3.541 \(\int x^2 (a+b x^2)^{5/2} (A+B x^2) \, dx\)

Optimal. Leaf size=188 \[ -\frac {a^4 (10 A b-3 a B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{256 b^{5/2}}+\frac {a^3 x \sqrt {a+b x^2} (10 A b-3 a B)}{256 b^2}+\frac {a^2 x^3 \sqrt {a+b x^2} (10 A b-3 a B)}{128 b}+\frac {a x^3 \left (a+b x^2\right )^{3/2} (10 A b-3 a B)}{96 b}+\frac {x^3 \left (a+b x^2\right )^{5/2} (10 A b-3 a B)}{80 b}+\frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b} \]

[Out]

1/96*a*(10*A*b-3*B*a)*x^3*(b*x^2+a)^(3/2)/b+1/80*(10*A*b-3*B*a)*x^3*(b*x^2+a)^(5/2)/b+1/10*B*x^3*(b*x^2+a)^(7/
2)/b-1/256*a^4*(10*A*b-3*B*a)*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))/b^(5/2)+1/256*a^3*(10*A*b-3*B*a)*x*(b*x^2+a)^
(1/2)/b^2+1/128*a^2*(10*A*b-3*B*a)*x^3*(b*x^2+a)^(1/2)/b

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Rubi [A]  time = 0.09, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {459, 279, 321, 217, 206} \[ \frac {a^3 x \sqrt {a+b x^2} (10 A b-3 a B)}{256 b^2}-\frac {a^4 (10 A b-3 a B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{256 b^{5/2}}+\frac {a^2 x^3 \sqrt {a+b x^2} (10 A b-3 a B)}{128 b}+\frac {a x^3 \left (a+b x^2\right )^{3/2} (10 A b-3 a B)}{96 b}+\frac {x^3 \left (a+b x^2\right )^{5/2} (10 A b-3 a B)}{80 b}+\frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*x^2)^(5/2)*(A + B*x^2),x]

[Out]

(a^3*(10*A*b - 3*a*B)*x*Sqrt[a + b*x^2])/(256*b^2) + (a^2*(10*A*b - 3*a*B)*x^3*Sqrt[a + b*x^2])/(128*b) + (a*(
10*A*b - 3*a*B)*x^3*(a + b*x^2)^(3/2))/(96*b) + ((10*A*b - 3*a*B)*x^3*(a + b*x^2)^(5/2))/(80*b) + (B*x^3*(a +
b*x^2)^(7/2))/(10*b) - (a^4*(10*A*b - 3*a*B)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(256*b^(5/2))

