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

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

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Rubi [A]  time = 0.06, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {78, 47, 50, 63, 205} \begin {gather*} \frac {x^{5/2} (3 A b-7 a B)}{4 a b^2 (a+b x)}-\frac {5 x^{3/2} (3 A b-7 a B)}{12 a b^3}+\frac {5 \sqrt {x} (3 A b-7 a B)}{4 b^4}-\frac {5 \sqrt {a} (3 A b-7 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{9/2}}+\frac {x^{7/2} (A b-a B)}{2 a b (a+b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [C]  time = 0.03, size = 61, normalized size = 0.41 \begin {gather*} \frac {x^{7/2} \left (\frac {7 a^2 (A b-a B)}{(a+b x)^2}+(7 a B-3 A b) \, _2F_1\left (2,\frac {7}{2};\frac {9}{2};-\frac {b x}{a}\right )\right )}{14 a^3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(7/2)*((7*a^2*(A*b - a*B))/(a + b*x)^2 + (-3*A*b + 7*a*B)*Hypergeometric2F1[2, 7/2, 9/2, -((b*x)/a)]))/(14*
a^3*b)

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IntegrateAlgebraic [A]  time = 0.19, size = 122, normalized size = 0.83 \begin {gather*} \frac {5 \left (7 a^{3/2} B-3 \sqrt {a} A b\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{9/2}}+\frac {\sqrt {x} \left (-105 a^3 B+45 a^2 A b-175 a^2 b B x+75 a A b^2 x-56 a b^2 B x^2+24 A b^3 x^2+8 b^3 B x^3\right )}{12 b^4 (a+b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.12, size = 349, normalized size = 2.37 \begin {gather*} \left [-\frac {15 \, {\left (7 \, B a^{3} - 3 \, A a^{2} b + {\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} x^{2} + 2 \, {\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} x\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x - 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) - 2 \, {\left (8 \, B b^{3} x^{3} - 105 \, B a^{3} + 45 \, A a^{2} b - 8 \, {\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} x^{2} - 25 \, {\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} x\right )} \sqrt {x}}{24 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}, \frac {15 \, {\left (7 \, B a^{3} - 3 \, A a^{2} b + {\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} x^{2} + 2 \, {\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} x\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) + {\left (8 \, B b^{3} x^{3} - 105 \, B a^{3} + 45 \, A a^{2} b - 8 \, {\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} x^{2} - 25 \, {\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} x\right )} \sqrt {x}}{12 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/24*(15*(7*B*a^3 - 3*A*a^2*b + (7*B*a*b^2 - 3*A*b^3)*x^2 + 2*(7*B*a^2*b - 3*A*a*b^2)*x)*sqrt(-a/b)*log((b*x
- 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x + a)) - 2*(8*B*b^3*x^3 - 105*B*a^3 + 45*A*a^2*b - 8*(7*B*a*b^2 - 3*A*b^3)*
x^2 - 25*(7*B*a^2*b - 3*A*a*b^2)*x)*sqrt(x))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4), 1/12*(15*(7*B*a^3 - 3*A*a^2*b +
(7*B*a*b^2 - 3*A*b^3)*x^2 + 2*(7*B*a^2*b - 3*A*a*b^2)*x)*sqrt(a/b)*arctan(b*sqrt(x)*sqrt(a/b)/a) + (8*B*b^3*x^
3 - 105*B*a^3 + 45*A*a^2*b - 8*(7*B*a*b^2 - 3*A*b^3)*x^2 - 25*(7*B*a^2*b - 3*A*a*b^2)*x)*sqrt(x))/(b^6*x^2 + 2
*a*b^5*x + a^2*b^4)]

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giac [A]  time = 1.33, size = 119, normalized size = 0.81 \begin {gather*} \frac {5 \, {\left (7 \, B a^{2} - 3 \, A a b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{4}} - \frac {13 \, B a^{2} b x^{\frac {3}{2}} - 9 \, A a b^{2} x^{\frac {3}{2}} + 11 \, B a^{3} \sqrt {x} - 7 \, A a^{2} b \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} b^{4}} + \frac {2 \, {\left (B b^{6} x^{\frac {3}{2}} - 9 \, B a b^{5} \sqrt {x} + 3 \, A b^{6} \sqrt {x}\right )}}{3 \, b^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/4*(7*B*a^2 - 3*A*a*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^4) - 1/4*(13*B*a^2*b*x^(3/2) - 9*A*a*b^2*x^(3
/2) + 11*B*a^3*sqrt(x) - 7*A*a^2*b*sqrt(x))/((b*x + a)^2*b^4) + 2/3*(B*b^6*x^(3/2) - 9*B*a*b^5*sqrt(x) + 3*A*b
^6*sqrt(x))/b^9

