3.2.87 \(\int \frac {2+3 x^2}{x^5 \sqrt {3+5 x^2+x^4}} \, dx\) [187]

Optimal. Leaf size=83 \[ -\frac {\sqrt {3+5 x^2+x^4}}{6 x^4}-\frac {\sqrt {3+5 x^2+x^4}}{12 x^2}+\frac {1}{8} \sqrt {3} \tanh ^{-1}\left (\frac {6+5 x^2}{2 \sqrt {3} \sqrt {3+5 x^2+x^4}}\right ) \]

[Out]

1/8*arctanh(1/6*(5*x^2+6)*3^(1/2)/(x^4+5*x^2+3)^(1/2))*3^(1/2)-1/6*(x^4+5*x^2+3)^(1/2)/x^4-1/12*(x^4+5*x^2+3)^
(1/2)/x^2

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Rubi [A]
time = 0.04, antiderivative size = 83, 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, 848, 820, 738, 212} \begin {gather*} -\frac {\sqrt {x^4+5 x^2+3}}{12 x^2}-\frac {\sqrt {x^4+5 x^2+3}}{6 x^4}+\frac {1}{8} \sqrt {3} \tanh ^{-1}\left (\frac {5 x^2+6}{2 \sqrt {3} \sqrt {x^4+5 x^2+3}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*Sqrt[3 + 5*x^2 + x^4]/x^4 - Sqrt[3 + 5*x^2 + x^4]/(12*x^2) + (Sqrt[3]*ArcTanh[(6 + 5*x^2)/(2*Sqrt[3]*Sqrt
[3 + 5*x^2 + x^4])])/8

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 820

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2))), x] - Dist[
(b*(e*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(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] && EqQ[S
implify[m + 2*p + 3], 0]

Rule 848

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(e*f - d*g)*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/((m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Dist[1/((m
+ 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*Simp[(c*d*f - f*b*e + a*e*g)*(m + 1)
+ b*(d*g - e*f)*(p + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] &&
NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, -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^5 \sqrt {3+5 x^2+x^4}} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {2+3 x}{x^3 \sqrt {3+5 x+x^2}} \, dx,x,x^2\right )\\ &=-\frac {\sqrt {3+5 x^2+x^4}}{6 x^4}-\frac {1}{12} \text {Subst}\left (\int \frac {-3+2 x}{x^2 \sqrt {3+5 x+x^2}} \, dx,x,x^2\right )\\ &=-\frac {\sqrt {3+5 x^2+x^4}}{6 x^4}-\frac {\sqrt {3+5 x^2+x^4}}{12 x^2}-\frac {3}{8} \text {Subst}\left (\int \frac {1}{x \sqrt {3+5 x+x^2}} \, dx,x,x^2\right )\\ &=-\frac {\sqrt {3+5 x^2+x^4}}{6 x^4}-\frac {\sqrt {3+5 x^2+x^4}}{12 x^2}+\frac {3}{4} \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 {\sqrt {3+5 x^2+x^4}}{6 x^4}-\frac {\sqrt {3+5 x^2+x^4}}{12 x^2}+\frac {1}{8} \sqrt {3} \tanh ^{-1}\left (\frac {6+5 x^2}{2 \sqrt {3} \sqrt {3+5 x^2+x^4}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.16, size = 63, normalized size = 0.76 \begin {gather*} -\frac {\left (2+x^2\right ) \sqrt {3+5 x^2+x^4}}{12 x^4}-\frac {1}{4} \sqrt {3} \tanh ^{-1}\left (\frac {x^2-\sqrt {3+5 x^2+x^4}}{\sqrt {3}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*((2 + x^2)*Sqrt[3 + 5*x^2 + x^4])/x^4 - (Sqrt[3]*ArcTanh[(x^2 - Sqrt[3 + 5*x^2 + x^4])/Sqrt[3]])/4

