3.245 \(\int \frac {(c+\frac {d}{x})^3}{\sqrt {a+\frac {b}{x}}} \, dx\)

Optimal. Leaf size=126 \[ -\frac {c^2 (b c-6 a d) \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {d \sqrt {a+\frac {b}{x}} \left (2 \left (-2 a^2 d^2+9 a b c d+3 b^2 c^2\right )+\frac {b d (2 a d+3 b c)}{x}\right )}{3 a b^2}+\frac {c x \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^2}{a} \]

[Out]

-c^2*(-6*a*d+b*c)*arctanh((a+b/x)^(1/2)/a^(1/2))/a^(3/2)-1/3*d*(-4*a^2*d^2+18*a*b*c*d+6*b^2*c^2+b*d*(2*a*d+3*b
*c)/x)*(a+b/x)^(1/2)/a/b^2+c*(c+d/x)^2*x*(a+b/x)^(1/2)/a

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Rubi [A]  time = 0.09, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {375, 98, 147, 63, 208} \[ -\frac {d \sqrt {a+\frac {b}{x}} \left (2 \left (-2 a^2 d^2+9 a b c d+3 b^2 c^2\right )+\frac {b d (2 a d+3 b c)}{x}\right )}{3 a b^2}-\frac {c^2 (b c-6 a d) \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{a^{3/2}}+\frac {c x \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^2}{a} \]

Antiderivative was successfully verified.

[In]

Int[(c + d/x)^3/Sqrt[a + b/x],x]

[Out]

-(d*Sqrt[a + b/x]*(2*(3*b^2*c^2 + 9*a*b*c*d - 2*a^2*d^2) + (b*d*(3*b*c + 2*a*d))/x))/(3*a*b^2) + (c*Sqrt[a + b
/x]*(c + d/x)^2*x)/a - (c^2*(b*c - 6*a*d)*ArcTanh[Sqrt[a + b/x]/Sqrt[a]])/a^(3/2)

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 98

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

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

Rule 208

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

Rule 375

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> -Subst[Int[((a + b/x^n)^p*(c +
d/x^n)^q)/x^2, x], x, 1/x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && ILtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\left (c+\frac {d}{x}\right )^3}{\sqrt {a+\frac {b}{x}}} \, dx &=-\operatorname {Subst}\left (\int \frac {(c+d x)^3}{x^2 \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {c \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^2 x}{a}+\frac {\operatorname {Subst}\left (\int \frac {(c+d x) \left (\frac {1}{2} c (b c-6 a d)-\frac {1}{2} d (3 b c+2 a d) x\right )}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )}{a}\\ &=-\frac {d \sqrt {a+\frac {b}{x}} \left (2 \left (3 b^2 c^2+9 a b c d-2 a^2 d^2\right )+\frac {b d (3 b c+2 a d)}{x}\right )}{3 a b^2}+\frac {c \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^2 x}{a}+\frac {\left (c^2 (b c-6 a d)\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )}{2 a}\\ &=-\frac {d \sqrt {a+\frac {b}{x}} \left (2 \left (3 b^2 c^2+9 a b c d-2 a^2 d^2\right )+\frac {b d (3 b c+2 a d)}{x}\right )}{3 a b^2}+\frac {c \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^2 x}{a}+\frac {\left (c^2 (b c-6 a d)\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+\frac {b}{x}}\right )}{a b}\\ &=-\frac {d \sqrt {a+\frac {b}{x}} \left (2 \left (3 b^2 c^2+9 a b c d-2 a^2 d^2\right )+\frac {b d (3 b c+2 a d)}{x}\right )}{3 a b^2}+\frac {c \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^2 x}{a}-\frac {c^2 (b c-6 a d) \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{a^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 95, normalized size = 0.75 \[ \frac {c^2 (6 a d-b c) \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{a^{3/2}}+\frac {\sqrt {a+\frac {b}{x}} \left (4 a^2 d^3 x-2 a b d^2 (9 c x+d)+3 b^2 c^3 x^2\right )}{3 a b^2 x} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d/x)^3/Sqrt[a + b/x],x]

[Out]

(Sqrt[a + b/x]*(4*a^2*d^3*x + 3*b^2*c^3*x^2 - 2*a*b*d^2*(d + 9*c*x)))/(3*a*b^2*x) + (c^2*(-(b*c) + 6*a*d)*ArcT
anh[Sqrt[a + b/x]/Sqrt[a]])/a^(3/2)

