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

Optimal. Leaf size=130 \[ \frac {(5 a B+A b) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{8 a^{3/2} b^{7/2}}-\frac {\sqrt {x} (5 a B+A b)}{8 a b^3 (a+b x)}-\frac {x^{3/2} (5 a B+A b)}{12 a b^2 (a+b x)^2}+\frac {x^{5/2} (A b-a B)}{3 a b (a+b x)^3} \]

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Rubi [A]  time = 0.06, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {27, 78, 47, 63, 205} \begin {gather*} \frac {(5 a B+A b) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{8 a^{3/2} b^{7/2}}-\frac {x^{3/2} (5 a B+A b)}{12 a b^2 (a+b x)^2}-\frac {\sqrt {x} (5 a B+A b)}{8 a b^3 (a+b x)}+\frac {x^{5/2} (A b-a B)}{3 a b (a+b x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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^{3/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac {x^{3/2} (A+B x)}{(a+b x)^4} \, dx\\ &=\frac {(A b-a B) x^{5/2}}{3 a b (a+b x)^3}+\frac {(A b+5 a B) \int \frac {x^{3/2}}{(a+b x)^3} \, dx}{6 a b}\\ &=\frac {(A b-a B) x^{5/2}}{3 a b (a+b x)^3}-\frac {(A b+5 a B) x^{3/2}}{12 a b^2 (a+b x)^2}+\frac {(A b+5 a B) \int \frac {\sqrt {x}}{(a+b x)^2} \, dx}{8 a b^2}\\ &=\frac {(A b-a B) x^{5/2}}{3 a b (a+b x)^3}-\frac {(A b+5 a B) x^{3/2}}{12 a b^2 (a+b x)^2}-\frac {(A b+5 a B) \sqrt {x}}{8 a b^3 (a+b x)}+\frac {(A b+5 a B) \int \frac {1}{\sqrt {x} (a+b x)} \, dx}{16 a b^3}\\ &=\frac {(A b-a B) x^{5/2}}{3 a b (a+b x)^3}-\frac {(A b+5 a B) x^{3/2}}{12 a b^2 (a+b x)^2}-\frac {(A b+5 a B) \sqrt {x}}{8 a b^3 (a+b x)}+\frac {(A b+5 a B) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {x}\right )}{8 a b^3}\\ &=\frac {(A b-a B) x^{5/2}}{3 a b (a+b x)^3}-\frac {(A b+5 a B) x^{3/2}}{12 a b^2 (a+b x)^2}-\frac {(A b+5 a B) \sqrt {x}}{8 a b^3 (a+b x)}+\frac {(A b+5 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{8 a^{3/2} b^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 105, normalized size = 0.81 \begin {gather*} \frac {(5 a B+A b) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{8 a^{3/2} b^{7/2}}-\frac {\sqrt {x} \left (15 a^3 B+a^2 b (3 A+40 B x)+a b^2 x (8 A+33 B x)-3 A b^3 x^2\right )}{24 a b^3 (a+b x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/24*(Sqrt[x]*(15*a^3*B - 3*A*b^3*x^2 + a*b^2*x*(8*A + 33*B*x) + a^2*b*(3*A + 40*B*x)))/(a*b^3*(a + b*x)^3) +
((A*b + 5*a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(8*a^(3/2)*b^(7/2))

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IntegrateAlgebraic [A]  time = 0.22, size = 111, normalized size = 0.85 \begin {gather*} \frac {(5 a B+A b) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{8 a^{3/2} b^{7/2}}-\frac {\sqrt {x} \left (15 a^3 B+3 a^2 A b+40 a^2 b B x+8 a A b^2 x+33 a b^2 B x^2-3 A b^3 x^2\right )}{24 a b^3 (a+b x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/24*(Sqrt[x]*(3*a^2*A*b + 15*a^3*B + 8*a*A*b^2*x + 40*a^2*b*B*x - 3*A*b^3*x^2 + 33*a*b^2*B*x^2))/(a*b^3*(a +
b*x)^3) + ((A*b + 5*a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(8*a^(3/2)*b^(7/2))

