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

Optimal. Leaf size=146 \[ -\frac {5 b (7 A b-4 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 a^{9/2}}+\frac {5 b (7 A b-4 a B)}{8 a^4 \sqrt {a+b x^2}}+\frac {5 b (7 A b-4 a B)}{24 a^3 \left (a+b x^2\right )^{3/2}}+\frac {7 A b-4 a B}{8 a^2 x^2 \left (a+b x^2\right )^{3/2}}-\frac {A}{4 a x^4 \left (a+b x^2\right )^{3/2}} \]

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Rubi [A]  time = 0.11, antiderivative size = 150, normalized size of antiderivative = 1.03, 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 = {446, 78, 51, 63, 208} \begin {gather*} \frac {5 \sqrt {a+b x^2} (7 A b-4 a B)}{8 a^4 x^2}-\frac {5 (7 A b-4 a B)}{12 a^3 x^2 \sqrt {a+b x^2}}-\frac {7 A b-4 a B}{12 a^2 x^2 \left (a+b x^2\right )^{3/2}}-\frac {5 b (7 A b-4 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 a^{9/2}}-\frac {A}{4 a x^4 \left (a+b x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-A/(4*a*x^4*(a + b*x^2)^(3/2)) - (7*A*b - 4*a*B)/(12*a^2*x^2*(a + b*x^2)^(3/2)) - (5*(7*A*b - 4*a*B))/(12*a^3*
x^2*Sqrt[a + b*x^2]) + (5*(7*A*b - 4*a*B)*Sqrt[a + b*x^2])/(8*a^4*x^2) - (5*b*(7*A*b - 4*a*B)*ArcTanh[Sqrt[a +
b*x^2]/Sqrt[a]])/(8*a^(9/2))

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 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 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]

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]]

Rubi steps

\begin {align*} \int \frac {A+B x^2}{x^5 \left (a+b x^2\right )^{5/2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {A+B x}{x^3 (a+b x)^{5/2}} \, dx,x,x^2\right )\\ &=-\frac {A}{4 a x^4 \left (a+b x^2\right )^{3/2}}+\frac {\left (-\frac {7 A b}{2}+2 a B\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 (a+b x)^{5/2}} \, dx,x,x^2\right )}{4 a}\\ &=-\frac {A}{4 a x^4 \left (a+b x^2\right )^{3/2}}-\frac {7 A b-4 a B}{12 a^2 x^2 \left (a+b x^2\right )^{3/2}}-\frac {(5 (7 A b-4 a B)) \operatorname {Subst}\left (\int \frac {1}{x^2 (a+b x)^{3/2}} \, dx,x,x^2\right )}{24 a^2}\\ &=-\frac {A}{4 a x^4 \left (a+b x^2\right )^{3/2}}-\frac {7 A b-4 a B}{12 a^2 x^2 \left (a+b x^2\right )^{3/2}}-\frac {5 (7 A b-4 a B)}{12 a^3 x^2 \sqrt {a+b x^2}}-\frac {(5 (7 A b-4 a B)) \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,x^2\right )}{8 a^3}\\ &=-\frac {A}{4 a x^4 \left (a+b x^2\right )^{3/2}}-\frac {7 A b-4 a B}{12 a^2 x^2 \left (a+b x^2\right )^{3/2}}-\frac {5 (7 A b-4 a B)}{12 a^3 x^2 \sqrt {a+b x^2}}+\frac {5 (7 A b-4 a B) \sqrt {a+b x^2}}{8 a^4 x^2}+\frac {(5 b (7 A b-4 a B)) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right )}{16 a^4}\\ &=-\frac {A}{4 a x^4 \left (a+b x^2\right )^{3/2}}-\frac {7 A b-4 a B}{12 a^2 x^2 \left (a+b x^2\right )^{3/2}}-\frac {5 (7 A b-4 a B)}{12 a^3 x^2 \sqrt {a+b x^2}}+\frac {5 (7 A b-4 a B) \sqrt {a+b x^2}}{8 a^4 x^2}+\frac {(5 (7 A b-4 a B)) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^2}\right )}{8 a^4}\\ &=-\frac {A}{4 a x^4 \left (a+b x^2\right )^{3/2}}-\frac {7 A b-4 a B}{12 a^2 x^2 \left (a+b x^2\right )^{3/2}}-\frac {5 (7 A b-4 a B)}{12 a^3 x^2 \sqrt {a+b x^2}}+\frac {5 (7 A b-4 a B) \sqrt {a+b x^2}}{8 a^4 x^2}-\frac {5 b (7 A b-4 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 a^{9/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 60, normalized size = 0.41 \begin {gather*} \frac {b x^4 (7 A b-4 a B) \, _2F_1\left (-\frac {3}{2},2;-\frac {1}{2};\frac {b x^2}{a}+1\right )-3 a^2 A}{12 a^3 x^4 \left (a+b x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a^2*A + b*(7*A*b - 4*a*B)*x^4*Hypergeometric2F1[-3/2, 2, -1/2, 1 + (b*x^2)/a])/(12*a^3*x^4*(a + b*x^2)^(3/
2))

