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

Optimal. Leaf size=103 \[ -\frac {(5 b c-2 a d) \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{a^{7/2}}+\frac {5 b c-2 a d}{a^3 \sqrt {a+\frac {b}{x}}}+\frac {5 b c-2 a d}{3 a^2 \left (a+\frac {b}{x}\right )^{3/2}}+\frac {c x}{a \left (a+\frac {b}{x}\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {375, 78, 51, 63, 208} \[ \frac {5 b c-2 a d}{a^3 \sqrt {a+\frac {b}{x}}}+\frac {5 b c-2 a d}{3 a^2 \left (a+\frac {b}{x}\right )^{3/2}}-\frac {(5 b c-2 a d) \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{a^{7/2}}+\frac {c x}{a \left (a+\frac {b}{x}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d/x)/(a + b/x)^(5/2),x]

[Out]

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

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 \frac {c+\frac {d}{x}}{\left (a+\frac {b}{x}\right )^{5/2}} \, dx &=-\operatorname {Subst}\left (\int \frac {c+d x}{x^2 (a+b x)^{5/2}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {c x}{a \left (a+\frac {b}{x}\right )^{3/2}}-\frac {\left (-\frac {5 b c}{2}+a d\right ) \operatorname {Subst}\left (\int \frac {1}{x (a+b x)^{5/2}} \, dx,x,\frac {1}{x}\right )}{a}\\ &=\frac {5 b c-2 a d}{3 a^2 \left (a+\frac {b}{x}\right )^{3/2}}+\frac {c x}{a \left (a+\frac {b}{x}\right )^{3/2}}+\frac {(5 b c-2 a d) \operatorname {Subst}\left (\int \frac {1}{x (a+b x)^{3/2}} \, dx,x,\frac {1}{x}\right )}{2 a^2}\\ &=\frac {5 b c-2 a d}{3 a^2 \left (a+\frac {b}{x}\right )^{3/2}}+\frac {5 b c-2 a d}{a^3 \sqrt {a+\frac {b}{x}}}+\frac {c x}{a \left (a+\frac {b}{x}\right )^{3/2}}+\frac {(5 b c-2 a d) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )}{2 a^3}\\ &=\frac {5 b c-2 a d}{3 a^2 \left (a+\frac {b}{x}\right )^{3/2}}+\frac {5 b c-2 a d}{a^3 \sqrt {a+\frac {b}{x}}}+\frac {c x}{a \left (a+\frac {b}{x}\right )^{3/2}}+\frac {(5 b c-2 a d) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+\frac {b}{x}}\right )}{a^3 b}\\ &=\frac {5 b c-2 a d}{3 a^2 \left (a+\frac {b}{x}\right )^{3/2}}+\frac {5 b c-2 a d}{a^3 \sqrt {a+\frac {b}{x}}}+\frac {c x}{a \left (a+\frac {b}{x}\right )^{3/2}}-\frac {(5 b c-2 a d) \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{a^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 60, normalized size = 0.58 \[ \frac {x \left ((5 b c-2 a d) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {b}{a x}+1\right )+3 a c x\right )}{3 a^2 \sqrt {a+\frac {b}{x}} (a x+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(3*a*c*x + (5*b*c - 2*a*d)*Hypergeometric2F1[-3/2, 1, -1/2, 1 + b/(a*x)]))/(3*a^2*Sqrt[a + b/x]*(b + a*x))

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fricas [A]  time = 0.92, size = 331, normalized size = 3.21 \[ \left [-\frac {3 \, {\left (5 \, b^{3} c - 2 \, a b^{2} d + {\left (5 \, a^{2} b c - 2 \, a^{3} d\right )} x^{2} + 2 \, {\left (5 \, a b^{2} c - 2 \, a^{2} b d\right )} x\right )} \sqrt {a} \log \left (2 \, a x + 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) - 2 \, {\left (3 \, a^{3} c x^{3} + 4 \, {\left (5 \, a^{2} b c - 2 \, a^{3} d\right )} x^{2} + 3 \, {\left (5 \, a b^{2} c - 2 \, a^{2} b d\right )} x\right )} \sqrt {\frac {a x + b}{x}}}{6 \, {\left (a^{6} x^{2} + 2 \, a^{5} b x + a^{4} b^{2}\right )}}, \frac {3 \, {\left (5 \, b^{3} c - 2 \, a b^{2} d + {\left (5 \, a^{2} b c - 2 \, a^{3} d\right )} x^{2} + 2 \, {\left (5 \, a b^{2} c - 2 \, a^{2} b d\right )} x\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) + {\left (3 \, a^{3} c x^{3} + 4 \, {\left (5 \, a^{2} b c - 2 \, a^{3} d\right )} x^{2} + 3 \, {\left (5 \, a b^{2} c - 2 \, a^{2} b d\right )} x\right )} \sqrt {\frac {a x + b}{x}}}{3 \, {\left (a^{6} x^{2} + 2 \, a^{5} b x + a^{4} b^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6*(3*(5*b^3*c - 2*a*b^2*d + (5*a^2*b*c - 2*a^3*d)*x^2 + 2*(5*a*b^2*c - 2*a^2*b*d)*x)*sqrt(a)*log(2*a*x + 2
*sqrt(a)*x*sqrt((a*x + b)/x) + b) - 2*(3*a^3*c*x^3 + 4*(5*a^2*b*c - 2*a^3*d)*x^2 + 3*(5*a*b^2*c - 2*a^2*b*d)*x
)*sqrt((a*x + b)/x))/(a^6*x^2 + 2*a^5*b*x + a^4*b^2), 1/3*(3*(5*b^3*c - 2*a*b^2*d + (5*a^2*b*c - 2*a^3*d)*x^2
+ 2*(5*a*b^2*c - 2*a^2*b*d)*x)*sqrt(-a)*arctan(sqrt(-a)*sqrt((a*x + b)/x)/a) + (3*a^3*c*x^3 + 4*(5*a^2*b*c - 2
*a^3*d)*x^2 + 3*(5*a*b^2*c - 2*a^2*b*d)*x)*sqrt((a*x + b)/x))/(a^6*x^2 + 2*a^5*b*x + a^4*b^2)]

