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

Optimal. Leaf size=192 \[ -\frac {a^4 (10 A b-7 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{128 b^{9/2}}+\frac {a^3 \sqrt {x} \sqrt {a+b x} (10 A b-7 a B)}{128 b^4}-\frac {a^2 x^{3/2} \sqrt {a+b x} (10 A b-7 a B)}{192 b^3}+\frac {a x^{5/2} \sqrt {a+b x} (10 A b-7 a B)}{240 b^2}+\frac {x^{7/2} \sqrt {a+b x} (10 A b-7 a B)}{40 b}+\frac {B x^{7/2} (a+b x)^{3/2}}{5 b} \]

[Out]

1/5*B*x^(7/2)*(b*x+a)^(3/2)/b-1/128*a^4*(10*A*b-7*B*a)*arctanh(b^(1/2)*x^(1/2)/(b*x+a)^(1/2))/b^(9/2)-1/192*a^
2*(10*A*b-7*B*a)*x^(3/2)*(b*x+a)^(1/2)/b^3+1/240*a*(10*A*b-7*B*a)*x^(5/2)*(b*x+a)^(1/2)/b^2+1/40*(10*A*b-7*B*a
)*x^(7/2)*(b*x+a)^(1/2)/b+1/128*a^3*(10*A*b-7*B*a)*x^(1/2)*(b*x+a)^(1/2)/b^4

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {80, 50, 63, 217, 206} \[ -\frac {a^2 x^{3/2} \sqrt {a+b x} (10 A b-7 a B)}{192 b^3}+\frac {a^3 \sqrt {x} \sqrt {a+b x} (10 A b-7 a B)}{128 b^4}-\frac {a^4 (10 A b-7 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{128 b^{9/2}}+\frac {a x^{5/2} \sqrt {a+b x} (10 A b-7 a B)}{240 b^2}+\frac {x^{7/2} \sqrt {a+b x} (10 A b-7 a B)}{40 b}+\frac {B x^{7/2} (a+b x)^{3/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(10*A*b - 7*a*B)*Sqrt[x]*Sqrt[a + b*x])/(128*b^4) - (a^2*(10*A*b - 7*a*B)*x^(3/2)*Sqrt[a + b*x])/(192*b^3
) + (a*(10*A*b - 7*a*B)*x^(5/2)*Sqrt[a + b*x])/(240*b^2) + ((10*A*b - 7*a*B)*x^(7/2)*Sqrt[a + b*x])/(40*b) + (
B*x^(7/2)*(a + b*x)^(3/2))/(5*b) - (a^4*(10*A*b - 7*a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(128*b^(9/2
))

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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps

\begin {align*} \int x^{5/2} \sqrt {a+b x} (A+B x) \, dx &=\frac {B x^{7/2} (a+b x)^{3/2}}{5 b}+\frac {\left (5 A b-\frac {7 a B}{2}\right ) \int x^{5/2} \sqrt {a+b x} \, dx}{5 b}\\ &=\frac {(10 A b-7 a B) x^{7/2} \sqrt {a+b x}}{40 b}+\frac {B x^{7/2} (a+b x)^{3/2}}{5 b}+\frac {(a (10 A b-7 a B)) \int \frac {x^{5/2}}{\sqrt {a+b x}} \, dx}{80 b}\\ &=\frac {a (10 A b-7 a B) x^{5/2} \sqrt {a+b x}}{240 b^2}+\frac {(10 A b-7 a B) x^{7/2} \sqrt {a+b x}}{40 b}+\frac {B x^{7/2} (a+b x)^{3/2}}{5 b}-\frac {\left (a^2 (10 A b-7 a B)\right ) \int \frac {x^{3/2}}{\sqrt {a+b x}} \, dx}{96 b^2}\\ &=-\frac {a^2 (10 A b-7 a B) x^{3/2} \sqrt {a+b x}}{192 b^3}+\frac {a (10 A b-7 a B) x^{5/2} \sqrt {a+b x}}{240 b^2}+\frac {(10 A b-7 a B) x^{7/2} \sqrt {a+b x}}{40 b}+\frac {B x^{7/2} (a+b x)^{3/2}}{5 b}+\frac {\left (a^3 (10 A b-7 a B)\right ) \int \frac {\sqrt {x}}{\sqrt {a+b x}} \, dx}{128 b^3}\\ &=\frac {a^3 (10 A b-7 a B) \sqrt {x} \sqrt {a+b x}}{128 b^4}-\frac {a^2 (10 A b-7 a B) x^{3/2} \sqrt {a+b x}}{192 b^3}+\frac {a (10 A b-7 a B) x^{5/2} \sqrt {a+b x}}{240 b^2}+\frac {(10 A b-7 a B) x^{7/2} \sqrt {a+b x}}{40 b}+\frac {B x^{7/2} (a+b x)^{3/2}}{5 b}-\frac {\left (a^4 (10 A b-7 a B)\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx}{256 b^4}\\ &=\frac {a^3 (10 A b-7 a B) \sqrt {x} \sqrt {a+b x}}{128 b^4}-\frac {a^2 (10 A b-7 a B) x^{3/2} \sqrt {a+b x}}{192 b^3}+\frac {a (10 A b-7 a B) x^{5/2} \sqrt {a+b x}}{240 b^2}+\frac {(10 A b-7 a B) x^{7/2} \sqrt {a+b x}}{40 b}+\frac {B x^{7/2} (a+b x)^{3/2}}{5 b}-\frac {\left (a^4 (10 A b-7 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right )}{128 b^4}\\ &=\frac {a^3 (10 A b-7 a B) \sqrt {x} \sqrt {a+b x}}{128 b^4}-\frac {a^2 (10 A b-7 a B) x^{3/2} \sqrt {a+b x}}{192 b^3}+\frac {a (10 A b-7 a B) x^{5/2} \sqrt {a+b x}}{240 b^2}+\frac {(10 A b-7 a B) x^{7/2} \sqrt {a+b x}}{40 b}+\frac {B x^{7/2} (a+b x)^{3/2}}{5 b}-\frac {\left (a^4 (10 A b-7 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a+b x}}\right )}{128 b^4}\\ &=\frac {a^3 (10 A b-7 a B) \sqrt {x} \sqrt {a+b x}}{128 b^4}-\frac {a^2 (10 A b-7 a B) x^{3/2} \sqrt {a+b x}}{192 b^3}+\frac {a (10 A b-7 a B) x^{5/2} \sqrt {a+b x}}{240 b^2}+\frac {(10 A b-7 a B) x^{7/2} \sqrt {a+b x}}{40 b}+\frac {B x^{7/2} (a+b x)^{3/2}}{5 b}-\frac {a^4 (10 A b-7 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{128 b^{9/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.32, size = 146, normalized size = 0.76 \[ \frac {\sqrt {a+b x} \left (\frac {15 a^{7/2} (7 a B-10 A b) \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {\frac {b x}{a}+1}}+\sqrt {b} \sqrt {x} \left (-105 a^4 B+10 a^3 b (15 A+7 B x)-4 a^2 b^2 x (25 A+14 B x)+16 a b^3 x^2 (5 A+3 B x)+96 b^4 x^3 (5 A+4 B x)\right )\right )}{1920 b^{9/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*(Sqrt[b]*Sqrt[x]*(-105*a^4*B + 16*a*b^3*x^2*(5*A + 3*B*x) + 96*b^4*x^3*(5*A + 4*B*x) + 10*a^3*b
*(15*A + 7*B*x) - 4*a^2*b^2*x*(25*A + 14*B*x)) + (15*a^(7/2)*(-10*A*b + 7*a*B)*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[
a]])/Sqrt[1 + (b*x)/a]))/(1920*b^(9/2))

