3.825 \(\int \frac{A+B x}{x^{7/2} (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=302 \[ -\frac{7 b (a+b x) (9 A b-5 a B)}{4 a^5 \sqrt{x} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{7 (a+b x) (9 A b-5 a B)}{12 a^4 x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{7 (a+b x) (9 A b-5 a B)}{20 a^3 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{A b-a B}{2 a b x^{5/2} (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{9 A b-5 a B}{4 a^2 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{7 b^{3/2} (a+b x) (9 A b-5 a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 a^{11/2} \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

(9*A*b - 5*a*B)/(4*a^2*b*x^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (A*b - a*B)/(2*a*b*x^(5/2)*(a + b*x)*Sqrt[a^
2 + 2*a*b*x + b^2*x^2]) - (7*(9*A*b - 5*a*B)*(a + b*x))/(20*a^3*b*x^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (7*
(9*A*b - 5*a*B)*(a + b*x))/(12*a^4*x^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (7*b*(9*A*b - 5*a*B)*(a + b*x))/(4
*a^5*Sqrt[x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (7*b^(3/2)*(9*A*b - 5*a*B)*(a + b*x)*ArcTan[(Sqrt[b]*Sqrt[x])/Sq
rt[a]])/(4*a^(11/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

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Rubi [A]  time = 0.146108, antiderivative size = 302, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {770, 78, 51, 63, 205} \[ -\frac{7 b (a+b x) (9 A b-5 a B)}{4 a^5 \sqrt{x} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{7 (a+b x) (9 A b-5 a B)}{12 a^4 x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{7 (a+b x) (9 A b-5 a B)}{20 a^3 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{A b-a B}{2 a b x^{5/2} (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{9 A b-5 a B}{4 a^2 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{7 b^{3/2} (a+b x) (9 A b-5 a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 a^{11/2} \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^(7/2)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)),x]

[Out]

(9*A*b - 5*a*B)/(4*a^2*b*x^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (A*b - a*B)/(2*a*b*x^(5/2)*(a + b*x)*Sqrt[a^
2 + 2*a*b*x + b^2*x^2]) - (7*(9*A*b - 5*a*B)*(a + b*x))/(20*a^3*b*x^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (7*
(9*A*b - 5*a*B)*(a + b*x))/(12*a^4*x^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (7*b*(9*A*b - 5*a*B)*(a + b*x))/(4
*a^5*Sqrt[x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (7*b^(3/2)*(9*A*b - 5*a*B)*(a + b*x)*ArcTan[(Sqrt[b]*Sqrt[x])/Sq
rt[a]])/(4*a^(11/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

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}{x^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac{\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac{A+B x}{x^{7/2} \left (a b+b^2 x\right )^3} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{A b-a B}{2 a b x^{5/2} (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left ((9 A b-5 a B) \left (a b+b^2 x\right )\right ) \int \frac{1}{x^{7/2} \left (a b+b^2 x\right )^2} \, dx}{4 a \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{9 A b-5 a B}{4 a^2 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{A b-a B}{2 a b x^{5/2} (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (7 (9 A b-5 a B) \left (a b+b^2 x\right )\right ) \int \frac{1}{x^{7/2} \left (a b+b^2 x\right )} \, dx}{8 a^2 b \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{9 A b-5 a B}{4 a^2 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{A b-a B}{2 a b x^{5/2} (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{7 (9 A b-5 a B) (a+b x)}{20 a^3 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (7 (9 A b-5 a B) \left (a b+b^2 x\right )\right ) \int \frac{1}{x^{5/2} \left (a b+b^2 x\right )} \, dx}{8 a^3 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{9 A b-5 a B}{4 a^2 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{A b-a B}{2 a b x^{5/2} (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{7 (9 A b-5 a B) (a+b x)}{20 a^3 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{7 (9 A b-5 a B) (a+b x)}{12 a^4 x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (7 b (9 A b-5 a B) \left (a b+b^2 x\right )\right ) \int \frac{1}{x^{3/2} \left (a b+b^2 x\right )} \, dx}{8 a^4 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{9 A b-5 a B}{4 a^2 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{A b-a B}{2 a b x^{5/2} (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{7 (9 A b-5 a B) (a+b x)}{20 a^3 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{7 (9 A b-5 a B) (a+b x)}{12 a^4 x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{7 b (9 A b-5 a B) (a+b x)}{4 a^5 \sqrt{x} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (7 b^2 (9 A b-5 a B) \left (a b+b^2 x\right )\right ) \int \frac{1}{\sqrt{x} \left (a b+b^2 x\right )} \, dx}{8 a^5 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{9 A b-5 a B}{4 a^2 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{A b-a B}{2 a b x^{5/2} (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{7 (9 A b-5 a B) (a+b x)}{20 a^3 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{7 (9 A b-5 a B) (a+b x)}{12 a^4 x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{7 b (9 A b-5 a B) (a+b x)}{4 a^5 \sqrt{x} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (7 b^2 (9 A b-5 a B) \left (a b+b^2 x\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a b+b^2 x^2} \, dx,x,\sqrt{x}\right )}{4 a^5 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{9 A b-5 a B}{4 a^2 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{A b-a B}{2 a b x^{5/2} (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{7 (9 A b-5 a B) (a+b x)}{20 a^3 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{7 (9 A b-5 a B) (a+b x)}{12 a^4 x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{7 b (9 A b-5 a B) (a+b x)}{4 a^5 \sqrt{x} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{7 b^{3/2} (9 A b-5 a B) (a+b x) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 a^{11/2} \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0329658, size = 79, normalized size = 0.26 \[ \frac{5 a^2 (A b-a B)+(a+b x)^2 (5 a B-9 A b) \, _2F_1\left (-\frac{5}{2},2;-\frac{3}{2};-\frac{b x}{a}\right )}{10 a^3 b x^{5/2} (a+b x) \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^(7/2)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)),x]

