\(\int \frac {(a+b x)^{5/2} (A+B x)}{x^5} \, dx\) [419]

Optimal result

Integrand size = 18, antiderivative size = 142 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^5} \, dx=\frac {5 b^2 (A b-8 a B) \sqrt {a+b x}}{64 a x}+\frac {5 b (A b-8 a B) (a+b x)^{3/2}}{96 a x^2}+\frac {(A b-8 a B) (a+b x)^{5/2}}{24 a x^3}-\frac {A (a+b x)^{7/2}}{4 a x^4}+\frac {5 b^3 (A b-8 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{3/2}} \]

[Out]

5/96*b*(A*b-8*B*a)*(b*x+a)^(3/2)/a/x^2+1/24*(A*b-8*B*a)*(b*x+a)^(5/2)/a/x^3-1/4*A*(b*x+a)^(7/2)/a/x^4+5/64*b^3
*(A*b-8*B*a)*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(3/2)+5/64*b^2*(A*b-8*B*a)*(b*x+a)^(1/2)/a/x

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {79, 43, 65, 214} \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^5} \, dx=\frac {5 b^3 (A b-8 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{3/2}}+\frac {5 b^2 \sqrt {a+b x} (A b-8 a B)}{64 a x}+\frac {(a+b x)^{5/2} (A b-8 a B)}{24 a x^3}+\frac {5 b (a+b x)^{3/2} (A b-8 a B)}{96 a x^2}-\frac {A (a+b x)^{7/2}}{4 a x^4} \]

[In]

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

[Out]

(5*b^2*(A*b - 8*a*B)*Sqrt[a + b*x])/(64*a*x) + (5*b*(A*b - 8*a*B)*(a + b*x)^(3/2))/(96*a*x^2) + ((A*b - 8*a*B)
*(a + b*x)^(5/2))/(24*a*x^3) - (A*(a + b*x)^(7/2))/(4*a*x^4) + (5*b^3*(A*b - 8*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt
[a]])/(64*a^(3/2))

Rule 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {A (a+b x)^{7/2}}{4 a x^4}+\frac {\left (-\frac {A b}{2}+4 a B\right ) \int \frac {(a+b x)^{5/2}}{x^4} \, dx}{4 a} \\ & = \frac {(A b-8 a B) (a+b x)^{5/2}}{24 a x^3}-\frac {A (a+b x)^{7/2}}{4 a x^4}-\frac {(5 b (A b-8 a B)) \int \frac {(a+b x)^{3/2}}{x^3} \, dx}{48 a} \\ & = \frac {5 b (A b-8 a B) (a+b x)^{3/2}}{96 a x^2}+\frac {(A b-8 a B) (a+b x)^{5/2}}{24 a x^3}-\frac {A (a+b x)^{7/2}}{4 a x^4}-\frac {\left (5 b^2 (A b-8 a B)\right ) \int \frac {\sqrt {a+b x}}{x^2} \, dx}{64 a} \\ & = \frac {5 b^2 (A b-8 a B) \sqrt {a+b x}}{64 a x}+\frac {5 b (A b-8 a B) (a+b x)^{3/2}}{96 a x^2}+\frac {(A b-8 a B) (a+b x)^{5/2}}{24 a x^3}-\frac {A (a+b x)^{7/2}}{4 a x^4}-\frac {\left (5 b^3 (A b-8 a B)\right ) \int \frac {1}{x \sqrt {a+b x}} \, dx}{128 a} \\ & = \frac {5 b^2 (A b-8 a B) \sqrt {a+b x}}{64 a x}+\frac {5 b (A b-8 a B) (a+b x)^{3/2}}{96 a x^2}+\frac {(A b-8 a B) (a+b x)^{5/2}}{24 a x^3}-\frac {A (a+b x)^{7/2}}{4 a x^4}-\frac {\left (5 b^2 (A b-8 a B)\right ) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{64 a} \\ & = \frac {5 b^2 (A b-8 a B) \sqrt {a+b x}}{64 a x}+\frac {5 b (A b-8 a B) (a+b x)^{3/2}}{96 a x^2}+\frac {(A b-8 a B) (a+b x)^{5/2}}{24 a x^3}-\frac {A (a+b x)^{7/2}}{4 a x^4}+\frac {5 b^3 (A b-8 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.78 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^5} \, dx=-\frac {\sqrt {a+b x} \left (15 A b^3 x^3+16 a^3 (3 A+4 B x)+8 a^2 b x (17 A+26 B x)+2 a b^2 x^2 (59 A+132 B x)\right )}{192 a x^4}+\frac {5 b^3 (A b-8 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{3/2}} \]