________________________________________________________________________________________

fricas [A]  time = 0.90, size = 295, normalized size = 1.54 \[ \left [-\frac {15 \, {\left (7 \, B a^{5} - 10 \, A a^{4} b\right )} \sqrt {b} \log \left (2 \, b x - 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (384 \, B b^{5} x^{4} - 105 \, B a^{4} b + 150 \, A a^{3} b^{2} + 48 \, {\left (B a b^{4} + 10 \, A b^{5}\right )} x^{3} - 8 \, {\left (7 \, B a^{2} b^{3} - 10 \, A a b^{4}\right )} x^{2} + 10 \, {\left (7 \, B a^{3} b^{2} - 10 \, A a^{2} b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{3840 \, b^{5}}, -\frac {15 \, {\left (7 \, B a^{5} - 10 \, A a^{4} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (384 \, B b^{5} x^{4} - 105 \, B a^{4} b + 150 \, A a^{3} b^{2} + 48 \, {\left (B a b^{4} + 10 \, A b^{5}\right )} x^{3} - 8 \, {\left (7 \, B a^{2} b^{3} - 10 \, A a b^{4}\right )} x^{2} + 10 \, {\left (7 \, B a^{3} b^{2} - 10 \, A a^{2} b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{1920 \, b^{5}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)*(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[-1/3840*(15*(7*B*a^5 - 10*A*a^4*b)*sqrt(b)*log(2*b*x - 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*(384*B*b^5*x^
4 - 105*B*a^4*b + 150*A*a^3*b^2 + 48*(B*a*b^4 + 10*A*b^5)*x^3 - 8*(7*B*a^2*b^3 - 10*A*a*b^4)*x^2 + 10*(7*B*a^3
*b^2 - 10*A*a^2*b^3)*x)*sqrt(b*x + a)*sqrt(x))/b^5, -1/1920*(15*(7*B*a^5 - 10*A*a^4*b)*sqrt(-b)*arctan(sqrt(b*
x + a)*sqrt(-b)/(b*sqrt(x))) - (384*B*b^5*x^4 - 105*B*a^4*b + 150*A*a^3*b^2 + 48*(B*a*b^4 + 10*A*b^5)*x^3 - 8*
(7*B*a^2*b^3 - 10*A*a*b^4)*x^2 + 10*(7*B*a^3*b^2 - 10*A*a^2*b^3)*x)*sqrt(b*x + a)*sqrt(x))/b^5]

________________________________________________________________________________________

giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)*(b*x+a)^(1/2),x, algorithm="giac")

[Out]

Timed out

________________________________________________________________________________________

maple [A]  time = 0.02, size = 260, normalized size = 1.35 \[ -\frac {\sqrt {b x +a}\, \left (-768 \sqrt {\left (b x +a \right ) x}\, B \,b^{\frac {9}{2}} x^{4}-960 \sqrt {\left (b x +a \right ) x}\, A \,b^{\frac {9}{2}} x^{3}-96 \sqrt {\left (b x +a \right ) x}\, B a \,b^{\frac {7}{2}} x^{3}-160 \sqrt {\left (b x +a \right ) x}\, A a \,b^{\frac {7}{2}} x^{2}+112 \sqrt {\left (b x +a \right ) x}\, B \,a^{2} b^{\frac {5}{2}} x^{2}+150 A \,a^{4} b \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )-105 B \,a^{5} \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )+200 \sqrt {\left (b x +a \right ) x}\, A \,a^{2} b^{\frac {5}{2}} x -140 \sqrt {\left (b x +a \right ) x}\, B \,a^{3} b^{\frac {3}{2}} x -300 \sqrt {\left (b x +a \right ) x}\, A \,a^{3} b^{\frac {3}{2}}+210 \sqrt {\left (b x +a \right ) x}\, B \,a^{4} \sqrt {b}\right ) \sqrt {x}}{3840 \sqrt {\left (b x +a \right ) x}\, b^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(B*x+A)*(b*x+a)^(1/2),x)

[Out]

-1/3840*x^(1/2)*(b*x+a)^(1/2)/b^(9/2)*(-768*B*x^4*b^(9/2)*((b*x+a)*x)^(1/2)-960*A*x^3*b^(9/2)*((b*x+a)*x)^(1/2
)-96*B*x^3*a*b^(7/2)*((b*x+a)*x)^(1/2)-160*A*x^2*a*b^(7/2)*((b*x+a)*x)^(1/2)+112*B*x^2*a^2*b^(5/2)*((b*x+a)*x)
^(1/2)+200*A*((b*x+a)*x)^(1/2)*b^(5/2)*x*a^2-140*B*((b*x+a)*x)^(1/2)*b^(3/2)*x*a^3+150*A*ln(1/2*(2*((b*x+a)*x)
^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*a^4*b-300*A*((b*x+a)*x)^(1/2)*b^(3/2)*a^3-105*B*ln(1/2*(2*((b*x+a)*x)^(1/2)*b
^(1/2)+2*b*x+a)/b^(1/2))*a^5+210*B*((b*x+a)*x)^(1/2)*b^(1/2)*a^4)/((b*x+a)*x)^(1/2)

