3.2.97 \(\int \frac {2+3 x^2}{x (3+5 x^2+x^4)^{3/2}} \, dx\) [197]

Optimal. Leaf size=66 \[ \frac {-7-8 x^2}{39 \sqrt {3+5 x^2+x^4}}-\frac {\tanh ^{-1}\left (\frac {6+5 x^2}{2 \sqrt {3} \sqrt {3+5 x^2+x^4}}\right )}{3 \sqrt {3}} \]

[Out]

-1/9*arctanh(1/6*(5*x^2+6)*3^(1/2)/(x^4+5*x^2+3)^(1/2))*3^(1/2)+1/39*(-8*x^2-7)/(x^4+5*x^2+3)^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1265, 836, 12, 738, 212} \begin {gather*} -\frac {8 x^2+7}{39 \sqrt {x^4+5 x^2+3}}-\frac {\tanh ^{-1}\left (\frac {5 x^2+6}{2 \sqrt {3} \sqrt {x^4+5 x^2+3}}\right )}{3 \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/39*(7 + 8*x^2)/Sqrt[3 + 5*x^2 + x^4] - ArcTanh[(6 + 5*x^2)/(2*Sqrt[3]*Sqrt[3 + 5*x^2 + x^4])]/(3*Sqrt[3])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 836

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2*a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*e))*x)
*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2))), x] + Dist[1/((p + 1)*(b^2 - 4*a*
c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2
*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d*m + b*e*m) - b*d*(3*c*d -
b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, b,
c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] ||
IntegerQ[p] || IntegersQ[2*m, 2*p])

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 \left (3+5 x^2+x^4\right )^{3/2}} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {2+3 x}{x \left (3+5 x+x^2\right )^{3/2}} \, dx,x,x^2\right )\\ &=-\frac {7+8 x^2}{39 \sqrt {3+5 x^2+x^4}}-\frac {1}{39} \text {Subst}\left (\int -\frac {13}{x \sqrt {3+5 x+x^2}} \, dx,x,x^2\right )\\ &=-\frac {7+8 x^2}{39 \sqrt {3+5 x^2+x^4}}+\frac {1}{3} \text {Subst}\left (\int \frac {1}{x \sqrt {3+5 x+x^2}} \, dx,x,x^2\right )\\ &=-\frac {7+8 x^2}{39 \sqrt {3+5 x^2+x^4}}-\frac {2}{3} \text {Subst}\left (\int \frac {1}{12-x^2} \, dx,x,\frac {6+5 x^2}{\sqrt {3+5 x^2+x^4}}\right )\\ &=-\frac {7+8 x^2}{39 \sqrt {3+5 x^2+x^4}}-\frac {\tanh ^{-1}\left (\frac {6+5 x^2}{2 \sqrt {3} \sqrt {3+5 x^2+x^4}}\right )}{3 \sqrt {3}}\\ \end {align*}

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Mathematica [A]
time = 0.20, size = 62, normalized size = 0.94 \begin {gather*} \frac {-7-8 x^2}{39 \sqrt {3+5 x^2+x^4}}+\frac {2 \tanh ^{-1}\left (\frac {x^2-\sqrt {3+5 x^2+x^4}}{\sqrt {3}}\right )}{3 \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-7 - 8*x^2)/(39*Sqrt[3 + 5*x^2 + x^4]) + (2*ArcTanh[(x^2 - Sqrt[3 + 5*x^2 + x^4])/Sqrt[3]])/(3*Sqrt[3])

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Maple [A]
time = 0.20, size = 67, normalized size = 1.02

