3.770 \(\int \frac{A+B x}{\sqrt{x} (a^2+2 a b x+b^2 x^2)^2} \, dx\)

Optimal. Leaf size=130 \[ \frac{(a B+5 A b) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{8 a^{7/2} b^{3/2}}+\frac{\sqrt{x} (a B+5 A b)}{8 a^3 b (a+b x)}+\frac{\sqrt{x} (a B+5 A b)}{12 a^2 b (a+b x)^2}+\frac{\sqrt{x} (A b-a B)}{3 a b (a+b x)^3} \]

[Out]

((A*b - a*B)*Sqrt[x])/(3*a*b*(a + b*x)^3) + ((5*A*b + a*B)*Sqrt[x])/(12*a^2*b*(a + b*x)^2) + ((5*A*b + a*B)*Sq
rt[x])/(8*a^3*b*(a + b*x)) + ((5*A*b + a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(8*a^(7/2)*b^(3/2))

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Rubi [A]  time = 0.0568772, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {27, 78, 51, 63, 205} \[ \frac{(a B+5 A b) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{8 a^{7/2} b^{3/2}}+\frac{\sqrt{x} (a B+5 A b)}{8 a^3 b (a+b x)}+\frac{\sqrt{x} (a B+5 A b)}{12 a^2 b (a+b x)^2}+\frac{\sqrt{x} (A b-a B)}{3 a b (a+b x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A*b - a*B)*Sqrt[x])/(3*a*b*(a + b*x)^3) + ((5*A*b + a*B)*Sqrt[x])/(12*a^2*b*(a + b*x)^2) + ((5*A*b + a*B)*Sq
rt[x])/(8*a^3*b*(a + b*x)) + ((5*A*b + a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(8*a^(7/2)*b^(3/2))

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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 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 205

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

Rubi steps

\begin{align*} \int \frac{A+B x}{\sqrt{x} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac{A+B x}{\sqrt{x} (a+b x)^4} \, dx\\ &=\frac{(A b-a B) \sqrt{x}}{3 a b (a+b x)^3}+\frac{(5 A b+a B) \int \frac{1}{\sqrt{x} (a+b x)^3} \, dx}{6 a b}\\ &=\frac{(A b-a B) \sqrt{x}}{3 a b (a+b x)^3}+\frac{(5 A b+a B) \sqrt{x}}{12 a^2 b (a+b x)^2}+\frac{(5 A b+a B) \int \frac{1}{\sqrt{x} (a+b x)^2} \, dx}{8 a^2 b}\\ &=\frac{(A b-a B) \sqrt{x}}{3 a b (a+b x)^3}+\frac{(5 A b+a B) \sqrt{x}}{12 a^2 b (a+b x)^2}+\frac{(5 A b+a B) \sqrt{x}}{8 a^3 b (a+b x)}+\frac{(5 A b+a B) \int \frac{1}{\sqrt{x} (a+b x)} \, dx}{16 a^3 b}\\ &=\frac{(A b-a B) \sqrt{x}}{3 a b (a+b x)^3}+\frac{(5 A b+a B) \sqrt{x}}{12 a^2 b (a+b x)^2}+\frac{(5 A b+a B) \sqrt{x}}{8 a^3 b (a+b x)}+\frac{(5 A b+a B) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\sqrt{x}\right )}{8 a^3 b}\\ &=\frac{(A b-a B) \sqrt{x}}{3 a b (a+b x)^3}+\frac{(5 A b+a B) \sqrt{x}}{12 a^2 b (a+b x)^2}+\frac{(5 A b+a B) \sqrt{x}}{8 a^3 b (a+b x)}+\frac{(5 A b+a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{8 a^{7/2} b^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.0261596, size = 59, normalized size = 0.45 \[ \frac{\sqrt{x} \left (\frac{a^3 (A b-a B)}{(a+b x)^3}+(a B+5 A b) \, _2F_1\left (\frac{1}{2},3;\frac{3}{2};-\frac{b x}{a}\right )\right )}{3 a^4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x]*((a^3*(A*b - a*B))/(a + b*x)^3 + (5*A*b + a*B)*Hypergeometric2F1[1/2, 3, 3/2, -((b*x)/a)]))/(3*a^4*b)

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Maple [A]  time = 0.015, size = 112, normalized size = 0.9 \begin{align*} 2\,{\frac{1}{ \left ( bx+a \right ) ^{3}} \left ( 1/16\,{\frac{ \left ( 5\,Ab+aB \right ) b{x}^{5/2}}{{a}^{3}}}+1/6\,{\frac{ \left ( 5\,Ab+aB \right ){x}^{3/2}}{{a}^{2}}}+1/16\,{\frac{ \left ( 11\,Ab-aB \right ) \sqrt{x}}{ab}} \right ) }+{\frac{5\,A}{8\,{a}^{3}}\arctan \left ({b\sqrt{x}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{B}{8\,{a}^{2}b}\arctan \left ({b\sqrt{x}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2/x^(1/2),x)

