3.537 \(\int \frac{(a+b x^2)^{3/2} (A+B x^2)}{x^{11}} \, dx\)

Optimal. Leaf size=184 \[ -\frac{3 b^3 \sqrt{a+b x^2} (A b-2 a B)}{256 a^3 x^2}+\frac{b^2 \sqrt{a+b x^2} (A b-2 a B)}{128 a^2 x^4}+\frac{3 b^4 (A b-2 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{256 a^{7/2}}+\frac{b \sqrt{a+b x^2} (A b-2 a B)}{32 a x^6}+\frac{\left (a+b x^2\right )^{3/2} (A b-2 a B)}{16 a x^8}-\frac{A \left (a+b x^2\right )^{5/2}}{10 a x^{10}} \]

[Out]

(b*(A*b - 2*a*B)*Sqrt[a + b*x^2])/(32*a*x^6) + (b^2*(A*b - 2*a*B)*Sqrt[a + b*x^2])/(128*a^2*x^4) - (3*b^3*(A*b
- 2*a*B)*Sqrt[a + b*x^2])/(256*a^3*x^2) + ((A*b - 2*a*B)*(a + b*x^2)^(3/2))/(16*a*x^8) - (A*(a + b*x^2)^(5/2)
)/(10*a*x^10) + (3*b^4*(A*b - 2*a*B)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(256*a^(7/2))

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Rubi [A]  time = 0.145562, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {446, 78, 47, 51, 63, 208} \[ -\frac{3 b^3 \sqrt{a+b x^2} (A b-2 a B)}{256 a^3 x^2}+\frac{b^2 \sqrt{a+b x^2} (A b-2 a B)}{128 a^2 x^4}+\frac{3 b^4 (A b-2 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{256 a^{7/2}}+\frac{b \sqrt{a+b x^2} (A b-2 a B)}{32 a x^6}+\frac{\left (a+b x^2\right )^{3/2} (A b-2 a B)}{16 a x^8}-\frac{A \left (a+b x^2\right )^{5/2}}{10 a x^{10}} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^(3/2)*(A + B*x^2))/x^11,x]

[Out]

(b*(A*b - 2*a*B)*Sqrt[a + b*x^2])/(32*a*x^6) + (b^2*(A*b - 2*a*B)*Sqrt[a + b*x^2])/(128*a^2*x^4) - (3*b^3*(A*b
- 2*a*B)*Sqrt[a + b*x^2])/(256*a^3*x^2) + ((A*b - 2*a*B)*(a + b*x^2)^(3/2))/(16*a*x^8) - (A*(a + b*x^2)^(5/2)
)/(10*a*x^10) + (3*b^4*(A*b - 2*a*B)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(256*a^(7/2))