method	result	size
risch	\(-\frac {8 x^{2}+7}{39 \sqrt {x^{4}+5 x^{2}+3}}-\frac {\arctanh \left (\frac {\left (5 x^{2}+6\right ) \sqrt {3}}{6 \sqrt {x^{4}+5 x^{2}+3}}\right ) \sqrt {3}}{9}\)	\(53\)
default	\(-\frac {4 \left (2 x^{2}+5\right )}{39 \sqrt {x^{4}+5 x^{2}+3}}+\frac {1}{3 \sqrt {x^{4}+5 x^{2}+3}}-\frac {\arctanh \left (\frac {\left (5 x^{2}+6\right ) \sqrt {3}}{6 \sqrt {x^{4}+5 x^{2}+3}}\right ) \sqrt {3}}{9}\)	\(67\)
elliptic	\(-\frac {4 \left (2 x^{2}+5\right )}{39 \sqrt {x^{4}+5 x^{2}+3}}+\frac {1}{3 \sqrt {x^{4}+5 x^{2}+3}}-\frac {\arctanh \left (\frac {\left (5 x^{2}+6\right ) \sqrt {3}}{6 \sqrt {x^{4}+5 x^{2}+3}}\right ) \sqrt {3}}{9}\)	\(67\)
trager	\(-\frac {8 x^{2}+7}{39 \sqrt {x^{4}+5 x^{2}+3}}-\frac {\RootOf \left (\textit {\_Z}^{2}-3\right ) \ln \left (-\frac {5 \RootOf \left (\textit {\_Z}^{2}-3\right ) x^{2}+6 \RootOf \left (\textit {\_Z}^{2}-3\right )+6 \sqrt {x^{4}+5 x^{2}+3}}{x^{2}}\right )}{9}\)	\(71\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/39*(2*x^2+5)/(x^4+5*x^2+3)^(1/2)+1/3/(x^4+5*x^2+3)^(1/2)-1/9*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 = 0.51, size = 65, normalized size = 0.98 \begin {gather*} -\frac {8 \, x^{2}}{39 \, \sqrt {x^{4} + 5 \, x^{2} + 3}} - \frac {1}{9} \, \sqrt {3} \log \left (\frac {2 \, \sqrt {3} \sqrt {x^{4} + 5 \, x^{2} + 3}}{x^{2}} + \frac {6}{x^{2}} + 5\right ) - \frac {7}{39 \, \sqrt {x^{4} + 5 \, x^{2} + 3}} \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)^(3/2),x, algorithm="maxima")

[Out]

-8/39*x^2/sqrt(x^4 + 5*x^2 + 3) - 1/9*sqrt(3)*log(2*sqrt(3)*sqrt(x^4 + 5*x^2 + 3)/x^2 + 6/x^2 + 5) - 7/39/sqrt
(x^4 + 5*x^2 + 3)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 107 vs. \(2 (52) = 104\).
time = 0.34, size = 107, normalized size = 1.62 \begin {gather*} -\frac {24 \, x^{4} - 13 \, \sqrt {3} {\left (x^{4} + 5 \, x^{2} + 3\right )} \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 ) + 120 \, x^{2} + 3 \, \sqrt {x^{4} + 5 \, x^{2} + 3} {\left (8 \, x^{2} + 7\right )} + 72}{117 \, {\left (x^{4} + 5 \, x^{2} + 3\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)^(3/2),x, algorithm="fricas")

[Out]

-1/117*(24*x^4 - 13*sqrt(3)*(x^4 + 5*x^2 + 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) + 120*x^2 + 3*sqrt(x^4 + 5*x^2 + 3)*(8*x^2 + 7) + 72)/(x^4 + 5*x^2 + 3)

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

[Out]

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

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Giac [A]
time = 3.61, size = 78, normalized size = 1.18 \begin {gather*} -\frac {1}{9} \, \sqrt {3} \log \left (-x^{2} + \sqrt {3} + \sqrt {x^{4} + 5 \, x^{2} + 3}\right ) + \frac {1}{9} \, \sqrt {3} \log \left (-x^{2} - \sqrt {3} + \sqrt {x^{4} + 5 \, x^{2} + 3}\right ) - \frac {8 \, x^{2} + 7}{39 \, \sqrt {x^{4} + 5 \, x^{2} + 3}} \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)^(3/2),x, algorithm="giac")

[Out]

-1/9*sqrt(3)*log(-x^2 + sqrt(3) + sqrt(x^4 + 5*x^2 + 3)) + 1/9*sqrt(3)*log(-x^2 - sqrt(3) + sqrt(x^4 + 5*x^2 +
3)) - 1/39*(8*x^2 + 7)/sqrt(x^4 + 5*x^2 + 3)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {3\,x^2+2}{x\,{\left (x^4+5\,x^2+3\right )}^{3/2}} \,d x \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)^(3/2)),x)

[Out]

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

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