\(\int \frac {(A+B x) (a+b x^2)^{3/2}}{x^2} \, dx\) [13]

Optimal result

Integrand size = 20, antiderivative size = 108 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x^2} \, dx=\frac {1}{2} (2 a B+3 A b x) \sqrt {a+b x^2}-\frac {(3 A-B x) \left (a+b x^2\right )^{3/2}}{3 x}+\frac {3}{2} a A \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-a^{3/2} B \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {827, 829, 858, 223, 212, 272, 65, 214} \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x^2} \, dx=a^{3/2} (-B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )+\frac {3}{2} a A \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {\left (a+b x^2\right )^{3/2} (3 A-B x)}{3 x}+\frac {1}{2} \sqrt {a+b x^2} (2 a B+3 A b x) \]

[In]

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

[Out]

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

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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 223

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 272

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

Rule 827

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m
+ 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*((a + c*x^2)^p/(e^2*(m + 1)*(m + 2*p + 2))), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 829

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m
+ 1)*(c*e*f*(m + 2*p + 2) - g*c*d*(2*p + 1) + g*c*e*(m + 2*p + 1)*x)*((a + c*x^2)^p/(c*e^2*(m + 2*p + 1)*(m +
2*p + 2))), x] + Dist[2*(p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2))), Int[(d + e*x)^m*(a + c*x^2)^(p - 1)*Simp[f*a
*c*e^2*(m + 2*p + 2) + a*c*d*e*g*m - (c^2*f*d*e*(m + 2*p + 2) - g*(c^2*d^2*(2*p + 1) + a*c*e^2*(m + 2*p + 1)))
*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !R
ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*
m, 2*p])

Rule 858

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

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

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.03 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x^2} \, dx=\frac {\sqrt {a+b x^2} \left (b x^2 (3 A+2 B x)+a (-6 A+8 B x)\right )}{6 x}+2 a^{3/2} B \text {arctanh}\left (\frac {\sqrt {b} x-\sqrt {a+b x^2}}{\sqrt {a}}\right )-\frac {3}{2} a A \sqrt {b} \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right ) \]

[In]

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

[Out]

(Sqrt[a + b*x^2]*(b*x^2*(3*A + 2*B*x) + a*(-6*A + 8*B*x)))/(6*x) + 2*a^(3/2)*B*ArcTanh[(Sqrt[b]*x - Sqrt[a + b
*x^2])/Sqrt[a]] - (3*a*A*Sqrt[b]*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/2

Maple [A] (verified)

Time = 3.41 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.05

method	result	size
risch	\(-\frac {a A \sqrt {b \,x^{2}+a}}{x}+\frac {B b \,x^{2} \sqrt {b \,x^{2}+a}}{3}+\frac {4 a B \sqrt {b \,x^{2}+a}}{3}+\frac {b A x \sqrt {b \,x^{2}+a}}{2}+\frac {3 a \sqrt {b}\, A \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2}-a^{\frac {3}{2}} B \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\)	\(113\)
default	\(B \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )+A \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{a x}+\frac {4 b \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{a}\right )\)	\(133\)

[In]

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

[Out]

-a*A*(b*x^2+a)^(1/2)/x+1/3*B*b*x^2*(b*x^2+a)^(1/2)+4/3*a*B*(b*x^2+a)^(1/2)+1/2*b*A*x*(b*x^2+a)^(1/2)+3/2*a*b^(
1/2)*A*ln(x*b^(1/2)+(b*x^2+a)^(1/2))-a^(3/2)*B*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 411, normalized size of antiderivative = 3.81 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x^2} \, dx=\left [\frac {9 \, A a \sqrt {b} x \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 6 \, B a^{\frac {3}{2}} x \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (2 \, B b x^{3} + 3 \, A b x^{2} + 8 \, B a x - 6 \, A a\right )} \sqrt {b x^{2} + a}}{12 \, x}, -\frac {9 \, A a \sqrt {-b} x \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - 3 \, B a^{\frac {3}{2}} x \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - {\left (2 \, B b x^{3} + 3 \, A b x^{2} + 8 \, B a x - 6 \, A a\right )} \sqrt {b x^{2} + a}}{6 \, x}, \frac {12 \, B \sqrt {-a} a x \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + 9 \, A a \sqrt {b} x \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (2 \, B b x^{3} + 3 \, A b x^{2} + 8 \, B a x - 6 \, A a\right )} \sqrt {b x^{2} + a}}{12 \, x}, -\frac {9 \, A a \sqrt {-b} x \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - 6 \, B \sqrt {-a} a x \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) - {\left (2 \, B b x^{3} + 3 \, A b x^{2} + 8 \, B a x - 6 \, A a\right )} \sqrt {b x^{2} + a}}{6 \, x}\right ] \]