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fricas [A]  time = 0.46, size = 411, normalized size = 3.16 \begin {gather*} \left [-\frac {3 \, {\left (5 \, B a^{4} + A a^{3} b + {\left (5 \, B a b^{3} + A b^{4}\right )} x^{3} + 3 \, {\left (5 \, B a^{2} b^{2} + A a b^{3}\right )} x^{2} + 3 \, {\left (5 \, 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 (15 \, B a^{4} b + 3 \, A a^{3} b^{2} + 3 \, {\left (11 \, B a^{2} b^{3} - A a b^{4}\right )} x^{2} + 8 \, {\left (5 \, B a^{3} b^{2} + A a^{2} b^{3}\right )} x\right )} \sqrt {x}}{48 \, {\left (a^{2} b^{7} x^{3} + 3 \, a^{3} b^{6} x^{2} + 3 \, a^{4} b^{5} x + a^{5} b^{4}\right )}}, -\frac {3 \, {\left (5 \, B a^{4} + A a^{3} b + {\left (5 \, B a b^{3} + A b^{4}\right )} x^{3} + 3 \, {\left (5 \, B a^{2} b^{2} + A a b^{3}\right )} x^{2} + 3 \, {\left (5 \, 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 (15 \, B a^{4} b + 3 \, A a^{3} b^{2} + 3 \, {\left (11 \, B a^{2} b^{3} - A a b^{4}\right )} x^{2} + 8 \, {\left (5 \, B a^{3} b^{2} + A a^{2} b^{3}\right )} x\right )} \sqrt {x}}{24 \, {\left (a^{2} b^{7} x^{3} + 3 \, a^{3} b^{6} x^{2} + 3 \, a^{4} b^{5} x + a^{5} b^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/48*(3*(5*B*a^4 + A*a^3*b + (5*B*a*b^3 + A*b^4)*x^3 + 3*(5*B*a^2*b^2 + A*a*b^3)*x^2 + 3*(5*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*(15*B*a^4*b + 3*A*a^3*b^2 + 3*(11*B*a^2
*b^3 - A*a*b^4)*x^2 + 8*(5*B*a^3*b^2 + A*a^2*b^3)*x)*sqrt(x))/(a^2*b^7*x^3 + 3*a^3*b^6*x^2 + 3*a^4*b^5*x + a^5
*b^4), -1/24*(3*(5*B*a^4 + A*a^3*b + (5*B*a*b^3 + A*b^4)*x^3 + 3*(5*B*a^2*b^2 + A*a*b^3)*x^2 + 3*(5*B*a^3*b +
A*a^2*b^2)*x)*sqrt(a*b)*arctan(sqrt(a*b)/(b*sqrt(x))) + (15*B*a^4*b + 3*A*a^3*b^2 + 3*(11*B*a^2*b^3 - A*a*b^4)
*x^2 + 8*(5*B*a^3*b^2 + A*a^2*b^3)*x)*sqrt(x))/(a^2*b^7*x^3 + 3*a^3*b^6*x^2 + 3*a^4*b^5*x + a^5*b^4)]

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giac [A]  time = 0.16, size = 107, normalized size = 0.82 \begin {gather*} \frac {{\left (5 \, B a + A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a b^{3}} - \frac {33 \, B a b^{2} x^{\frac {5}{2}} - 3 \, A b^{3} x^{\frac {5}{2}} + 40 \, B a^{2} b x^{\frac {3}{2}} + 8 \, A a b^{2} x^{\frac {3}{2}} + 15 \, B a^{3} \sqrt {x} + 3 \, A a^{2} b \sqrt {x}}{24 \, {\left (b x + a\right )}^{3} a b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.10, size = 111, normalized size = 0.85 \begin {gather*} \frac {A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, a \,b^{2}}+\frac {5 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, b^{3}}+\frac {\frac {\left (A b -11 B a \right ) x^{\frac {5}{2}}}{8 a b}-\frac {\left (A b +5 B a \right ) x^{\frac {3}{2}}}{3 b^{2}}-\frac {\left (A b +5 B a \right ) a \sqrt {x}}{8 b^{3}}}{\left (b x +a \right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(1/16*(A*b-11*B*a)/a/b*x^(5/2)-1/6/b^2*(A*b+5*B*a)*x^(3/2)-1/16*(A*b+5*B*a)*a/b^3*x^(1/2))/(b*x+a)^3+1/8/a/b
^2/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x^(1/2))*A+5/8/b^3/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x^(1/2))*B

