5*x^2 + x^4])])/32

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fricas [A]  time = 0.66, size = 51, normalized size = 0.69 \[ \frac {1}{16} \, {\left (8 \, x^{4} + 18 \, x^{2} - 31\right )} \sqrt {x^{4} + 5 \, x^{2} + 3} - \frac {143}{32} \, \log \left (-2 \, x^{2} + 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} - 5\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(8*x^4 + 18*x^2 - 31)*sqrt(x^4 + 5*x^2 + 3) - 143/32*log(-2*x^2 + 2*sqrt(x^4 + 5*x^2 + 3) - 5)

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giac [A]  time = 0.34, size = 74, normalized size = 1.00 \[ \frac {1}{16} \, \sqrt {x^{4} + 5 \, x^{2} + 3} {\left (2 \, {\left (4 \, x^{2} + 5\right )} x^{2} - 51\right )} + \frac {1}{4} \, \sqrt {x^{4} + 5 \, x^{2} + 3} {\left (2 \, x^{2} + 5\right )} - \frac {143}{32} \, \log \left (2 \, x^{2} - 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} + 5\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*sqrt(x^4 + 5*x^2 + 3)*(2*(4*x^2 + 5)*x^2 - 51) + 1/4*sqrt(x^4 + 5*x^2 + 3)*(2*x^2 + 5) - 143/32*log(2*x^2
- 2*sqrt(x^4 + 5*x^2 + 3) + 5)

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maple [A]  time = 0.01, size = 57, normalized size = 0.77 \[ \frac {143 \ln \left (x^{2}+\frac {5}{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{32}+\frac {\left (x^{4}+5 x^{2}+3\right )^{\frac {3}{2}}}{2}-\frac {11 \left (2 x^{2}+5\right ) \sqrt {x^{4}+5 x^{2}+3}}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(x^4+5*x^2+3)^(3/2)-11/16*(2*x^2+5)*(x^4+5*x^2+3)^(1/2)+143/32*ln(x^2+5/2+(x^4+5*x^2+3)^(1/2))

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maxima [A]  time = 0.65, size = 70, normalized size = 0.95 \[ -\frac {11}{8} \, \sqrt {x^{4} + 5 \, x^{2} + 3} x^{2} + \frac {1}{2} \, {\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac {3}{2}} - \frac {55}{16} \, \sqrt {x^{4} + 5 \, x^{2} + 3} + \frac {143}{32} \, \log \left (2 \, x^{2} + 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} + 5\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-11/8*sqrt(x^4 + 5*x^2 + 3)*x^2 + 1/2*(x^4 + 5*x^2 + 3)^(3/2) - 55/16*sqrt(x^4 + 5*x^2 + 3) + 143/32*log(2*x^2
+ 2*sqrt(x^4 + 5*x^2 + 3) + 5)

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mupad [B]  time = 0.29, size = 67, normalized size = 0.91 \[ \frac {143\,\ln \left (\sqrt {x^4+5\,x^2+3}+x^2+\frac {5}{2}\right )}{32}+\left (\frac {x^2}{2}+\frac {5}{4}\right )\,\sqrt {x^4+5\,x^2+3}+\frac {\sqrt {x^4+5\,x^2+3}\,\left (8\,x^4+10\,x^2-51\right )}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(143*log((5*x^2 + x^4 + 3)^(1/2) + x^2 + 5/2))/32 + (x^2/2 + 5/4)*(5*x^2 + x^4 + 3)^(1/2) + ((5*x^2 + x^4 + 3)
^(1/2)*(10*x^2 + 8*x^4 - 51))/16

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \left (3 x^{2} + 2\right ) \sqrt {x^{4} + 5 x^{2} + 3}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*(3*x**2 + 2)*sqrt(x**4 + 5*x**2 + 3), x)

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