[In]

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

[Out]

[1/12*(9*A*a*sqrt(b)*x*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 6*B*a^(3/2)*x*log(-(b*x^2 - 2*sqrt(b*
x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(2*B*b*x^3 + 3*A*b*x^2 + 8*B*a*x - 6*A*a)*sqrt(b*x^2 + a))/x, -1/6*(9*A*a*sqr
t(-b)*x*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - 3*B*a^(3/2)*x*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2)
- (2*B*b*x^3 + 3*A*b*x^2 + 8*B*a*x - 6*A*a)*sqrt(b*x^2 + a))/x, 1/12*(12*B*sqrt(-a)*a*x*arctan(sqrt(-a)/sqrt(
b*x^2 + a)) + 9*A*a*sqrt(b)*x*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(2*B*b*x^3 + 3*A*b*x^2 + 8*B
*a*x - 6*A*a)*sqrt(b*x^2 + a))/x, -1/6*(9*A*a*sqrt(-b)*x*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - 6*B*sqrt(-a)*a*x
*arctan(sqrt(-a)/sqrt(b*x^2 + a)) - (2*B*b*x^3 + 3*A*b*x^2 + 8*B*a*x - 6*A*a)*sqrt(b*x^2 + a))/x]

Sympy [A] (verification not implemented)

Time = 2.28 (sec) , antiderivative size = 243, normalized size of antiderivative = 2.25 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x^2} \, dx=- \frac {A a^{\frac {3}{2}}}{x \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {A \sqrt {a} b x}{\sqrt {1 + \frac {b x^{2}}{a}}} + A a \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )} + A b \left (\begin {cases} \frac {a \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {x \sqrt {a + b x^{2}}}{2} & \text {for}\: b \neq 0 \\\sqrt {a} x & \text {otherwise} \end {cases}\right ) - B a^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )} + \frac {B a^{2}}{\sqrt {b} x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {B a \sqrt {b} x}{\sqrt {\frac {a}{b x^{2}} + 1}} + B b \left (\begin {cases} \frac {a \sqrt {a + b x^{2}}}{3 b} + \frac {x^{2} \sqrt {a + b x^{2}}}{3} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{2}}{2} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.81 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x^2} \, dx=\frac {3}{2} \, \sqrt {b x^{2} + a} A b x + \frac {3}{2} \, A a \sqrt {b} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - B a^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \frac {1}{3} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B + \sqrt {b x^{2} + a} B a - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A}{x} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.15 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x^2} \, dx=\frac {2 \, B a^{2} \arctan \left (-\frac {\sqrt {b} x - \sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {3}{2} \, A a \sqrt {b} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right ) + \frac {2 \, A a^{2} \sqrt {b}}{{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a} + \frac {1}{6} \, \sqrt {b x^{2} + a} {\left (8 \, B a + {\left (2 \, B b x + 3 \, A b\right )} x\right )} \]

[In]

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

[Out]

2*B*a^2*arctan(-(sqrt(b)*x - sqrt(b*x^2 + a))/sqrt(-a))/sqrt(-a) - 3/2*A*a*sqrt(b)*log(abs(-sqrt(b)*x + sqrt(b
*x^2 + a))) + 2*A*a^2*sqrt(b)/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a) + 1/6*sqrt(b*x^2 + a)*(8*B*a + (2*B*b*x +
3*A*b)*x)

Mupad [B] (verification not implemented)

Time = 6.73 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.80 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x^2} \, dx=\frac {B\,{\left (b\,x^2+a\right )}^{3/2}}{3}-B\,a^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )+B\,a\,\sqrt {b\,x^2+a}-\frac {A\,{\left (b\,x^2+a\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},-\frac {1}{2};\ \frac {1}{2};\ -\frac {b\,x^2}{a}\right )}{x\,{\left (\frac {b\,x^2}{a}+1\right )}^{3/2}} \]

[In]

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

[Out]

(B*(a + b*x^2)^(3/2))/3 - B*a^(3/2)*atanh((a + b*x^2)^(1/2)/a^(1/2)) + B*a*(a + b*x^2)^(1/2) - (A*(a + b*x^2)^
(3/2)*hypergeom([-3/2, -1/2], 1/2, -(b*x^2)/a))/(x*((b*x^2)/a + 1)^(3/2))