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

[Out]

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

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maxima [A]  time = 2.01, size = 124, normalized size = 0.84 \begin {gather*} -\frac {{\left (13 \, B a^{2} b - 9 \, A a b^{2}\right )} x^{\frac {3}{2}} + {\left (11 \, B a^{3} - 7 \, A a^{2} b\right )} \sqrt {x}}{4 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} + \frac {5 \, {\left (7 \, B a^{2} - 3 \, A a b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{4}} + \frac {2 \, {\left (B b x^{\frac {3}{2}} - 3 \, {\left (3 \, B a - A b\right )} \sqrt {x}\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*x+a)^3,x, algorithm="maxima")

[Out]

-1/4*((13*B*a^2*b - 9*A*a*b^2)*x^(3/2) + (11*B*a^3 - 7*A*a^2*b)*sqrt(x))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4) + 5/4
*(7*B*a^2 - 3*A*a*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^4) + 2/3*(B*b*x^(3/2) - 3*(3*B*a - A*b)*sqrt(x))
/b^4

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mupad [B]  time = 0.42, size = 143, normalized size = 0.97 \begin {gather*} \frac {x^{3/2}\,\left (\frac {9\,A\,a\,b^2}{4}-\frac {13\,B\,a^2\,b}{4}\right )-\sqrt {x}\,\left (\frac {11\,B\,a^3}{4}-\frac {7\,A\,a^2\,b}{4}\right )}{a^2\,b^4+2\,a\,b^5\,x+b^6\,x^2}+\sqrt {x}\,\left (\frac {2\,A}{b^3}-\frac {6\,B\,a}{b^4}\right )+\frac {2\,B\,x^{3/2}}{3\,b^3}+\frac {5\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {b}\,\sqrt {x}\,\left (3\,A\,b-7\,B\,a\right )}{7\,B\,a^2-3\,A\,a\,b}\right )\,\left (3\,A\,b-7\,B\,a\right )}{4\,b^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^(3/2)*((9*A*a*b^2)/4 - (13*B*a^2*b)/4) - x^(1/2)*((11*B*a^3)/4 - (7*A*a^2*b)/4))/(a^2*b^4 + b^6*x^2 + 2*a*b
^5*x) + x^(1/2)*((2*A)/b^3 - (6*B*a)/b^4) + (2*B*x^(3/2))/(3*b^3) + (5*a^(1/2)*atan((a^(1/2)*b^(1/2)*x^(1/2)*(
3*A*b - 7*B*a))/(7*B*a^2 - 3*A*a*b))*(3*A*b - 7*B*a))/(4*b^(9/2))

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sympy [A]  time = 85.13, size = 1773, normalized size = 12.06