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IntegrateAlgebraic [A]  time = 0.19, size = 126, normalized size = 0.86 \begin {gather*} \frac {5 \left (4 a b B-7 A b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 a^{9/2}}+\frac {-6 a^3 A-12 a^3 B x^2+21 a^2 A b x^2-80 a^2 b B x^4+140 a A b^2 x^4-60 a b^2 B x^6+105 A b^3 x^6}{24 a^4 x^4 \left (a+b x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*a^3*A + 21*a^2*A*b*x^2 - 12*a^3*B*x^2 + 140*a*A*b^2*x^4 - 80*a^2*b*B*x^4 + 105*A*b^3*x^6 - 60*a*b^2*B*x^6)
/(24*a^4*x^4*(a + b*x^2)^(3/2)) + (5*(-7*A*b^2 + 4*a*b*B)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(8*a^(9/2))

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fricas [A]  time = 1.06, size = 407, normalized size = 2.79 \begin {gather*} \left [-\frac {15 \, {\left ({\left (4 \, B a b^{3} - 7 \, A b^{4}\right )} x^{8} + 2 \, {\left (4 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{6} + {\left (4 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} x^{4}\right )} \sqrt {a} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (15 \, {\left (4 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{6} + 6 \, A a^{4} + 20 \, {\left (4 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} x^{4} + 3 \, {\left (4 \, B a^{4} - 7 \, A a^{3} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{48 \, {\left (a^{5} b^{2} x^{8} + 2 \, a^{6} b x^{6} + a^{7} x^{4}\right )}}, -\frac {15 \, {\left ({\left (4 \, B a b^{3} - 7 \, A b^{4}\right )} x^{8} + 2 \, {\left (4 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{6} + {\left (4 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} x^{4}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (15 \, {\left (4 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{6} + 6 \, A a^{4} + 20 \, {\left (4 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} x^{4} + 3 \, {\left (4 \, B a^{4} - 7 \, A a^{3} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{24 \, {\left (a^{5} b^{2} x^{8} + 2 \, a^{6} b x^{6} + a^{7} x^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/48*(15*((4*B*a*b^3 - 7*A*b^4)*x^8 + 2*(4*B*a^2*b^2 - 7*A*a*b^3)*x^6 + (4*B*a^3*b - 7*A*a^2*b^2)*x^4)*sqrt(
a)*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(15*(4*B*a^2*b^2 - 7*A*a*b^3)*x^6 + 6*A*a^4 + 20*(4
*B*a^3*b - 7*A*a^2*b^2)*x^4 + 3*(4*B*a^4 - 7*A*a^3*b)*x^2)*sqrt(b*x^2 + a))/(a^5*b^2*x^8 + 2*a^6*b*x^6 + a^7*x
^4), -1/24*(15*((4*B*a*b^3 - 7*A*b^4)*x^8 + 2*(4*B*a^2*b^2 - 7*A*a*b^3)*x^6 + (4*B*a^3*b - 7*A*a^2*b^2)*x^4)*s
qrt(-a)*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (15*(4*B*a^2*b^2 - 7*A*a*b^3)*x^6 + 6*A*a^4 + 20*(4*B*a^3*b - 7*A*a
^2*b^2)*x^4 + 3*(4*B*a^4 - 7*A*a^3*b)*x^2)*sqrt(b*x^2 + a))/(a^5*b^2*x^8 + 2*a^6*b*x^6 + a^7*x^4)]

