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

Optimal. Leaf size=143 \[ -\frac {\left (a+b x^2\right )^{5/2} (4 a B+3 A b)}{8 a x^2}+\frac {5 b \left (a+b x^2\right )^{3/2} (4 a B+3 A b)}{24 a}+\frac {5}{8} b \sqrt {a+b x^2} (4 a B+3 A b)-\frac {5}{8} \sqrt {a} b (4 a B+3 A b) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )-\frac {A \left (a+b x^2\right )^{7/2}}{4 a x^4} \]

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Rubi [A]  time = 0.10, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {446, 78, 47, 50, 63, 208} \begin {gather*} -\frac {\left (a+b x^2\right )^{5/2} (4 a B+3 A b)}{8 a x^2}+\frac {5 b \left (a+b x^2\right )^{3/2} (4 a B+3 A b)}{24 a}+\frac {5}{8} b \sqrt {a+b x^2} (4 a B+3 A b)-\frac {5}{8} \sqrt {a} b (4 a B+3 A b) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )-\frac {A \left (a+b x^2\right )^{7/2}}{4 a x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*b*(3*A*b + 4*a*B)*Sqrt[a + b*x^2])/8 + (5*b*(3*A*b + 4*a*B)*(a + b*x^2)^(3/2))/(24*a) - ((3*A*b + 4*a*B)*(a
+ b*x^2)^(5/2))/(8*a*x^2) - (A*(a + b*x^2)^(7/2))/(4*a*x^4) - (5*Sqrt[a]*b*(3*A*b + 4*a*B)*ArcTanh[Sqrt[a + b
*x^2]/Sqrt[a]])/8

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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^5} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(a+b x)^{5/2} (A+B x)}{x^3} \, dx,x,x^2\right )\\ &=-\frac {A \left (a+b x^2\right )^{7/2}}{4 a x^4}+\frac {\left (\frac {3 A b}{2}+2 a B\right ) \operatorname {Subst}\left (\int \frac {(a+b x)^{5/2}}{x^2} \, dx,x,x^2\right )}{4 a}\\ &=-\frac {(3 A b+4 a B) \left (a+b x^2\right )^{5/2}}{8 a x^2}-\frac {A \left (a+b x^2\right )^{7/2}}{4 a x^4}+\frac {(5 b (3 A b+4 a B)) \operatorname {Subst}\left (\int \frac {(a+b x)^{3/2}}{x} \, dx,x,x^2\right )}{16 a}\\ &=\frac {5 b (3 A b+4 a B) \left (a+b x^2\right )^{3/2}}{24 a}-\frac {(3 A b+4 a B) \left (a+b x^2\right )^{5/2}}{8 a x^2}-\frac {A \left (a+b x^2\right )^{7/2}}{4 a x^4}+\frac {1}{16} (5 b (3 A b+4 a B)) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x} \, dx,x,x^2\right )\\ &=\frac {5}{8} b (3 A b+4 a B) \sqrt {a+b x^2}+\frac {5 b (3 A b+4 a B) \left (a+b x^2\right )^{3/2}}{24 a}-\frac {(3 A b+4 a B) \left (a+b x^2\right )^{5/2}}{8 a x^2}-\frac {A \left (a+b x^2\right )^{7/2}}{4 a x^4}+\frac {1}{16} (5 a b (3 A b+4 a B)) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right )\\ &=\frac {5}{8} b (3 A b+4 a B) \sqrt {a+b x^2}+\frac {5 b (3 A b+4 a B) \left (a+b x^2\right )^{3/2}}{24 a}-\frac {(3 A b+4 a B) \left (a+b x^2\right )^{5/2}}{8 a x^2}-\frac {A \left (a+b x^2\right )^{7/2}}{4 a x^4}+\frac {1}{8} (5 a (3 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 )\\ &=\frac {5}{8} b (3 A b+4 a B) \sqrt {a+b x^2}+\frac {5 b (3 A b+4 a B) \left (a+b x^2\right )^{3/2}}{24 a}-\frac {(3 A b+4 a B) \left (a+b x^2\right )^{5/2}}{8 a x^2}-\frac {A \left (a+b x^2\right )^{7/2}}{4 a x^4}-\frac {5}{8} \sqrt {a} b (3 A b+4 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 60, normalized size = 0.42 \begin {gather*} \frac {\left (a+b x^2\right )^{7/2} \left (b x^4 (4 a B+3 A b) \, _2F_1\left (2,\frac {7}{2};\frac {9}{2};\frac {b x^2}{a}+1\right )-7 a^2 A\right )}{28 a^3 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^2)^(7/2)*(-7*a^2*A + b*(3*A*b + 4*a*B)*x^4*Hypergeometric2F1[2, 7/2, 9/2, 1 + (b*x^2)/a]))/(28*a^3*x
^4)

