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

Optimal. Leaf size=153 \[ \frac {5 (A b-7 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{8 \sqrt {a} b^{9/2}}-\frac {5 \sqrt {x} (A b-7 a B)}{8 a b^4}+\frac {5 x^{3/2} (A b-7 a B)}{24 a b^3 (a+b x)}+\frac {x^{5/2} (A b-7 a B)}{12 a b^2 (a+b x)^2}+\frac {x^{7/2} (A b-a B)}{3 a b (a+b x)^3} \]

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {27, 78, 47, 50, 63, 205} \begin {gather*} \frac {x^{5/2} (A b-7 a B)}{12 a b^2 (a+b x)^2}+\frac {5 x^{3/2} (A b-7 a B)}{24 a b^3 (a+b x)}-\frac {5 \sqrt {x} (A b-7 a B)}{8 a b^4}+\frac {5 (A b-7 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{8 \sqrt {a} b^{9/2}}+\frac {x^{7/2} (A b-a B)}{3 a b (a+b x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*(A*b - 7*a*B)*Sqrt[x])/(8*a*b^4) + ((A*b - a*B)*x^(7/2))/(3*a*b*(a + b*x)^3) + ((A*b - 7*a*B)*x^(5/2))/(12
*a*b^2*(a + b*x)^2) + (5*(A*b - 7*a*B)*x^(3/2))/(24*a*b^3*(a + b*x)) + (5*(A*b - 7*a*B)*ArcTan[(Sqrt[b]*Sqrt[x
])/Sqrt[a]])/(8*Sqrt[a]*b^(9/2))

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 47

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},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

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 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 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 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 {x^{5/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac {x^{5/2} (A+B x)}{(a+b x)^4} \, dx\\ &=\frac {(A b-a B) x^{7/2}}{3 a b (a+b x)^3}-\frac {\left (\frac {A b}{2}-\frac {7 a B}{2}\right ) \int \frac {x^{5/2}}{(a+b x)^3} \, dx}{3 a b}\\ &=\frac {(A b-a B) x^{7/2}}{3 a b (a+b x)^3}+\frac {(A b-7 a B) x^{5/2}}{12 a b^2 (a+b x)^2}-\frac {(5 (A b-7 a B)) \int \frac {x^{3/2}}{(a+b x)^2} \, dx}{24 a b^2}\\ &=\frac {(A b-a B) x^{7/2}}{3 a b (a+b x)^3}+\frac {(A b-7 a B) x^{5/2}}{12 a b^2 (a+b x)^2}+\frac {5 (A b-7 a B) x^{3/2}}{24 a b^3 (a+b x)}-\frac {(5 (A b-7 a B)) \int \frac {\sqrt {x}}{a+b x} \, dx}{16 a b^3}\\ &=-\frac {5 (A b-7 a B) \sqrt {x}}{8 a b^4}+\frac {(A b-a B) x^{7/2}}{3 a b (a+b x)^3}+\frac {(A b-7 a B) x^{5/2}}{12 a b^2 (a+b x)^2}+\frac {5 (A b-7 a B) x^{3/2}}{24 a b^3 (a+b x)}+\frac {(5 (A b-7 a B)) \int \frac {1}{\sqrt {x} (a+b x)} \, dx}{16 b^4}\\ &=-\frac {5 (A b-7 a B) \sqrt {x}}{8 a b^4}+\frac {(A b-a B) x^{7/2}}{3 a b (a+b x)^3}+\frac {(A b-7 a B) x^{5/2}}{12 a b^2 (a+b x)^2}+\frac {5 (A b-7 a B) x^{3/2}}{24 a b^3 (a+b x)}+\frac {(5 (A b-7 a B)) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {x}\right )}{8 b^4}\\ &=-\frac {5 (A b-7 a B) \sqrt {x}}{8 a b^4}+\frac {(A b-a B) x^{7/2}}{3 a b (a+b x)^3}+\frac {(A b-7 a B) x^{5/2}}{12 a b^2 (a+b x)^2}+\frac {5 (A b-7 a B) x^{3/2}}{24 a b^3 (a+b x)}+\frac {5 (A b-7 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{8 \sqrt {a} b^{9/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.03, size = 61, normalized size = 0.40 \begin {gather*} \frac {x^{7/2} \left (\frac {7 a^3 (A b-a B)}{(a+b x)^3}+(7 a B-A b) \, _2F_1\left (3,\frac {7}{2};\frac {9}{2};-\frac {b x}{a}\right )\right )}{21 a^4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(7/2)*((7*a^3*(A*b - a*B))/(a + b*x)^3 + (-(A*b) + 7*a*B)*Hypergeometric2F1[3, 7/2, 9/2, -((b*x)/a)]))/(21*
a^4*b)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.23, size = 118, normalized size = 0.77 \begin {gather*} \frac {\sqrt {x} \left (105 a^3 B-15 a^2 A b+280 a^2 b B x-40 a A b^2 x+231 a b^2 B x^2-33 A b^3 x^2+48 b^3 B x^3\right )}{24 b^4 (a+b x)^3}-\frac {5 (7 a B-A b) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{8 \sqrt {a} b^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x]*(-15*a^2*A*b + 105*a^3*B - 40*a*A*b^2*x + 280*a^2*b*B*x - 33*A*b^3*x^2 + 231*a*b^2*B*x^2 + 48*b^3*B*x
^3))/(24*b^4*(a + b*x)^3) - (5*(-(A*b) + 7*a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(8*Sqrt[a]*b^(9/2))