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*(2*A*sqrt(x) + 2*B*x**(3/2)/3), Eq(a, 0) & Eq(b, 0)), ((2*A*sqrt(x) + 2*B*x**(3/2)/3)/b**3, Eq(
a, 0)), ((2*A*x**(7/2)/7 + 2*B*x**(9/2)/9)/a**3, Eq(b, 0)), (90*I*A*a**(5/2)*b**2*sqrt(x)*sqrt(1/b)/(24*I*a**(
5/2)*b**5*sqrt(1/b) + 48*I*a**(3/2)*b**6*x*sqrt(1/b) + 24*I*sqrt(a)*b**7*x**2*sqrt(1/b)) + 150*I*A*a**(3/2)*b*
*3*x**(3/2)*sqrt(1/b)/(24*I*a**(5/2)*b**5*sqrt(1/b) + 48*I*a**(3/2)*b**6*x*sqrt(1/b) + 24*I*sqrt(a)*b**7*x**2*
sqrt(1/b)) + 48*I*A*sqrt(a)*b**4*x**(5/2)*sqrt(1/b)/(24*I*a**(5/2)*b**5*sqrt(1/b) + 48*I*a**(3/2)*b**6*x*sqrt(
1/b) + 24*I*sqrt(a)*b**7*x**2*sqrt(1/b)) - 45*A*a**3*b*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(24*I*a**(5/2)*b**5
*sqrt(1/b) + 48*I*a**(3/2)*b**6*x*sqrt(1/b) + 24*I*sqrt(a)*b**7*x**2*sqrt(1/b)) + 45*A*a**3*b*log(I*sqrt(a)*sq
rt(1/b) + sqrt(x))/(24*I*a**(5/2)*b**5*sqrt(1/b) + 48*I*a**(3/2)*b**6*x*sqrt(1/b) + 24*I*sqrt(a)*b**7*x**2*sqr
t(1/b)) - 90*A*a**2*b**2*x*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(24*I*a**(5/2)*b**5*sqrt(1/b) + 48*I*a**(3/2)*b
**6*x*sqrt(1/b) + 24*I*sqrt(a)*b**7*x**2*sqrt(1/b)) + 90*A*a**2*b**2*x*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(24*
I*a**(5/2)*b**5*sqrt(1/b) + 48*I*a**(3/2)*b**6*x*sqrt(1/b) + 24*I*sqrt(a)*b**7*x**2*sqrt(1/b)) - 45*A*a*b**3*x
**2*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(24*I*a**(5/2)*b**5*sqrt(1/b) + 48*I*a**(3/2)*b**6*x*sqrt(1/b) + 24*I*
sqrt(a)*b**7*x**2*sqrt(1/b)) + 45*A*a*b**3*x**2*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(24*I*a**(5/2)*b**5*sqrt(1/
b) + 48*I*a**(3/2)*b**6*x*sqrt(1/b) + 24*I*sqrt(a)*b**7*x**2*sqrt(1/b)) - 210*I*B*a**(7/2)*b*sqrt(x)*sqrt(1/b)
/(24*I*a**(5/2)*b**5*sqrt(1/b) + 48*I*a**(3/2)*b**6*x*sqrt(1/b) + 24*I*sqrt(a)*b**7*x**2*sqrt(1/b)) - 350*I*B*
a**(5/2)*b**2*x**(3/2)*sqrt(1/b)/(24*I*a**(5/2)*b**5*sqrt(1/b) + 48*I*a**(3/2)*b**6*x*sqrt(1/b) + 24*I*sqrt(a)
*b**7*x**2*sqrt(1/b)) - 112*I*B*a**(3/2)*b**3*x**(5/2)*sqrt(1/b)/(24*I*a**(5/2)*b**5*sqrt(1/b) + 48*I*a**(3/2)
*b**6*x*sqrt(1/b) + 24*I*sqrt(a)*b**7*x**2*sqrt(1/b)) + 16*I*B*sqrt(a)*b**4*x**(7/2)*sqrt(1/b)/(24*I*a**(5/2)*
b**5*sqrt(1/b) + 48*I*a**(3/2)*b**6*x*sqrt(1/b) + 24*I*sqrt(a)*b**7*x**2*sqrt(1/b)) + 105*B*a**4*log(-I*sqrt(a
)*sqrt(1/b) + sqrt(x))/(24*I*a**(5/2)*b**5*sqrt(1/b) + 48*I*a**(3/2)*b**6*x*sqrt(1/b) + 24*I*sqrt(a)*b**7*x**2
*sqrt(1/b)) - 105*B*a**4*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(24*I*a**(5/2)*b**5*sqrt(1/b) + 48*I*a**(3/2)*b**6
*x*sqrt(1/b) + 24*I*sqrt(a)*b**7*x**2*sqrt(1/b)) + 210*B*a**3*b*x*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(24*I*a*
*(5/2)*b**5*sqrt(1/b) + 48*I*a**(3/2)*b**6*x*sqrt(1/b) + 24*I*sqrt(a)*b**7*x**2*sqrt(1/b)) - 210*B*a**3*b*x*lo
g(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(24*I*a**(5/2)*b**5*sqrt(1/b) + 48*I*a**(3/2)*b**6*x*sqrt(1/b) + 24*I*sqrt(a)
*b**7*x**2*sqrt(1/b)) + 105*B*a**2*b**2*x**2*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(24*I*a**(5/2)*b**5*sqrt(1/b)
+ 48*I*a**(3/2)*b**6*x*sqrt(1/b) + 24*I*sqrt(a)*b**7*x**2*sqrt(1/b)) - 105*B*a**2*b**2*x**2*log(I*sqrt(a)*sqr
t(1/b) + sqrt(x))/(24*I*a**(5/2)*b**5*sqrt(1/b) + 48*I*a**(3/2)*b**6*x*sqrt(1/b) + 24*I*sqrt(a)*b**7*x**2*sqrt
(1/b)), True))

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