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rubi steps

\begin {align*} \int x^2 \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx &=\frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}-\frac {(-10 A b+3 a B) \int x^2 \left (a+b x^2\right )^{5/2} \, dx}{10 b}\\ &=\frac {(10 A b-3 a B) x^3 \left (a+b x^2\right )^{5/2}}{80 b}+\frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}+\frac {(a (10 A b-3 a B)) \int x^2 \left (a+b x^2\right )^{3/2} \, dx}{16 b}\\ &=\frac {a (10 A b-3 a B) x^3 \left (a+b x^2\right )^{3/2}}{96 b}+\frac {(10 A b-3 a B) x^3 \left (a+b x^2\right )^{5/2}}{80 b}+\frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}+\frac {\left (a^2 (10 A b-3 a B)\right ) \int x^2 \sqrt {a+b x^2} \, dx}{32 b}\\ &=\frac {a^2 (10 A b-3 a B) x^3 \sqrt {a+b x^2}}{128 b}+\frac {a (10 A b-3 a B) x^3 \left (a+b x^2\right )^{3/2}}{96 b}+\frac {(10 A b-3 a B) x^3 \left (a+b x^2\right )^{5/2}}{80 b}+\frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}+\frac {\left (a^3 (10 A b-3 a B)\right ) \int \frac {x^2}{\sqrt {a+b x^2}} \, dx}{128 b}\\ &=\frac {a^3 (10 A b-3 a B) x \sqrt {a+b x^2}}{256 b^2}+\frac {a^2 (10 A b-3 a B) x^3 \sqrt {a+b x^2}}{128 b}+\frac {a (10 A b-3 a B) x^3 \left (a+b x^2\right )^{3/2}}{96 b}+\frac {(10 A b-3 a B) x^3 \left (a+b x^2\right )^{5/2}}{80 b}+\frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}-\frac {\left (a^4 (10 A b-3 a B)\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx}{256 b^2}\\ &=\frac {a^3 (10 A b-3 a B) x \sqrt {a+b x^2}}{256 b^2}+\frac {a^2 (10 A b-3 a B) x^3 \sqrt {a+b x^2}}{128 b}+\frac {a (10 A b-3 a B) x^3 \left (a+b x^2\right )^{3/2}}{96 b}+\frac {(10 A b-3 a B) x^3 \left (a+b x^2\right )^{5/2}}{80 b}+\frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}-\frac {\left (a^4 (10 A b-3 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )}{256 b^2}\\ &=\frac {a^3 (10 A b-3 a B) x \sqrt {a+b x^2}}{256 b^2}+\frac {a^2 (10 A b-3 a B) x^3 \sqrt {a+b x^2}}{128 b}+\frac {a (10 A b-3 a B) x^3 \left (a+b x^2\right )^{3/2}}{96 b}+\frac {(10 A b-3 a B) x^3 \left (a+b x^2\right )^{5/2}}{80 b}+\frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}-\frac {a^4 (10 A b-3 a B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{256 b^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.31, size = 151, normalized size = 0.80 \[ \frac {\sqrt {a+b x^2} \left (\frac {15 a^{7/2} (3 a B-10 A b) \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {\frac {b x^2}{a}+1}}+\sqrt {b} x \left (-45 a^4 B+30 a^3 b \left (5 A+B x^2\right )+4 a^2 b^2 x^2 \left (295 A+186 B x^2\right )+16 a b^3 x^4 \left (85 A+63 B x^2\right )+96 b^4 x^6 \left (5 A+4 B x^2\right )\right )\right )}{3840 b^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*x^2)^(5/2)*(A + B*x^2),x]

[Out]

(Sqrt[a + b*x^2]*(Sqrt[b]*x*(-45*a^4*B + 30*a^3*b*(5*A + B*x^2) + 96*b^4*x^6*(5*A + 4*B*x^2) + 16*a*b^3*x^4*(8
5*A + 63*B*x^2) + 4*a^2*b^2*x^2*(295*A + 186*B*x^2)) + (15*a^(7/2)*(-10*A*b + 3*a*B)*ArcSinh[(Sqrt[b]*x)/Sqrt[
a]])/Sqrt[1 + (b*x^2)/a]))/(3840*b^(5/2))

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fricas [A]  time = 0.82, size = 308, normalized size = 1.64 \[ \left [-\frac {15 \, {\left (3 \, B a^{5} - 10 \, A a^{4} b\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (384 \, B b^{5} x^{9} + 48 \, {\left (21 \, B a b^{4} + 10 \, A b^{5}\right )} x^{7} + 8 \, {\left (93 \, B a^{2} b^{3} + 170 \, A a b^{4}\right )} x^{5} + 10 \, {\left (3 \, B a^{3} b^{2} + 118 \, A a^{2} b^{3}\right )} x^{3} - 15 \, {\left (3 \, B a^{4} b - 10 \, A a^{3} b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{7680 \, b^{3}}, -\frac {15 \, {\left (3 \, B a^{5} - 10 \, A a^{4} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (384 \, B b^{5} x^{9} + 48 \, {\left (21 \, B a b^{4} + 10 \, A b^{5}\right )} x^{7} + 8 \, {\left (93 \, B a^{2} b^{3} + 170 \, A a b^{4}\right )} x^{5} + 10 \, {\left (3 \, B a^{3} b^{2} + 118 \, A a^{2} b^{3}\right )} x^{3} - 15 \, {\left (3 \, B a^{4} b - 10 \, A a^{3} b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{3840 \, b^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x^2+a)^(5/2)*(B*x^2+A),x, algorithm="fricas")

[Out]