[In]

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

[Out]

-1/192*(Sqrt[a + b*x]*(15*A*b^3*x^3 + 16*a^3*(3*A + 4*B*x) + 8*a^2*b*x*(17*A + 26*B*x) + 2*a*b^2*x^2*(59*A + 1
32*B*x)))/(a*x^4) + (5*b^3*(A*b - 8*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(64*a^(3/2))

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.71

method	result	size
pseudoelliptic	\(-\frac {17 \left (-\frac {15 b^{3} x^{4} \left (A b -8 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{136}+\sqrt {b x +a}\, \left (\frac {59 \left (\frac {132 B x}{59}+A \right ) x^{2} b^{2} a^{\frac {3}{2}}}{68}+b x \left (\frac {26 B x}{17}+A \right ) a^{\frac {5}{2}}+\frac {2 \left (4 B x +3 A \right ) a^{\frac {7}{2}}}{17}+\frac {15 A \sqrt {a}\, b^{3} x^{3}}{136}\right )\right )}{24 a^{\frac {3}{2}} x^{4}}\)	\(101\)
risch	\(-\frac {\sqrt {b x +a}\, \left (15 A \,b^{3} x^{3}+264 B a \,b^{2} x^{3}+118 a A \,b^{2} x^{2}+208 B \,a^{2} b \,x^{2}+136 a^{2} A b x +64 a^{3} B x +48 a^{3} A \right )}{192 x^{4} a}+\frac {5 b^{3} \left (A b -8 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{64 a^{\frac {3}{2}}}\)	\(106\)
derivativedivides	\(2 b^{3} \left (-\frac {\frac {\left (5 A b +88 B a \right ) \left (b x +a \right )^{\frac {7}{2}}}{128 a}+\left (-\frac {73 B a}{48}+\frac {73 A b}{384}\right ) \left (b x +a \right )^{\frac {5}{2}}-\frac {55 a \left (A b -8 B a \right ) \left (b x +a \right )^{\frac {3}{2}}}{384}+\left (-\frac {5}{16} a^{3} B +\frac {5}{128} a^{2} b A \right ) \sqrt {b x +a}}{b^{4} x^{4}}+\frac {5 \left (A b -8 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{128 a^{\frac {3}{2}}}\right )\)	\(119\)
default	\(2 b^{3} \left (-\frac {\frac {\left (5 A b +88 B a \right ) \left (b x +a \right )^{\frac {7}{2}}}{128 a}+\left (-\frac {73 B a}{48}+\frac {73 A b}{384}\right ) \left (b x +a \right )^{\frac {5}{2}}-\frac {55 a \left (A b -8 B a \right ) \left (b x +a \right )^{\frac {3}{2}}}{384}+\left (-\frac {5}{16} a^{3} B +\frac {5}{128} a^{2} b A \right ) \sqrt {b x +a}}{b^{4} x^{4}}+\frac {5 \left (A b -8 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{128 a^{\frac {3}{2}}}\right )\)	\(119\)

[In]

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

[Out]

-17/24*(-15/136*b^3*x^4*(A*b-8*B*a)*arctanh((b*x+a)^(1/2)/a^(1/2))+(b*x+a)^(1/2)*(59/68*(132/59*B*x+A)*x^2*b^2
*a^(3/2)+b*x*(26/17*B*x+A)*a^(5/2)+2/17*(4*B*x+3*A)*a^(7/2)+15/136*A*a^(1/2)*b^3*x^3))/a^(3/2)/x^4

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.83 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^5} \, dx=\left [-\frac {15 \, {\left (8 \, B a b^{3} - A b^{4}\right )} \sqrt {a} x^{4} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (48 \, A a^{4} + 3 \, {\left (88 \, B a^{2} b^{2} + 5 \, A a b^{3}\right )} x^{3} + 2 \, {\left (104 \, B a^{3} b + 59 \, A a^{2} b^{2}\right )} x^{2} + 8 \, {\left (8 \, B a^{4} + 17 \, A a^{3} b\right )} x\right )} \sqrt {b x + a}}{384 \, a^{2} x^{4}}, \frac {15 \, {\left (8 \, B a b^{3} - A b^{4}\right )} \sqrt {-a} x^{4} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) - {\left (48 \, A a^{4} + 3 \, {\left (88 \, B a^{2} b^{2} + 5 \, A a b^{3}\right )} x^{3} + 2 \, {\left (104 \, B a^{3} b + 59 \, A a^{2} b^{2}\right )} x^{2} + 8 \, {\left (8 \, B a^{4} + 17 \, A a^{3} b\right )} x\right )} \sqrt {b x + a}}{192 \, a^{2} x^{4}}\right ] \]