________________________________________________________________________________________

fricas [A]  time = 0.44, size = 437, normalized size = 2.86 \begin {gather*} \left [\frac {15 \, {\left (7 \, B a^{4} - A a^{3} b + {\left (7 \, B a b^{3} - A b^{4}\right )} x^{3} + 3 \, {\left (7 \, B a^{2} b^{2} - A a b^{3}\right )} x^{2} + 3 \, {\left (7 \, B a^{3} b - 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 (48 \, B a b^{4} x^{3} + 105 \, B a^{4} b - 15 \, A a^{3} b^{2} + 33 \, {\left (7 \, B a^{2} b^{3} - A a b^{4}\right )} x^{2} + 40 \, {\left (7 \, B a^{3} b^{2} - A a^{2} b^{3}\right )} x\right )} \sqrt {x}}{48 \, {\left (a b^{8} x^{3} + 3 \, a^{2} b^{7} x^{2} + 3 \, a^{3} b^{6} x + a^{4} b^{5}\right )}}, \frac {15 \, {\left (7 \, B a^{4} - A a^{3} b + {\left (7 \, B a b^{3} - A b^{4}\right )} x^{3} + 3 \, {\left (7 \, B a^{2} b^{2} - A a b^{3}\right )} x^{2} + 3 \, {\left (7 \, B a^{3} b - A a^{2} b^{2}\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) + {\left (48 \, B a b^{4} x^{3} + 105 \, B a^{4} b - 15 \, A a^{3} b^{2} + 33 \, {\left (7 \, B a^{2} b^{3} - A a b^{4}\right )} x^{2} + 40 \, {\left (7 \, B a^{3} b^{2} - A a^{2} b^{3}\right )} x\right )} \sqrt {x}}{24 \, {\left (a b^{8} x^{3} + 3 \, a^{2} b^{7} x^{2} + 3 \, a^{3} b^{6} x + a^{4} b^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(15*(7*B*a^4 - A*a^3*b + (7*B*a*b^3 - A*b^4)*x^3 + 3*(7*B*a^2*b^2 - A*a*b^3)*x^2 + 3*(7*B*a^3*b - A*a^2*
b^2)*x)*sqrt(-a*b)*log((b*x - a - 2*sqrt(-a*b)*sqrt(x))/(b*x + a)) + 2*(48*B*a*b^4*x^3 + 105*B*a^4*b - 15*A*a^
3*b^2 + 33*(7*B*a^2*b^3 - A*a*b^4)*x^2 + 40*(7*B*a^3*b^2 - A*a^2*b^3)*x)*sqrt(x))/(a*b^8*x^3 + 3*a^2*b^7*x^2 +
3*a^3*b^6*x + a^4*b^5), 1/24*(15*(7*B*a^4 - A*a^3*b + (7*B*a*b^3 - A*b^4)*x^3 + 3*(7*B*a^2*b^2 - A*a*b^3)*x^2
+ 3*(7*B*a^3*b - A*a^2*b^2)*x)*sqrt(a*b)*arctan(sqrt(a*b)/(b*sqrt(x))) + (48*B*a*b^4*x^3 + 105*B*a^4*b - 15*A
*a^3*b^2 + 33*(7*B*a^2*b^3 - A*a*b^4)*x^2 + 40*(7*B*a^3*b^2 - A*a^2*b^3)*x)*sqrt(x))/(a*b^8*x^3 + 3*a^2*b^7*x^
2 + 3*a^3*b^6*x + a^4*b^5)]

