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

Optimal. Leaf size=157 \[ -\frac {5 (7 A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{8 a^{9/2} \sqrt {b}}-\frac {5 (7 A b-a B)}{8 a^4 b \sqrt {x}}+\frac {5 (7 A b-a B)}{24 a^3 b \sqrt {x} (a+b x)}+\frac {7 A b-a B}{12 a^2 b \sqrt {x} (a+b x)^2}+\frac {A b-a B}{3 a b \sqrt {x} (a+b x)^3} \]

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \begin {gather*} -\frac {5 (7 A b-a B)}{8 a^4 b \sqrt {x}}+\frac {5 (7 A b-a B)}{24 a^3 b \sqrt {x} (a+b x)}+\frac {7 A b-a B}{12 a^2 b \sqrt {x} (a+b x)^2}-\frac {5 (7 A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{8 a^{9/2} \sqrt {b}}+\frac {A b-a B}{3 a b \sqrt {x} (a+b x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

________________________________________________________________________________________

Mathematica [C]  time = 0.04, size = 59, normalized size = 0.38 \begin {gather*} \frac {\frac {a^3 (A b-a B)}{(a+b x)^3}+(a B-7 A b) \, _2F_1\left (-\frac {1}{2},3;\frac {1}{2};-\frac {b x}{a}\right )}{3 a^4 b \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a^3*(A*b - a*B))/(a + b*x)^3 + (-7*A*b + a*B)*Hypergeometric2F1[-1/2, 3, 1/2, -((b*x)/a)])/(3*a^4*b*Sqrt[x])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.21, size = 120, normalized size = 0.76 \begin {gather*} \frac {5 (a B-7 A b) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{8 a^{9/2} \sqrt {b}}+\frac {-48 a^3 A+33 a^3 B x-231 a^2 A b x+40 a^2 b B x^2-280 a A b^2 x^2+15 a b^2 B x^3-105 A b^3 x^3}{24 a^4 \sqrt {x} (a+b x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.45, size = 445, normalized size = 2.83 \begin {gather*} \left [\frac {15 \, {\left ({\left (B a b^{3} - 7 \, A b^{4}\right )} x^{4} + 3 \, {\left (B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{3} + 3 \, {\left (B a^{3} b - 7 \, A a^{2} b^{2}\right )} x^{2} + {\left (B a^{4} - 7 \, A a^{3} b\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 \, A a^{4} b - 15 \, {\left (B a^{2} b^{3} - 7 \, A a b^{4}\right )} x^{3} - 40 \, {\left (B a^{3} b^{2} - 7 \, A a^{2} b^{3}\right )} x^{2} - 33 \, {\left (B a^{4} b - 7 \, A a^{3} b^{2}\right )} x\right )} \sqrt {x}}{48 \, {\left (a^{5} b^{4} x^{4} + 3 \, a^{6} b^{3} x^{3} + 3 \, a^{7} b^{2} x^{2} + a^{8} b x\right )}}, -\frac {15 \, {\left ({\left (B a b^{3} - 7 \, A b^{4}\right )} x^{4} + 3 \, {\left (B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{3} + 3 \, {\left (B a^{3} b - 7 \, A a^{2} b^{2}\right )} x^{2} + {\left (B a^{4} - 7 \, A a^{3} b\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) + {\left (48 \, A a^{4} b - 15 \, {\left (B a^{2} b^{3} - 7 \, A a b^{4}\right )} x^{3} - 40 \, {\left (B a^{3} b^{2} - 7 \, A a^{2} b^{3}\right )} x^{2} - 33 \, {\left (B a^{4} b - 7 \, A a^{3} b^{2}\right )} x\right )} \sqrt {x}}{24 \, {\left (a^{5} b^{4} x^{4} + 3 \, a^{6} b^{3} x^{3} + 3 \, a^{7} b^{2} x^{2} + a^{8} b x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.16, size = 110, normalized size = 0.70 \begin {gather*} \frac {5 \, {\left (B a - 7 \, A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{4}} - \frac {2 \, A}{a^{4} \sqrt {x}} + \frac {15 \, B a b^{2} x^{\frac {5}{2}} - 57 \, A b^{3} x^{\frac {5}{2}} + 40 \, B a^{2} b x^{\frac {3}{2}} - 136 \, A a b^{2} x^{\frac {3}{2}} + 33 \, B a^{3} \sqrt {x} - 87 \, A a^{2} b \sqrt {x}}{24 \, {\left (b x + a\right )}^{3} a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/8*(B*a - 7*A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^4) - 2*A/(a^4*sqrt(x)) + 1/24*(15*B*a*b^2*x^(5/2) -
57*A*b^3*x^(5/2) + 40*B*a^2*b*x^(3/2) - 136*A*a*b^2*x^(3/2) + 33*B*a^3*sqrt(x) - 87*A*a^2*b*sqrt(x))/((b*x +
a)^3*a^4)