________________________________________________________________________________________

maxima [A]  time = 0.92, size = 242, normalized size = 1.26 \[ \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} B x^{2}}{5 \, b} - \frac {7 \, \sqrt {b x^{2} + a x} B a^{3} x}{64 \, b^{3}} - \frac {7 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} B a x}{40 \, b^{2}} + \frac {5 \, \sqrt {b x^{2} + a x} A a^{2} x}{32 \, b^{2}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} A x}{4 \, b} + \frac {7 \, B a^{5} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{256 \, b^{\frac {9}{2}}} - \frac {5 \, A a^{4} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{128 \, b^{\frac {7}{2}}} - \frac {7 \, \sqrt {b x^{2} + a x} B a^{4}}{128 \, b^{4}} + \frac {7 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} B a^{2}}{48 \, b^{3}} + \frac {5 \, \sqrt {b x^{2} + a x} A a^{3}}{64 \, b^{3}} - \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} A a}{24 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)*(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

1/5*(b*x^2 + a*x)^(3/2)*B*x^2/b - 7/64*sqrt(b*x^2 + a*x)*B*a^3*x/b^3 - 7/40*(b*x^2 + a*x)^(3/2)*B*a*x/b^2 + 5/
32*sqrt(b*x^2 + a*x)*A*a^2*x/b^2 + 1/4*(b*x^2 + a*x)^(3/2)*A*x/b + 7/256*B*a^5*log(2*b*x + a + 2*sqrt(b*x^2 +
a*x)*sqrt(b))/b^(9/2) - 5/128*A*a^4*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(7/2) - 7/128*sqrt(b*x^2 +
a*x)*B*a^4/b^4 + 7/48*(b*x^2 + a*x)^(3/2)*B*a^2/b^3 + 5/64*sqrt(b*x^2 + a*x)*A*a^3/b^3 - 5/24*(b*x^2 + a*x)^(3
/2)*A*a/b^2

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^{5/2}\,\left (A+B\,x\right )\,\sqrt {a+b\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(A + B*x)*(a + b*x)^(1/2),x)

[Out]

int(x^(5/2)*(A + B*x)*(a + b*x)^(1/2), x)

________________________________________________________________________________________

sympy [C]  time = 48.22, size = 2370, normalized size = 12.34 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)*(B*x+A)*(b*x+a)**(1/2),x)

[Out]