[In]

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

[Out]

[-1/384*(15*(8*B*a*b^3 - A*b^4)*sqrt(a)*x^4*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(48*A*a^4 + 3*(88
*B*a^2*b^2 + 5*A*a*b^3)*x^3 + 2*(104*B*a^3*b + 59*A*a^2*b^2)*x^2 + 8*(8*B*a^4 + 17*A*a^3*b)*x)*sqrt(b*x + a))/
(a^2*x^4), 1/192*(15*(8*B*a*b^3 - A*b^4)*sqrt(-a)*x^4*arctan(sqrt(b*x + a)*sqrt(-a)/a) - (48*A*a^4 + 3*(88*B*a
^2*b^2 + 5*A*a*b^3)*x^3 + 2*(104*B*a^3*b + 59*A*a^2*b^2)*x^2 + 8*(8*B*a^4 + 17*A*a^3*b)*x)*sqrt(b*x + a))/(a^2
*x^4)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 326 vs. \(2 (129) = 258\).

Time = 124.43 (sec) , antiderivative size = 326, normalized size of antiderivative = 2.30 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^5} \, dx=- \frac {A a^{3}}{4 \sqrt {b} x^{\frac {9}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {23 A a^{2} \sqrt {b}}{24 x^{\frac {7}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {127 A a b^{\frac {3}{2}}}{96 x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {133 A b^{\frac {5}{2}}}{192 x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {5 A b^{\frac {7}{2}}}{64 a \sqrt {x} \sqrt {\frac {a}{b x} + 1}} + \frac {5 A b^{4} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{64 a^{\frac {3}{2}}} - \frac {B a^{3}}{3 \sqrt {b} x^{\frac {7}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {17 B a^{2} \sqrt {b}}{12 x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {35 B a b^{\frac {3}{2}}}{24 x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {B b^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}}{\sqrt {x}} - \frac {3 B b^{\frac {5}{2}}}{8 \sqrt {x} \sqrt {\frac {a}{b x} + 1}} - \frac {5 B b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{8 \sqrt {a}} \]

[In]

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

[Out]

-A*a**3/(4*sqrt(b)*x**(9/2)*sqrt(a/(b*x) + 1)) - 23*A*a**2*sqrt(b)/(24*x**(7/2)*sqrt(a/(b*x) + 1)) - 127*A*a*b
**(3/2)/(96*x**(5/2)*sqrt(a/(b*x) + 1)) - 133*A*b**(5/2)/(192*x**(3/2)*sqrt(a/(b*x) + 1)) - 5*A*b**(7/2)/(64*a
*sqrt(x)*sqrt(a/(b*x) + 1)) + 5*A*b**4*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/(64*a**(3/2)) - B*a**3/(3*sqrt(b)*x**(
7/2)*sqrt(a/(b*x) + 1)) - 17*B*a**2*sqrt(b)/(12*x**(5/2)*sqrt(a/(b*x) + 1)) - 35*B*a*b**(3/2)/(24*x**(3/2)*sqr
t(a/(b*x) + 1)) - B*b**(5/2)*sqrt(a/(b*x) + 1)/sqrt(x) - 3*B*b**(5/2)/(8*sqrt(x)*sqrt(a/(b*x) + 1)) - 5*B*b**3
*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/(8*sqrt(a))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.37 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^5} \, dx=-\frac {1}{384} \, b^{4} {\left (\frac {2 \, {\left (3 \, {\left (88 \, B a + 5 \, A b\right )} {\left (b x + a\right )}^{\frac {7}{2}} - 73 \, {\left (8 \, B a^{2} - A a b\right )} {\left (b x + a\right )}^{\frac {5}{2}} + 55 \, {\left (8 \, B a^{3} - A a^{2} b\right )} {\left (b x + a\right )}^{\frac {3}{2}} - 15 \, {\left (8 \, B a^{4} - A a^{3} b\right )} \sqrt {b x + a}\right )}}{{\left (b x + a\right )}^{4} a b - 4 \, {\left (b x + a\right )}^{3} a^{2} b + 6 \, {\left (b x + a\right )}^{2} a^{3} b - 4 \, {\left (b x + a\right )} a^{4} b + a^{5} b} - \frac {15 \, {\left (8 \, B a - A b\right )} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {3}{2}} b}\right )} \]

