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

Optimal. Leaf size=192 \[ \frac {3 a^4 (2 A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{128 b^{7/2}}-\frac {3 a^3 \sqrt {x} \sqrt {a+b x} (2 A b-a B)}{128 b^3}+\frac {a^2 x^{3/2} \sqrt {a+b x} (2 A b-a B)}{64 b^2}+\frac {a x^{5/2} \sqrt {a+b x} (2 A b-a B)}{16 b}+\frac {x^{5/2} (a+b x)^{3/2} (2 A b-a B)}{8 b}+\frac {B x^{5/2} (a+b x)^{5/2}}{5 b} \]

[Out]

1/8*(2*A*b-B*a)*x^(5/2)*(b*x+a)^(3/2)/b+1/5*B*x^(5/2)*(b*x+a)^(5/2)/b+3/128*a^4*(2*A*b-B*a)*arctanh(b^(1/2)*x^
(1/2)/(b*x+a)^(1/2))/b^(7/2)+1/64*a^2*(2*A*b-B*a)*x^(3/2)*(b*x+a)^(1/2)/b^2+1/16*a*(2*A*b-B*a)*x^(5/2)*(b*x+a)
^(1/2)/b-3/128*a^3*(2*A*b-B*a)*x^(1/2)*(b*x+a)^(1/2)/b^3

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Rubi [A]  time = 0.08, 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} \[ \frac {a^2 x^{3/2} \sqrt {a+b x} (2 A b-a B)}{64 b^2}-\frac {3 a^3 \sqrt {x} \sqrt {a+b x} (2 A b-a B)}{128 b^3}+\frac {3 a^4 (2 A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{128 b^{7/2}}+\frac {a x^{5/2} \sqrt {a+b x} (2 A b-a B)}{16 b}+\frac {x^{5/2} (a+b x)^{3/2} (2 A b-a B)}{8 b}+\frac {B x^{5/2} (a+b x)^{5/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a^3*(2*A*b - a*B)*Sqrt[x]*Sqrt[a + b*x])/(128*b^3) + (a^2*(2*A*b - a*B)*x^(3/2)*Sqrt[a + b*x])/(64*b^2) +
(a*(2*A*b - a*B)*x^(5/2)*Sqrt[a + b*x])/(16*b) + ((2*A*b - a*B)*x^(5/2)*(a + b*x)^(3/2))/(8*b) + (B*x^(5/2)*(a
+ b*x)^(5/2))/(5*b) + (3*a^4*(2*A*b - a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(128*b^(7/2))