2*A*a**2*Piecewise((a**(3/2)*sqrt(a + b*x)/(8*sqrt(b)*sqrt(b*x/a)) - 3*sqrt(a)*(a + b*x)**(3/2)/(8*sqrt(b)*sqr
t(b*x/a)) - a**2*acosh(sqrt(a + b*x)/sqrt(a))/(8*sqrt(b)) + (a + b*x)**(5/2)/(4*sqrt(a)*sqrt(b)*sqrt(b*x/a)),
Abs(1 + b*x/a) > 1), (-I*a**(3/2)*sqrt(a + b*x)/(8*sqrt(b)*sqrt(-b*x/a)) + 3*I*sqrt(a)*(a + b*x)**(3/2)/(8*sqr
t(b)*sqrt(-b*x/a)) + I*a**2*asin(sqrt(a + b*x)/sqrt(a))/(8*sqrt(b)) - I*(a + b*x)**(5/2)/(4*sqrt(a)*sqrt(b)*sq
rt(-b*x/a)), True))/b**3 - 4*A*a*Piecewise((a**(5/2)*sqrt(a + b*x)/(16*sqrt(b)*sqrt(b*x/a)) - a**(3/2)*(a + b*
x)**(3/2)/(48*sqrt(b)*sqrt(b*x/a)) - 5*sqrt(a)*(a + b*x)**(5/2)/(24*sqrt(b)*sqrt(b*x/a)) - a**3*acosh(sqrt(a +
b*x)/sqrt(a))/(16*sqrt(b)) + (a + b*x)**(7/2)/(6*sqrt(a)*sqrt(b)*sqrt(b*x/a)), Abs(1 + b*x/a) > 1), (-I*a**(5
/2)*sqrt(a + b*x)/(16*sqrt(b)*sqrt(-b*x/a)) + I*a**(3/2)*(a + b*x)**(3/2)/(48*sqrt(b)*sqrt(-b*x/a)) + 5*I*sqrt
(a)*(a + b*x)**(5/2)/(24*sqrt(b)*sqrt(-b*x/a)) + I*a**3*asin(sqrt(a + b*x)/sqrt(a))/(16*sqrt(b)) - I*(a + b*x)
**(7/2)/(6*sqrt(a)*sqrt(b)*sqrt(-b*x/a)), True))/b**3 + 2*A*Piecewise((5*a**(7/2)*sqrt(a + b*x)/(128*sqrt(b)*s
qrt(b*x/a)) - 5*a**(5/2)*(a + b*x)**(3/2)/(384*sqrt(b)*sqrt(b*x/a)) - a**(3/2)*(a + b*x)**(5/2)/(192*sqrt(b)*s
qrt(b*x/a)) - 7*sqrt(a)*(a + b*x)**(7/2)/(48*sqrt(b)*sqrt(b*x/a)) - 5*a**4*acosh(sqrt(a + b*x)/sqrt(a))/(128*s
qrt(b)) + (a + b*x)**(9/2)/(8*sqrt(a)*sqrt(b)*sqrt(b*x/a)), Abs(1 + b*x/a) > 1), (-5*I*a**(7/2)*sqrt(a + b*x)/
(128*sqrt(b)*sqrt(-b*x/a)) + 5*I*a**(5/2)*(a + b*x)**(3/2)/(384*sqrt(b)*sqrt(-b*x/a)) + I*a**(3/2)*(a + b*x)**
(5/2)/(192*sqrt(b)*sqrt(-b*x/a)) + 7*I*sqrt(a)*(a + b*x)**(7/2)/(48*sqrt(b)*sqrt(-b*x/a)) + 5*I*a**4*asin(sqrt
(a + b*x)/sqrt(a))/(128*sqrt(b)) - I*(a + b*x)**(9/2)/(8*sqrt(a)*sqrt(b)*sqrt(-b*x/a)), True))/b**3 - 2*B*a**3
*Piecewise((a**(3/2)*sqrt(a + b*x)/(8*sqrt(b)*sqrt(b*x/a)) - 3*sqrt(a)*(a + b*x)**(3/2)/(8*sqrt(b)*sqrt(b*x/a)
) - a**2*acosh(sqrt(a + b*x)/sqrt(a))/(8*sqrt(b)) + (a + b*x)**(5/2)/(4*sqrt(a)*sqrt(b)*sqrt(b*x/a)), Abs(1 +
b*x/a) > 1), (-I*a**(3/2)*sqrt(a + b*x)/(8*sqrt(b)*sqrt(-b*x/a)) + 3*I*sqrt(a)*(a + b*x)**(3/2)/(8*sqrt(b)*sqr
t(-b*x/a)) + I*a**2*asin(sqrt(a + b*x)/sqrt(a))/(8*sqrt(b)) - I*(a + b*x)**(5/2)/(4*sqrt(a)*sqrt(b)*sqrt(-b*x/
