3.4.55 \(\int \frac {A+B x}{x^{5/2} (a+b x)^3} \, dx\)

Optimal. Leaf size=147 \[ \frac {5 \sqrt {b} (7 A b-3 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{9/2}}+\frac {5 (7 A b-3 a B)}{4 a^4 \sqrt {x}}-\frac {5 (7 A b-3 a B)}{12 a^3 b x^{3/2}}+\frac {7 A b-3 a B}{4 a^2 b x^{3/2} (a+b x)}+\frac {A b-a B}{2 a b x^{3/2} (a+b x)^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {78, 51, 63, 205} \begin {gather*} \frac {7 A b-3 a B}{4 a^2 b x^{3/2} (a+b x)}-\frac {5 (7 A b-3 a B)}{12 a^3 b x^{3/2}}+\frac {5 (7 A b-3 a B)}{4 a^4 \sqrt {x}}+\frac {5 \sqrt {b} (7 A b-3 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{9/2}}+\frac {A b-a B}{2 a b x^{3/2} (a+b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^(5/2)*(a + b*x)^3),x]

[Out]

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

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

________________________________________________________________________________________

Mathematica [C]  time = 0.03, size = 61, normalized size = 0.41 \begin {gather*} \frac {\frac {3 a^2 (A b-a B)}{(a+b x)^2}+(3 a B-7 A b) \, _2F_1\left (-\frac {3}{2},2;-\frac {1}{2};-\frac {b x}{a}\right )}{6 a^3 b x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^(5/2)*(a + b*x)^3),x]

[Out]

((3*a^2*(A*b - a*B))/(a + b*x)^2 + (-7*A*b + 3*a*B)*Hypergeometric2F1[-3/2, 2, -1/2, -((b*x)/a)])/(6*a^3*b*x^(
3/2))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.18, size = 125, normalized size = 0.85 \begin {gather*} \frac {-8 a^3 A-24 a^3 B x+56 a^2 A b x-75 a^2 b B x^2+175 a A b^2 x^2-45 a b^2 B x^3+105 A b^3 x^3}{12 a^4 x^{3/2} (a+b x)^2}-\frac {5 \left (3 a \sqrt {b} B-7 A b^{3/2}\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x)/(x^(5/2)*(a + b*x)^3),x]

[Out]

(-8*a^3*A + 56*a^2*A*b*x - 24*a^3*B*x + 175*a*A*b^2*x^2 - 75*a^2*b*B*x^2 + 105*A*b^3*x^3 - 45*a*b^2*B*x^3)/(12
*a^4*x^(3/2)*(a + b*x)^2) - (5*(-7*A*b^(3/2) + 3*a*Sqrt[b]*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*a^(9/2))

________________________________________________________________________________________

fricas [A]  time = 1.07, size = 380, normalized size = 2.59 \begin {gather*} \left [-\frac {15 \, {\left ({\left (3 \, B a b^{2} - 7 \, A b^{3}\right )} x^{4} + 2 \, {\left (3 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{3} + {\left (3 \, B a^{3} - 7 \, A a^{2} b\right )} x^{2}\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 (8 \, A a^{3} + 15 \, {\left (3 \, B a b^{2} - 7 \, A b^{3}\right )} x^{3} + 25 \, {\left (3 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{2} + 8 \, {\left (3 \, B a^{3} - 7 \, A a^{2} b\right )} x\right )} \sqrt {x}}{24 \, {\left (a^{4} b^{2} x^{4} + 2 \, a^{5} b x^{3} + a^{6} x^{2}\right )}}, \frac {15 \, {\left ({\left (3 \, B a b^{2} - 7 \, A b^{3}\right )} x^{4} + 2 \, {\left (3 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{3} + {\left (3 \, B a^{3} - 7 \, A a^{2} b\right )} x^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - {\left (8 \, A a^{3} + 15 \, {\left (3 \, B a b^{2} - 7 \, A b^{3}\right )} x^{3} + 25 \, {\left (3 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{2} + 8 \, {\left (3 \, B a^{3} - 7 \, A a^{2} b\right )} x\right )} \sqrt {x}}{12 \, {\left (a^{4} b^{2} x^{4} + 2 \, a^{5} b x^{3} + a^{6} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(5/2)/(b*x+a)^3,x, algorithm="fricas")

