3.240 \(\int (a+\frac {b}{x})^{5/2} (c+\frac {d}{x}) \, dx\)

Optimal. Leaf size=125 \[ a^{3/2} (2 a d+5 b c) \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )-\frac {\left (a+\frac {b}{x}\right )^{5/2} (2 a d+5 b c)}{5 a}-\frac {1}{3} \left (a+\frac {b}{x}\right )^{3/2} (2 a d+5 b c)-a \sqrt {a+\frac {b}{x}} (2 a d+5 b c)+\frac {c x \left (a+\frac {b}{x}\right )^{7/2}}{a} \]

[Out]

-1/3*(2*a*d+5*b*c)*(a+b/x)^(3/2)-1/5*(2*a*d+5*b*c)*(a+b/x)^(5/2)/a+c*(a+b/x)^(7/2)*x/a+a^(3/2)*(2*a*d+5*b*c)*a
rctanh((a+b/x)^(1/2)/a^(1/2))-a*(2*a*d+5*b*c)*(a+b/x)^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {375, 78, 50, 63, 208} \[ a^{3/2} (2 a d+5 b c) \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )-\frac {\left (a+\frac {b}{x}\right )^{5/2} (2 a d+5 b c)}{5 a}-\frac {1}{3} \left (a+\frac {b}{x}\right )^{3/2} (2 a d+5 b c)-a \sqrt {a+\frac {b}{x}} (2 a d+5 b c)+\frac {c x \left (a+\frac {b}{x}\right )^{7/2}}{a} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^(5/2)*(c + d/x),x]

[Out]

-(a*(5*b*c + 2*a*d)*Sqrt[a + b/x]) - ((5*b*c + 2*a*d)*(a + b/x)^(3/2))/3 - ((5*b*c + 2*a*d)*(a + b/x)^(5/2))/(
5*a) + (c*(a + b/x)^(7/2)*x)/a + a^(3/2)*(5*b*c + 2*a*d)*ArcTanh[Sqrt[a + b/x]/Sqrt[a]]

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 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 \left (a+\frac {b}{x}\right )^{5/2} \left (c+\frac {d}{x}\right ) \, dx &=-\operatorname {Subst}\left (\int \frac {(a+b x)^{5/2} (c+d x)}{x^2} \, dx,x,\frac {1}{x}\right )\\ &=\frac {c \left (a+\frac {b}{x}\right )^{7/2} x}{a}-\frac {\left (\frac {5 b c}{2}+a d\right ) \operatorname {Subst}\left (\int \frac {(a+b x)^{5/2}}{x} \, dx,x,\frac {1}{x}\right )}{a}\\ &=-\frac {(5 b c+2 a d) \left (a+\frac {b}{x}\right )^{5/2}}{5 a}+\frac {c \left (a+\frac {b}{x}\right )^{7/2} x}{a}-\frac {1}{2} (5 b c+2 a d) \operatorname {Subst}\left (\int \frac {(a+b x)^{3/2}}{x} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {1}{3} (5 b c+2 a d) \left (a+\frac {b}{x}\right )^{3/2}-\frac {(5 b c+2 a d) \left (a+\frac {b}{x}\right )^{5/2}}{5 a}+\frac {c \left (a+\frac {b}{x}\right )^{7/2} x}{a}-\frac {1}{2} (a (5 b c+2 a d)) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x} \, dx,x,\frac {1}{x}\right )\\ &=-a (5 b c+2 a d) \sqrt {a+\frac {b}{x}}-\frac {1}{3} (5 b c+2 a d) \left (a+\frac {b}{x}\right )^{3/2}-\frac {(5 b c+2 a d) \left (a+\frac {b}{x}\right )^{5/2}}{5 a}+\frac {c \left (a+\frac {b}{x}\right )^{7/2} x}{a}-\frac {1}{2} \left (a^2 (5 b c+2 a d)\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )\\ &=-a (5 b c+2 a d) \sqrt {a+\frac {b}{x}}-\frac {1}{3} (5 b c+2 a d) \left (a+\frac {b}{x}\right )^{3/2}-\frac {(5 b c+2 a d) \left (a+\frac {b}{x}\right )^{5/2}}{5 a}+\frac {c \left (a+\frac {b}{x}\right )^{7/2} x}{a}-\frac {\left (a^2 (5 b c+2 a d)\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+\frac {b}{x}}\right )}{b}\\ &=-a (5 b c+2 a d) \sqrt {a+\frac {b}{x}}-\frac {1}{3} (5 b c+2 a d) \left (a+\frac {b}{x}\right )^{3/2}-\frac {(5 b c+2 a d) \left (a+\frac {b}{x}\right )^{5/2}}{5 a}+\frac {c \left (a+\frac {b}{x}\right )^{7/2} x}{a}+a^{3/2} (5 b c+2 a d) \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 94, normalized size = 0.75 \[ a^{3/2} (2 a d+5 b c) \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )+\frac {\sqrt {a+\frac {b}{x}} \left (a^2 x^2 (15 c x-46 d)-2 a b x (35 c x+11 d)-2 b^2 (5 c x+3 d)\right )}{15 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^(5/2)*(c + d/x),x]

