3.2.85 \(\int \frac {2+3 x^2}{x \sqrt {3+5 x^2+x^4}} \, dx\) [185]

Optimal. Leaf size=69 \[ \frac {3}{2} \tanh ^{-1}\left (\frac {5+2 x^2}{2 \sqrt {3+5 x^2+x^4}}\right )-\frac {\tanh ^{-1}\left (\frac {6+5 x^2}{2 \sqrt {3} \sqrt {3+5 x^2+x^4}}\right )}{\sqrt {3}} \]

[Out]

3/2*arctanh(1/2*(2*x^2+5)/(x^4+5*x^2+3)^(1/2))-1/3*arctanh(1/6*(5*x^2+6)*3^(1/2)/(x^4+5*x^2+3)^(1/2))*3^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1265, 857, 635, 212, 738} \begin {gather*} \frac {3}{2} \tanh ^{-1}\left (\frac {2 x^2+5}{2 \sqrt {x^4+5 x^2+3}}\right )-\frac {\tanh ^{-1}\left (\frac {5 x^2+6}{2 \sqrt {3} \sqrt {x^4+5 x^2+3}}\right )}{\sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*ArcTanh[(5 + 2*x^2)/(2*Sqrt[3 + 5*x^2 + x^4])])/2 - ArcTanh[(6 + 5*x^2)/(2*Sqrt[3]*Sqrt[3 + 5*x^2 + x^4])]/
Sqrt[3]

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 738

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

Rule 857

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 1265

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(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] &&
IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \frac {2+3 x^2}{x \sqrt {3+5 x^2+x^4}} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {2+3 x}{x \sqrt {3+5 x+x^2}} \, dx,x,x^2\right )\\ &=\frac {3}{2} \text {Subst}\left (\int \frac {1}{\sqrt {3+5 x+x^2}} \, dx,x,x^2\right )+\text {Subst}\left (\int \frac {1}{x \sqrt {3+5 x+x^2}} \, dx,x,x^2\right )\\ &=-\left (2 \text {Subst}\left (\int \frac {1}{12-x^2} \, dx,x,\frac {6+5 x^2}{\sqrt {3+5 x^2+x^4}}\right )\right )+3 \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} \tanh ^{-1}\left (\frac {5+2 x^2}{2 \sqrt {3+5 x^2+x^4}}\right )-\frac {\tanh ^{-1}\left (\frac {6+5 x^2}{2 \sqrt {3} \sqrt {3+5 x^2+x^4}}\right )}{\sqrt {3}}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 63, normalized size = 0.91 \begin {gather*} \frac {2 \tanh ^{-1}\left (\frac {x^2-\sqrt {3+5 x^2+x^4}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {3}{2} \log \left (-5-2 x^2+2 \sqrt {3+5 x^2+x^4}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.25, size = 52, normalized size = 0.75

method	result	size
default	\(\frac {3 \ln \left (x^{2}+\frac {5}{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{2}-\frac {\arctanh \left (\frac {\left (5 x^{2}+6\right ) \sqrt {3}}{6 \sqrt {x^{4}+5 x^{2}+3}}\right ) \sqrt {3}}{3}\)	\(52\)
elliptic	\(\frac {3 \ln \left (x^{2}+\frac {5}{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{2}-\frac {\arctanh \left (\frac {\left (5 x^{2}+6\right ) \sqrt {3}}{6 \sqrt {x^{4}+5 x^{2}+3}}\right ) \sqrt {3}}{3}\)	\(52\)
trager	\(-\frac {\RootOf \left (\textit {\_Z}^{2}-3\right ) \ln \left (-\frac {5 \RootOf \left (\textit {\_Z}^{2}-3\right ) x^{2}+6 \RootOf \left (\textit {\_Z}^{2}-3\right )+6 \sqrt {x^{4}+5 x^{2}+3}}{x^{2}}\right )}{3}-\frac {3 \ln \left (2 x^{2}-2 \sqrt {x^{4}+5 x^{2}+3}+5\right )}{2}\)	\(74\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*ln(x^2+5/2+(x^4+5*x^2+3)^(1/2))-1/3*arctanh(1/6*(5*x^2+6)*3^(1/2)/(x^4+5*x^2+3)^(1/2))*3^(1/2)

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Maxima [A]
time = 0.54, size = 58, normalized size = 0.84 \begin {gather*} -\frac {1}{3} \, \sqrt {3} \log \left (\frac {2 \, \sqrt {3} \sqrt {x^{4} + 5 \, x^{2} + 3}}{x^{2}} + \frac {6}{x^{2}} + 5\right ) + \frac {3}{2} \, \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((3*x^2+2)/x/(x^4+5*x^2+3)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.37, size = 75, normalized size = 1.09 \begin {gather*} \frac {1}{3} \, \sqrt {3} \log \left (\frac {25 \, x^{2} - 2 \, \sqrt {3} {\left (5 \, x^{2} + 6\right )} - 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} {\left (5 \, \sqrt {3} - 6\right )} + 30}{x^{2}}\right ) - \frac {3}{2} \, \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((3*x^2+2)/x/(x^4+5*x^2+3)^(1/2),x, algorithm="fricas")

[Out]

1/3*sqrt(3)*log((25*x^2 - 2*sqrt(3)*(5*x^2 + 6) - 2*sqrt(x^4 + 5*x^2 + 3)*(5*sqrt(3) - 6) + 30)/x^2) - 3/2*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 {3 x^{2} + 2}{x \sqrt {x^{4} + 5 x^{2} + 3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 3.84, size = 78, normalized size = 1.13 \begin {gather*} \frac {1}{3} \, \sqrt {3} \log \left (\frac {x^{2} + \sqrt {3} - \sqrt {x^{4} + 5 \, x^{2} + 3}}{x^{2} - \sqrt {3} - \sqrt {x^{4} + 5 \, x^{2} + 3}}\right ) - \frac {3}{2} \, \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((3*x^2+2)/x/(x^4+5*x^2+3)^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 1.01, size = 56, normalized size = 0.81 \begin {gather*} \frac {3\,\ln \left (\sqrt {x^4+5\,x^2+3}+x^2+\frac {5}{2}\right )}{2}-\frac {\sqrt {3}\,\left (\ln \left (\frac {1}{x^2}\right )+\ln \left (2\,\sqrt {3}\,\sqrt {x^4+5\,x^2+3}+5\,x^2+6\right )\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*log((5*x^2 + x^4 + 3)^(1/2) + x^2 + 5/2))/2 - (3^(1/2)*(log(1/x^2) + log(2*3^(1/2)*(5*x^2 + x^4 + 3)^(1/2)
+ 5*x^2 + 6)))/3

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