[-1/7680*(15*(3*B*a^5 - 10*A*a^4*b)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(384*B*b^5*x^9
+ 48*(21*B*a*b^4 + 10*A*b^5)*x^7 + 8*(93*B*a^2*b^3 + 170*A*a*b^4)*x^5 + 10*(3*B*a^3*b^2 + 118*A*a^2*b^3)*x^3
- 15*(3*B*a^4*b - 10*A*a^3*b^2)*x)*sqrt(b*x^2 + a))/b^3, -1/3840*(15*(3*B*a^5 - 10*A*a^4*b)*sqrt(-b)*arctan(sq
rt(-b)*x/sqrt(b*x^2 + a)) - (384*B*b^5*x^9 + 48*(21*B*a*b^4 + 10*A*b^5)*x^7 + 8*(93*B*a^2*b^3 + 170*A*a*b^4)*x
^5 + 10*(3*B*a^3*b^2 + 118*A*a^2*b^3)*x^3 - 15*(3*B*a^4*b - 10*A*a^3*b^2)*x)*sqrt(b*x^2 + a))/b^3]

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giac [A]  time = 0.41, size = 165, normalized size = 0.88 \[ \frac {1}{3840} \, {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, B b^{2} x^{2} + \frac {21 \, B a b^{9} + 10 \, A b^{10}}{b^{8}}\right )} x^{2} + \frac {93 \, B a^{2} b^{8} + 170 \, A a b^{9}}{b^{8}}\right )} x^{2} + \frac {5 \, {\left (3 \, B a^{3} b^{7} + 118 \, A a^{2} b^{8}\right )}}{b^{8}}\right )} x^{2} - \frac {15 \, {\left (3 \, B a^{4} b^{6} - 10 \, A a^{3} b^{7}\right )}}{b^{8}}\right )} \sqrt {b x^{2} + a} x - \frac {{\left (3 \, B a^{5} - 10 \, A a^{4} b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{256 \, b^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x^2+a)^(5/2)*(B*x^2+A),x, algorithm="giac")

[Out]

1/3840*(2*(4*(6*(8*B*b^2*x^2 + (21*B*a*b^9 + 10*A*b^10)/b^8)*x^2 + (93*B*a^2*b^8 + 170*A*a*b^9)/b^8)*x^2 + 5*(
3*B*a^3*b^7 + 118*A*a^2*b^8)/b^8)*x^2 - 15*(3*B*a^4*b^6 - 10*A*a^3*b^7)/b^8)*sqrt(b*x^2 + a)*x - 1/256*(3*B*a^
5 - 10*A*a^4*b)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/2)

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maple [A]  time = 0.01, size = 215, normalized size = 1.14 \[ -\frac {5 A \,a^{4} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{128 b^{\frac {3}{2}}}+\frac {3 B \,a^{5} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{256 b^{\frac {5}{2}}}-\frac {5 \sqrt {b \,x^{2}+a}\, A \,a^{3} x}{128 b}+\frac {3 \sqrt {b \,x^{2}+a}\, B \,a^{4} x}{256 b^{2}}-\frac {5 \left (b \,x^{2}+a \right )^{\frac {3}{2}} A \,a^{2} x}{192 b}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} B \,a^{3} x}{128 b^{2}}+\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} B \,x^{3}}{10 b}-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} A a x}{48 b}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} B \,a^{2} x}{160 b^{2}}+\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} A x}{8 b}-\frac {3 \left (b \,x^{2}+a \right )^{\frac {7}{2}} B a x}{80 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(b*x^2+a)^(5/2)*(B*x^2+A),x)

[Out]

1/10*B*x^3*(b*x^2+a)^(7/2)/b-3/80*B*a/b^2*x*(b*x^2+a)^(7/2)+1/160*B*a^2/b^2*x*(b*x^2+a)^(5/2)+1/128*B*a^3/b^2*
x*(b*x^2+a)^(3/2)+3/256*B*a^4/b^2*x*(b*x^2+a)^(1/2)+3/256*B*a^5/b^(5/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))+1/8*A*x*
(b*x^2+a)^(7/2)/b-1/48*A*a/b*x*(b*x^2+a)^(5/2)-5/192*A*a^2/b*x*(b*x^2+a)^(3/2)-5/128*A*a^3/b*x*(b*x^2+a)^(1/2)
-5/128*A*a^4/b^(3/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))

