3.2.84 \(\int \frac {x (2+3 x^2)}{\sqrt {3+5 x^2+x^4}} \, dx\) [184]

Optimal. Leaf size=49 \[ \frac {3}{2} \sqrt {3+5 x^2+x^4}-\frac {11}{4} \tanh ^{-1}\left (\frac {5+2 x^2}{2 \sqrt {3+5 x^2+x^4}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {1261, 654, 635, 212} \begin {gather*} \frac {3}{2} \sqrt {x^4+5 x^2+3}-\frac {11}{4} \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]

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

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 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 \frac {x \left (2+3 x^2\right )}{\sqrt {3+5 x^2+x^4}} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {2+3 x}{\sqrt {3+5 x+x^2}} \, dx,x,x^2\right )\\ &=\frac {3}{2} \sqrt {3+5 x^2+x^4}-\frac {11}{4} \text {Subst}\left (\int \frac {1}{\sqrt {3+5 x+x^2}} \, dx,x,x^2\right )\\ &=\frac {3}{2} \sqrt {3+5 x^2+x^4}-\frac {11}{2} \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 {3}{2} \sqrt {3+5 x^2+x^4}-\frac {11}{4} \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 = 47, normalized size = 0.96 \begin {gather*} \frac {3}{2} \sqrt {3+5 x^2+x^4}+\frac {11}{4} \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]

(3*Sqrt[3 + 5*x^2 + x^4])/2 + (11*Log[-5 - 2*x^2 + 2*Sqrt[3 + 5*x^2 + x^4]])/4

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Maple [A]
time = 0.10, size = 36, normalized size = 0.73

method	result	size
default	\(\frac {3 \sqrt {x^{4}+5 x^{2}+3}}{2}-\frac {11 \ln \left (x^{2}+\frac {5}{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{4}\)	\(36\)
risch	\(\frac {3 \sqrt {x^{4}+5 x^{2}+3}}{2}-\frac {11 \ln \left (x^{2}+\frac {5}{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{4}\)	\(36\)
elliptic	\(\frac {3 \sqrt {x^{4}+5 x^{2}+3}}{2}-\frac {11 \ln \left (x^{2}+\frac {5}{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{4}\)	\(36\)
trager	\(\frac {3 \sqrt {x^{4}+5 x^{2}+3}}{2}+\frac {11 \ln \left (-2 x^{2}+2 \sqrt {x^{4}+5 x^{2}+3}-5\right )}{4}\)	\(40\)

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]

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

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Maxima [A]
time = 0.28, size = 39, normalized size = 0.80 \begin {gather*} \frac {3}{2} \, \sqrt {x^{4} + 5 \, x^{2} + 3} - \frac {11}{4} \, \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]

3/2*sqrt(x^4 + 5*x^2 + 3) - 11/4*log(2*x^2 + 2*sqrt(x^4 + 5*x^2 + 3) + 5)

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Fricas [A]
time = 0.41, size = 39, normalized size = 0.80 \begin {gather*} \frac {3}{2} \, \sqrt {x^{4} + 5 \, x^{2} + 3} + \frac {11}{4} \, \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]

3/2*sqrt(x^4 + 5*x^2 + 3) + 11/4*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 \frac {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 = 3.80, size = 39, normalized size = 0.80 \begin {gather*} \frac {3}{2} \, \sqrt {x^{4} + 5 \, x^{2} + 3} + \frac {11}{4} \, \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]

3/2*sqrt(x^4 + 5*x^2 + 3) + 11/4*log(2*x^2 - 2*sqrt(x^4 + 5*x^2 + 3) + 5)

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Mupad [B]
time = 0.52, size = 35, normalized size = 0.71 \begin {gather*} \frac {3\,\sqrt {x^4+5\,x^2+3}}{2}-\frac {11\,\ln \left (\sqrt {x^4+5\,x^2+3}+x^2+\frac {5}{2}\right )}{4} \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]

(3*(5*x^2 + x^4 + 3)^(1/2))/2 - (11*log((5*x^2 + x^4 + 3)^(1/2) + x^2 + 5/2))/4

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