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IntegrateAlgebraic [A]  time = 0.17, size = 112, normalized size = 0.78 \begin {gather*} \frac {\sqrt {a+b x^2} \left (-6 a^2 A-12 a^2 B x^2-27 a A b x^2+56 a b B x^4+24 A b^2 x^4+8 b^2 B x^6\right )}{24 x^4}-\frac {5}{8} \left (4 a^{3/2} b B+3 \sqrt {a} A b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*(-6*a^2*A - 27*a*A*b*x^2 - 12*a^2*B*x^2 + 24*A*b^2*x^4 + 56*a*b*B*x^4 + 8*b^2*B*x^6))/(24*x^4
) - (5*(3*Sqrt[a]*A*b^2 + 4*a^(3/2)*b*B)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/8

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fricas [A]  time = 0.94, size = 221, normalized size = 1.55 \begin {gather*} \left [\frac {15 \, {\left (4 \, B a b + 3 \, A b^{2}\right )} \sqrt {a} x^{4} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (8 \, B b^{2} x^{6} + 8 \, {\left (7 \, B a b + 3 \, A b^{2}\right )} x^{4} - 6 \, A a^{2} - 3 \, {\left (4 \, B a^{2} + 9 \, A a b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{48 \, x^{4}}, \frac {15 \, {\left (4 \, B a b + 3 \, A b^{2}\right )} \sqrt {-a} x^{4} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (8 \, B b^{2} x^{6} + 8 \, {\left (7 \, B a b + 3 \, A b^{2}\right )} x^{4} - 6 \, A a^{2} - 3 \, {\left (4 \, B a^{2} + 9 \, A a b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{24 \, x^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(15*(4*B*a*b + 3*A*b^2)*sqrt(a)*x^4*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(8*B*b^2*x^6
+ 8*(7*B*a*b + 3*A*b^2)*x^4 - 6*A*a^2 - 3*(4*B*a^2 + 9*A*a*b)*x^2)*sqrt(b*x^2 + a))/x^4, 1/24*(15*(4*B*a*b +
3*A*b^2)*sqrt(-a)*x^4*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (8*B*b^2*x^6 + 8*(7*B*a*b + 3*A*b^2)*x^4 - 6*A*a^2 -
3*(4*B*a^2 + 9*A*a*b)*x^2)*sqrt(b*x^2 + a))/x^4]

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giac [A]  time = 0.35, size = 171, normalized size = 1.20 \begin {gather*} \frac {8 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B b^{2} + 48 \, \sqrt {b x^{2} + a} B a b^{2} + 24 \, \sqrt {b x^{2} + a} A b^{3} + \frac {15 \, {\left (4 \, B a^{2} b^{2} + 3 \, A a b^{3}\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {3 \, {\left (4 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{2} b^{2} - 4 \, \sqrt {b x^{2} + a} B a^{3} b^{2} + 9 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a b^{3} - 7 \, \sqrt {b x^{2} + a} A a^{2} b^{3}\right )}}{b^{2} x^{4}}}{24 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(8*(b*x^2 + a)^(3/2)*B*b^2 + 48*sqrt(b*x^2 + a)*B*a*b^2 + 24*sqrt(b*x^2 + a)*A*b^3 + 15*(4*B*a^2*b^2 + 3*
A*a*b^3)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/sqrt(-a) - 3*(4*(b*x^2 + a)^(3/2)*B*a^2*b^2 - 4*sqrt(b*x^2 + a)*B*a^
3*b^2 + 9*(b*x^2 + a)^(3/2)*A*a*b^3 - 7*sqrt(b*x^2 + a)*A*a^2*b^3)/(b^2*x^4))/b

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maple [A]  time = 0.01, size = 213, normalized size = 1.49 \begin {gather*} -\frac {15 A \sqrt {a}\, b^{2} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{8}-\frac {5 B \,a^{\frac {3}{2}} b \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{2}+\frac {15 \sqrt {b \,x^{2}+a}\, A \,b^{2}}{8}+\frac {5 \sqrt {b \,x^{2}+a}\, B a b}{2}+\frac {5 \left (b \,x^{2}+a \right )^{\frac {3}{2}} A \,b^{2}}{8 a}+\frac {5 \left (b \,x^{2}+a \right )^{\frac {3}{2}} B b}{6}+\frac {3 \left (b \,x^{2}+a \right )^{\frac {5}{2}} A \,b^{2}}{8 a^{2}}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} B b}{2 a}-\frac {3 \left (b \,x^{2}+a \right )^{\frac {7}{2}} A b}{8 a^{2} x^{2}}-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} B}{2 a \,x^{2}}-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} A}{4 a \,x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*B/a/x^2*(b*x^2+a)^(7/2)+1/2*B*b/a*(b*x^2+a)^(5/2)+5/6*B*b*(b*x^2+a)^(3/2)-5/2*B*b*a^(3/2)*ln((2*a+2*(b*x^
2+a)^(1/2)*a^(1/2))/x)+5/2*B*b*a*(b*x^2+a)^(1/2)-1/4*A*(b*x^2+a)^(7/2)/a/x^4-3/8*A*b/a^2/x^2*(b*x^2+a)^(7/2)+3
/8*A*b^2/a^2*(b*x^2+a)^(5/2)+5/8*A*b^2/a*(b*x^2+a)^(3/2)-15/8*A*b^2*a^(1/2)*ln((2*a+2*(b*x^2+a)^(1/2)*a^(1/2))
/x)+15/8*A*b^2*(b*x^2+a)^(1/2)