________________________________________________________________________________________

giac [A]  time = 0.17, size = 111, normalized size = 0.73 \begin {gather*} \frac {2 \, B \sqrt {x}}{b^{4}} - \frac {5 \, {\left (7 \, B a - A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{4}} + \frac {87 \, B a b^{2} x^{\frac {5}{2}} - 33 \, A b^{3} x^{\frac {5}{2}} + 136 \, B a^{2} b x^{\frac {3}{2}} - 40 \, A a b^{2} x^{\frac {3}{2}} + 57 \, B a^{3} \sqrt {x} - 15 \, A a^{2} b \sqrt {x}}{24 \, {\left (b x + a\right )}^{3} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*B*sqrt(x)/b^4 - 5/8*(7*B*a - A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^4) + 1/24*(87*B*a*b^2*x^(5/2) - 3
3*A*b^3*x^(5/2) + 136*B*a^2*b*x^(3/2) - 40*A*a*b^2*x^(3/2) + 57*B*a^3*sqrt(x) - 15*A*a^2*b*sqrt(x))/((b*x + a)
^3*b^4)

________________________________________________________________________________________

maple [A]  time = 0.07, size = 163, normalized size = 1.07 \begin {gather*} -\frac {11 A \,x^{\frac {5}{2}}}{8 \left (b x +a \right )^{3} b}+\frac {29 B a \,x^{\frac {5}{2}}}{8 \left (b x +a \right )^{3} b^{2}}-\frac {5 A a \,x^{\frac {3}{2}}}{3 \left (b x +a \right )^{3} b^{2}}+\frac {17 B \,a^{2} x^{\frac {3}{2}}}{3 \left (b x +a \right )^{3} b^{3}}-\frac {5 A \,a^{2} \sqrt {x}}{8 \left (b x +a \right )^{3} b^{3}}+\frac {19 B \,a^{3} \sqrt {x}}{8 \left (b x +a \right )^{3} b^{4}}+\frac {5 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, b^{3}}-\frac {35 B a \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, b^{4}}+\frac {2 B \sqrt {x}}{b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*B/b^4*x^(1/2)+29/8/b^2/(b*x+a)^3*x^(5/2)*B*a-11/8/b/(b*x+a)^3*x^(5/2)*A-5/3/b^2/(b*x+a)^3*A*x^(3/2)*a+17/3/b
^3/(b*x+a)^3*B*x^(3/2)*a^2+19/8/b^4/(b*x+a)^3*x^(1/2)*B*a^3-5/8/b^3/(b*x+a)^3*x^(1/2)*A*a^2+5/8/b^3/(a*b)^(1/2
)*arctan(1/(a*b)^(1/2)*b*x^(1/2))*A-35/8/b^4/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x^(1/2))*B*a