________________________________________________________________________________________

maple [A]  time = 0.09, size = 163, normalized size = 1.04 \begin {gather*} -\frac {19 A \,b^{3} x^{\frac {5}{2}}}{8 \left (b x +a \right )^{3} a^{4}}+\frac {5 B \,b^{2} x^{\frac {5}{2}}}{8 \left (b x +a \right )^{3} a^{3}}-\frac {17 A \,b^{2} x^{\frac {3}{2}}}{3 \left (b x +a \right )^{3} a^{3}}+\frac {5 B b \,x^{\frac {3}{2}}}{3 \left (b x +a \right )^{3} a^{2}}-\frac {29 A b \sqrt {x}}{8 \left (b x +a \right )^{3} a^{2}}+\frac {11 B \sqrt {x}}{8 \left (b x +a \right )^{3} a}-\frac {35 A b \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, a^{4}}+\frac {5 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, a^{3}}-\frac {2 A}{a^{4} \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 1.15, size = 132, normalized size = 0.84 \begin {gather*} -\frac {48 \, A a^{3} - 15 \, {\left (B a b^{2} - 7 \, A b^{3}\right )} x^{3} - 40 \, {\left (B a^{2} b - 7 \, A a b^{2}\right )} x^{2} - 33 \, {\left (B a^{3} - 7 \, A a^{2} b\right )} x}{24 \, {\left (a^{4} b^{3} x^{\frac {7}{2}} + 3 \, a^{5} b^{2} x^{\frac {5}{2}} + 3 \, a^{6} b x^{\frac {3}{2}} + a^{7} \sqrt {x}\right )}} + \frac {5 \, {\left (B a - 7 \, A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.28, size = 147, normalized size = 0.94 \begin {gather*} -\frac {\frac {2\,A}{a}+\frac {11\,x\,\left (7\,A\,b-B\,a\right )}{8\,a^2}+\frac {5\,b^2\,x^3\,\left (7\,A\,b-B\,a\right )}{8\,a^4}+\frac {5\,b\,x^2\,\left (7\,A\,b-B\,a\right )}{3\,a^3}}{a^3\,\sqrt {x}+b^3\,x^{7/2}+3\,a^2\,b\,x^{3/2}+3\,a\,b^2\,x^{5/2}}-\frac {5\,\mathrm {atan}\left (\frac {5\,\sqrt {b}\,\sqrt {x}\,\left (7\,A\,b-B\,a\right )}{\sqrt {a}\,\left (35\,A\,b-5\,B\,a\right )}\right )\,\left (7\,A\,b-B\,a\right )}{8\,a^{9/2}\,\sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((2*A)/a + (11*x*(7*A*b - B*a))/(8*a^2) + (5*b^2*x^3*(7*A*b - B*a))/(8*a^4) + (5*b*x^2*(7*A*b - B*a))/(3*a^3
))/(a^3*x^(1/2) + b^3*x^(7/2) + 3*a^2*b*x^(3/2) + 3*a*b^2*x^(5/2)) - (5*atan((5*b^(1/2)*x^(1/2)*(7*A*b - B*a))
/(a^(1/2)*(35*A*b - 5*B*a)))*(7*A*b - B*a))/(8*a^(9/2)*b^(1/2))