[Out]

(Sqrt[a + b/x]*(-2*b^2*(3*d + 5*c*x) + a^2*x^2*(-46*d + 15*c*x) - 2*a*b*x*(11*d + 35*c*x)))/(15*x^2) + a^(3/2)
*(5*b*c + 2*a*d)*ArcTanh[Sqrt[a + b/x]/Sqrt[a]]

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fricas [A]  time = 0.96, size = 222, normalized size = 1.78 \[ \left [\frac {15 \, {\left (5 \, a b c + 2 \, a^{2} d\right )} \sqrt {a} x^{2} \log \left (2 \, a x + 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) + 2 \, {\left (15 \, a^{2} c x^{3} - 6 \, b^{2} d - 2 \, {\left (35 \, a b c + 23 \, a^{2} d\right )} x^{2} - 2 \, {\left (5 \, b^{2} c + 11 \, a b d\right )} x\right )} \sqrt {\frac {a x + b}{x}}}{30 \, x^{2}}, -\frac {15 \, {\left (5 \, a b c + 2 \, a^{2} d\right )} \sqrt {-a} x^{2} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) - {\left (15 \, a^{2} c x^{3} - 6 \, b^{2} d - 2 \, {\left (35 \, a b c + 23 \, a^{2} d\right )} x^{2} - 2 \, {\left (5 \, b^{2} c + 11 \, a b d\right )} x\right )} \sqrt {\frac {a x + b}{x}}}{15 \, x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^(5/2)*(c+d/x),x, algorithm="fricas")

[Out]

[1/30*(15*(5*a*b*c + 2*a^2*d)*sqrt(a)*x^2*log(2*a*x + 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + 2*(15*a^2*c*x^3 - 6
*b^2*d - 2*(35*a*b*c + 23*a^2*d)*x^2 - 2*(5*b^2*c + 11*a*b*d)*x)*sqrt((a*x + b)/x))/x^2, -1/15*(15*(5*a*b*c +
2*a^2*d)*sqrt(-a)*x^2*arctan(sqrt(-a)*sqrt((a*x + b)/x)/a) - (15*a^2*c*x^3 - 6*b^2*d - 2*(35*a*b*c + 23*a^2*d)
*x^2 - 2*(5*b^2*c + 11*a*b*d)*x)*sqrt((a*x + b)/x))/x^2]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^(5/2)*(c+d/x),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(x)]Warning, choosing root of [1,0,%%%{-2,[1,2,0]%%%}+%%%{-2,[1,0,0]%%%}+%%%{-2,[0,1,1]%%%},0,%%%{1,[2,4,0]%%
%}+%%%{-2,[2,2,0]%%%}+%%%{1,[2,0,0]%%%}+%%%{2,[1,3,1]%%%}+%%%{-2,[1,1,1]%%%}+%%%{1,[0,2,2]%%%}] at parameters
values [86,-97,-82]Warning, choosing root of [1,0,%%%{-2,[1,2,0]%%%}+%%%{-2,[1,0,0]%%%}+%%%{-2,[0,1,1]%%%},0,%
%%{1,[2,4,0]%%%}+%%%{-2,[2,2,0]%%%}+%%%{1,[2,0,0]%%%}+%%%{2,[1,3,1]%%%}+%%%{-2,[1,1,1]%%%}+%%%{1,[0,2,2]%%%}]
at parameters values [7,-27,26]Warning, choosing root of [1,0,%%%{-4,[1,0,0]%%%}+%%%{-2,[0,1,1]%%%},0,%%%{1,[0
,2,2]%%%}] at parameters values [18.6420984049,-49,-86]Warning, choosing root of [1,0,%%%{-4,[1,0,0]%%%}+%%%{-
2,[0,1,1]%%%},0,%%%{1,[0,2,2]%%%}] at parameters values [78.6493344628,22,42]Warning, choosing root of [1,0,%%
%{-2,[1,0,1]%%%}+%%%{-4,[0,1,0]%%%},0,%%%{1,[2,0,2]%%%}] at parameters values [-13,74.7709350525,24]Sign error
(%%%{-b,0%%%}+%%%{2*sqrt(a)*sqrt(b),1/2%%%}+%%%{-2*a,1%%%}+%%%{a*sqrt(a)*sqrt(b)/b,3/2%%%}+%%%{-a^2*sqrt(a)*s
qrt(b)/(4*b^2),5/2%%%}+%%%{undef,7/2%%%})Limit: Max order reached or unable to make series expansion Error: Ba
d Argument Value