________________________________________________________________________________________

maxima [A]  time = 1.20, size = 136, normalized size = 0.89 \begin {gather*} \frac {3 \, {\left (29 \, B a b^{2} - 11 \, A b^{3}\right )} x^{\frac {5}{2}} + 8 \, {\left (17 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{\frac {3}{2}} + 3 \, {\left (19 \, B a^{3} - 5 \, A a^{2} b\right )} \sqrt {x}}{24 \, {\left (b^{7} x^{3} + 3 \, a b^{6} x^{2} + 3 \, a^{2} b^{5} x + a^{3} b^{4}\right )}} + \frac {2 \, B \sqrt {x}}{b^{4}} - \frac {5 \, {\left (7 \, B a - A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(29*B*a*b^2 - 11*A*b^3)*x^(5/2) + 8*(17*B*a^2*b - 5*A*a*b^2)*x^(3/2) + 3*(19*B*a^3 - 5*A*a^2*b)*sqrt(x
))/(b^7*x^3 + 3*a*b^6*x^2 + 3*a^2*b^5*x + a^3*b^4) + 2*B*sqrt(x)/b^4 - 5/8*(7*B*a - A*b)*arctan(b*sqrt(x)/sqrt
(a*b))/(sqrt(a*b)*b^4)

________________________________________________________________________________________

mupad [B]  time = 1.25, size = 131, normalized size = 0.86 \begin {gather*} \frac {2\,B\,\sqrt {x}}{b^4}-\frac {x^{3/2}\,\left (\frac {5\,A\,a\,b^2}{3}-\frac {17\,B\,a^2\,b}{3}\right )-\sqrt {x}\,\left (\frac {19\,B\,a^3}{8}-\frac {5\,A\,a^2\,b}{8}\right )+x^{5/2}\,\left (\frac {11\,A\,b^3}{8}-\frac {29\,B\,a\,b^2}{8}\right )}{a^3\,b^4+3\,a^2\,b^5\,x+3\,a\,b^6\,x^2+b^7\,x^3}+\frac {5\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (A\,b-7\,B\,a\right )}{8\,\sqrt {a}\,b^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*B*x^(1/2))/b^4 - (x^(3/2)*((5*A*a*b^2)/3 - (17*B*a^2*b)/3) - x^(1/2)*((19*B*a^3)/8 - (5*A*a^2*b)/8) + x^(5/
2)*((11*A*b^3)/8 - (29*B*a*b^2)/8))/(a^3*b^4 + b^7*x^3 + 3*a^2*b^5*x + 3*a*b^6*x^2) + (5*atan((b^(1/2)*x^(1/2)
)/a^(1/2))*(A*b - 7*B*a))/(8*a^(1/2)*b^(9/2))