________________________________________________________________________________________

sympy [A]  time = 144.73, size = 2917, normalized size = 18.58

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*(-2*A/(9*x**(9/2)) - 2*B/(7*x**(7/2))), Eq(a, 0) & Eq(b, 0)), ((-2*A/sqrt(x) + 2*B*sqrt(x))/a**
4, Eq(b, 0)), ((-2*A/(9*x**(9/2)) - 2*B/(7*x**(7/2)))/b**4, Eq(a, 0)), (-96*I*A*a**(7/2)*b*sqrt(1/b)/(48*I*a**
(15/2)*b*sqrt(x)*sqrt(1/b) + 144*I*a**(13/2)*b**2*x**(3/2)*sqrt(1/b) + 144*I*a**(11/2)*b**3*x**(5/2)*sqrt(1/b)
+ 48*I*a**(9/2)*b**4*x**(7/2)*sqrt(1/b)) - 462*I*A*a**(5/2)*b**2*x*sqrt(1/b)/(48*I*a**(15/2)*b*sqrt(x)*sqrt(1
/b) + 144*I*a**(13/2)*b**2*x**(3/2)*sqrt(1/b) + 144*I*a**(11/2)*b**3*x**(5/2)*sqrt(1/b) + 48*I*a**(9/2)*b**4*x
**(7/2)*sqrt(1/b)) - 560*I*A*a**(3/2)*b**3*x**2*sqrt(1/b)/(48*I*a**(15/2)*b*sqrt(x)*sqrt(1/b) + 144*I*a**(13/2
)*b**2*x**(3/2)*sqrt(1/b) + 144*I*a**(11/2)*b**3*x**(5/2)*sqrt(1/b) + 48*I*a**(9/2)*b**4*x**(7/2)*sqrt(1/b)) -
210*I*A*sqrt(a)*b**4*x**3*sqrt(1/b)/(48*I*a**(15/2)*b*sqrt(x)*sqrt(1/b) + 144*I*a**(13/2)*b**2*x**(3/2)*sqrt(
1/b) + 144*I*a**(11/2)*b**3*x**(5/2)*sqrt(1/b) + 48*I*a**(9/2)*b**4*x**(7/2)*sqrt(1/b)) - 105*A*a**3*b*sqrt(x)
*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(15/2)*b*sqrt(x)*sqrt(1/b) + 144*I*a**(13/2)*b**2*x**(3/2)*sqrt(
1/b) + 144*I*a**(11/2)*b**3*x**(5/2)*sqrt(1/b) + 48*I*a**(9/2)*b**4*x**(7/2)*sqrt(1/b)) + 105*A*a**3*b*sqrt(x)
*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(15/2)*b*sqrt(x)*sqrt(1/b) + 144*I*a**(13/2)*b**2*x**(3/2)*sqrt(1
/b) + 144*I*a**(11/2)*b**3*x**(5/2)*sqrt(1/b) + 48*I*a**(9/2)*b**4*x**(7/2)*sqrt(1/b)) - 315*A*a**2*b**2*x**(3
/2)*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(15/2)*b*sqrt(x)*sqrt(1/b) + 144*I*a**(13/2)*b**2*x**(3/2)*sq
rt(1/b) + 144*I*a**(11/2)*b**3*x**(5/2)*sqrt(1/b) + 48*I*a**(9/2)*b**4*x**(7/2)*sqrt(1/b)) + 315*A*a**2*b**2*x
**(3/2)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(15/2)*b*sqrt(x)*sqrt(1/b) + 144*I*a**(13/2)*b**2*x**(3/2)
*sqrt(1/b) + 144*I*a**(11/2)*b**3*x**(5/2)*sqrt(1/b) + 48*I*a**(9/2)*b**4*x**(7/2)*sqrt(1/b)) - 315*A*a*b**3*x
**(5/2)*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(15/2)*b*sqrt(x)*sqrt(1/b) + 144*I*a**(13/2)*b**2*x**(3/2
)*sqrt(1/b) + 144*I*a**(11/2)*b**3*x**(5/2)*sqrt(1/b) + 48*I*a**(9/2)*b**4*x**(7/2)*sqrt(1/b)) + 315*A*a*b**3*
x**(5/2)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(15/2)*b*sqrt(x)*sqrt(1/b) + 144*I*a**(13/2)*b**2*x**(3/2
)*sqrt(1/b) + 144*I*a**(11/2)*b**3*x**(5/2)*sqrt(1/b) + 48*I*a**(9/2)*b**4*x**(7/2)*sqrt(1/b)) - 105*A*b**4*x*
*(7/2)*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(15/2)*b*sqrt(x)*sqrt(1/b) + 144*I*a**(13/2)*b**2*x**(3/2)
*sqrt(1/b) + 144*I*a**(11/2)*b**3*x**(5/2)*sqrt(1/b) + 48*I*a**(9/2)*b**4*x**(7/2)*sqrt(1/b)) + 105*A*b**4*x**