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fricas [A]  time = 0.98, size = 233, normalized size = 1.85 \[ \left [-\frac {3 \, {\left (b^{3} c^{3} - 6 \, a b^{2} c^{2} d\right )} \sqrt {a} x \log \left (2 \, a x + 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) - 2 \, {\left (3 \, a b^{2} c^{3} x^{2} - 2 \, a^{2} b d^{3} - 2 \, {\left (9 \, a^{2} b c d^{2} - 2 \, a^{3} d^{3}\right )} x\right )} \sqrt {\frac {a x + b}{x}}}{6 \, a^{2} b^{2} x}, \frac {3 \, {\left (b^{3} c^{3} - 6 \, a b^{2} c^{2} d\right )} \sqrt {-a} x \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) + {\left (3 \, a b^{2} c^{3} x^{2} - 2 \, a^{2} b d^{3} - 2 \, {\left (9 \, a^{2} b c d^{2} - 2 \, a^{3} d^{3}\right )} x\right )} \sqrt {\frac {a x + b}{x}}}{3 \, a^{2} b^{2} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d/x)^3/(a+b/x)^(1/2),x, algorithm="fricas")

[Out]

[-1/6*(3*(b^3*c^3 - 6*a*b^2*c^2*d)*sqrt(a)*x*log(2*a*x + 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) - 2*(3*a*b^2*c^3*x
^2 - 2*a^2*b*d^3 - 2*(9*a^2*b*c*d^2 - 2*a^3*d^3)*x)*sqrt((a*x + b)/x))/(a^2*b^2*x), 1/3*(3*(b^3*c^3 - 6*a*b^2*
c^2*d)*sqrt(-a)*x*arctan(sqrt(-a)*sqrt((a*x + b)/x)/a) + (3*a*b^2*c^3*x^2 - 2*a^2*b*d^3 - 2*(9*a^2*b*c*d^2 - 2
*a^3*d^3)*x)*sqrt((a*x + b)/x))/(a^2*b^2*x)]

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giac [A]  time = 0.20, size = 158, normalized size = 1.25 \[ -\frac {\frac {3 \, b^{2} c^{3} \sqrt {\frac {a x + b}{x}}}{{\left (a - \frac {a x + b}{x}\right )} a} - \frac {3 \, {\left (b^{2} c^{3} - 6 \, a b c^{2} d\right )} \arctan \left (\frac {\sqrt {\frac {a x + b}{x}}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {2 \, {\left (9 \, b^{3} c d^{2} \sqrt {\frac {a x + b}{x}} - 3 \, a b^{2} d^{3} \sqrt {\frac {a x + b}{x}} + \frac {{\left (a x + b\right )} b^{2} d^{3} \sqrt {\frac {a x + b}{x}}}{x}\right )}}{b^{3}}}{3 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d/x)^3/(a+b/x)^(1/2),x, algorithm="giac")

[Out]

-1/3*(3*b^2*c^3*sqrt((a*x + b)/x)/((a - (a*x + b)/x)*a) - 3*(b^2*c^3 - 6*a*b*c^2*d)*arctan(sqrt((a*x + b)/x)/s
qrt(-a))/(sqrt(-a)*a) + 2*(9*b^3*c*d^2*sqrt((a*x + b)/x) - 3*a*b^2*d^3*sqrt((a*x + b)/x) + (a*x + b)*b^2*d^3*s
qrt((a*x + b)/x)/x)/b^3)/b

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maple [B]  time = 0.06, size = 535, normalized size = 4.25 \[ \frac {\sqrt {\frac {a x +b}{x}}\, \left (3 a^{3} b \,d^{3} x^{3} \ln \left (\frac {2 a x +b +2 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}}{2 \sqrt {a}}\right )-3 a^{3} b \,d^{3} x^{3} \ln \left (\frac {2 a x +b +2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}}{2 \sqrt {a}}\right )-9 a^{2} b^{2} c \,d^{2} x^{3} \ln \left (\frac {2 a x +b +2 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}}{2 \sqrt {a}}\right )+9 a^{2} b^{2} c \,d^{2} x^{3} \ln \left (\frac {2 a x +b +2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}}{2 \sqrt {a}}\right )+9 a \,b^{3} c^{2} d \,x^{3} \ln \left (\frac {2 a x +b +2 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}}{2 \sqrt {a}}\right )+9 a \,b^{3} c^{2} d \,x^{3} \ln \left (\frac {2 a x +b +2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}}{2 \sqrt {a}}\right )-3 b^{4} c^{3} x^{3} \ln \left (\frac {2 a x +b +2 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}}{2 \sqrt {a}}\right )-6 \sqrt {a \,x^{2}+b x}\, a^{\frac {7}{2}} d^{3} x^{3}-6 \sqrt {\left (a x +b \right ) x}\, a^{\frac {7}{2}} d^{3} x^{3}+18 \sqrt {a \,x^{2}+b x}\, a^{\frac {5}{2}} b c \,d^{2} x^{3}+18 \sqrt {\left (a x +b \right ) x}\, a^{\frac {5}{2}} b c \,d^{2} x^{3}+18 \sqrt {a \,x^{2}+b x}\, a^{\frac {3}{2}} b^{2} c^{2} d \,x^{3}-18 \sqrt {\left (a x +b \right ) x}\, a^{\frac {3}{2}} b^{2} c^{2} d \,x^{3}+6 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}\, b^{3} c^{3} x^{3}+12 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {5}{2}} d^{3} x -36 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {3}{2}} b c \,d^{2} x -4 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {3}{2}} b \,d^{3}\right )}{6 \sqrt {\left (a x +b \right ) x}\, a^{\frac {3}{2}} b^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d/x)^3/(a+b/x)^(1/2),x)