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maxima [A]  time = 1.10, size = 130, normalized size = 1.00 \begin {gather*} -\frac {3 \, {\left (11 \, B a b^{2} - A b^{3}\right )} x^{\frac {5}{2}} + 8 \, {\left (5 \, B a^{2} b + A a b^{2}\right )} x^{\frac {3}{2}} + 3 \, {\left (5 \, B a^{3} + A a^{2} b\right )} \sqrt {x}}{24 \, {\left (a b^{6} x^{3} + 3 \, a^{2} b^{5} x^{2} + 3 \, a^{3} b^{4} x + a^{4} b^{3}\right )}} + \frac {{\left (5 \, B a + A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.25, size = 112, normalized size = 0.86 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (A\,b+5\,B\,a\right )}{8\,a^{3/2}\,b^{7/2}}-\frac {\frac {x^{3/2}\,\left (A\,b+5\,B\,a\right )}{3\,b^2}-\frac {x^{5/2}\,\left (A\,b-11\,B\,a\right )}{8\,a\,b}+\frac {a\,\sqrt {x}\,\left (A\,b+5\,B\,a\right )}{8\,b^3}}{a^3+3\,a^2\,b\,x+3\,a\,b^2\,x^2+b^3\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 84.21, size = 2547, normalized size = 19.59

