\(\int \frac {(a+b x^2)^{5/2} (A+B x^2)}{x^3} \, dx\) [546]

Optimal result

Integrand size = 22, antiderivative size = 135 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^3} \, dx=\frac {1}{2} a (5 A b+2 a B) \sqrt {a+b x^2}+\frac {1}{6} (5 A b+2 a B) \left (a+b x^2\right )^{3/2}+\frac {(5 A b+2 a B) \left (a+b x^2\right )^{5/2}}{10 a}-\frac {A \left (a+b x^2\right )^{7/2}}{2 a x^2}-\frac {1}{2} a^{3/2} (5 A b+2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]

[Out]

1/6*(5*A*b+2*B*a)*(b*x^2+a)^(3/2)+1/10*(5*A*b+2*B*a)*(b*x^2+a)^(5/2)/a-1/2*A*(b*x^2+a)^(7/2)/a/x^2-1/2*a^(3/2)
*(5*A*b+2*B*a)*arctanh((b*x^2+a)^(1/2)/a^(1/2))+1/2*a*(5*A*b+2*B*a)*(b*x^2+a)^(1/2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {457, 79, 52, 65, 214} \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^3} \, dx=-\frac {1}{2} a^{3/2} (2 a B+5 A b) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )+\frac {\left (a+b x^2\right )^{5/2} (2 a B+5 A b)}{10 a}+\frac {1}{6} \left (a+b x^2\right )^{3/2} (2 a B+5 A b)+\frac {1}{2} a \sqrt {a+b x^2} (2 a B+5 A b)-\frac {A \left (a+b x^2\right )^{7/2}}{2 a x^2} \]

[In]

Int[((a + b*x^2)^(5/2)*(A + B*x^2))/x^3,x]

[Out]

(a*(5*A*b + 2*a*B)*Sqrt[a + b*x^2])/2 + ((5*A*b + 2*a*B)*(a + b*x^2)^(3/2))/6 + ((5*A*b + 2*a*B)*(a + b*x^2)^(
5/2))/(10*a) - (A*(a + b*x^2)^(7/2))/(2*a*x^2) - (a^(3/2)*(5*A*b + 2*a*B)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/2

Rule 52

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 65

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 79

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] || L
tQ[p, n]))))

Rule 214

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

Rule 457

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*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {(a+b x)^{5/2} (A+B x)}{x^2} \, dx,x,x^2\right ) \\ & = -\frac {A \left (a+b x^2\right )^{7/2}}{2 a x^2}+\frac {\left (\frac {5 A b}{2}+a B\right ) \text {Subst}\left (\int \frac {(a+b x)^{5/2}}{x} \, dx,x,x^2\right )}{2 a} \\ & = \frac {(5 A b+2 a B) \left (a+b x^2\right )^{5/2}}{10 a}-\frac {A \left (a+b x^2\right )^{7/2}}{2 a x^2}+\frac {1}{4} (5 A b+2 a B) \text {Subst}\left (\int \frac {(a+b x)^{3/2}}{x} \, dx,x,x^2\right ) \\ & = \frac {1}{6} (5 A b+2 a B) \left (a+b x^2\right )^{3/2}+\frac {(5 A b+2 a B) \left (a+b x^2\right )^{5/2}}{10 a}-\frac {A \left (a+b x^2\right )^{7/2}}{2 a x^2}+\frac {1}{4} (a (5 A b+2 a B)) \text {Subst}\left (\int \frac {\sqrt {a+b x}}{x} \, dx,x,x^2\right ) \\ & = \frac {1}{2} a (5 A b+2 a B) \sqrt {a+b x^2}+\frac {1}{6} (5 A b+2 a B) \left (a+b x^2\right )^{3/2}+\frac {(5 A b+2 a B) \left (a+b x^2\right )^{5/2}}{10 a}-\frac {A \left (a+b x^2\right )^{7/2}}{2 a x^2}+\frac {1}{4} \left (a^2 (5 A b+2 a B)\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right ) \\ & = \frac {1}{2} a (5 A b+2 a B) \sqrt {a+b x^2}+\frac {1}{6} (5 A b+2 a B) \left (a+b x^2\right )^{3/2}+\frac {(5 A b+2 a B) \left (a+b x^2\right )^{5/2}}{10 a}-\frac {A \left (a+b x^2\right )^{7/2}}{2 a x^2}+\frac {\left (a^2 (5 A b+2 a B)\right ) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^2}\right )}{2 b} \\ & = \frac {1}{2} a (5 A b+2 a B) \sqrt {a+b x^2}+\frac {1}{6} (5 A b+2 a B) \left (a+b x^2\right )^{3/2}+\frac {(5 A b+2 a B) \left (a+b x^2\right )^{5/2}}{10 a}-\frac {A \left (a+b x^2\right )^{7/2}}{2 a x^2}-\frac {1}{2} a^{3/2} (5 A b+2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.78 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^3} \, dx=\frac {\sqrt {a+b x^2} \left (-15 a^2 A+70 a A b x^2+46 a^2 B x^2+10 A b^2 x^4+22 a b B x^4+6 b^2 B x^6\right )}{30 x^2}-\frac {1}{2} a^{3/2} (5 A b+2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]