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maxima [A]  time = 1.06, size = 200, normalized size = 1.06 \[ \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} B x^{3}}{10 \, b} - \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} B a x}{80 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B a^{2} x}{160 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{3} x}{128 \, b^{2}} + \frac {3 \, \sqrt {b x^{2} + a} B a^{4} x}{256 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} A x}{8 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A a x}{48 \, b} - \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a^{2} x}{192 \, b} - \frac {5 \, \sqrt {b x^{2} + a} A a^{3} x}{128 \, b} + \frac {3 \, B a^{5} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{256 \, b^{\frac {5}{2}}} - \frac {5 \, A a^{4} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{128 \, b^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x^2+a)^(5/2)*(B*x^2+A),x, algorithm="maxima")

[Out]

1/10*(b*x^2 + a)^(7/2)*B*x^3/b - 3/80*(b*x^2 + a)^(7/2)*B*a*x/b^2 + 1/160*(b*x^2 + a)^(5/2)*B*a^2*x/b^2 + 1/12
8*(b*x^2 + a)^(3/2)*B*a^3*x/b^2 + 3/256*sqrt(b*x^2 + a)*B*a^4*x/b^2 + 1/8*(b*x^2 + a)^(7/2)*A*x/b - 1/48*(b*x^
2 + a)^(5/2)*A*a*x/b - 5/192*(b*x^2 + a)^(3/2)*A*a^2*x/b - 5/128*sqrt(b*x^2 + a)*A*a^3*x/b + 3/256*B*a^5*arcsi
nh(b*x/sqrt(a*b))/b^(5/2) - 5/128*A*a^4*arcsinh(b*x/sqrt(a*b))/b^(3/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,\left (B\,x^2+A\right )\,{\left (b\,x^2+a\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(A + B*x^2)*(a + b*x^2)^(5/2),x)

[Out]

int(x^2*(A + B*x^2)*(a + b*x^2)^(5/2), x)

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sympy [B]  time = 59.84, size = 348, normalized size = 1.85 \[ \frac {5 A a^{\frac {7}{2}} x}{128 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {133 A a^{\frac {5}{2}} x^{3}}{384 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {127 A a^{\frac {3}{2}} b x^{5}}{192 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {23 A \sqrt {a} b^{2} x^{7}}{48 \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {5 A a^{4} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{128 b^{\frac {3}{2}}} + \frac {A b^{3} x^{9}}{8 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {3 B a^{\frac {9}{2}} x}{256 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {B a^{\frac {7}{2}} x^{3}}{256 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {129 B a^{\frac {5}{2}} x^{5}}{640 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {73 B a^{\frac {3}{2}} b x^{7}}{160 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {29 B \sqrt {a} b^{2} x^{9}}{80 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 B a^{5} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{256 b^{\frac {5}{2}}} + \frac {B b^{3} x^{11}}{10 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(b*x**2+a)**(5/2)*(B*x**2+A),x)

[Out]

5*A*a**(7/2)*x/(128*b*sqrt(1 + b*x**2/a)) + 133*A*a**(5/2)*x**3/(384*sqrt(1 + b*x**2/a)) + 127*A*a**(3/2)*b*x*
*5/(192*sqrt(1 + b*x**2/a)) + 23*A*sqrt(a)*b**2*x**7/(48*sqrt(1 + b*x**2/a)) - 5*A*a**4*asinh(sqrt(b)*x/sqrt(a
))/(128*b**(3/2)) + A*b**3*x**9/(8*sqrt(a)*sqrt(1 + b*x**2/a)) - 3*B*a**(9/2)*x/(256*b**2*sqrt(1 + b*x**2/a))
- B*a**(7/2)*x**3/(256*b*sqrt(1 + b*x**2/a)) + 129*B*a**(5/2)*x**5/(640*sqrt(1 + b*x**2/a)) + 73*B*a**(3/2)*b*
x**7/(160*sqrt(1 + b*x**2/a)) + 29*B*sqrt(a)*b**2*x**9/(80*sqrt(1 + b*x**2/a)) + 3*B*a**5*asinh(sqrt(b)*x/sqrt
(a))/(256*b**(5/2)) + B*b**3*x**11/(10*sqrt(a)*sqrt(1 + b*x**2/a))

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