[Out]

[-1/24*(15*((3*B*a*b^2 - 7*A*b^3)*x^4 + 2*(3*B*a^2*b - 7*A*a*b^2)*x^3 + (3*B*a^3 - 7*A*a^2*b)*x^2)*sqrt(-b/a)*
log((b*x + 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) + 2*(8*A*a^3 + 15*(3*B*a*b^2 - 7*A*b^3)*x^3 + 25*(3*B*a^2*b
- 7*A*a*b^2)*x^2 + 8*(3*B*a^3 - 7*A*a^2*b)*x)*sqrt(x))/(a^4*b^2*x^4 + 2*a^5*b*x^3 + a^6*x^2), 1/12*(15*((3*B*a
*b^2 - 7*A*b^3)*x^4 + 2*(3*B*a^2*b - 7*A*a*b^2)*x^3 + (3*B*a^3 - 7*A*a^2*b)*x^2)*sqrt(b/a)*arctan(a*sqrt(b/a)/
(b*sqrt(x))) - (8*A*a^3 + 15*(3*B*a*b^2 - 7*A*b^3)*x^3 + 25*(3*B*a^2*b - 7*A*a*b^2)*x^2 + 8*(3*B*a^3 - 7*A*a^2
*b)*x)*sqrt(x))/(a^4*b^2*x^4 + 2*a^5*b*x^3 + a^6*x^2)]

________________________________________________________________________________________

giac [A]  time = 1.21, size = 108, normalized size = 0.73 \begin {gather*} -\frac {5 \, {\left (3 \, B a b - 7 \, A b^{2}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{4}} - \frac {2 \, {\left (3 \, B a x - 9 \, A b x + A a\right )}}{3 \, a^{4} x^{\frac {3}{2}}} - \frac {7 \, B a b^{2} x^{\frac {3}{2}} - 11 \, A b^{3} x^{\frac {3}{2}} + 9 \, B a^{2} b \sqrt {x} - 13 \, A a b^{2} \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(5/2)/(b*x+a)^3,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.02, size = 152, normalized size = 1.03 \begin {gather*} \frac {11 A \,b^{3} x^{\frac {3}{2}}}{4 \left (b x +a \right )^{2} a^{4}}-\frac {7 B \,b^{2} x^{\frac {3}{2}}}{4 \left (b x +a \right )^{2} a^{3}}+\frac {13 A \,b^{2} \sqrt {x}}{4 \left (b x +a \right )^{2} a^{3}}-\frac {9 B b \sqrt {x}}{4 \left (b x +a \right )^{2} a^{2}}+\frac {35 A \,b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}\, a^{4}}-\frac {15 B b \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}\, a^{3}}+\frac {6 A b}{a^{4} \sqrt {x}}-\frac {2 B}{a^{3} \sqrt {x}}-\frac {2 A}{3 a^{3} x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^(5/2)/(b*x+a)^3,x)

[Out]

11/4/a^4*b^3/(b*x+a)^2*x^(3/2)*A-7/4/a^3*b^2/(b*x+a)^2*x^(3/2)*B+13/4/a^3*b^2/(b*x+a)^2*A*x^(1/2)-9/4/a^2*b/(b
*x+a)^2*B*x^(1/2)+35/4/a^4*b^2/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x^(1/2))*A-15/4/a^3*b/(a*b)^(1/2)*arctan(1/(
a*b)^(1/2)*b*x^(1/2))*B-2/3*A/a^3/x^(3/2)+6/a^4/x^(1/2)*A*b-2/a^3/x^(1/2)*B

________________________________________________________________________________________

maxima [A]  time = 2.01, size = 128, normalized size = 0.87 \begin {gather*} -\frac {8 \, A a^{3} + 15 \, {\left (3 \, B a b^{2} - 7 \, A b^{3}\right )} x^{3} + 25 \, {\left (3 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{2} + 8 \, {\left (3 \, B a^{3} - 7 \, A a^{2} b\right )} x}{12 \, {\left (a^{4} b^{2} x^{\frac {7}{2}} + 2 \, a^{5} b x^{\frac {5}{2}} + a^{6} x^{\frac {3}{2}}\right )}} - \frac {5 \, {\left (3 \, B a b - 7 \, A b^{2}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(5/2)/(b*x+a)^3,x, algorithm="maxima")