[In]

Integrate[((a + b*x^2)^(5/2)*(A + B*x^2))/x^3,x]

[Out]

(Sqrt[a + b*x^2]*(-15*a^2*A + 70*a*A*b*x^2 + 46*a^2*B*x^2 + 10*A*b^2*x^4 + 22*a*b*B*x^4 + 6*b^2*B*x^6))/(30*x^
2) - (a^(3/2)*(5*A*b + 2*a*B)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/2

Maple [A] (verified)

Time = 2.86 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.76

method	result	size
pseudoelliptic	\(\frac {\left (-\frac {15}{2} a^{2} b A -3 a^{3} B \right ) x^{2} \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right )+\left (7 x^{2} \left (\frac {11 x^{2} B}{35}+A \right ) b \,a^{\frac {3}{2}}+\left (\frac {23 x^{2} B}{5}-\frac {3 A}{2}\right ) a^{\frac {5}{2}}+b^{2} x^{4} \sqrt {a}\, \left (\frac {3 x^{2} B}{5}+A \right )\right ) \sqrt {b \,x^{2}+a}}{3 \sqrt {a}\, x^{2}}\)	\(102\)
risch	\(-\frac {a^{2} A \sqrt {b \,x^{2}+a}}{2 x^{2}}+\frac {B \,b^{2} x^{4} \sqrt {b \,x^{2}+a}}{5}+\frac {11 B a b \,x^{2} \sqrt {b \,x^{2}+a}}{15}+\frac {23 B \,a^{2} \sqrt {b \,x^{2}+a}}{15}+\frac {A \,b^{2} x^{2} \sqrt {b \,x^{2}+a}}{3}+\frac {7 a b A \sqrt {b \,x^{2}+a}}{3}-\frac {5 A \,a^{\frac {3}{2}} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right ) b}{2}-B \,a^{\frac {5}{2}} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\)	\(161\)
default	\(B \left (\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5}+a \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )\right )+A \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{2 a \,x^{2}}+\frac {5 b \left (\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5}+a \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )\right )}{2 a}\right )\)	\(162\)

[In]

int((b*x^2+a)^(5/2)*(B*x^2+A)/x^3,x,method=_RETURNVERBOSE)

[Out]

1/3*((-15/2*a^2*b*A-3*a^3*B)*x^2*arctanh((b*x^2+a)^(1/2)/a^(1/2))+(7*x^2*(11/35*x^2*B+A)*b*a^(3/2)+(23/5*x^2*B
-3/2*A)*a^(5/2)+b^2*x^4*a^(1/2)*(3/5*x^2*B+A))*(b*x^2+a)^(1/2))/a^(1/2)/x^2

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.64 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^3} \, dx=\left [\frac {15 \, {\left (2 \, B a^{2} + 5 \, A a b\right )} \sqrt {a} x^{2} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (6 \, B b^{2} x^{6} + 2 \, {\left (11 \, B a b + 5 \, A b^{2}\right )} x^{4} - 15 \, A a^{2} + 2 \, {\left (23 \, B a^{2} + 35 \, A a b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{60 \, x^{2}}, \frac {15 \, {\left (2 \, B a^{2} + 5 \, A a b\right )} \sqrt {-a} x^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (6 \, B b^{2} x^{6} + 2 \, {\left (11 \, B a b + 5 \, A b^{2}\right )} x^{4} - 15 \, A a^{2} + 2 \, {\left (23 \, B a^{2} + 35 \, A a b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{30 \, x^{2}}\right ] \]

[In]

integrate((b*x^2+a)^(5/2)*(B*x^2+A)/x^3,x, algorithm="fricas")

[Out]

[1/60*(15*(2*B*a^2 + 5*A*a*b)*sqrt(a)*x^2*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(6*B*b^2*x^6
+ 2*(11*B*a*b + 5*A*b^2)*x^4 - 15*A*a^2 + 2*(23*B*a^2 + 35*A*a*b)*x^2)*sqrt(b*x^2 + a))/x^2, 1/30*(15*(2*B*a^
2 + 5*A*a*b)*sqrt(-a)*x^2*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (6*B*b^2*x^6 + 2*(11*B*a*b + 5*A*b^2)*x^4 - 15*A*
a^2 + 2*(23*B*a^2 + 35*A*a*b)*x^2)*sqrt(b*x^2 + a))/x^2]