[Out]

2*(1/16*(5*A*b+B*a)/a^3*b*x^(5/2)+1/6/a^2*(5*A*b+B*a)*x^(3/2)+1/16*(11*A*b-B*a)/a/b*x^(1/2))/(b*x+a)^3+5/8/a^3
/(a*b)^(1/2)*arctan(x^(1/2)*b/(a*b)^(1/2))*A+1/8/a^2/b/(a*b)^(1/2)*arctan(x^(1/2)*b/(a*b)^(1/2))*B

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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+A)/(b^2*x^2+2*a*b*x+a^2)^2/x^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.94058, size = 882, normalized size = 6.78 \begin{align*} \left [-\frac{3 \,{\left (B a^{4} + 5 \, A a^{3} b +{\left (B a b^{3} + 5 \, A b^{4}\right )} x^{3} + 3 \,{\left (B a^{2} b^{2} + 5 \, A a b^{3}\right )} x^{2} + 3 \,{\left (B a^{3} b + 5 \, A a^{2} b^{2}\right )} x\right )} \sqrt{-a b} \log \left (\frac{b x - a - 2 \, \sqrt{-a b} \sqrt{x}}{b x + a}\right ) + 2 \,{\left (3 \, B a^{4} b - 33 \, A a^{3} b^{2} - 3 \,{\left (B a^{2} b^{3} + 5 \, A a b^{4}\right )} x^{2} - 8 \,{\left (B a^{3} b^{2} + 5 \, A a^{2} b^{3}\right )} x\right )} \sqrt{x}}{48 \,{\left (a^{4} b^{5} x^{3} + 3 \, a^{5} b^{4} x^{2} + 3 \, a^{6} b^{3} x + a^{7} b^{2}\right )}}, -\frac{3 \,{\left (B a^{4} + 5 \, A a^{3} b +{\left (B a b^{3} + 5 \, A b^{4}\right )} x^{3} + 3 \,{\left (B a^{2} b^{2} + 5 \, A a b^{3}\right )} x^{2} + 3 \,{\left (B a^{3} b + 5 \, A a^{2} b^{2}\right )} x\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b}}{b \sqrt{x}}\right ) +{\left (3 \, B a^{4} b - 33 \, A a^{3} b^{2} - 3 \,{\left (B a^{2} b^{3} + 5 \, A a b^{4}\right )} x^{2} - 8 \,{\left (B a^{3} b^{2} + 5 \, A a^{2} b^{3}\right )} x\right )} \sqrt{x}}{24 \,{\left (a^{4} b^{5} x^{3} + 3 \, a^{5} b^{4} x^{2} + 3 \, a^{6} b^{3} x + a^{7} b^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2/x^(1/2),x, algorithm="fricas")

[Out]

[-1/48*(3*(B*a^4 + 5*A*a^3*b + (B*a*b^3 + 5*A*b^4)*x^3 + 3*(B*a^2*b^2 + 5*A*a*b^3)*x^2 + 3*(B*a^3*b + 5*A*a^2*
b^2)*x)*sqrt(-a*b)*log((b*x - a - 2*sqrt(-a*b)*sqrt(x))/(b*x + a)) + 2*(3*B*a^4*b - 33*A*a^3*b^2 - 3*(B*a^2*b^
3 + 5*A*a*b^4)*x^2 - 8*(B*a^3*b^2 + 5*A*a^2*b^3)*x)*sqrt(x))/(a^4*b^5*x^3 + 3*a^5*b^4*x^2 + 3*a^6*b^3*x + a^7*
b^2), -1/24*(3*(B*a^4 + 5*A*a^3*b + (B*a*b^3 + 5*A*b^4)*x^3 + 3*(B*a^2*b^2 + 5*A*a*b^3)*x^2 + 3*(B*a^3*b + 5*A
*a^2*b^2)*x)*sqrt(a*b)*arctan(sqrt(a*b)/(b*sqrt(x))) + (3*B*a^4*b - 33*A*a^3*b^2 - 3*(B*a^2*b^3 + 5*A*a*b^4)*x
^2 - 8*(B*a^3*b^2 + 5*A*a^2*b^3)*x)*sqrt(x))/(a^4*b^5*x^3 + 3*a^5*b^4*x^2 + 3*a^6*b^3*x + a^7*b^2)]

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Sympy [A]  time = 138.417, size = 2548, normalized size = 19.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b**2*x**2+2*a*b*x+a**2)**2/x**(1/2),x)