[Out]

(5*a^2*(A*b - a*B) + (-9*A*b + 5*a*B)*(a + b*x)^2*Hypergeometric2F1[-5/2, 2, -3/2, -((b*x)/a)])/(10*a^3*b*x^(5
/2)*(a + b*x)*Sqrt[(a + b*x)^2])

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Maple [A]  time = 0.022, size = 289, normalized size = 1. \begin{align*} -{\frac{bx+a}{60\,{a}^{5}} \left ( 945\,A\sqrt{ab}{x}^{4}{b}^{4}+40\,B\sqrt{ab}x{a}^{4}-1050\,B\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){x}^{7/2}{a}^{2}{b}^{3}+1890\,A\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){x}^{7/2}a{b}^{4}-525\,B\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){x}^{9/2}a{b}^{4}+24\,A\sqrt{ab}{a}^{4}-525\,B\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){x}^{5/2}{a}^{3}{b}^{2}+504\,A\sqrt{ab}{x}^{2}{a}^{2}{b}^{2}-280\,B\sqrt{ab}{x}^{2}{a}^{3}b-72\,A\sqrt{ab}x{a}^{3}b+945\,A\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){x}^{5/2}{a}^{2}{b}^{3}-875\,B\sqrt{ab}{x}^{3}{a}^{2}{b}^{2}+945\,A\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){x}^{9/2}{b}^{5}-525\,B\sqrt{ab}{x}^{4}a{b}^{3}+1575\,A\sqrt{ab}{x}^{3}a{b}^{3} \right ){\frac{1}{\sqrt{ab}}}{x}^{-{\frac{5}{2}}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^(7/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

-1/60*(945*A*(a*b)^(1/2)*x^4*b^4+40*B*(a*b)^(1/2)*x*a^4-1050*B*arctan(x^(1/2)*b/(a*b)^(1/2))*x^(7/2)*a^2*b^3+1
890*A*arctan(x^(1/2)*b/(a*b)^(1/2))*x^(7/2)*a*b^4-525*B*arctan(x^(1/2)*b/(a*b)^(1/2))*x^(9/2)*a*b^4+24*A*(a*b)
^(1/2)*a^4-525*B*arctan(x^(1/2)*b/(a*b)^(1/2))*x^(5/2)*a^3*b^2+504*A*(a*b)^(1/2)*x^2*a^2*b^2-280*B*(a*b)^(1/2)
*x^2*a^3*b-72*A*(a*b)^(1/2)*x*a^3*b+945*A*arctan(x^(1/2)*b/(a*b)^(1/2))*x^(5/2)*a^2*b^3-875*B*(a*b)^(1/2)*x^3*
a^2*b^2+945*A*arctan(x^(1/2)*b/(a*b)^(1/2))*x^(9/2)*b^5-525*B*(a*b)^(1/2)*x^4*a*b^3+1575*A*(a*b)^(1/2)*x^3*a*b
^3)*(b*x+a)/(a*b)^(1/2)/x^(5/2)/a^5/((b*x+a)^2)^(3/2)