[Out]

1/6*((a*x+b)/x)^(1/2)/x^2*(3*ln(1/2*(2*a*x+b+2*((a*x+b)*x)^(1/2)*a^(1/2))/a^(1/2))*x^3*a^3*b*d^3-9*ln(1/2*(2*a
*x+b+2*((a*x+b)*x)^(1/2)*a^(1/2))/a^(1/2))*x^3*a^2*b^2*c*d^2+9*ln(1/2*(2*a*x+b+2*((a*x+b)*x)^(1/2)*a^(1/2))/a^
(1/2))*x^3*a*b^3*c^2*d-3*ln(1/2*(2*a*x+b+2*((a*x+b)*x)^(1/2)*a^(1/2))/a^(1/2))*x^3*b^4*c^3-6*(a*x^2+b*x)^(1/2)
*a^(7/2)*x^3*d^3+18*(a*x^2+b*x)^(1/2)*a^(5/2)*x^3*b*c*d^2+18*(a*x^2+b*x)^(1/2)*a^(3/2)*x^3*b^2*c^2*d-3*ln(1/2*
(2*a*x+b+2*(a*x^2+b*x)^(1/2)*a^(1/2))/a^(1/2))*x^3*a^3*b*d^3+9*ln(1/2*(2*a*x+b+2*(a*x^2+b*x)^(1/2)*a^(1/2))/a^
(1/2))*x^3*a^2*b^2*c*d^2+9*ln(1/2*(2*a*x+b+2*(a*x^2+b*x)^(1/2)*a^(1/2))/a^(1/2))*x^3*a*b^3*c^2*d-6*a^(7/2)*((a
*x+b)*x)^(1/2)*x^3*d^3+18*a^(5/2)*((a*x+b)*x)^(1/2)*x^3*b*c*d^2-18*a^(3/2)*((a*x+b)*x)^(1/2)*x^3*b^2*c^2*d+6*a
^(1/2)*((a*x+b)*x)^(1/2)*x^3*b^3*c^3+12*(a*x^2+b*x)^(3/2)*a^(5/2)*x*d^3-36*(a*x^2+b*x)^(3/2)*a^(3/2)*x*b*c*d^2
-4*d^3*(a*x^2+b*x)^(3/2)*b*a^(3/2))/((a*x+b)*x)^(1/2)/b^3/a^(3/2)

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maxima [A]  time = 1.33, size = 166, normalized size = 1.32 \[ \frac {1}{2} \, c^{3} {\left (\frac {2 \, \sqrt {a + \frac {b}{x}} b}{{\left (a + \frac {b}{x}\right )} a - a^{2}} + \frac {b \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right )}{a^{\frac {3}{2}}}\right )} - \frac {2}{3} \, d^{3} {\left (\frac {{\left (a + \frac {b}{x}\right )}^{\frac {3}{2}}}{b^{2}} - \frac {3 \, \sqrt {a + \frac {b}{x}} a}{b^{2}}\right )} - \frac {3 \, c^{2} d \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right )}{\sqrt {a}} - \frac {6 \, \sqrt {a + \frac {b}{x}} c d^{2}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d/x)^3/(a+b/x)^(1/2),x, algorithm="maxima")

[Out]

1/2*c^3*(2*sqrt(a + b/x)*b/((a + b/x)*a - a^2) + b*log((sqrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a)))/a^
(3/2)) - 2/3*d^3*((a + b/x)^(3/2)/b^2 - 3*sqrt(a + b/x)*a/b^2) - 3*c^2*d*log((sqrt(a + b/x) - sqrt(a))/(sqrt(a
+ b/x) + sqrt(a)))/sqrt(a) - 6*sqrt(a + b/x)*c*d^2/b