[Out]

Piecewise((zoo*(-2*A/(7*x**(7/2)) - 2*B/(5*x**(5/2))), Eq(a, 0) & Eq(b, 0)), ((2*A*sqrt(x) + 2*B*x**(3/2)/3)/a
**4, Eq(b, 0)), ((-2*A/(7*x**(7/2)) - 2*B/(5*x**(5/2)))/b**4, Eq(a, 0)), (66*I*A*a**(5/2)*b**2*sqrt(x)*sqrt(1/
b)/(48*I*a**(13/2)*b**2*sqrt(1/b) + 144*I*a**(11/2)*b**3*x*sqrt(1/b) + 144*I*a**(9/2)*b**4*x**2*sqrt(1/b) + 48
*I*a**(7/2)*b**5*x**3*sqrt(1/b)) + 80*I*A*a**(3/2)*b**3*x**(3/2)*sqrt(1/b)/(48*I*a**(13/2)*b**2*sqrt(1/b) + 14
4*I*a**(11/2)*b**3*x*sqrt(1/b) + 144*I*a**(9/2)*b**4*x**2*sqrt(1/b) + 48*I*a**(7/2)*b**5*x**3*sqrt(1/b)) + 30*
I*A*sqrt(a)*b**4*x**(5/2)*sqrt(1/b)/(48*I*a**(13/2)*b**2*sqrt(1/b) + 144*I*a**(11/2)*b**3*x*sqrt(1/b) + 144*I*
a**(9/2)*b**4*x**2*sqrt(1/b) + 48*I*a**(7/2)*b**5*x**3*sqrt(1/b)) + 15*A*a**3*b*log(-I*sqrt(a)*sqrt(1/b) + sqr
t(x))/(48*I*a**(13/2)*b**2*sqrt(1/b) + 144*I*a**(11/2)*b**3*x*sqrt(1/b) + 144*I*a**(9/2)*b**4*x**2*sqrt(1/b) +
48*I*a**(7/2)*b**5*x**3*sqrt(1/b)) - 15*A*a**3*b*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(13/2)*b**2*sqrt
(1/b) + 144*I*a**(11/2)*b**3*x*sqrt(1/b) + 144*I*a**(9/2)*b**4*x**2*sqrt(1/b) + 48*I*a**(7/2)*b**5*x**3*sqrt(1
/b)) + 45*A*a**2*b**2*x*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(13/2)*b**2*sqrt(1/b) + 144*I*a**(11/2)*b
**3*x*sqrt(1/b) + 144*I*a**(9/2)*b**4*x**2*sqrt(1/b) + 48*I*a**(7/2)*b**5*x**3*sqrt(1/b)) - 45*A*a**2*b**2*x*l
og(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(13/2)*b**2*sqrt(1/b) + 144*I*a**(11/2)*b**3*x*sqrt(1/b) + 144*I*a*
*(9/2)*b**4*x**2*sqrt(1/b) + 48*I*a**(7/2)*b**5*x**3*sqrt(1/b)) + 45*A*a*b**3*x**2*log(-I*sqrt(a)*sqrt(1/b) +
sqrt(x))/(48*I*a**(13/2)*b**2*sqrt(1/b) + 144*I*a**(11/2)*b**3*x*sqrt(1/b) + 144*I*a**(9/2)*b**4*x**2*sqrt(1/b
) + 48*I*a**(7/2)*b**5*x**3*sqrt(1/b)) - 45*A*a*b**3*x**2*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(13/2)*b
**2*sqrt(1/b) + 144*I*a**(11/2)*b**3*x*sqrt(1/b) + 144*I*a**(9/2)*b**4*x**2*sqrt(1/b) + 48*I*a**(7/2)*b**5*x**
3*sqrt(1/b)) + 15*A*b**4*x**3*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(13/2)*b**2*sqrt(1/b) + 144*I*a**(1
1/2)*b**3*x*sqrt(1/b) + 144*I*a**(9/2)*b**4*x**2*sqrt(1/b) + 48*I*a**(7/2)*b**5*x**3*sqrt(1/b)) - 15*A*b**4*x*
*3*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(13/2)*b**2*sqrt(1/b) + 144*I*a**(11/2)*b**3*x*sqrt(1/b) + 144*
I*a**(9/2)*b**4*x**2*sqrt(1/b) + 48*I*a**(7/2)*b**5*x**3*sqrt(1/b)) - 6*I*B*a**(7/2)*b*sqrt(x)*sqrt(1/b)/(48*I
*a**(13/2)*b**2*sqrt(1/b) + 144*I*a**(11/2)*b**3*x*sqrt(1/b) + 144*I*a**(9/2)*b**4*x**2*sqrt(1/b) + 48*I*a**(7
/2)*b**5*x**3*sqrt(1/b)) + 16*I*B*a**(5/2)*b**2*x**(3/2)*sqrt(1/b)/(48*I*a**(13/2)*b**2*sqrt(1/b) + 144*I*a**(