Rule 446

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^{11}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2} (A+B x)}{x^6} \, dx,x,x^2\right )\\ &=-\frac{A \left (a+b x^2\right )^{5/2}}{10 a x^{10}}+\frac{\left (-\frac{5 A b}{2}+5 a B\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{x^5} \, dx,x,x^2\right )}{10 a}\\ &=\frac{(A b-2 a B) \left (a+b x^2\right )^{3/2}}{16 a x^8}-\frac{A \left (a+b x^2\right )^{5/2}}{10 a x^{10}}-\frac{(3 b (A b-2 a B)) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x^4} \, dx,x,x^2\right )}{32 a}\\ &=\frac{b (A b-2 a B) \sqrt{a+b x^2}}{32 a x^6}+\frac{(A b-2 a B) \left (a+b x^2\right )^{3/2}}{16 a x^8}-\frac{A \left (a+b x^2\right )^{5/2}}{10 a x^{10}}-\frac{\left (b^2 (A b-2 a B)\right ) \operatorname{Subst}\left (\int \frac{1}{x^3 \sqrt{a+b x}} \, dx,x,x^2\right )}{64 a}\\ &=\frac{b (A b-2 a B) \sqrt{a+b x^2}}{32 a x^6}+\frac{b^2 (A b-2 a B) \sqrt{a+b x^2}}{128 a^2 x^4}+\frac{(A b-2 a B) \left (a+b x^2\right )^{3/2}}{16 a x^8}-\frac{A \left (a+b x^2\right )^{5/2}}{10 a x^{10}}+\frac{\left (3 b^3 (A b-2 a B)\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b x}} \, dx,x,x^2\right )}{256 a^2}\\ &=\frac{b (A b-2 a B) \sqrt{a+b x^2}}{32 a x^6}+\frac{b^2 (A b-2 a B) \sqrt{a+b x^2}}{128 a^2 x^4}-\frac{3 b^3 (A b-2 a B) \sqrt{a+b x^2}}{256 a^3 x^2}+\frac{(A b-2 a B) \left (a+b x^2\right )^{3/2}}{16 a x^8}-\frac{A \left (a+b x^2\right )^{5/2}}{10 a x^{10}}-\frac{\left (3 b^4 (A b-2 a B)\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^2\right )}{512 a^3}\\ &=\frac{b (A b-2 a B) \sqrt{a+b x^2}}{32 a x^6}+\frac{b^2 (A b-2 a B) \sqrt{a+b x^2}}{128 a^2 x^4}-\frac{3 b^3 (A b-2 a B) \sqrt{a+b x^2}}{256 a^3 x^2}+\frac{(A b-2 a B) \left (a+b x^2\right )^{3/2}}{16 a x^8}-\frac{A \left (a+b x^2\right )^{5/2}}{10 a x^{10}}-\frac{\left (3 b^3 (A b-2 a B)\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^2}\right )}{256 a^3}\\ &=\frac{b (A b-2 a B) \sqrt{a+b x^2}}{32 a x^6}+\frac{b^2 (A b-2 a B) \sqrt{a+b x^2}}{128 a^2 x^4}-\frac{3 b^3 (A b-2 a B) \sqrt{a+b x^2}}{256 a^3 x^2}+\frac{(A b-2 a B) \left (a+b x^2\right )^{3/2}}{16 a x^8}-\frac{A \left (a+b x^2\right )^{5/2}}{10 a x^{10}}+\frac{3 b^4 (A b-2 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{256 a^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0267115, size = 61, normalized size = 0.33 \[ -\frac{\left (a+b x^2\right )^{5/2} \left (a^5 A+b^4 x^{10} (2 a B-A b) \, _2F_1\left (\frac{5}{2},5;\frac{7}{2};\frac{b x^2}{a}+1\right )\right )}{10 a^6 x^{10}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^(3/2)*(A + B*x^2))/x^11,x]

[Out]

-((a + b*x^2)^(5/2)*(a^5*A + b^4*(-(A*b) + 2*a*B)*x^10*Hypergeometric2F1[5/2, 5, 7/2, 1 + (b*x^2)/a]))/(10*a^6
*x^10)

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Maple [B]  time = 0.027, size = 317, normalized size = 1.7 \begin{align*} -{\frac{A}{10\,a{x}^{10}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{Ab}{16\,{a}^{2}{x}^{8}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{A{b}^{2}}{32\,{a}^{3}{x}^{6}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{A{b}^{3}}{128\,{a}^{4}{x}^{4}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{A{b}^{4}}{256\,{a}^{5}{x}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{A{b}^{5}}{256\,{a}^{5}} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{3\,A{b}^{5}}{256}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{7}{2}}}}-{\frac{3\,A{b}^{5}}{256\,{a}^{4}}\sqrt{b{x}^{2}+a}}-{\frac{B}{8\,a{x}^{8}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{Bb}{16\,{a}^{2}{x}^{6}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{B{b}^{2}}{64\,{a}^{3}{x}^{4}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{B{b}^{3}}{128\,{a}^{4}{x}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{B{b}^{4}}{128\,{a}^{4}} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{3\,B{b}^{4}}{128}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{5}{2}}}}+{\frac{3\,B{b}^{4}}{128\,{a}^{3}}\sqrt{b{x}^{2}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(3/2)*(B*x^2+A)/x^11,x)