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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)/x^(7/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.28022, size = 950, normalized size = 3.15 \begin{align*} \left [-\frac{105 \,{\left ({\left (5 \, B a b^{3} - 9 \, A b^{4}\right )} x^{5} + 2 \,{\left (5 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{4} +{\left (5 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{3}\right )} \sqrt{-\frac{b}{a}} \log \left (\frac{b x - 2 \, a \sqrt{x} \sqrt{-\frac{b}{a}} - a}{b x + a}\right ) + 2 \,{\left (24 \, A a^{4} - 105 \,{\left (5 \, B a b^{3} - 9 \, A b^{4}\right )} x^{4} - 175 \,{\left (5 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{3} - 56 \,{\left (5 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{2} + 8 \,{\left (5 \, B a^{4} - 9 \, A a^{3} b\right )} x\right )} \sqrt{x}}{120 \,{\left (a^{5} b^{2} x^{5} + 2 \, a^{6} b x^{4} + a^{7} x^{3}\right )}}, -\frac{105 \,{\left ({\left (5 \, B a b^{3} - 9 \, A b^{4}\right )} x^{5} + 2 \,{\left (5 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{4} +{\left (5 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{3}\right )} \sqrt{\frac{b}{a}} \arctan \left (\frac{a \sqrt{\frac{b}{a}}}{b \sqrt{x}}\right ) +{\left (24 \, A a^{4} - 105 \,{\left (5 \, B a b^{3} - 9 \, A b^{4}\right )} x^{4} - 175 \,{\left (5 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{3} - 56 \,{\left (5 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{2} + 8 \,{\left (5 \, B a^{4} - 9 \, A a^{3} b\right )} x\right )} \sqrt{x}}{60 \,{\left (a^{5} b^{2} x^{5} + 2 \, a^{6} b x^{4} + a^{7} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(7/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

[-1/120*(105*((5*B*a*b^3 - 9*A*b^4)*x^5 + 2*(5*B*a^2*b^2 - 9*A*a*b^3)*x^4 + (5*B*a^3*b - 9*A*a^2*b^2)*x^3)*sqr
t(-b/a)*log((b*x - 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) + 2*(24*A*a^4 - 105*(5*B*a*b^3 - 9*A*b^4)*x^4 - 175*
(5*B*a^2*b^2 - 9*A*a*b^3)*x^3 - 56*(5*B*a^3*b - 9*A*a^2*b^2)*x^2 + 8*(5*B*a^4 - 9*A*a^3*b)*x)*sqrt(x))/(a^5*b^
2*x^5 + 2*a^6*b*x^4 + a^7*x^3), -1/60*(105*((5*B*a*b^3 - 9*A*b^4)*x^5 + 2*(5*B*a^2*b^2 - 9*A*a*b^3)*x^4 + (5*B
*a^3*b - 9*A*a^2*b^2)*x^3)*sqrt(b/a)*arctan(a*sqrt(b/a)/(b*sqrt(x))) + (24*A*a^4 - 105*(5*B*a*b^3 - 9*A*b^4)*x
^4 - 175*(5*B*a^2*b^2 - 9*A*a*b^3)*x^3 - 56*(5*B*a^3*b - 9*A*a^2*b^2)*x^2 + 8*(5*B*a^4 - 9*A*a^3*b)*x)*sqrt(x)
)/(a^5*b^2*x^5 + 2*a^6*b*x^4 + a^7*x^3)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**(7/2)/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.19422, size = 215, normalized size = 0.71 \begin{align*} \frac{7 \,{\left (5 \, B a b^{2} - 9 \, A b^{3}\right )} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{4 \, \sqrt{a b} a^{5} \mathrm{sgn}\left (b x + a\right )} + \frac{11 \, B a b^{3} x^{\frac{3}{2}} - 15 \, A b^{4} x^{\frac{3}{2}} + 13 \, B a^{2} b^{2} \sqrt{x} - 17 \, A a b^{3} \sqrt{x}}{4 \,{\left (b x + a\right )}^{2} a^{5} \mathrm{sgn}\left (b x + a\right )} + \frac{2 \,{\left (45 \, B a b x^{2} - 90 \, A b^{2} x^{2} - 5 \, B a^{2} x + 15 \, A a b x - 3 \, A a^{2}\right )}}{15 \, a^{5} x^{\frac{5}{2}} \mathrm{sgn}\left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(7/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

7/4*(5*B*a*b^2 - 9*A*b^3)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^5*sgn(b*x + a)) + 1/4*(11*B*a*b^3*x^(3/2) -
15*A*b^4*x^(3/2) + 13*B*a^2*b^2*sqrt(x) - 17*A*a*b^3*sqrt(x))/((b*x + a)^2*a^5*sgn(b*x + a)) + 2/15*(45*B*a*b
*x^2 - 90*A*b^2*x^2 - 5*B*a^2*x + 15*A*a*b*x - 3*A*a^2)/(a^5*x^(5/2)*sgn(b*x + a))