________________________________________________________________________________________

sympy [A]  time = 151.18, size = 2649, normalized size = 17.31

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*(-2*A/sqrt(x) + 2*B*sqrt(x)), Eq(a, 0) & Eq(b, 0)), ((-2*A/sqrt(x) + 2*B*sqrt(x))/b**4, Eq(a, 0
)), ((2*A*x**(7/2)/7 + 2*B*x**(9/2)/9)/a**4, Eq(b, 0)), (-30*I*A*a**(5/2)*b**2*sqrt(x)*sqrt(1/b)/(48*I*a**(7/2
)*b**5*sqrt(1/b) + 144*I*a**(5/2)*b**6*x*sqrt(1/b) + 144*I*a**(3/2)*b**7*x**2*sqrt(1/b) + 48*I*sqrt(a)*b**8*x*
*3*sqrt(1/b)) - 80*I*A*a**(3/2)*b**3*x**(3/2)*sqrt(1/b)/(48*I*a**(7/2)*b**5*sqrt(1/b) + 144*I*a**(5/2)*b**6*x*
sqrt(1/b) + 144*I*a**(3/2)*b**7*x**2*sqrt(1/b) + 48*I*sqrt(a)*b**8*x**3*sqrt(1/b)) - 66*I*A*sqrt(a)*b**4*x**(5
/2)*sqrt(1/b)/(48*I*a**(7/2)*b**5*sqrt(1/b) + 144*I*a**(5/2)*b**6*x*sqrt(1/b) + 144*I*a**(3/2)*b**7*x**2*sqrt(
1/b) + 48*I*sqrt(a)*b**8*x**3*sqrt(1/b)) + 15*A*a**3*b*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(7/2)*b**5
*sqrt(1/b) + 144*I*a**(5/2)*b**6*x*sqrt(1/b) + 144*I*a**(3/2)*b**7*x**2*sqrt(1/b) + 48*I*sqrt(a)*b**8*x**3*sqr
t(1/b)) - 15*A*a**3*b*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(7/2)*b**5*sqrt(1/b) + 144*I*a**(5/2)*b**6*x
*sqrt(1/b) + 144*I*a**(3/2)*b**7*x**2*sqrt(1/b) + 48*I*sqrt(a)*b**8*x**3*sqrt(1/b)) + 45*A*a**2*b**2*x*log(-I*
sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(7/2)*b**5*sqrt(1/b) + 144*I*a**(5/2)*b**6*x*sqrt(1/b) + 144*I*a**(3/2)*
b**7*x**2*sqrt(1/b) + 48*I*sqrt(a)*b**8*x**3*sqrt(1/b)) - 45*A*a**2*b**2*x*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/
(48*I*a**(7/2)*b**5*sqrt(1/b) + 144*I*a**(5/2)*b**6*x*sqrt(1/b) + 144*I*a**(3/2)*b**7*x**2*sqrt(1/b) + 48*I*sq
rt(a)*b**8*x**3*sqrt(1/b)) + 45*A*a*b**3*x**2*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(7/2)*b**5*sqrt(1/b
) + 144*I*a**(5/2)*b**6*x*sqrt(1/b) + 144*I*a**(3/2)*b**7*x**2*sqrt(1/b) + 48*I*sqrt(a)*b**8*x**3*sqrt(1/b)) -
45*A*a*b**3*x**2*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(7/2)*b**5*sqrt(1/b) + 144*I*a**(5/2)*b**6*x*sqr
t(1/b) + 144*I*a**(3/2)*b**7*x**2*sqrt(1/b) + 48*I*sqrt(a)*b**8*x**3*sqrt(1/b)) + 15*A*b**4*x**3*log(-I*sqrt(a
)*sqrt(1/b) + sqrt(x))/(48*I*a**(7/2)*b**5*sqrt(1/b) + 144*I*a**(5/2)*b**6*x*sqrt(1/b) + 144*I*a**(3/2)*b**7*x
**2*sqrt(1/b) + 48*I*sqrt(a)*b**8*x**3*sqrt(1/b)) - 15*A*b**4*x**3*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a*
*(7/2)*b**5*sqrt(1/b) + 144*I*a**(5/2)*b**6*x*sqrt(1/b) + 144*I*a**(3/2)*b**7*x**2*sqrt(1/b) + 48*I*sqrt(a)*b*