(7/2)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(15/2)*b*sqrt(x)*sqrt(1/b) + 144*I*a**(13/2)*b**2*x**(3/2)*s
qrt(1/b) + 144*I*a**(11/2)*b**3*x**(5/2)*sqrt(1/b) + 48*I*a**(9/2)*b**4*x**(7/2)*sqrt(1/b)) + 66*I*B*a**(7/2)*
b*x*sqrt(1/b)/(48*I*a**(15/2)*b*sqrt(x)*sqrt(1/b) + 144*I*a**(13/2)*b**2*x**(3/2)*sqrt(1/b) + 144*I*a**(11/2)*
b**3*x**(5/2)*sqrt(1/b) + 48*I*a**(9/2)*b**4*x**(7/2)*sqrt(1/b)) + 80*I*B*a**(5/2)*b**2*x**2*sqrt(1/b)/(48*I*a
**(15/2)*b*sqrt(x)*sqrt(1/b) + 144*I*a**(13/2)*b**2*x**(3/2)*sqrt(1/b) + 144*I*a**(11/2)*b**3*x**(5/2)*sqrt(1/
b) + 48*I*a**(9/2)*b**4*x**(7/2)*sqrt(1/b)) + 30*I*B*a**(3/2)*b**3*x**3*sqrt(1/b)/(48*I*a**(15/2)*b*sqrt(x)*sq
rt(1/b) + 144*I*a**(13/2)*b**2*x**(3/2)*sqrt(1/b) + 144*I*a**(11/2)*b**3*x**(5/2)*sqrt(1/b) + 48*I*a**(9/2)*b*
*4*x**(7/2)*sqrt(1/b)) + 15*B*a**4*sqrt(x)*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(15/2)*b*sqrt(x)*sqrt(
1/b) + 144*I*a**(13/2)*b**2*x**(3/2)*sqrt(1/b) + 144*I*a**(11/2)*b**3*x**(5/2)*sqrt(1/b) + 48*I*a**(9/2)*b**4*
x**(7/2)*sqrt(1/b)) - 15*B*a**4*sqrt(x)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(15/2)*b*sqrt(x)*sqrt(1/b)
+ 144*I*a**(13/2)*b**2*x**(3/2)*sqrt(1/b) + 144*I*a**(11/2)*b**3*x**(5/2)*sqrt(1/b) + 48*I*a**(9/2)*b**4*x**(
7/2)*sqrt(1/b)) + 45*B*a**3*b*x**(3/2)*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(15/2)*b*sqrt(x)*sqrt(1/b)
+ 144*I*a**(13/2)*b**2*x**(3/2)*sqrt(1/b) + 144*I*a**(11/2)*b**3*x**(5/2)*sqrt(1/b) + 48*I*a**(9/2)*b**4*x**(
7/2)*sqrt(1/b)) - 45*B*a**3*b*x**(3/2)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(15/2)*b*sqrt(x)*sqrt(1/b)
+ 144*I*a**(13/2)*b**2*x**(3/2)*sqrt(1/b) + 144*I*a**(11/2)*b**3*x**(5/2)*sqrt(1/b) + 48*I*a**(9/2)*b**4*x**(7
/2)*sqrt(1/b)) + 45*B*a**2*b**2*x**(5/2)*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(15/2)*b*sqrt(x)*sqrt(1/
b) + 144*I*a**(13/2)*b**2*x**(3/2)*sqrt(1/b) + 144*I*a**(11/2)*b**3*x**(5/2)*sqrt(1/b) + 48*I*a**(9/2)*b**4*x*
*(7/2)*sqrt(1/b)) - 45*B*a**2*b**2*x**(5/2)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(15/2)*b*sqrt(x)*sqrt(
1/b) + 144*I*a**(13/2)*b**2*x**(3/2)*sqrt(1/b) + 144*I*a**(11/2)*b**3*x**(5/2)*sqrt(1/b) + 48*I*a**(9/2)*b**4*
x**(7/2)*sqrt(1/b)) + 15*B*a*b**3*x**(7/2)*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(15/2)*b*sqrt(x)*sqrt(
1/b) + 144*I*a**(13/2)*b**2*x**(3/2)*sqrt(1/b) + 144*I*a**(11/2)*b**3*x**(5/2)*sqrt(1/b) + 48*I*a**(9/2)*b**4*
x**(7/2)*sqrt(1/b)) - 15*B*a*b**3*x**(7/2)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(15/2)*b*sqrt(x)*sqrt(1
/b) + 144*I*a**(13/2)*b**2*x**(3/2)*sqrt(1/b) + 144*I*a**(11/2)*b**3*x**(5/2)*sqrt(1/b) + 48*I*a**(9/2)*b**4*x
**(7/2)*sqrt(1/b)), True))

________________________________________________________________________________________