11/2)*b**3*x*sqrt(1/b) + 144*I*a**(9/2)*b**4*x**2*sqrt(1/b) + 48*I*a**(7/2)*b**5*x**3*sqrt(1/b)) + 6*I*B*a**(3
/2)*b**3*x**(5/2)*sqrt(1/b)/(48*I*a**(13/2)*b**2*sqrt(1/b) + 144*I*a**(11/2)*b**3*x*sqrt(1/b) + 144*I*a**(9/2)
*b**4*x**2*sqrt(1/b) + 48*I*a**(7/2)*b**5*x**3*sqrt(1/b)) + 3*B*a**4*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I
*a**(13/2)*b**2*sqrt(1/b) + 144*I*a**(11/2)*b**3*x*sqrt(1/b) + 144*I*a**(9/2)*b**4*x**2*sqrt(1/b) + 48*I*a**(7
/2)*b**5*x**3*sqrt(1/b)) - 3*B*a**4*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(13/2)*b**2*sqrt(1/b) + 144*I*
a**(11/2)*b**3*x*sqrt(1/b) + 144*I*a**(9/2)*b**4*x**2*sqrt(1/b) + 48*I*a**(7/2)*b**5*x**3*sqrt(1/b)) + 9*B*a**
3*b*x*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(13/2)*b**2*sqrt(1/b) + 144*I*a**(11/2)*b**3*x*sqrt(1/b) +
144*I*a**(9/2)*b**4*x**2*sqrt(1/b) + 48*I*a**(7/2)*b**5*x**3*sqrt(1/b)) - 9*B*a**3*b*x*log(I*sqrt(a)*sqrt(1/b)
+ sqrt(x))/(48*I*a**(13/2)*b**2*sqrt(1/b) + 144*I*a**(11/2)*b**3*x*sqrt(1/b) + 144*I*a**(9/2)*b**4*x**2*sqrt(
1/b) + 48*I*a**(7/2)*b**5*x**3*sqrt(1/b)) + 9*B*a**2*b**2*x**2*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(1
3/2)*b**2*sqrt(1/b) + 144*I*a**(11/2)*b**3*x*sqrt(1/b) + 144*I*a**(9/2)*b**4*x**2*sqrt(1/b) + 48*I*a**(7/2)*b*
*5*x**3*sqrt(1/b)) - 9*B*a**2*b**2*x**2*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(13/2)*b**2*sqrt(1/b) + 14
4*I*a**(11/2)*b**3*x*sqrt(1/b) + 144*I*a**(9/2)*b**4*x**2*sqrt(1/b) + 48*I*a**(7/2)*b**5*x**3*sqrt(1/b)) + 3*B
*a*b**3*x**3*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(13/2)*b**2*sqrt(1/b) + 144*I*a**(11/2)*b**3*x*sqrt(
1/b) + 144*I*a**(9/2)*b**4*x**2*sqrt(1/b) + 48*I*a**(7/2)*b**5*x**3*sqrt(1/b)) - 3*B*a*b**3*x**3*log(I*sqrt(a)
*sqrt(1/b) + sqrt(x))/(48*I*a**(13/2)*b**2*sqrt(1/b) + 144*I*a**(11/2)*b**3*x*sqrt(1/b) + 144*I*a**(9/2)*b**4*
x**2*sqrt(1/b) + 48*I*a**(7/2)*b**5*x**3*sqrt(1/b)), True))

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Giac [A]  time = 1.17665, size = 144, normalized size = 1.11 \begin{align*} \frac{{\left (B a + 5 \, A b\right )} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{8 \, \sqrt{a b} a^{3} b} + \frac{3 \, B a b^{2} x^{\frac{5}{2}} + 15 \, A b^{3} x^{\frac{5}{2}} + 8 \, B a^{2} b x^{\frac{3}{2}} + 40 \, A a b^{2} x^{\frac{3}{2}} - 3 \, B a^{3} \sqrt{x} + 33 \, A a^{2} b \sqrt{x}}{24 \,{\left (b x + a\right )}^{3} a^{3} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2/x^(1/2),x, algorithm="giac")

[Out]

1/8*(B*a + 5*A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3*b) + 1/24*(3*B*a*b^2*x^(5/2) + 15*A*b^3*x^(5/2) +
8*B*a^2*b*x^(3/2) + 40*A*a*b^2*x^(3/2) - 3*B*a^3*sqrt(x) + 33*A*a^2*b*sqrt(x))/((b*x + a)^3*a^3*b)