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giac [A]  time = 0.41, size = 165, normalized size = 1.13 \begin {gather*} -\frac {5 \, {\left (4 \, B a b - 7 \, A b^{2}\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{8 \, \sqrt {-a} a^{4}} - \frac {6 \, {\left (b x^{2} + a\right )} B a b + B a^{2} b - 9 \, {\left (b x^{2} + a\right )} A b^{2} - A a b^{2}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{4}} - \frac {4 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a b - 4 \, \sqrt {b x^{2} + a} B a^{2} b - 11 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{2} + 13 \, \sqrt {b x^{2} + a} A a b^{2}}{8 \, a^{4} b^{2} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/8*(4*B*a*b - 7*A*b^2)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^4) - 1/3*(6*(b*x^2 + a)*B*a*b + B*a^2*b
- 9*(b*x^2 + a)*A*b^2 - A*a*b^2)/((b*x^2 + a)^(3/2)*a^4) - 1/8*(4*(b*x^2 + a)^(3/2)*B*a*b - 4*sqrt(b*x^2 + a)*
B*a^2*b - 11*(b*x^2 + a)^(3/2)*A*b^2 + 13*sqrt(b*x^2 + a)*A*a*b^2)/(a^4*b^2*x^4)

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maple [A]  time = 0.01, size = 187, normalized size = 1.28 \begin {gather*} \frac {35 A \,b^{2}}{24 \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{3}}-\frac {5 B b}{6 \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{2}}-\frac {35 A \,b^{2} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{8 a^{\frac {9}{2}}}+\frac {5 B b \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{2 a^{\frac {7}{2}}}+\frac {35 A \,b^{2}}{8 \sqrt {b \,x^{2}+a}\, a^{4}}-\frac {5 B b}{2 \sqrt {b \,x^{2}+a}\, a^{3}}+\frac {7 A b}{8 \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{2} x^{2}}-\frac {B}{2 \left (b \,x^{2}+a \right )^{\frac {3}{2}} a \,x^{2}}-\frac {A}{4 \left (b \,x^{2}+a \right )^{\frac {3}{2}} a \,x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*A/a/x^4/(b*x^2+a)^(3/2)+7/8*A*b/a^2/x^2/(b*x^2+a)^(3/2)+35/24*A*b^2/a^3/(b*x^2+a)^(3/2)+35/8*A*b^2/a^4/(b
*x^2+a)^(1/2)-35/8*A*b^2/a^(9/2)*ln((2*a+2*(b*x^2+a)^(1/2)*a^(1/2))/x)-1/2*B/a/x^2/(b*x^2+a)^(3/2)-5/6*B*b/a^2
/(b*x^2+a)^(3/2)-5/2*B*b/a^3/(b*x^2+a)^(1/2)+5/2*B*b/a^(7/2)*ln((2*a+2*(b*x^2+a)^(1/2)*a^(1/2))/x)

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maxima [A]  time = 1.10, size = 164, normalized size = 1.12 \begin {gather*} \frac {5 \, B b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{2 \, a^{\frac {7}{2}}} - \frac {35 \, A b^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{8 \, a^{\frac {9}{2}}} - \frac {5 \, B b}{2 \, \sqrt {b x^{2} + a} a^{3}} - \frac {5 \, B b}{6 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2}} + \frac {35 \, A b^{2}}{8 \, \sqrt {b x^{2} + a} a^{4}} + \frac {35 \, A b^{2}}{24 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{3}} - \frac {B}{2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a x^{2}} + \frac {7 \, A b}{8 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} x^{2}} - \frac {A}{4 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/2*B*b*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(7/2) - 35/8*A*b^2*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(9/2) - 5/2*B*b/(sq
rt(b*x^2 + a)*a^3) - 5/6*B*b/((b*x^2 + a)^(3/2)*a^2) + 35/8*A*b^2/(sqrt(b*x^2 + a)*a^4) + 35/24*A*b^2/((b*x^2
+ a)^(3/2)*a^3) - 1/2*B/((b*x^2 + a)^(3/2)*a*x^2) + 7/8*A*b/((b*x^2 + a)^(3/2)*a^2*x^2) - 1/4*A/((b*x^2 + a)^(
3/2)*a*x^4)