a)), True))/b**4 + 6*B*a**2*Piecewise((a**(5/2)*sqrt(a + b*x)/(16*sqrt(b)*sqrt(b*x/a)) - a**(3/2)*(a + b*x)**(
3/2)/(48*sqrt(b)*sqrt(b*x/a)) - 5*sqrt(a)*(a + b*x)**(5/2)/(24*sqrt(b)*sqrt(b*x/a)) - a**3*acosh(sqrt(a + b*x)
/sqrt(a))/(16*sqrt(b)) + (a + b*x)**(7/2)/(6*sqrt(a)*sqrt(b)*sqrt(b*x/a)), Abs(1 + b*x/a) > 1), (-I*a**(5/2)*s
qrt(a + b*x)/(16*sqrt(b)*sqrt(-b*x/a)) + I*a**(3/2)*(a + b*x)**(3/2)/(48*sqrt(b)*sqrt(-b*x/a)) + 5*I*sqrt(a)*(
a + b*x)**(5/2)/(24*sqrt(b)*sqrt(-b*x/a)) + I*a**3*asin(sqrt(a + b*x)/sqrt(a))/(16*sqrt(b)) - I*(a + b*x)**(7/
2)/(6*sqrt(a)*sqrt(b)*sqrt(-b*x/a)), True))/b**4 - 6*B*a*Piecewise((5*a**(7/2)*sqrt(a + b*x)/(128*sqrt(b)*sqrt
(b*x/a)) - 5*a**(5/2)*(a + b*x)**(3/2)/(384*sqrt(b)*sqrt(b*x/a)) - a**(3/2)*(a + b*x)**(5/2)/(192*sqrt(b)*sqrt
(b*x/a)) - 7*sqrt(a)*(a + b*x)**(7/2)/(48*sqrt(b)*sqrt(b*x/a)) - 5*a**4*acosh(sqrt(a + b*x)/sqrt(a))/(128*sqrt
(b)) + (a + b*x)**(9/2)/(8*sqrt(a)*sqrt(b)*sqrt(b*x/a)), Abs(1 + b*x/a) > 1), (-5*I*a**(7/2)*sqrt(a + b*x)/(12
8*sqrt(b)*sqrt(-b*x/a)) + 5*I*a**(5/2)*(a + b*x)**(3/2)/(384*sqrt(b)*sqrt(-b*x/a)) + I*a**(3/2)*(a + b*x)**(5/
2)/(192*sqrt(b)*sqrt(-b*x/a)) + 7*I*sqrt(a)*(a + b*x)**(7/2)/(48*sqrt(b)*sqrt(-b*x/a)) + 5*I*a**4*asin(sqrt(a
+ b*x)/sqrt(a))/(128*sqrt(b)) - I*(a + b*x)**(9/2)/(8*sqrt(a)*sqrt(b)*sqrt(-b*x/a)), True))/b**4 + 2*B*Piecewi
se((7*a**(9/2)*sqrt(a + b*x)/(256*sqrt(b)*sqrt(b*x/a)) - 7*a**(7/2)*(a + b*x)**(3/2)/(768*sqrt(b)*sqrt(b*x/a))
- 7*a**(5/2)*(a + b*x)**(5/2)/(1920*sqrt(b)*sqrt(b*x/a)) - a**(3/2)*(a + b*x)**(7/2)/(480*sqrt(b)*sqrt(b*x/a)
) - 9*sqrt(a)*(a + b*x)**(9/2)/(80*sqrt(b)*sqrt(b*x/a)) - 7*a**5*acosh(sqrt(a + b*x)/sqrt(a))/(256*sqrt(b)) +
(a + b*x)**(11/2)/(10*sqrt(a)*sqrt(b)*sqrt(b*x/a)), Abs(1 + b*x/a) > 1), (-7*I*a**(9/2)*sqrt(a + b*x)/(256*sqr
t(b)*sqrt(-b*x/a)) + 7*I*a**(7/2)*(a + b*x)**(3/2)/(768*sqrt(b)*sqrt(-b*x/a)) + 7*I*a**(5/2)*(a + b*x)**(5/2)/
(1920*sqrt(b)*sqrt(-b*x/a)) + I*a**(3/2)*(a + b*x)**(7/2)/(480*sqrt(b)*sqrt(-b*x/a)) + 9*I*sqrt(a)*(a + b*x)**
(9/2)/(80*sqrt(b)*sqrt(-b*x/a)) + 7*I*a**5*asin(sqrt(a + b*x)/sqrt(a))/(256*sqrt(b)) - I*(a + b*x)**(11/2)/(10
*sqrt(a)*sqrt(b)*sqrt(-b*x/a)), True))/b**4

________________________________________________________________________________________