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maple [B]  time = 0.06, size = 253, normalized size = 2.02 \[ -\frac {\sqrt {\frac {a x +b}{x}}\, \left (-30 a^{3} b d \,x^{4} \ln \left (\frac {2 a x +b +2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}}{2 \sqrt {a}}\right )-75 a^{2} b^{2} c \,x^{4} \ln \left (\frac {2 a x +b +2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}}{2 \sqrt {a}}\right )-60 \sqrt {a \,x^{2}+b x}\, a^{\frac {7}{2}} d \,x^{4}-150 \sqrt {a \,x^{2}+b x}\, a^{\frac {5}{2}} b c \,x^{4}+60 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {5}{2}} d \,x^{2}+120 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {3}{2}} b c \,x^{2}+32 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {3}{2}} b d x +20 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} \sqrt {a}\, b^{2} c x +12 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} \sqrt {a}\, b^{2} d \right )}{30 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}\, b \,x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x)^(5/2)*(c+d/x),x)

[Out]

-1/30*((a*x+b)/x)^(1/2)/x^3/b*(-60*(a*x^2+b*x)^(1/2)*a^(7/2)*x^4*d-150*(a*x^2+b*x)^(1/2)*a^(5/2)*x^4*b*c-30*ln
(1/2*(2*a*x+b+2*(a*x^2+b*x)^(1/2)*a^(1/2))/a^(1/2))*x^4*a^3*b*d-75*ln(1/2*(2*a*x+b+2*(a*x^2+b*x)^(1/2)*a^(1/2)
)/a^(1/2))*x^4*a^2*b^2*c+60*(a*x^2+b*x)^(3/2)*a^(5/2)*x^2*d+120*(a*x^2+b*x)^(3/2)*a^(3/2)*x^2*b*c+32*(a*x^2+b*
x)^(3/2)*a^(3/2)*x*b*d+20*(a*x^2+b*x)^(3/2)*a^(1/2)*x*b^2*c+12*(a*x^2+b*x)^(3/2)*a^(1/2)*b^2*d)/((a*x+b)*x)^(1
/2)/a^(1/2)

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maxima [A]  time = 1.51, size = 161, normalized size = 1.29 \[ \frac {1}{6} \, {\left (6 \, \sqrt {a + \frac {b}{x}} a^{2} x - 15 \, a^{\frac {3}{2}} b \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right ) - 4 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} b - 24 \, \sqrt {a + \frac {b}{x}} a b\right )} c - \frac {1}{15} \, {\left (15 \, a^{\frac {5}{2}} \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right ) + 6 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} + 10 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a + 30 \, \sqrt {a + \frac {b}{x}} a^{2}\right )} d \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^(5/2)*(c+d/x),x, algorithm="maxima")

[Out]

1/6*(6*sqrt(a + b/x)*a^2*x - 15*a^(3/2)*b*log((sqrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a))) - 4*(a + b/
x)^(3/2)*b - 24*sqrt(a + b/x)*a*b)*c - 1/15*(15*a^(5/2)*log((sqrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a)
)) + 6*(a + b/x)^(5/2) + 10*(a + b/x)^(3/2)*a + 30*sqrt(a + b/x)*a^2)*d

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mupad [B]  time = 3.48, size = 99, normalized size = 0.79 \[ -\frac {2\,d\,{\left (a+\frac {b}{x}\right )}^{5/2}}{5}-2\,a^2\,d\,\sqrt {a+\frac {b}{x}}-\frac {2\,a\,d\,{\left (a+\frac {b}{x}\right )}^{3/2}}{3}-\frac {2\,c\,x\,{\left (a+\frac {b}{x}\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{2},-\frac {3}{2};\ -\frac {1}{2};\ -\frac {a\,x}{b}\right )}{3\,{\left (\frac {a\,x}{b}+1\right )}^{5/2}}-a^{5/2}\,d\,\mathrm {atan}\left (\frac {\sqrt {a+\frac {b}{x}}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,2{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x)^(5/2)*(c + d/x),x)

[Out]

- (2*d*(a + b/x)^(5/2))/5 - 2*a^2*d*(a + b/x)^(1/2) - a^(5/2)*d*atan(((a + b/x)^(1/2)*1i)/a^(1/2))*2i - (2*a*d
*(a + b/x)^(3/2))/3 - (2*c*x*(a + b/x)^(5/2)*hypergeom([-5/2, -3/2], -1/2, -(a*x)/b))/(3*((a*x)/b + 1)^(5/2))

