3.2.44 \(\int x (2+3 x^2) \sqrt {3+5 x^2+x^4} \, dx\) [144]

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

[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 = 0.03, 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 = {1261, 654, 626, 635, 212} \begin {gather*} \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 ) \end {gather*}

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 212

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

Rule 626

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 635

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 654

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 1261

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} \text {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} \text {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} \text {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} \text {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 = 0.08, size = 59, normalized size = 0.80 \begin {gather*} \frac {1}{16} \sqrt {3+5 x^2+x^4} \left (-31+18 x^2+8 x^4\right )-\frac {143}{32} \log \left (-5-2 x^2+2 \sqrt {3+5 x^2+x^4}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[3 + 5*x^2 + x^4]*(-31 + 18*x^2 + 8*x^4))/16 - (143*Log[-5 - 2*x^2 + 2*Sqrt[3 + 5*x^2 + x^4]])/32

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Maple [A]
time = 0.10, size = 57, normalized size = 0.77

method	result	size
risch	\(\frac {\left (8 x^{4}+18 x^{2}-31\right ) \sqrt {x^{4}+5 x^{2}+3}}{16}+\frac {143 \ln \left (x^{2}+\frac {5}{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{32}\)	\(48\)
trager	\(\left (\frac {1}{2} x^{4}+\frac {9}{8} x^{2}-\frac {31}{16}\right ) \sqrt {x^{4}+5 x^{2}+3}-\frac {143 \ln \left (-2 x^{2}+2 \sqrt {x^{4}+5 x^{2}+3}-5\right )}{32}\)	\(51\)
default	\(\frac {\left (x^{4}+5 x^{2}+3\right )^{\frac {3}{2}}}{2}-\frac {11 \left (2 x^{2}+5\right ) \sqrt {x^{4}+5 x^{2}+3}}{16}+\frac {143 \ln \left (x^{2}+\frac {5}{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{32}\)	\(57\)
elliptic	\(\frac {x^{4} \sqrt {x^{4}+5 x^{2}+3}}{2}+\frac {9 x^{2} \sqrt {x^{4}+5 x^{2}+3}}{8}-\frac {31 \sqrt {x^{4}+5 x^{2}+3}}{16}+\frac {143 \ln \left (x^{2}+\frac {5}{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{32}\)	\(70\)

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,method=_RETURNVERBOSE)

[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 = 0.30, size = 70, normalized size = 0.95 \begin {gather*} -\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 ) \end {gather*}

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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Fricas [A]
time = 0.35, size = 51, normalized size = 0.69 \begin {gather*} \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 ) \end {gather*}

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

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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Giac [A]
time = 4.37, size = 74, normalized size = 1.00 \begin {gather*} \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 ) \end {gather*}

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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Mupad [B]
time = 0.29, size = 67, normalized size = 0.91 \begin {gather*} \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} \end {gather*}

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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