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*(-2*A/(3*x**(3/2)) - 2*B/sqrt(x)), Eq(a, 0) & Eq(b, 0)), ((2*A*x**(5/2)/5 + 2*B*x**(7/2)/7)/a**
4, Eq(b, 0)), ((-2*A/(3*x**(3/2)) - 2*B/sqrt(x))/b**4, Eq(a, 0)), (-6*I*A*a**(5/2)*b**2*sqrt(x)*sqrt(1/b)/(48*
I*a**(9/2)*b**4*sqrt(1/b) + 144*I*a**(7/2)*b**5*x*sqrt(1/b) + 144*I*a**(5/2)*b**6*x**2*sqrt(1/b) + 48*I*a**(3/
2)*b**7*x**3*sqrt(1/b)) - 16*I*A*a**(3/2)*b**3*x**(3/2)*sqrt(1/b)/(48*I*a**(9/2)*b**4*sqrt(1/b) + 144*I*a**(7/
2)*b**5*x*sqrt(1/b) + 144*I*a**(5/2)*b**6*x**2*sqrt(1/b) + 48*I*a**(3/2)*b**7*x**3*sqrt(1/b)) + 6*I*A*sqrt(a)*
b**4*x**(5/2)*sqrt(1/b)/(48*I*a**(9/2)*b**4*sqrt(1/b) + 144*I*a**(7/2)*b**5*x*sqrt(1/b) + 144*I*a**(5/2)*b**6*
x**2*sqrt(1/b) + 48*I*a**(3/2)*b**7*x**3*sqrt(1/b)) + 3*A*a**3*b*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**
(9/2)*b**4*sqrt(1/b) + 144*I*a**(7/2)*b**5*x*sqrt(1/b) + 144*I*a**(5/2)*b**6*x**2*sqrt(1/b) + 48*I*a**(3/2)*b*
*7*x**3*sqrt(1/b)) - 3*A*a**3*b*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(9/2)*b**4*sqrt(1/b) + 144*I*a**(7
/2)*b**5*x*sqrt(1/b) + 144*I*a**(5/2)*b**6*x**2*sqrt(1/b) + 48*I*a**(3/2)*b**7*x**3*sqrt(1/b)) + 9*A*a**2*b**2
*x*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(9/2)*b**4*sqrt(1/b) + 144*I*a**(7/2)*b**5*x*sqrt(1/b) + 144*I
*a**(5/2)*b**6*x**2*sqrt(1/b) + 48*I*a**(3/2)*b**7*x**3*sqrt(1/b)) - 9*A*a**2*b**2*x*log(I*sqrt(a)*sqrt(1/b) +
sqrt(x))/(48*I*a**(9/2)*b**4*sqrt(1/b) + 144*I*a**(7/2)*b**5*x*sqrt(1/b) + 144*I*a**(5/2)*b**6*x**2*sqrt(1/b)
+ 48*I*a**(3/2)*b**7*x**3*sqrt(1/b)) + 9*A*a*b**3*x**2*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(9/2)*b**
4*sqrt(1/b) + 144*I*a**(7/2)*b**5*x*sqrt(1/b) + 144*I*a**(5/2)*b**6*x**2*sqrt(1/b) + 48*I*a**(3/2)*b**7*x**3*s
qrt(1/b)) - 9*A*a*b**3*x**2*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(9/2)*b**4*sqrt(1/b) + 144*I*a**(7/2)*
b**5*x*sqrt(1/b) + 144*I*a**(5/2)*b**6*x**2*sqrt(1/b) + 48*I*a**(3/2)*b**7*x**3*sqrt(1/b)) + 3*A*b**4*x**3*log
(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(9/2)*b**4*sqrt(1/b) + 144*I*a**(7/2)*b**5*x*sqrt(1/b) + 144*I*a**(5
/2)*b**6*x**2*sqrt(1/b) + 48*I*a**(3/2)*b**7*x**3*sqrt(1/b)) - 3*A*b**4*x**3*log(I*sqrt(a)*sqrt(1/b) + sqrt(x)
)/(48*I*a**(9/2)*b**4*sqrt(1/b) + 144*I*a**(7/2)*b**5*x*sqrt(1/b) + 144*I*a**(5/2)*b**6*x**2*sqrt(1/b) + 48*I*
a**(3/2)*b**7*x**3*sqrt(1/b)) - 30*I*B*a**(7/2)*b*sqrt(x)*sqrt(1/b)/(48*I*a**(9/2)*b**4*sqrt(1/b) + 144*I*a**(
7/2)*b**5*x*sqrt(1/b) + 144*I*a**(5/2)*b**6*x**2*sqrt(1/b) + 48*I*a**(3/2)*b**7*x**3*sqrt(1/b)) - 80*I*B*a**(5
/2)*b**2*x**(3/2)*sqrt(1/b)/(48*I*a**(9/2)*b**4*sqrt(1/b) + 144*I*a**(7/2)*b**5*x*sqrt(1/b) + 144*I*a**(5/2)*b
**6*x**2*sqrt(1/b) + 48*I*a**(3/2)*b**7*x**3*sqrt(1/b)) - 66*I*B*a**(3/2)*b**3*x**(5/2)*sqrt(1/b)/(48*I*a**(9/
2)*b**4*sqrt(1/b) + 144*I*a**(7/2)*b**5*x*sqrt(1/b) + 144*I*a**(5/2)*b**6*x**2*sqrt(1/b) + 48*I*a**(3/2)*b**7*
x**3*sqrt(1/b)) + 15*B*a**4*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(9/2)*b**4*sqrt(1/b) + 144*I*a**(7/2)
*b**5*x*sqrt(1/b) + 144*I*a**(5/2)*b**6*x**2*sqrt(1/b) + 48*I*a**(3/2)*b**7*x**3*sqrt(1/b)) - 15*B*a**4*log(I*
sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(9/2)*b**4*sqrt(1/b) + 144*I*a**(7/2)*b**5*x*sqrt(1/b) + 144*I*a**(5/2)*
b**6*x**2*sqrt(1/b) + 48*I*a**(3/2)*b**7*x**3*sqrt(1/b)) + 45*B*a**3*b*x*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(
48*I*a**(9/2)*b**4*sqrt(1/b) + 144*I*a**(7/2)*b**5*x*sqrt(1/b) + 144*I*a**(5/2)*b**6*x**2*sqrt(1/b) + 48*I*a**
(3/2)*b**7*x**3*sqrt(1/b)) - 45*B*a**3*b*x*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(9/2)*b**4*sqrt(1/b) +
144*I*a**(7/2)*b**5*x*sqrt(1/b) + 144*I*a**(5/2)*b**6*x**2*sqrt(1/b) + 48*I*a**(3/2)*b**7*x**3*sqrt(1/b)) + 45
*B*a**2*b**2*x**2*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(9/2)*b**4*sqrt(1/b) + 144*I*a**(7/2)*b**5*x*sq
rt(1/b) + 144*I*a**(5/2)*b**6*x**2*sqrt(1/b) + 48*I*a**(3/2)*b**7*x**3*sqrt(1/b)) - 45*B*a**2*b**2*x**2*log(I*
sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(9/2)*b**4*sqrt(1/b) + 144*I*a**(7/2)*b**5*x*sqrt(1/b) + 144*I*a**(5/2)*
b**6*x**2*sqrt(1/b) + 48*I*a**(3/2)*b**7*x**3*sqrt(1/b)) + 15*B*a*b**3*x**3*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x)
)/(48*I*a**(9/2)*b**4*sqrt(1/b) + 144*I*a**(7/2)*b**5*x*sqrt(1/b) + 144*I*a**(5/2)*b**6*x**2*sqrt(1/b) + 48*I*
a**(3/2)*b**7*x**3*sqrt(1/b)) - 15*B*a*b**3*x**3*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(48*I*a**(9/2)*b**4*sqrt(1
/b) + 144*I*a**(7/2)*b**5*x*sqrt(1/b) + 144*I*a**(5/2)*b**6*x**2*sqrt(1/b) + 48*I*a**(3/2)*b**7*x**3*sqrt(1/b)
), True))

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