3.147 \(\int \frac {(2+3 x^2) \sqrt {3+5 x^2+x^4}}{x^5} \, dx\)

Optimal. Leaf size=99 \[ -\frac {\sqrt {x^4+5 x^2+3} \left (23 x^2+6\right )}{12 x^4}+\frac {3}{2} \tanh ^{-1}\left (\frac {2 x^2+5}{2 \sqrt {x^4+5 x^2+3}}\right )-\frac {77 \tanh ^{-1}\left (\frac {5 x^2+6}{2 \sqrt {3} \sqrt {x^4+5 x^2+3}}\right )}{24 \sqrt {3}} \]

[Out]

3/2*arctanh(1/2*(2*x^2+5)/(x^4+5*x^2+3)^(1/2))-77/72*arctanh(1/6*(5*x^2+6)*3^(1/2)/(x^4+5*x^2+3)^(1/2))*3^(1/2
)-1/12*(23*x^2+6)*(x^4+5*x^2+3)^(1/2)/x^4

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Rubi [A]  time = 0.08, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {1251, 810, 843, 621, 206, 724} \[ -\frac {\sqrt {x^4+5 x^2+3} \left (23 x^2+6\right )}{12 x^4}+\frac {3}{2} \tanh ^{-1}\left (\frac {2 x^2+5}{2 \sqrt {x^4+5 x^2+3}}\right )-\frac {77 \tanh ^{-1}\left (\frac {5 x^2+6}{2 \sqrt {3} \sqrt {x^4+5 x^2+3}}\right )}{24 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((6 + 23*x^2)*Sqrt[3 + 5*x^2 + x^4])/(12*x^4) + (3*ArcTanh[(5 + 2*x^2)/(2*Sqrt[3 + 5*x^2 + x^4])])/2 - (77*Ar
cTanh[(6 + 5*x^2)/(2*Sqrt[3]*Sqrt[3 + 5*x^2 + x^4])])/(24*Sqrt[3])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 621

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 724

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 810

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

Rule 843

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 1251

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 {\left (2+3 x^2\right ) \sqrt {3+5 x^2+x^4}}{x^5} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(2+3 x) \sqrt {3+5 x+x^2}}{x^3} \, dx,x,x^2\right )\\ &=-\frac {\left (6+23 x^2\right ) \sqrt {3+5 x^2+x^4}}{12 x^4}-\frac {1}{24} \operatorname {Subst}\left (\int \frac {-77-36 x}{x \sqrt {3+5 x+x^2}} \, dx,x,x^2\right )\\ &=-\frac {\left (6+23 x^2\right ) \sqrt {3+5 x^2+x^4}}{12 x^4}+\frac {3}{2} \operatorname {Subst}\left (\int \frac {1}{\sqrt {3+5 x+x^2}} \, dx,x,x^2\right )+\frac {77}{24} \operatorname {Subst}\left (\int \frac {1}{x \sqrt {3+5 x+x^2}} \, dx,x,x^2\right )\\ &=-\frac {\left (6+23 x^2\right ) \sqrt {3+5 x^2+x^4}}{12 x^4}+3 \operatorname {Subst}\left (\int \frac {1}{4-x^2} \, dx,x,\frac {5+2 x^2}{\sqrt {3+5 x^2+x^4}}\right )-\frac {77}{12} \operatorname {Subst}\left (\int \frac {1}{12-x^2} \, dx,x,\frac {6+5 x^2}{\sqrt {3+5 x^2+x^4}}\right )\\ &=-\frac {\left (6+23 x^2\right ) \sqrt {3+5 x^2+x^4}}{12 x^4}+\frac {3}{2} \tanh ^{-1}\left (\frac {5+2 x^2}{2 \sqrt {3+5 x^2+x^4}}\right )-\frac {77 \tanh ^{-1}\left (\frac {6+5 x^2}{2 \sqrt {3} \sqrt {3+5 x^2+x^4}}\right )}{24 \sqrt {3}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 97, normalized size = 0.98 \[ \frac {1}{72} \left (-\frac {6 \sqrt {x^4+5 x^2+3} \left (23 x^2+6\right )}{x^4}+108 \tanh ^{-1}\left (\frac {2 x^2+5}{2 \sqrt {x^4+5 x^2+3}}\right )-77 \sqrt {3} \tanh ^{-1}\left (\frac {5 x^2+6}{2 \sqrt {3} \sqrt {x^4+5 x^2+3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-6*(6 + 23*x^2)*Sqrt[3 + 5*x^2 + x^4])/x^4 + 108*ArcTanh[(5 + 2*x^2)/(2*Sqrt[3 + 5*x^2 + x^4])] - 77*Sqrt[3]
*ArcTanh[(6 + 5*x^2)/(2*Sqrt[3]*Sqrt[3 + 5*x^2 + x^4])])/72