[In]

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

[Out]

-1/384*b^4*(2*(3*(88*B*a + 5*A*b)*(b*x + a)^(7/2) - 73*(8*B*a^2 - A*a*b)*(b*x + a)^(5/2) + 55*(8*B*a^3 - A*a^2
*b)*(b*x + a)^(3/2) - 15*(8*B*a^4 - A*a^3*b)*sqrt(b*x + a))/((b*x + a)^4*a*b - 4*(b*x + a)^3*a^2*b + 6*(b*x +
a)^2*a^3*b - 4*(b*x + a)*a^4*b + a^5*b) - 15*(8*B*a - A*b)*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt
(a)))/(a^(3/2)*b))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.25 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^5} \, dx=\frac {\frac {15 \, {\left (8 \, B a b^{4} - A b^{5}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} - \frac {264 \, {\left (b x + a\right )}^{\frac {7}{2}} B a b^{4} - 584 \, {\left (b x + a\right )}^{\frac {5}{2}} B a^{2} b^{4} + 440 \, {\left (b x + a\right )}^{\frac {3}{2}} B a^{3} b^{4} - 120 \, \sqrt {b x + a} B a^{4} b^{4} + 15 \, {\left (b x + a\right )}^{\frac {7}{2}} A b^{5} + 73 \, {\left (b x + a\right )}^{\frac {5}{2}} A a b^{5} - 55 \, {\left (b x + a\right )}^{\frac {3}{2}} A a^{2} b^{5} + 15 \, \sqrt {b x + a} A a^{3} b^{5}}{a b^{4} x^{4}}}{192 \, b} \]

[In]

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

[Out]

1/192*(15*(8*B*a*b^4 - A*b^5)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a) - (264*(b*x + a)^(7/2)*B*a*b^4 - 584
*(b*x + a)^(5/2)*B*a^2*b^4 + 440*(b*x + a)^(3/2)*B*a^3*b^4 - 120*sqrt(b*x + a)*B*a^4*b^4 + 15*(b*x + a)^(7/2)*
A*b^5 + 73*(b*x + a)^(5/2)*A*a*b^5 - 55*(b*x + a)^(3/2)*A*a^2*b^5 + 15*sqrt(b*x + a)*A*a^3*b^5)/(a*b^4*x^4))/b

Mupad [B] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.25 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^5} \, dx=\frac {5\,b^3\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )\,\left (A\,b-8\,B\,a\right )}{64\,a^{3/2}}-\frac {\left (\frac {73\,A\,b^4}{192}-\frac {73\,B\,a\,b^3}{24}\right )\,{\left (a+b\,x\right )}^{5/2}+\left (\frac {5\,A\,a^2\,b^4}{64}-\frac {5\,B\,a^3\,b^3}{8}\right )\,\sqrt {a+b\,x}+\left (\frac {55\,B\,a^2\,b^3}{24}-\frac {55\,A\,a\,b^4}{192}\right )\,{\left (a+b\,x\right )}^{3/2}+\frac {\left (5\,A\,b^4+88\,B\,a\,b^3\right )\,{\left (a+b\,x\right )}^{7/2}}{64\,a}}{{\left (a+b\,x\right )}^4-4\,a^3\,\left (a+b\,x\right )-4\,a\,{\left (a+b\,x\right )}^3+6\,a^2\,{\left (a+b\,x\right )}^2+a^4} \]

[In]

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

[Out]

(5*b^3*atanh((a + b*x)^(1/2)/a^(1/2))*(A*b - 8*B*a))/(64*a^(3/2)) - (((73*A*b^4)/192 - (73*B*a*b^3)/24)*(a + b
*x)^(5/2) + ((5*A*a^2*b^4)/64 - (5*B*a^3*b^3)/8)*(a + b*x)^(1/2) + ((55*B*a^2*b^3)/24 - (55*A*a*b^4)/192)*(a +
b*x)^(3/2) + ((5*A*b^4 + 88*B*a*b^3)*(a + b*x)^(7/2))/(64*a))/((a + b*x)^4 - 4*a^3*(a + b*x) - 4*a*(a + b*x)^
3 + 6*a^2*(a + b*x)^2 + a^4)