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^{3/2} (a+b x)^{3/2} (A+B x) \, dx &=\frac {B x^{5/2} (a+b x)^{5/2}}{5 b}+\frac {\left (5 A b-\frac {5 a B}{2}\right ) \int x^{3/2} (a+b x)^{3/2} \, dx}{5 b}\\ &=\frac {(2 A b-a B) x^{5/2} (a+b x)^{3/2}}{8 b}+\frac {B x^{5/2} (a+b x)^{5/2}}{5 b}+\frac {(3 a (2 A b-a B)) \int x^{3/2} \sqrt {a+b x} \, dx}{16 b}\\ &=\frac {a (2 A b-a B) x^{5/2} \sqrt {a+b x}}{16 b}+\frac {(2 A b-a B) x^{5/2} (a+b x)^{3/2}}{8 b}+\frac {B x^{5/2} (a+b x)^{5/2}}{5 b}+\frac {\left (a^2 (2 A b-a B)\right ) \int \frac {x^{3/2}}{\sqrt {a+b x}} \, dx}{32 b}\\ &=\frac {a^2 (2 A b-a B) x^{3/2} \sqrt {a+b x}}{64 b^2}+\frac {a (2 A b-a B) x^{5/2} \sqrt {a+b x}}{16 b}+\frac {(2 A b-a B) x^{5/2} (a+b x)^{3/2}}{8 b}+\frac {B x^{5/2} (a+b x)^{5/2}}{5 b}-\frac {\left (3 a^3 (2 A b-a B)\right ) \int \frac {\sqrt {x}}{\sqrt {a+b x}} \, dx}{128 b^2}\\ &=-\frac {3 a^3 (2 A b-a B) \sqrt {x} \sqrt {a+b x}}{128 b^3}+\frac {a^2 (2 A b-a B) x^{3/2} \sqrt {a+b x}}{64 b^2}+\frac {a (2 A b-a B) x^{5/2} \sqrt {a+b x}}{16 b}+\frac {(2 A b-a B) x^{5/2} (a+b x)^{3/2}}{8 b}+\frac {B x^{5/2} (a+b x)^{5/2}}{5 b}+\frac {\left (3 a^4 (2 A b-a B)\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx}{256 b^3}\\ &=-\frac {3 a^3 (2 A b-a B) \sqrt {x} \sqrt {a+b x}}{128 b^3}+\frac {a^2 (2 A b-a B) x^{3/2} \sqrt {a+b x}}{64 b^2}+\frac {a (2 A b-a B) x^{5/2} \sqrt {a+b x}}{16 b}+\frac {(2 A b-a B) x^{5/2} (a+b x)^{3/2}}{8 b}+\frac {B x^{5/2} (a+b x)^{5/2}}{5 b}+\frac {\left (3 a^4 (2 A b-a B)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right )}{128 b^3}\\ &=-\frac {3 a^3 (2 A b-a B) \sqrt {x} \sqrt {a+b x}}{128 b^3}+\frac {a^2 (2 A b-a B) x^{3/2} \sqrt {a+b x}}{64 b^2}+\frac {a (2 A b-a B) x^{5/2} \sqrt {a+b x}}{16 b}+\frac {(2 A b-a B) x^{5/2} (a+b x)^{3/2}}{8 b}+\frac {B x^{5/2} (a+b x)^{5/2}}{5 b}+\frac {\left (3 a^4 (2 A b-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^3}\\ &=-\frac {3 a^3 (2 A b-a B) \sqrt {x} \sqrt {a+b x}}{128 b^3}+\frac {a^2 (2 A b-a B) x^{3/2} \sqrt {a+b x}}{64 b^2}+\frac {a (2 A b-a B) x^{5/2} \sqrt {a+b x}}{16 b}+\frac {(2 A b-a B) x^{5/2} (a+b x)^{3/2}}{8 b}+\frac {B x^{5/2} (a+b x)^{5/2}}{5 b}+\frac {3 a^4 (2 A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{128 b^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.32, size = 144, normalized size = 0.75 \[ \frac {\sqrt {a+b x} \left (\sqrt {b} \sqrt {x} \left (15 a^4 B-10 a^3 b (3 A+B x)+4 a^2 b^2 x (5 A+2 B x)+16 a b^3 x^2 (15 A+11 B x)+32 b^4 x^3 (5 A+4 B x)\right )-\frac {15 a^{7/2} (a B-2 A b) \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {\frac {b x}{a}+1}}\right )}{640 b^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*(Sqrt[b]*Sqrt[x]*(15*a^4*B - 10*a^3*b*(3*A + B*x) + 4*a^2*b^2*x*(5*A + 2*B*x) + 32*b^4*x^3*(5*A
+ 4*B*x) + 16*a*b^3*x^2*(15*A + 11*B*x)) - (15*a^(7/2)*(-2*A*b + a*B)*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/Sqr
t[1 + (b*x)/a]))/(640*b^(7/2))

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fricas [A]  time = 0.59, size = 290, normalized size = 1.51 \[ \left [-\frac {15 \, {\left (B a^{5} - 2 \, A a^{4} b\right )} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (128 \, B b^{5} x^{4} + 15 \, B a^{4} b - 30 \, A a^{3} b^{2} + 16 \, {\left (11 \, B a b^{4} + 10 \, A b^{5}\right )} x^{3} + 8 \, {\left (B a^{2} b^{3} + 30 \, A a b^{4}\right )} x^{2} - 10 \, {\left (B a^{3} b^{2} - 2 \, A a^{2} b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{1280 \, b^{4}}, \frac {15 \, {\left (B a^{5} - 2 \, A a^{4} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (128 \, B b^{5} x^{4} + 15 \, B a^{4} b - 30 \, A a^{3} b^{2} + 16 \, {\left (11 \, B a b^{4} + 10 \, A b^{5}\right )} x^{3} + 8 \, {\left (B a^{2} b^{3} + 30 \, A a b^{4}\right )} x^{2} - 10 \, {\left (B a^{3} b^{2} - 2 \, A a^{2} b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{640 \, b^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/1280*(15*(B*a^5 - 2*A*a^4*b)*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*(128*B*b^5*x^4 +
15*B*a^4*b - 30*A*a^3*b^2 + 16*(11*B*a*b^4 + 10*A*b^5)*x^3 + 8*(B*a^2*b^3 + 30*A*a*b^4)*x^2 - 10*(B*a^3*b^2 -
2*A*a^2*b^3)*x)*sqrt(b*x + a)*sqrt(x))/b^4, 1/640*(15*(B*a^5 - 2*A*a^4*b)*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(
-b)/(b*sqrt(x))) + (128*B*b^5*x^4 + 15*B*a^4*b - 30*A*a^3*b^2 + 16*(11*B*a*b^4 + 10*A*b^5)*x^3 + 8*(B*a^2*b^3
+ 30*A*a*b^4)*x^2 - 10*(B*a^3*b^2 - 2*A*a^2*b^3)*x)*sqrt(b*x + a)*sqrt(x))/b^4]

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

[Out]

