3.2.98 \(\int \frac {2+3 x^2}{x^3 (3+5 x^2+x^4)^{3/2}} \, dx\) [198]

Optimal. Leaf size=90 \[ \frac {-7-8 x^2}{39 x^2 \sqrt {3+5 x^2+x^4}}-\frac {2 \sqrt {3+5 x^2+x^4}}{39 x^2}+\frac {\tanh ^{-1}\left (\frac {6+5 x^2}{2 \sqrt {3} \sqrt {3+5 x^2+x^4}}\right )}{3 \sqrt {3}} \]

[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^2/(x^4+5*x^2+3)^(1/2)-2/39*(x
^4+5*x^2+3)^(1/2)/x^2

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Rubi [A]
time = 0.05, antiderivative size = 90, 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, 820, 738, 212} \begin {gather*} -\frac {8 x^2+7}{39 x^2 \sqrt {x^4+5 x^2+3}}-\frac {2 \sqrt {x^4+5 x^2+3}}{39 x^2}+\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*(3 + 5*x^2 + x^4)^(3/2)),x]

[Out]

-1/39*(7 + 8*x^2)/(x^2*Sqrt[3 + 5*x^2 + x^4]) - (2*Sqrt[3 + 5*x^2 + x^4])/(39*x^2) + ArcTanh[(6 + 5*x^2)/(2*Sq
rt[3]*Sqrt[3 + 5*x^2 + x^4])]/(3*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 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 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^3 \left (3+5 x^2+x^4\right )^{3/2}} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {2+3 x}{x^2 \left (3+5 x+x^2\right )^{3/2}} \, dx,x,x^2\right )\\ &=-\frac {7+8 x^2}{39 x^2 \sqrt {3+5 x^2+x^4}}-\frac {1}{39} \text {Subst}\left (\int \frac {-6+8 x}{x^2 \sqrt {3+5 x+x^2}} \, dx,x,x^2\right )\\ &=-\frac {7+8 x^2}{39 x^2 \sqrt {3+5 x^2+x^4}}-\frac {2 \sqrt {3+5 x^2+x^4}}{39 x^2}-\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 x^2 \sqrt {3+5 x^2+x^4}}-\frac {2 \sqrt {3+5 x^2+x^4}}{39 x^2}+\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 x^2 \sqrt {3+5 x^2+x^4}}-\frac {2 \sqrt {3+5 x^2+x^4}}{39 x^2}+\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 = 0.23, size = 70, normalized size = 0.78 \begin {gather*} \frac {-13-18 x^2-2 x^4}{39 x^2 \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*(3 + 5*x^2 + x^4)^(3/2)),x]

[Out]

(-13 - 18*x^2 - 2*x^4)/(39*x^2*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 = 84, normalized size = 0.93

method	result	size
risch	\(-\frac {2 x^{4}+18 x^{2}+13}{39 x^{2} \sqrt {x^{4}+5 x^{2}+3}}+\frac {\arctanh \left (\frac {\left (5 x^{2}+6\right ) \sqrt {3}}{6 \sqrt {x^{4}+5 x^{2}+3}}\right ) \sqrt {3}}{9}\)	\(61\)
trager	\(-\frac {2 x^{4}+18 x^{2}+13}{39 x^{2} \sqrt {x^{4}+5 x^{2}+3}}-\frac {\RootOf \left (\textit {\_Z}^{2}-3\right ) \ln \left (-\frac {-5 \RootOf \left (\textit {\_Z}^{2}-3\right ) x^{2}+6 \sqrt {x^{4}+5 x^{2}+3}-6 \RootOf \left (\textit {\_Z}^{2}-3\right )}{x^{2}}\right )}{9}\)	\(79\)
default	\(-\frac {1}{3 x^{2} \sqrt {x^{4}+5 x^{2}+3}}-\frac {1}{3 \sqrt {x^{4}+5 x^{2}+3}}-\frac {2 x^{2}+5}{39 \sqrt {x^{4}+5 x^{2}+3}}+\frac {\arctanh \left (\frac {\left (5 x^{2}+6\right ) \sqrt {3}}{6 \sqrt {x^{4}+5 x^{2}+3}}\right ) \sqrt {3}}{9}\)	\(84\)
elliptic	\(-\frac {1}{3 x^{2} \sqrt {x^{4}+5 x^{2}+3}}-\frac {1}{3 \sqrt {x^{4}+5 x^{2}+3}}-\frac {2 x^{2}+5}{39 \sqrt {x^{4}+5 x^{2}+3}}+\frac {\arctanh \left (\frac {\left (5 x^{2}+6\right ) \sqrt {3}}{6 \sqrt {x^{4}+5 x^{2}+3}}\right ) \sqrt {3}}{9}\)	\(84\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/x^2/(x^4+5*x^2+3)^(1/2)-1/3/(x^4+5*x^2+3)^(1/2)-1/39*(2*x^2+5)/(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 = 0.50, size = 82, normalized size = 0.91 \begin {gather*} -\frac {2 \, 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 {6}{13 \, \sqrt {x^{4} + 5 \, x^{2} + 3}} - \frac {1}{3 \, \sqrt {x^{4} + 5 \, x^{2} + 3} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/39*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) - 6/13/sqrt
(x^4 + 5*x^2 + 3) - 1/3/(sqrt(x^4 + 5*x^2 + 3)*x^2)

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Fricas [A]
time = 0.41, size = 124, normalized size = 1.38 \begin {gather*} -\frac {6 \, x^{6} + 30 \, x^{4} - 13 \, \sqrt {3} {\left (x^{6} + 5 \, x^{4} + 3 \, x^{2}\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 ) + 18 \, x^{2} + 3 \, {\left (2 \, x^{4} + 18 \, x^{2} + 13\right )} \sqrt {x^{4} + 5 \, x^{2} + 3}}{117 \, {\left (x^{6} + 5 \, x^{4} + 3 \, x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/117*(6*x^6 + 30*x^4 - 13*sqrt(3)*(x^6 + 5*x^4 + 3*x^2)*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) + 18*x^2 + 3*(2*x^4 + 18*x^2 + 13)*sqrt(x^4 + 5*x^2 + 3))/(x^6 + 5*x^4 +
3*x^2)

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

[Out]

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

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Giac [A]
time = 3.47, size = 122, normalized size = 1.36 \begin {gather*} -\frac {1}{9} \, \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 {7 \, x^{2} + 11}{117 \, \sqrt {x^{4} + 5 \, x^{2} + 3}} + \frac {5 \, x^{2} - 5 \, \sqrt {x^{4} + 5 \, x^{2} + 3} + 6}{9 \, {\left ({\left (x^{2} - \sqrt {x^{4} + 5 \, x^{2} + 3}\right )}^{2} - 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2+2)/x^3/(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))/(x^2 - sqrt(3) - sqrt(x^4 + 5*x^2 + 3))) + 1/117*(7*x
^2 + 11)/sqrt(x^4 + 5*x^2 + 3) + 1/9*(5*x^2 - 5*sqrt(x^4 + 5*x^2 + 3) + 6)/((x^2 - sqrt(x^4 + 5*x^2 + 3))^2 -
3)

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

[Out]

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

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