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sympy [A]  time = 82.77, size = 520, normalized size = 4.16 \[ \frac {4 a^{\frac {11}{2}} b^{\frac {7}{2}} d x^{3} \sqrt {\frac {a x}{b} + 1}}{15 a^{\frac {7}{2}} b^{3} x^{\frac {7}{2}} + 15 a^{\frac {5}{2}} b^{4} x^{\frac {5}{2}}} + \frac {2 a^{\frac {9}{2}} b^{\frac {9}{2}} d x^{2} \sqrt {\frac {a x}{b} + 1}}{15 a^{\frac {7}{2}} b^{3} x^{\frac {7}{2}} + 15 a^{\frac {5}{2}} b^{4} x^{\frac {5}{2}}} - \frac {8 a^{\frac {7}{2}} b^{\frac {11}{2}} d x \sqrt {\frac {a x}{b} + 1}}{15 a^{\frac {7}{2}} b^{3} x^{\frac {7}{2}} + 15 a^{\frac {5}{2}} b^{4} x^{\frac {5}{2}}} - \frac {6 a^{\frac {5}{2}} b^{\frac {13}{2}} d \sqrt {\frac {a x}{b} + 1}}{15 a^{\frac {7}{2}} b^{3} x^{\frac {7}{2}} + 15 a^{\frac {5}{2}} b^{4} x^{\frac {5}{2}}} + a^{\frac {3}{2}} b c \operatorname {asinh}{\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}} \right )} - \frac {4 a^{6} b^{3} d x^{\frac {7}{2}}}{15 a^{\frac {7}{2}} b^{3} x^{\frac {7}{2}} + 15 a^{\frac {5}{2}} b^{4} x^{\frac {5}{2}}} - \frac {4 a^{5} b^{4} d x^{\frac {5}{2}}}{15 a^{\frac {7}{2}} b^{3} x^{\frac {7}{2}} + 15 a^{\frac {5}{2}} b^{4} x^{\frac {5}{2}}} - \frac {2 a^{3} d \operatorname {atan}{\left (\frac {\sqrt {a + \frac {b}{x}}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + a^{2} \sqrt {b} c \sqrt {x} \sqrt {\frac {a x}{b} + 1} - \frac {4 a^{2} b c \operatorname {atan}{\left (\frac {\sqrt {a + \frac {b}{x}}}{\sqrt {- a}} \right )}}{\sqrt {- a}} - 2 a^{2} d \sqrt {a + \frac {b}{x}} - 4 a b c \sqrt {a + \frac {b}{x}} + 2 a b d \left (\begin {cases} - \frac {\sqrt {a}}{x} & \text {for}\: b = 0 \\- \frac {2 \left (a + \frac {b}{x}\right )^{\frac {3}{2}}}{3 b} & \text {otherwise} \end {cases}\right ) + b^{2} c \left (\begin {cases} - \frac {\sqrt {a}}{x} & \text {for}\: b = 0 \\- \frac {2 \left (a + \frac {b}{x}\right )^{\frac {3}{2}}}{3 b} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)**(5/2)*(c+d/x),x)

[Out]

4*a**(11/2)*b**(7/2)*d*x**3*sqrt(a*x/b + 1)/(15*a**(7/2)*b**3*x**(7/2) + 15*a**(5/2)*b**4*x**(5/2)) + 2*a**(9/
2)*b**(9/2)*d*x**2*sqrt(a*x/b + 1)/(15*a**(7/2)*b**3*x**(7/2) + 15*a**(5/2)*b**4*x**(5/2)) - 8*a**(7/2)*b**(11
/2)*d*x*sqrt(a*x/b + 1)/(15*a**(7/2)*b**3*x**(7/2) + 15*a**(5/2)*b**4*x**(5/2)) - 6*a**(5/2)*b**(13/2)*d*sqrt(
a*x/b + 1)/(15*a**(7/2)*b**3*x**(7/2) + 15*a**(5/2)*b**4*x**(5/2)) + a**(3/2)*b*c*asinh(sqrt(a)*sqrt(x)/sqrt(b
)) - 4*a**6*b**3*d*x**(7/2)/(15*a**(7/2)*b**3*x**(7/2) + 15*a**(5/2)*b**4*x**(5/2)) - 4*a**5*b**4*d*x**(5/2)/(
15*a**(7/2)*b**3*x**(7/2) + 15*a**(5/2)*b**4*x**(5/2)) - 2*a**3*d*atan(sqrt(a + b/x)/sqrt(-a))/sqrt(-a) + a**2
*sqrt(b)*c*sqrt(x)*sqrt(a*x/b + 1) - 4*a**2*b*c*atan(sqrt(a + b/x)/sqrt(-a))/sqrt(-a) - 2*a**2*d*sqrt(a + b/x)
- 4*a*b*c*sqrt(a + b/x) + 2*a*b*d*Piecewise((-sqrt(a)/x, Eq(b, 0)), (-2*(a + b/x)**(3/2)/(3*b), True)) + b**2
*c*Piecewise((-sqrt(a)/x, Eq(b, 0)), (-2*(a + b/x)**(3/2)/(3*b), True))

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