*8*x**3*sqrt(1/b)) + 210*I*B*a**(7/2)*b*sqrt(x)*sqrt(1/b)/(48*I*a**(7/2)*b**5*sqrt(1/b) + 144*I*a**(5/2)*b**6*
x*sqrt(1/b) + 144*I*a**(3/2)*b**7*x**2*sqrt(1/b) + 48*I*sqrt(a)*b**8*x**3*sqrt(1/b)) + 560*I*B*a**(5/2)*b**2*x
**(3/2)*sqrt(1/b)/(48*I*a**(7/2)*b**5*sqrt(1/b) + 144*I*a**(5/2)*b**6*x*sqrt(1/b) + 144*I*a**(3/2)*b**7*x**2*s
qrt(1/b) + 48*I*sqrt(a)*b**8*x**3*sqrt(1/b)) + 462*I*B*a**(3/2)*b**3*x**(5/2)*sqrt(1/b)/(48*I*a**(7/2)*b**5*sq
rt(1/b) + 144*I*a**(5/2)*b**6*x*sqrt(1/b) + 144*I*a**(3/2)*b**7*x**2*sqrt(1/b) + 48*I*sqrt(a)*b**8*x**3*sqrt(1
/b)) + 96*I*B*sqrt(a)*b**4*x**(7/2)*sqrt(1/b)/(48*I*a**(7/2)*b**5*sqrt(1/b) + 144*I*a**(5/2)*b**6*x*sqrt(1/b)
+ 144*I*a**(3/2)*b**7*x**2*sqrt(1/b) + 48*I*sqrt(a)*b**8*x**3*sqrt(1/b)) - 105*B*a**4*log(-I*sqrt(a)*sqrt(1/b)
+ sqrt(x))/(48*I*a**(7/2)*b**5*sqrt(1/b) + 144*I*a**(5/2)*b**6*x*sqrt(1/b) + 144*I*a**(3/2)*b**7*x**2*sqrt(1/
b) + 48*I*sqrt(a)*b**8*x**3*sqrt(1/b)) + 105*B*a**4*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(7/2)*b**5*sqr
t(1/b) + 144*I*a**(5/2)*b**6*x*sqrt(1/b) + 144*I*a**(3/2)*b**7*x**2*sqrt(1/b) + 48*I*sqrt(a)*b**8*x**3*sqrt(1/
b)) - 315*B*a**3*b*x*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(7/2)*b**5*sqrt(1/b) + 144*I*a**(5/2)*b**6*x
*sqrt(1/b) + 144*I*a**(3/2)*b**7*x**2*sqrt(1/b) + 48*I*sqrt(a)*b**8*x**3*sqrt(1/b)) + 315*B*a**3*b*x*log(I*sqr
t(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(7/2)*b**5*sqrt(1/b) + 144*I*a**(5/2)*b**6*x*sqrt(1/b) + 144*I*a**(3/2)*b**
7*x**2*sqrt(1/b) + 48*I*sqrt(a)*b**8*x**3*sqrt(1/b)) - 315*B*a**2*b**2*x**2*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x)
)/(48*I*a**(7/2)*b**5*sqrt(1/b) + 144*I*a**(5/2)*b**6*x*sqrt(1/b) + 144*I*a**(3/2)*b**7*x**2*sqrt(1/b) + 48*I*
sqrt(a)*b**8*x**3*sqrt(1/b)) + 315*B*a**2*b**2*x**2*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(7/2)*b**5*sqr
t(1/b) + 144*I*a**(5/2)*b**6*x*sqrt(1/b) + 144*I*a**(3/2)*b**7*x**2*sqrt(1/b) + 48*I*sqrt(a)*b**8*x**3*sqrt(1/
b)) - 105*B*a*b**3*x**3*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(7/2)*b**5*sqrt(1/b) + 144*I*a**(5/2)*b**
6*x*sqrt(1/b) + 144*I*a**(3/2)*b**7*x**2*sqrt(1/b) + 48*I*sqrt(a)*b**8*x**3*sqrt(1/b)) + 105*B*a*b**3*x**3*log
(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(7/2)*b**5*sqrt(1/b) + 144*I*a**(5/2)*b**6*x*sqrt(1/b) + 144*I*a**(3/
2)*b**7*x**2*sqrt(1/b) + 48*I*sqrt(a)*b**8*x**3*sqrt(1/b)), True))

________________________________________________________________________________________