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mupad [B]  time = 2.02, size = 176, normalized size = 1.21 \begin {gather*} \frac {35\,A\,b^2}{6\,a^3\,{\left (b\,x^2+a\right )}^{3/2}}-\frac {10\,B\,b}{3\,a^2\,{\left (b\,x^2+a\right )}^{3/2}}-\frac {35\,A\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{8\,a^{9/2}}-\frac {A}{4\,a\,x^4\,{\left (b\,x^2+a\right )}^{3/2}}-\frac {B}{2\,a\,x^2\,{\left (b\,x^2+a\right )}^{3/2}}+\frac {5\,B\,b\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{2\,a^{7/2}}+\frac {7\,A\,b}{8\,a^2\,x^2\,{\left (b\,x^2+a\right )}^{3/2}}+\frac {35\,A\,b^3\,x^2}{8\,a^4\,{\left (b\,x^2+a\right )}^{3/2}}-\frac {5\,B\,b^2\,x^2}{2\,a^3\,{\left (b\,x^2+a\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(35*A*b^2)/(6*a^3*(a + b*x^2)^(3/2)) - (10*B*b)/(3*a^2*(a + b*x^2)^(3/2)) - (35*A*b^2*atanh((a + b*x^2)^(1/2)/
a^(1/2)))/(8*a^(9/2)) - A/(4*a*x^4*(a + b*x^2)^(3/2)) - B/(2*a*x^2*(a + b*x^2)^(3/2)) + (5*B*b*atanh((a + b*x^
2)^(1/2)/a^(1/2)))/(2*a^(7/2)) + (7*A*b)/(8*a^2*x^2*(a + b*x^2)^(3/2)) + (35*A*b^3*x^2)/(8*a^4*(a + b*x^2)^(3/
2)) - (5*B*b^2*x^2)/(2*a^3*(a + b*x^2)^(3/2))

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sympy [B]  time = 139.24, size = 1323, normalized size = 9.06

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*(-6*a**(89/2)*b**75/(24*a**(93/2)*b**(151/2)*x**5*sqrt(a/(b*x**2) + 1) + 24*a**(91/2)*b**(153/2)*x**7*sqrt(a
/(b*x**2) + 1)) + 21*a**(87/2)*b**76*x**2/(24*a**(93/2)*b**(151/2)*x**5*sqrt(a/(b*x**2) + 1) + 24*a**(91/2)*b*
*(153/2)*x**7*sqrt(a/(b*x**2) + 1)) + 140*a**(85/2)*b**77*x**4/(24*a**(93/2)*b**(151/2)*x**5*sqrt(a/(b*x**2) +
1) + 24*a**(91/2)*b**(153/2)*x**7*sqrt(a/(b*x**2) + 1)) + 105*a**(83/2)*b**78*x**6/(24*a**(93/2)*b**(151/2)*x
**5*sqrt(a/(b*x**2) + 1) + 24*a**(91/2)*b**(153/2)*x**7*sqrt(a/(b*x**2) + 1)) - 105*a**42*b**(155/2)*x**5*sqrt
(a/(b*x**2) + 1)*asinh(sqrt(a)/(sqrt(b)*x))/(24*a**(93/2)*b**(151/2)*x**5*sqrt(a/(b*x**2) + 1) + 24*a**(91/2)*
b**(153/2)*x**7*sqrt(a/(b*x**2) + 1)) - 105*a**41*b**(157/2)*x**7*sqrt(a/(b*x**2) + 1)*asinh(sqrt(a)/(sqrt(b)*
x))/(24*a**(93/2)*b**(151/2)*x**5*sqrt(a/(b*x**2) + 1) + 24*a**(91/2)*b**(153/2)*x**7*sqrt(a/(b*x**2) + 1))) +
B*(-6*a**17*sqrt(1 + b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/
2)*b**3*x**8) - 46*a**16*b*x**2*sqrt(1 + b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**
2*x**6 + 12*a**(33/2)*b**3*x**8) - 15*a**16*b*x**2*log(b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36
*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) + 30*a**16*b*x**2*log(sqrt(1 + b*x**2/a) + 1)/(12*a**(39/2)*x**
2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) - 70*a**15*b**2*x**4*sqrt(1 + b*x**
2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) - 45*a**15*b*
*2*x**4*log(b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x*
*8) + 90*a**15*b**2*x**4*log(sqrt(1 + b*x**2/a) + 1)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b
**2*x**6 + 12*a**(33/2)*b**3*x**8) - 30*a**14*b**3*x**6*sqrt(1 + b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b
*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) - 45*a**14*b**3*x**6*log(b*x**2/a)/(12*a**(39/2)*x**2
+ 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) + 90*a**14*b**3*x**6*log(sqrt(1 + b*
x**2/a) + 1)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) - 15*
a**13*b**4*x**8*log(b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)
*b**3*x**8) + 30*a**13*b**4*x**8*log(sqrt(1 + b*x**2/a) + 1)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**
(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8))

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