[Out]

-1/10*A*(b*x^2+a)^(5/2)/a/x^10+1/16*A*b/a^2/x^8*(b*x^2+a)^(5/2)-1/32*A*b^2/a^3/x^6*(b*x^2+a)^(5/2)+1/128*A*b^3
/a^4/x^4*(b*x^2+a)^(5/2)+1/256*A*b^4/a^5/x^2*(b*x^2+a)^(5/2)-1/256*A*b^5/a^5*(b*x^2+a)^(3/2)+3/256*A*b^5/a^(7/
2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)-3/256*A*b^5/a^4*(b*x^2+a)^(1/2)-1/8*B/a/x^8*(b*x^2+a)^(5/2)+1/16*B*b/
a^2/x^6*(b*x^2+a)^(5/2)-1/64*B*b^2/a^3/x^4*(b*x^2+a)^(5/2)-1/128*B*b^3/a^4/x^2*(b*x^2+a)^(5/2)+1/128*B*b^4/a^4
*(b*x^2+a)^(3/2)-3/128*B*b^4/a^(5/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)+3/128*B*b^4/a^3*(b*x^2+a)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/x^11,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.0658, size = 722, normalized size = 3.92 \begin{align*} \left [-\frac{15 \,{\left (2 \, B a b^{4} - A b^{5}\right )} \sqrt{a} x^{10} \log \left (-\frac{b x^{2} + 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) - 2 \,{\left (15 \,{\left (2 \, B a^{2} b^{3} - A a b^{4}\right )} x^{8} - 10 \,{\left (2 \, B a^{3} b^{2} - A a^{2} b^{3}\right )} x^{6} - 128 \, A a^{5} - 8 \,{\left (30 \, B a^{4} b + A a^{3} b^{2}\right )} x^{4} - 16 \,{\left (10 \, B a^{5} + 11 \, A a^{4} b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{2560 \, a^{4} x^{10}}, \frac{15 \,{\left (2 \, B a b^{4} - A b^{5}\right )} \sqrt{-a} x^{10} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) +{\left (15 \,{\left (2 \, B a^{2} b^{3} - A a b^{4}\right )} x^{8} - 10 \,{\left (2 \, B a^{3} b^{2} - A a^{2} b^{3}\right )} x^{6} - 128 \, A a^{5} - 8 \,{\left (30 \, B a^{4} b + A a^{3} b^{2}\right )} x^{4} - 16 \,{\left (10 \, B a^{5} + 11 \, A a^{4} b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{1280 \, a^{4} x^{10}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/x^11,x, algorithm="fricas")

[Out]

[-1/2560*(15*(2*B*a*b^4 - A*b^5)*sqrt(a)*x^10*log(-(b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(15*(2*B
*a^2*b^3 - A*a*b^4)*x^8 - 10*(2*B*a^3*b^2 - A*a^2*b^3)*x^6 - 128*A*a^5 - 8*(30*B*a^4*b + A*a^3*b^2)*x^4 - 16*(
10*B*a^5 + 11*A*a^4*b)*x^2)*sqrt(b*x^2 + a))/(a^4*x^10), 1/1280*(15*(2*B*a*b^4 - A*b^5)*sqrt(-a)*x^10*arctan(s
qrt(-a)/sqrt(b*x^2 + a)) + (15*(2*B*a^2*b^3 - A*a*b^4)*x^8 - 10*(2*B*a^3*b^2 - A*a^2*b^3)*x^6 - 128*A*a^5 - 8*
(30*B*a^4*b + A*a^3*b^2)*x^4 - 16*(10*B*a^5 + 11*A*a^4*b)*x^2)*sqrt(b*x^2 + a))/(a^4*x^10)]