[Out]

-1/12*(8*A*a^3 + 15*(3*B*a*b^2 - 7*A*b^3)*x^3 + 25*(3*B*a^2*b - 7*A*a*b^2)*x^2 + 8*(3*B*a^3 - 7*A*a^2*b)*x)/(a
^4*b^2*x^(7/2) + 2*a^5*b*x^(5/2) + a^6*x^(3/2)) - 5/4*(3*B*a*b - 7*A*b^2)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*
b)*a^4)

________________________________________________________________________________________

mupad [B]  time = 0.45, size = 114, normalized size = 0.78 \begin {gather*} \frac {\frac {2\,x\,\left (7\,A\,b-3\,B\,a\right )}{3\,a^2}-\frac {2\,A}{3\,a}+\frac {5\,b^2\,x^3\,\left (7\,A\,b-3\,B\,a\right )}{4\,a^4}+\frac {25\,b\,x^2\,\left (7\,A\,b-3\,B\,a\right )}{12\,a^3}}{a^2\,x^{3/2}+b^2\,x^{7/2}+2\,a\,b\,x^{5/2}}+\frac {5\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (7\,A\,b-3\,B\,a\right )}{4\,a^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^(5/2)*(a + b*x)^3),x)

[Out]

((2*x*(7*A*b - 3*B*a))/(3*a^2) - (2*A)/(3*a) + (5*b^2*x^3*(7*A*b - 3*B*a))/(4*a^4) + (25*b*x^2*(7*A*b - 3*B*a)
)/(12*a^3))/(a^2*x^(3/2) + b^2*x^(7/2) + 2*a*b*x^(5/2)) + (5*b^(1/2)*atan((b^(1/2)*x^(1/2))/a^(1/2))*(7*A*b -
3*B*a))/(4*a^(9/2))

________________________________________________________________________________________

sympy [A]  time = 158.43, size = 1880, normalized size = 12.79

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**(5/2)/(b*x+a)**3,x)

[Out]