Timed out

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maple [A]  time = 0.01, size = 260, normalized size = 1.35 \[ \frac {\sqrt {b x +a}\, \left (256 \sqrt {\left (b x +a \right ) x}\, B \,b^{\frac {9}{2}} x^{4}+320 \sqrt {\left (b x +a \right ) x}\, A \,b^{\frac {9}{2}} x^{3}+352 \sqrt {\left (b x +a \right ) x}\, B a \,b^{\frac {7}{2}} x^{3}+480 \sqrt {\left (b x +a \right ) x}\, A a \,b^{\frac {7}{2}} x^{2}+16 \sqrt {\left (b x +a \right ) x}\, B \,a^{2} b^{\frac {5}{2}} x^{2}+30 A \,a^{4} b \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )-15 B \,a^{5} \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )+40 \sqrt {\left (b x +a \right ) x}\, A \,a^{2} b^{\frac {5}{2}} x -20 \sqrt {\left (b x +a \right ) x}\, B \,a^{3} b^{\frac {3}{2}} x -60 \sqrt {\left (b x +a \right ) x}\, A \,a^{3} b^{\frac {3}{2}}+30 \sqrt {\left (b x +a \right ) x}\, B \,a^{4} \sqrt {b}\right ) \sqrt {x}}{1280 \sqrt {\left (b x +a \right ) x}\, b^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1280*x^(1/2)*(b*x+a)^(1/2)/b^(7/2)*(256*((b*x+a)*x)^(1/2)*B*b^(9/2)*x^4+320*((b*x+a)*x)^(1/2)*A*b^(9/2)*x^3+
352*((b*x+a)*x)^(1/2)*B*a*b^(7/2)*x^3+480*((b*x+a)*x)^(1/2)*A*a*b^(7/2)*x^2+16*((b*x+a)*x)^(1/2)*B*a^2*b^(5/2)
*x^2+40*((b*x+a)*x)^(1/2)*A*a^2*b^(5/2)*x-20*((b*x+a)*x)^(1/2)*B*a^3*b^(3/2)*x+30*A*a^4*b*ln(1/2*(2*b*x+a+2*((
b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))-60*((b*x+a)*x)^(1/2)*A*a^3*b^(3/2)-15*B*a^5*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1
/2)*b^(1/2))/b^(1/2))+30*((b*x+a)*x)^(1/2)*B*a^4*b^(1/2))/((b*x+a)*x)^(1/2)

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maxima [A]  time = 0.92, size = 236, normalized size = 1.23 \[ \frac {1}{4} \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} A x + \frac {3 \, \sqrt {b x^{2} + a x} B a^{3} x}{64 \, b^{2}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} B a x}{8 \, b} - \frac {3 \, \sqrt {b x^{2} + a x} A a^{2} x}{32 \, b} - \frac {3 \, B a^{5} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{256 \, b^{\frac {7}{2}}} + \frac {3 \, A a^{4} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{128 \, b^{\frac {5}{2}}} + \frac {3 \, \sqrt {b x^{2} + a x} B a^{4}}{128 \, b^{3}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} B a^{2}}{16 \, b^{2}} - \frac {3 \, \sqrt {b x^{2} + a x} A a^{3}}{64 \, b^{2}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}} B}{5 \, b} + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} A a}{8 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(b*x^2 + a*x)^(3/2)*A*x + 3/64*sqrt(b*x^2 + a*x)*B*a^3*x/b^2 - 1/8*(b*x^2 + a*x)^(3/2)*B*a*x/b - 3/32*sqrt
(b*x^2 + a*x)*A*a^2*x/b - 3/256*B*a^5*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(7/2) + 3/128*A*a^4*log(2
*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(5/2) + 3/128*sqrt(b*x^2 + a*x)*B*a^4/b^3 - 1/16*(b*x^2 + a*x)^(3/2)
*B*a^2/b^2 - 3/64*sqrt(b*x^2 + a*x)*A*a^3/b^2 + 1/5*(b*x^2 + a*x)^(5/2)*B/b + 1/8*(b*x^2 + a*x)^(3/2)*A*a/b

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^{3/2}\,\left (A+B\,x\right )\,{\left (a+b\,x\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 76.84, size = 1856, normalized size = 9.67 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*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*s
qrt(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**2 + 2*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)/(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*sq
rt(b)) - I*(a + b*x)**(9/2)/(8*sqrt(a)*sqrt(b)*sqrt(-b*x/a)), True))/b**2 + 2*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)*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*as
in(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 - 4*
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)*s
qrt(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)*sqr
t(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)/(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*Piecewise((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)), A
bs(1 + b*x/a) > 1), (-7*I*a**(9/2)*sqrt(a + b*x)/(256*sqrt(b)*sqrt(-b*x/a)) + 7*I*a**(7/2)*(a + b*x)**(3/2)/(7
68*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**3

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