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Sympy [B]  time = 162.342, size = 345, normalized size = 1.88 \begin{align*} - \frac{A a^{2}}{10 \sqrt{b} x^{11} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{19 A a \sqrt{b}}{80 x^{9} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{23 A b^{\frac{3}{2}}}{160 x^{7} \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{A b^{\frac{5}{2}}}{640 a x^{5} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{A b^{\frac{7}{2}}}{256 a^{2} x^{3} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{3 A b^{\frac{9}{2}}}{256 a^{3} x \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{3 A b^{5} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{256 a^{\frac{7}{2}}} - \frac{B a^{2}}{8 \sqrt{b} x^{9} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{5 B a \sqrt{b}}{16 x^{7} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{13 B b^{\frac{3}{2}}}{64 x^{5} \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{B b^{\frac{5}{2}}}{128 a x^{3} \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{3 B b^{\frac{7}{2}}}{128 a^{2} x \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{3 B b^{4} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{128 a^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(3/2)*(B*x**2+A)/x**11,x)

[Out]

-A*a**2/(10*sqrt(b)*x**11*sqrt(a/(b*x**2) + 1)) - 19*A*a*sqrt(b)/(80*x**9*sqrt(a/(b*x**2) + 1)) - 23*A*b**(3/2
)/(160*x**7*sqrt(a/(b*x**2) + 1)) + A*b**(5/2)/(640*a*x**5*sqrt(a/(b*x**2) + 1)) - A*b**(7/2)/(256*a**2*x**3*s
qrt(a/(b*x**2) + 1)) - 3*A*b**(9/2)/(256*a**3*x*sqrt(a/(b*x**2) + 1)) + 3*A*b**5*asinh(sqrt(a)/(sqrt(b)*x))/(2
56*a**(7/2)) - B*a**2/(8*sqrt(b)*x**9*sqrt(a/(b*x**2) + 1)) - 5*B*a*sqrt(b)/(16*x**7*sqrt(a/(b*x**2) + 1)) - 1
3*B*b**(3/2)/(64*x**5*sqrt(a/(b*x**2) + 1)) + B*b**(5/2)/(128*a*x**3*sqrt(a/(b*x**2) + 1)) + 3*B*b**(7/2)/(128
*a**2*x*sqrt(a/(b*x**2) + 1)) - 3*B*b**4*asinh(sqrt(a)/(sqrt(b)*x))/(128*a**(5/2))

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Giac [A]  time = 1.12723, size = 286, normalized size = 1.55 \begin{align*} \frac{\frac{15 \,{\left (2 \, B a b^{5} - A b^{6}\right )} \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{3}} + \frac{30 \,{\left (b x^{2} + a\right )}^{\frac{9}{2}} B a b^{5} - 140 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}} B a^{2} b^{5} + 140 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} B a^{4} b^{5} - 30 \, \sqrt{b x^{2} + a} B a^{5} b^{5} - 15 \,{\left (b x^{2} + a\right )}^{\frac{9}{2}} A b^{6} + 70 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}} A a b^{6} - 128 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} A a^{2} b^{6} - 70 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} A a^{3} b^{6} + 15 \, \sqrt{b x^{2} + a} A a^{4} b^{6}}{a^{3} b^{5} x^{10}}}{1280 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/x^11,x, algorithm="giac")

[Out]

1/1280*(15*(2*B*a*b^5 - A*b^6)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^3) + (30*(b*x^2 + a)^(9/2)*B*a*b^5
- 140*(b*x^2 + a)^(7/2)*B*a^2*b^5 + 140*(b*x^2 + a)^(3/2)*B*a^4*b^5 - 30*sqrt(b*x^2 + a)*B*a^5*b^5 - 15*(b*x^
2 + a)^(9/2)*A*b^6 + 70*(b*x^2 + a)^(7/2)*A*a*b^6 - 128*(b*x^2 + a)^(5/2)*A*a^2*b^6 - 70*(b*x^2 + a)^(3/2)*A*a
^3*b^6 + 15*sqrt(b*x^2 + a)*A*a^4*b^6)/(a^3*b^5*x^10))/b