Sympy [A] (verification not implemented)

Time = 15.86 (sec) , antiderivative size = 330, normalized size of antiderivative = 2.44 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^3} \, dx=- \frac {5 A a^{\frac {3}{2}} b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{2} - \frac {A a^{2} \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{2 x} + \frac {2 A a^{2} \sqrt {b}}{x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {2 A a b^{\frac {3}{2}} x}{\sqrt {\frac {a}{b x^{2}} + 1}} + A b^{2} \left (\begin {cases} \frac {a \sqrt {a + b x^{2}}}{3 b} + \frac {x^{2} \sqrt {a + b x^{2}}}{3} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{2}}{2} & \text {otherwise} \end {cases}\right ) - B a^{\frac {5}{2}} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )} + \frac {B a^{3}}{\sqrt {b} x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {B a^{2} \sqrt {b} x}{\sqrt {\frac {a}{b x^{2}} + 1}} + 2 B a b \left (\begin {cases} \frac {a \sqrt {a + b x^{2}}}{3 b} + \frac {x^{2} \sqrt {a + b x^{2}}}{3} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{2}}{2} & \text {otherwise} \end {cases}\right ) + B b^{2} \left (\begin {cases} - \frac {2 a^{2} \sqrt {a + b x^{2}}}{15 b^{2}} + \frac {a x^{2} \sqrt {a + b x^{2}}}{15 b} + \frac {x^{4} \sqrt {a + b x^{2}}}{5} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{4}}{4} & \text {otherwise} \end {cases}\right ) \]

[In]

integrate((b*x**2+a)**(5/2)*(B*x**2+A)/x**3,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^3} \, dx=-B a^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) - \frac {5}{2} \, A a^{\frac {3}{2}} b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \frac {1}{5} \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B + \frac {1}{3} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a + \sqrt {b x^{2} + a} B a^{2} + \frac {5}{6} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A b}{2 \, a} + \frac {5}{2} \, \sqrt {b x^{2} + a} A a b - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} A}{2 \, a x^{2}} \]

[In]

integrate((b*x^2+a)^(5/2)*(B*x^2+A)/x^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.03 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^3} \, dx=\frac {6 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B b + 10 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a b + 30 \, \sqrt {b x^{2} + a} B a^{2} b + 10 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{2} + 60 \, \sqrt {b x^{2} + a} A a b^{2} - \frac {15 \, \sqrt {b x^{2} + a} A a^{2} b}{x^{2}} + \frac {15 \, {\left (2 \, B a^{3} b + 5 \, A a^{2} b^{2}\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}}}{30 \, b} \]

[In]

integrate((b*x^2+a)^(5/2)*(B*x^2+A)/x^3,x, algorithm="giac")

[Out]

1/30*(6*(b*x^2 + a)^(5/2)*B*b + 10*(b*x^2 + a)^(3/2)*B*a*b + 30*sqrt(b*x^2 + a)*B*a^2*b + 10*(b*x^2 + a)^(3/2)
*A*b^2 + 60*sqrt(b*x^2 + a)*A*a*b^2 - 15*sqrt(b*x^2 + a)*A*a^2*b/x^2 + 15*(2*B*a^3*b + 5*A*a^2*b^2)*arctan(sqr
t(b*x^2 + a)/sqrt(-a))/sqrt(-a))/b

Mupad [B] (verification not implemented)

Time = 6.14 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^3} \, dx=\frac {B\,{\left (b\,x^2+a\right )}^{5/2}}{5}+B\,a^2\,\sqrt {b\,x^2+a}+B\,a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,1{}\mathrm {i}+\frac {A\,b\,{\left (b\,x^2+a\right )}^{3/2}}{3}+\frac {B\,a\,{\left (b\,x^2+a\right )}^{3/2}}{3}+2\,A\,a\,b\,\sqrt {b\,x^2+a}-\frac {A\,a^2\,\sqrt {b\,x^2+a}}{2\,x^2}+\frac {A\,a^{3/2}\,b\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{2} \]

[In]

int(((A + B*x^2)*(a + b*x^2)^(5/2))/x^3,x)

[Out]

(B*(a + b*x^2)^(5/2))/5 + B*a^2*(a + b*x^2)^(1/2) + B*a^(5/2)*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*1i + (A*b*(
a + b*x^2)^(3/2))/3 + (B*a*(a + b*x^2)^(3/2))/3 + 2*A*a*b*(a + b*x^2)^(1/2) - (A*a^2*(a + b*x^2)^(1/2))/(2*x^2
) + (A*a^(3/2)*b*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*5i)/2