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fricas [A]  time = 0.65, size = 112, normalized size = 1.13 \[ \frac {77 \, \sqrt {3} x^{4} \log \left (\frac {25 \, x^{2} - 2 \, \sqrt {3} {\left (5 \, x^{2} + 6\right )} - 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} {\left (5 \, \sqrt {3} - 6\right )} + 30}{x^{2}}\right ) - 108 \, x^{4} \log \left (-2 \, x^{2} + 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} - 5\right ) - 138 \, x^{4} - 6 \, \sqrt {x^{4} + 5 \, x^{2} + 3} {\left (23 \, x^{2} + 6\right )}}{72 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/72*(77*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)
- 108*x^4*log(-2*x^2 + 2*sqrt(x^4 + 5*x^2 + 3) - 5) - 138*x^4 - 6*sqrt(x^4 + 5*x^2 + 3)*(23*x^2 + 6))/x^4

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giac [B]  time = 0.57, size = 169, normalized size = 1.71 \[ \frac {77}{72} \, \sqrt {3} \log \left (\frac {x^{2} + \sqrt {3} - \sqrt {x^{4} + 5 \, x^{2} + 3}}{x^{2} - \sqrt {3} - \sqrt {x^{4} + 5 \, x^{2} + 3}}\right ) + \frac {127 \, {\left (x^{2} - \sqrt {x^{4} + 5 \, x^{2} + 3}\right )}^{3} + 228 \, {\left (x^{2} - \sqrt {x^{4} + 5 \, x^{2} + 3}\right )}^{2} - 159 \, x^{2} + 159 \, \sqrt {x^{4} + 5 \, x^{2} + 3} - 324}{12 \, {\left ({\left (x^{2} - \sqrt {x^{4} + 5 \, x^{2} + 3}\right )}^{2} - 3\right )}^{2}} - \frac {3}{2} \, \log \left (2 \, x^{2} - 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} + 5\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

77/72*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*(127
*(x^2 - sqrt(x^4 + 5*x^2 + 3))^3 + 228*(x^2 - sqrt(x^4 + 5*x^2 + 3))^2 - 159*x^2 + 159*sqrt(x^4 + 5*x^2 + 3) -
324)/((x^2 - sqrt(x^4 + 5*x^2 + 3))^2 - 3)^2 - 3/2*log(2*x^2 - 2*sqrt(x^4 + 5*x^2 + 3) + 5)

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maple [A]  time = 0.02, size = 121, normalized size = 1.22 \[ -\frac {77 \sqrt {3}\, \arctanh \left (\frac {\left (5 x^{2}+6\right ) \sqrt {3}}{6 \sqrt {x^{4}+5 x^{2}+3}}\right )}{72}+\frac {3 \ln \left (x^{2}+\frac {5}{2}+\sqrt {x^{4}+5 x^{2}+3}\right )}{2}-\frac {13 \left (x^{4}+5 x^{2}+3\right )^{\frac {3}{2}}}{36 x^{2}}-\frac {\left (x^{4}+5 x^{2}+3\right )^{\frac {3}{2}}}{6 x^{4}}+\frac {77 \sqrt {x^{4}+5 x^{2}+3}}{72}+\frac {13 \left (2 x^{2}+5\right ) \sqrt {x^{4}+5 x^{2}+3}}{72} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/x^4*(x^4+5*x^2+3)^(3/2)-13/36*(x^4+5*x^2+3)^(3/2)/x^2+77/72*(x^4+5*x^2+3)^(1/2)-77/72*arctanh(1/6*(5*x^2+
6)*3^(1/2)/(x^4+5*x^2+3)^(1/2))*3^(1/2)+13/72*(2*x^2+5)*(x^4+5*x^2+3)^(1/2)+3/2*ln(x^2+5/2+(x^4+5*x^2+3)^(1/2)
)

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maxima [A]  time = 1.43, size = 106, normalized size = 1.07 \[ -\frac {77}{72} \, \sqrt {3} \log \left (\frac {2 \, \sqrt {3} \sqrt {x^{4} + 5 \, x^{2} + 3}}{x^{2}} + \frac {6}{x^{2}} + 5\right ) + \frac {1}{6} \, \sqrt {x^{4} + 5 \, x^{2} + 3} - \frac {13 \, \sqrt {x^{4} + 5 \, x^{2} + 3}}{12 \, x^{2}} - \frac {{\left (x^{4} + 5 \, x^{2} + 3\right )}^{\frac {3}{2}}}{6 \, x^{4}} + \frac {3}{2} \, \log \left (2 \, x^{2} + 2 \, \sqrt {x^{4} + 5 \, x^{2} + 3} + 5\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-77/72*sqrt(3)*log(2*sqrt(3)*sqrt(x^4 + 5*x^2 + 3)/x^2 + 6/x^2 + 5) + 1/6*sqrt(x^4 + 5*x^2 + 3) - 13/12*sqrt(x
^4 + 5*x^2 + 3)/x^2 - 1/6*(x^4 + 5*x^2 + 3)^(3/2)/x^4 + 3/2*log(2*x^2 + 2*sqrt(x^4 + 5*x^2 + 3) + 5)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (3\,x^2+2\right )\,\sqrt {x^4+5\,x^2+3}}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (3 x^{2} + 2\right ) \sqrt {x^{4} + 5 x^{2} + 3}}{x^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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