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Maple [A]
time = 0.21, size = 66, normalized size = 0.80

method	result	size
risch	\(-\frac {x^{6}+7 x^{4}+13 x^{2}+6}{12 x^{4} \sqrt {x^{4}+5 x^{2}+3}}+\frac {\arctanh \left (\frac {\left (5 x^{2}+6\right ) \sqrt {3}}{6 \sqrt {x^{4}+5 x^{2}+3}}\right ) \sqrt {3}}{8}\)	\(64\)
default	\(\frac {\arctanh \left (\frac {\left (5 x^{2}+6\right ) \sqrt {3}}{6 \sqrt {x^{4}+5 x^{2}+3}}\right ) \sqrt {3}}{8}-\frac {\sqrt {x^{4}+5 x^{2}+3}}{6 x^{4}}-\frac {\sqrt {x^{4}+5 x^{2}+3}}{12 x^{2}}\)	\(66\)
elliptic	\(\frac {\arctanh \left (\frac {\left (5 x^{2}+6\right ) \sqrt {3}}{6 \sqrt {x^{4}+5 x^{2}+3}}\right ) \sqrt {3}}{8}-\frac {\sqrt {x^{4}+5 x^{2}+3}}{6 x^{4}}-\frac {\sqrt {x^{4}+5 x^{2}+3}}{12 x^{2}}\)	\(66\)
trager	\(-\frac {\left (x^{2}+2\right ) \sqrt {x^{4}+5 x^{2}+3}}{12 x^{4}}-\frac {\RootOf \left (\textit {\_Z}^{2}-3\right ) \ln \left (-\frac {-5 \RootOf \left (\textit {\_Z}^{2}-3\right ) x^{2}+6 \sqrt {x^{4}+5 x^{2}+3}-6 \RootOf \left (\textit {\_Z}^{2}-3\right )}{x^{2}}\right )}{8}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*arctanh(1/6*(5*x^2+6)*3^(1/2)/(x^4+5*x^2+3)^(1/2))*3^(1/2)-1/6*(x^4+5*x^2+3)^(1/2)/x^4-1/12*(x^4+5*x^2+3)^
(1/2)/x^2

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Maxima [A]
time = 0.49, size = 68, normalized size = 0.82 \begin {gather*} \frac {1}{8} \, \sqrt {3} \log \left (\frac {2 \, \sqrt {3} \sqrt {x^{4} + 5 \, x^{2} + 3}}{x^{2}} + \frac {6}{x^{2}} + 5\right ) - \frac {\sqrt {x^{4} + 5 \, x^{2} + 3}}{12 \, x^{2}} - \frac {\sqrt {x^{4} + 5 \, x^{2} + 3}}{6 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(3)*log(2*sqrt(3)*sqrt(x^4 + 5*x^2 + 3)/x^2 + 6/x^2 + 5) - 1/12*sqrt(x^4 + 5*x^2 + 3)/x^2 - 1/6*sqrt(x
^4 + 5*x^2 + 3)/x^4

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Fricas [A]
time = 0.35, size = 83, normalized size = 1.00 \begin {gather*} \frac {3 \, \sqrt {3} x^{4} \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 ) - 2 \, x^{4} - 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} {\left (x^{2} + 2\right )}}{24 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*sqrt(3)*x^4*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) -
2*x^4 - 2*sqrt(x^4 + 5*x^2 + 3)*(x^2 + 2))/x^4

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {3 x^{2} + 2}{x^{5} \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**5/(x**4+5*x**2+3)**(1/2),x)

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 145 vs. \(2 (65) = 130\).
time = 3.72, size = 145, normalized size = 1.75 \begin {gather*} -\frac {1}{8} \, \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 {9 \, {\left (x^{2} - \sqrt {x^{4} + 5 \, x^{2} + 3}\right )}^{3} + 36 \, {\left (x^{2} - \sqrt {x^{4} + 5 \, x^{2} + 3}\right )}^{2} + 47 \, x^{2} - 47 \, \sqrt {x^{4} + 5 \, x^{2} + 3} + 12}{12 \, {\left ({\left (x^{2} - \sqrt {x^{4} + 5 \, x^{2} + 3}\right )}^{2} - 3\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*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))) + 1/12*(9*(x
^2 - sqrt(x^4 + 5*x^2 + 3))^3 + 36*(x^2 - sqrt(x^4 + 5*x^2 + 3))^2 + 47*x^2 - 47*sqrt(x^4 + 5*x^2 + 3) + 12)/(
(x^2 - sqrt(x^4 + 5*x^2 + 3))^2 - 3)^2

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {3\,x^2+2}{x^5\,\sqrt {x^4+5\,x^2+3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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