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maxima [A]  time = 1.06, size = 190, normalized size = 1.33 \begin {gather*} -\frac {5}{2} \, B a^{\frac {3}{2}} b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) - \frac {15}{8} \, A \sqrt {a} b^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \frac {5}{6} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B b + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B b}{2 \, a} + \frac {5}{2} \, \sqrt {b x^{2} + a} B a b + \frac {15}{8} \, \sqrt {b x^{2} + a} A b^{2} + \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A b^{2}}{8 \, a^{2}} + \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{2}}{8 \, a} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} B}{2 \, a x^{2}} - \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} A b}{8 \, a^{2} x^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} A}{4 \, a x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/2*B*a^(3/2)*b*arcsinh(a/(sqrt(a*b)*abs(x))) - 15/8*A*sqrt(a)*b^2*arcsinh(a/(sqrt(a*b)*abs(x))) + 5/6*(b*x^2
+ a)^(3/2)*B*b + 1/2*(b*x^2 + a)^(5/2)*B*b/a + 5/2*sqrt(b*x^2 + a)*B*a*b + 15/8*sqrt(b*x^2 + a)*A*b^2 + 3/8*(
b*x^2 + a)^(5/2)*A*b^2/a^2 + 5/8*(b*x^2 + a)^(3/2)*A*b^2/a - 1/2*(b*x^2 + a)^(7/2)*B/(a*x^2) - 3/8*(b*x^2 + a)
^(7/2)*A*b/(a^2*x^2) - 1/4*(b*x^2 + a)^(7/2)*A/(a*x^4)

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mupad [B]  time = 2.59, size = 144, normalized size = 1.01 \begin {gather*} A\,b^2\,\sqrt {b\,x^2+a}+\frac {B\,b\,{\left (b\,x^2+a\right )}^{3/2}}{3}+2\,B\,a\,b\,\sqrt {b\,x^2+a}+\frac {A\,\sqrt {a}\,b^2\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,15{}\mathrm {i}}{8}-\frac {9\,A\,a\,{\left (b\,x^2+a\right )}^{3/2}}{8\,x^4}+\frac {7\,A\,a^2\,\sqrt {b\,x^2+a}}{8\,x^4}-\frac {B\,a^2\,\sqrt {b\,x^2+a}}{2\,x^2}+\frac {B\,a^{3/2}\,b\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*b^2*(a + b*x^2)^(1/2) + (B*b*(a + b*x^2)^(3/2))/3 + 2*B*a*b*(a + b*x^2)^(1/2) + (A*a^(1/2)*b^2*atan(((a + b*
x^2)^(1/2)*1i)/a^(1/2))*15i)/8 - (9*A*a*(a + b*x^2)^(3/2))/(8*x^4) + (7*A*a^2*(a + b*x^2)^(1/2))/(8*x^4) - (B*
a^2*(a + b*x^2)^(1/2))/(2*x^2) + (B*a^(3/2)*b*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*5i)/2

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sympy [A]  time = 165.19, size = 279, normalized size = 1.95 \begin {gather*} - \frac {15 A \sqrt {a} b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{8} - \frac {A a^{3}}{4 \sqrt {b} x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {3 A a^{2} \sqrt {b}}{8 x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {A a b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{x} + \frac {7 A a b^{\frac {3}{2}}}{8 x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {A b^{\frac {5}{2}} x}{\sqrt {\frac {a}{b x^{2}} + 1}} - \frac {5 B a^{\frac {3}{2}} b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{2} - \frac {B a^{2} \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{2 x} + \frac {2 B a^{2} \sqrt {b}}{x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {2 B a b^{\frac {3}{2}} x}{\sqrt {\frac {a}{b x^{2}} + 1}} + B b^{2} \left (\begin {cases} \frac {\sqrt {a} x^{2}}{2} & \text {for}\: b = 0 \\\frac {\left (a + b x^{2}\right )^{\frac {3}{2}}}{3 b} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15*A*sqrt(a)*b**2*asinh(sqrt(a)/(sqrt(b)*x))/8 - A*a**3/(4*sqrt(b)*x**5*sqrt(a/(b*x**2) + 1)) - 3*A*a**2*sqrt
(b)/(8*x**3*sqrt(a/(b*x**2) + 1)) - A*a*b**(3/2)*sqrt(a/(b*x**2) + 1)/x + 7*A*a*b**(3/2)/(8*x*sqrt(a/(b*x**2)
+ 1)) + A*b**(5/2)*x/sqrt(a/(b*x**2) + 1) - 5*B*a**(3/2)*b*asinh(sqrt(a)/(sqrt(b)*x))/2 - B*a**2*sqrt(b)*sqrt(
a/(b*x**2) + 1)/(2*x) + 2*B*a**2*sqrt(b)/(x*sqrt(a/(b*x**2) + 1)) + 2*B*a*b**(3/2)*x/sqrt(a/(b*x**2) + 1) + B*
b**2*Piecewise((sqrt(a)*x**2/2, Eq(b, 0)), ((a + b*x**2)**(3/2)/(3*b), True))

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