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mupad [B]  time = 1.73, size = 107, normalized size = 0.85 \[ \sqrt {a+\frac {b}{x}}\,\left (\frac {6\,a\,d^3-6\,b\,c\,d^2}{b^2}-\frac {4\,a\,d^3}{b^2}\right )-\frac {2\,d^3\,{\left (a+\frac {b}{x}\right )}^{3/2}}{3\,b^2}+\frac {c^3\,x\,\sqrt {a+\frac {b}{x}}}{a}-\frac {c^2\,\mathrm {atan}\left (\frac {\sqrt {a+\frac {b}{x}}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,\left (6\,a\,d-b\,c\right )\,1{}\mathrm {i}}{a^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d/x)^3/(a + b/x)^(1/2),x)

[Out]

(a + b/x)^(1/2)*((6*a*d^3 - 6*b*c*d^2)/b^2 - (4*a*d^3)/b^2) - (2*d^3*(a + b/x)^(3/2))/(3*b^2) + (c^3*x*(a + b/
x)^(1/2))/a - (c^2*atan(((a + b/x)^(1/2)*1i)/a^(1/2))*(6*a*d - b*c)*1i)/a^(3/2)

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sympy [A]  time = 90.00, size = 386, normalized size = 3.06 \[ \frac {4 a^{\frac {7}{2}} b^{\frac {3}{2}} d^{3} x^{2} \sqrt {\frac {a x}{b} + 1}}{3 a^{\frac {5}{2}} b^{3} x^{\frac {5}{2}} + 3 a^{\frac {3}{2}} b^{4} x^{\frac {3}{2}}} + \frac {2 a^{\frac {5}{2}} b^{\frac {5}{2}} d^{3} x \sqrt {\frac {a x}{b} + 1}}{3 a^{\frac {5}{2}} b^{3} x^{\frac {5}{2}} + 3 a^{\frac {3}{2}} b^{4} x^{\frac {3}{2}}} - \frac {2 a^{\frac {3}{2}} b^{\frac {7}{2}} d^{3} \sqrt {\frac {a x}{b} + 1}}{3 a^{\frac {5}{2}} b^{3} x^{\frac {5}{2}} + 3 a^{\frac {3}{2}} b^{4} x^{\frac {3}{2}}} - \frac {4 a^{4} b d^{3} x^{\frac {5}{2}}}{3 a^{\frac {5}{2}} b^{3} x^{\frac {5}{2}} + 3 a^{\frac {3}{2}} b^{4} x^{\frac {3}{2}}} - \frac {4 a^{3} b^{2} d^{3} x^{\frac {3}{2}}}{3 a^{\frac {5}{2}} b^{3} x^{\frac {5}{2}} + 3 a^{\frac {3}{2}} b^{4} x^{\frac {3}{2}}} + 3 c d^{2} \left (\begin {cases} - \frac {1}{\sqrt {a} x} & \text {for}\: b = 0 \\- \frac {2 \sqrt {a + \frac {b}{x}}}{b} & \text {otherwise} \end {cases}\right ) + \frac {\sqrt {b} c^{3} \sqrt {x} \sqrt {\frac {a x}{b} + 1}}{a} - \frac {6 c^{2} d \operatorname {atan}{\left (\frac {1}{\sqrt {- \frac {1}{a}} \sqrt {a + \frac {b}{x}}} \right )}}{a \sqrt {- \frac {1}{a}}} - \frac {b c^{3} \operatorname {asinh}{\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}} \right )}}{a^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d/x)**3/(a+b/x)**(1/2),x)

[Out]

4*a**(7/2)*b**(3/2)*d**3*x**2*sqrt(a*x/b + 1)/(3*a**(5/2)*b**3*x**(5/2) + 3*a**(3/2)*b**4*x**(3/2)) + 2*a**(5/
2)*b**(5/2)*d**3*x*sqrt(a*x/b + 1)/(3*a**(5/2)*b**3*x**(5/2) + 3*a**(3/2)*b**4*x**(3/2)) - 2*a**(3/2)*b**(7/2)
*d**3*sqrt(a*x/b + 1)/(3*a**(5/2)*b**3*x**(5/2) + 3*a**(3/2)*b**4*x**(3/2)) - 4*a**4*b*d**3*x**(5/2)/(3*a**(5/
2)*b**3*x**(5/2) + 3*a**(3/2)*b**4*x**(3/2)) - 4*a**3*b**2*d**3*x**(3/2)/(3*a**(5/2)*b**3*x**(5/2) + 3*a**(3/2
)*b**4*x**(3/2)) + 3*c*d**2*Piecewise((-1/(sqrt(a)*x), Eq(b, 0)), (-2*sqrt(a + b/x)/b, True)) + sqrt(b)*c**3*s
qrt(x)*sqrt(a*x/b + 1)/a - 6*c**2*d*atan(1/(sqrt(-1/a)*sqrt(a + b/x)))/(a*sqrt(-1/a)) - b*c**3*asinh(sqrt(a)*s
qrt(x)/sqrt(b))/a**(3/2)

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