Piecewise((zoo*(-2*A/(9*x**(9/2)) - 2*B/(7*x**(7/2))), Eq(a, 0) & Eq(b, 0)), ((-2*A/(9*x**(9/2)) - 2*B/(7*x**(
7/2)))/b**3, Eq(a, 0)), ((-2*A/(3*x**(3/2)) - 2*B/sqrt(x))/a**3, Eq(b, 0)), (-16*I*A*a**(7/2)*sqrt(1/b)/(24*I*
a**(13/2)*x**(3/2)*sqrt(1/b) + 48*I*a**(11/2)*b*x**(5/2)*sqrt(1/b) + 24*I*a**(9/2)*b**2*x**(7/2)*sqrt(1/b)) +
112*I*A*a**(5/2)*b*x*sqrt(1/b)/(24*I*a**(13/2)*x**(3/2)*sqrt(1/b) + 48*I*a**(11/2)*b*x**(5/2)*sqrt(1/b) + 24*I
*a**(9/2)*b**2*x**(7/2)*sqrt(1/b)) + 350*I*A*a**(3/2)*b**2*x**2*sqrt(1/b)/(24*I*a**(13/2)*x**(3/2)*sqrt(1/b) +
48*I*a**(11/2)*b*x**(5/2)*sqrt(1/b) + 24*I*a**(9/2)*b**2*x**(7/2)*sqrt(1/b)) + 210*I*A*sqrt(a)*b**3*x**3*sqrt
(1/b)/(24*I*a**(13/2)*x**(3/2)*sqrt(1/b) + 48*I*a**(11/2)*b*x**(5/2)*sqrt(1/b) + 24*I*a**(9/2)*b**2*x**(7/2)*s
qrt(1/b)) + 105*A*a**2*b*x**(3/2)*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(24*I*a**(13/2)*x**(3/2)*sqrt(1/b) + 48*
I*a**(11/2)*b*x**(5/2)*sqrt(1/b) + 24*I*a**(9/2)*b**2*x**(7/2)*sqrt(1/b)) - 105*A*a**2*b*x**(3/2)*log(I*sqrt(a
)*sqrt(1/b) + sqrt(x))/(24*I*a**(13/2)*x**(3/2)*sqrt(1/b) + 48*I*a**(11/2)*b*x**(5/2)*sqrt(1/b) + 24*I*a**(9/2
)*b**2*x**(7/2)*sqrt(1/b)) + 210*A*a*b**2*x**(5/2)*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(24*I*a**(13/2)*x**(3/2
)*sqrt(1/b) + 48*I*a**(11/2)*b*x**(5/2)*sqrt(1/b) + 24*I*a**(9/2)*b**2*x**(7/2)*sqrt(1/b)) - 210*A*a*b**2*x**(
5/2)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(24*I*a**(13/2)*x**(3/2)*sqrt(1/b) + 48*I*a**(11/2)*b*x**(5/2)*sqrt(1/
b) + 24*I*a**(9/2)*b**2*x**(7/2)*sqrt(1/b)) + 105*A*b**3*x**(7/2)*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(24*I*a*
*(13/2)*x**(3/2)*sqrt(1/b) + 48*I*a**(11/2)*b*x**(5/2)*sqrt(1/b) + 24*I*a**(9/2)*b**2*x**(7/2)*sqrt(1/b)) - 10
5*A*b**3*x**(7/2)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(24*I*a**(13/2)*x**(3/2)*sqrt(1/b) + 48*I*a**(11/2)*b*x**
(5/2)*sqrt(1/b) + 24*I*a**(9/2)*b**2*x**(7/2)*sqrt(1/b)) - 48*I*B*a**(7/2)*x*sqrt(1/b)/(24*I*a**(13/2)*x**(3/2
)*sqrt(1/b) + 48*I*a**(11/2)*b*x**(5/2)*sqrt(1/b) + 24*I*a**(9/2)*b**2*x**(7/2)*sqrt(1/b)) - 150*I*B*a**(5/2)*
b*x**2*sqrt(1/b)/(24*I*a**(13/2)*x**(3/2)*sqrt(1/b) + 48*I*a**(11/2)*b*x**(5/2)*sqrt(1/b) + 24*I*a**(9/2)*b**2
*x**(7/2)*sqrt(1/b)) - 90*I*B*a**(3/2)*b**2*x**3*sqrt(1/b)/(24*I*a**(13/2)*x**(3/2)*sqrt(1/b) + 48*I*a**(11/2)
*b*x**(5/2)*sqrt(1/b) + 24*I*a**(9/2)*b**2*x**(7/2)*sqrt(1/b)) - 45*B*a**3*x**(3/2)*log(-I*sqrt(a)*sqrt(1/b) +
sqrt(x))/(24*I*a**(13/2)*x**(3/2)*sqrt(1/b) + 48*I*a**(11/2)*b*x**(5/2)*sqrt(1/b) + 24*I*a**(9/2)*b**2*x**(7/
2)*sqrt(1/b)) + 45*B*a**3*x**(3/2)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(24*I*a**(13/2)*x**(3/2)*sqrt(1/b) + 48*
I*a**(11/2)*b*x**(5/2)*sqrt(1/b) + 24*I*a**(9/2)*b**2*x**(7/2)*sqrt(1/b)) - 90*B*a**2*b*x**(5/2)*log(-I*sqrt(a
)*sqrt(1/b) + sqrt(x))/(24*I*a**(13/2)*x**(3/2)*sqrt(1/b) + 48*I*a**(11/2)*b*x**(5/2)*sqrt(1/b) + 24*I*a**(9/2
)*b**2*x**(7/2)*sqrt(1/b)) + 90*B*a**2*b*x**(5/2)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(24*I*a**(13/2)*x**(3/2)*
sqrt(1/b) + 48*I*a**(11/2)*b*x**(5/2)*sqrt(1/b) + 24*I*a**(9/2)*b**2*x**(7/2)*sqrt(1/b)) - 45*B*a*b**2*x**(7/2
)*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(24*I*a**(13/2)*x**(3/2)*sqrt(1/b) + 48*I*a**(11/2)*b*x**(5/2)*sqrt(1/b)
+ 24*I*a**(9/2)*b**2*x**(7/2)*sqrt(1/b)) + 45*B*a*b**2*x**(7/2)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(24*I*a**(
13/2)*x**(3/2)*sqrt(1/b) + 48*I*a**(11/2)*b*x**(5/2)*sqrt(1/b) + 24*I*a**(9/2)*b**2*x**(7/2)*sqrt(1/b)), True)
)

________________________________________________________________________________________