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giac [A]  time = 0.22, size = 145, normalized size = 1.41 \[ -\frac {\frac {3 \, b^{2} c \sqrt {\frac {a x + b}{x}}}{{\left (a - \frac {a x + b}{x}\right )} a^{3}} - \frac {3 \, {\left (5 \, b^{2} c - 2 \, a b d\right )} \arctan \left (\frac {\sqrt {\frac {a x + b}{x}}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3}} - \frac {2 \, {\left (a b^{2} c - a^{2} b d + \frac {6 \, {\left (a x + b\right )} b^{2} c}{x} - \frac {3 \, {\left (a x + b\right )} a b d}{x}\right )} x}{{\left (a x + b\right )} a^{3} \sqrt {\frac {a x + b}{x}}}}{3 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(3*b^2*c*sqrt((a*x + b)/x)/((a - (a*x + b)/x)*a^3) - 3*(5*b^2*c - 2*a*b*d)*arctan(sqrt((a*x + b)/x)/sqrt(
-a))/(sqrt(-a)*a^3) - 2*(a*b^2*c - a^2*b*d + 6*(a*x + b)*b^2*c/x - 3*(a*x + b)*a*b*d/x)*x/((a*x + b)*a^3*sqrt(
(a*x + b)/x)))/b

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maple [B]  time = 0.06, size = 541, normalized size = 5.25 \[ \frac {\sqrt {\frac {a x +b}{x}}\, \left (6 a^{4} b d \,x^{3} \ln \left (\frac {2 a x +b +2 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}}{2 \sqrt {a}}\right )-15 a^{3} b^{2} c \,x^{3} \ln \left (\frac {2 a x +b +2 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}}{2 \sqrt {a}}\right )+18 a^{3} b^{2} d \,x^{2} \ln \left (\frac {2 a x +b +2 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}}{2 \sqrt {a}}\right )-45 a^{2} b^{3} c \,x^{2} \ln \left (\frac {2 a x +b +2 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}}{2 \sqrt {a}}\right )-12 \sqrt {\left (a x +b \right ) x}\, a^{\frac {9}{2}} d \,x^{3}+30 \sqrt {\left (a x +b \right ) x}\, a^{\frac {7}{2}} b c \,x^{3}+18 a^{2} b^{3} d x \ln \left (\frac {2 a x +b +2 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}}{2 \sqrt {a}}\right )-45 a \,b^{4} c x \ln \left (\frac {2 a x +b +2 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}}{2 \sqrt {a}}\right )-36 \sqrt {\left (a x +b \right ) x}\, a^{\frac {7}{2}} b d \,x^{2}+90 \sqrt {\left (a x +b \right ) x}\, a^{\frac {5}{2}} b^{2} c \,x^{2}+6 a \,b^{4} d \ln \left (\frac {2 a x +b +2 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}}{2 \sqrt {a}}\right )-15 b^{5} c \ln \left (\frac {2 a x +b +2 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}}{2 \sqrt {a}}\right )-36 \sqrt {\left (a x +b \right ) x}\, a^{\frac {5}{2}} b^{2} d x +90 \sqrt {\left (a x +b \right ) x}\, a^{\frac {3}{2}} b^{3} c x +12 \left (\left (a x +b \right ) x \right )^{\frac {3}{2}} a^{\frac {7}{2}} d x -24 \left (\left (a x +b \right ) x \right )^{\frac {3}{2}} a^{\frac {5}{2}} b c x -12 \sqrt {\left (a x +b \right ) x}\, a^{\frac {3}{2}} b^{3} d +30 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}\, b^{4} c +8 \left (\left (a x +b \right ) x \right )^{\frac {3}{2}} a^{\frac {5}{2}} b d -20 \left (\left (a x +b \right ) x \right )^{\frac {3}{2}} a^{\frac {3}{2}} b^{2} c \right ) x}{6 \sqrt {\left (a x +b \right ) x}\, \left (a x +b \right )^{3} a^{\frac {7}{2}} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*((a*x+b)/x)^(1/2)*x/a^(7/2)*(6*ln(1/2*(2*a*x+b+2*((a*x+b)*x)^(1/2)*a^(1/2))/a^(1/2))*x^3*a^4*b*d-15*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^2*c-12*a^(9/2)*((a*x+b)*x)^(1/2)*x^3*d+30*a^(7/2)*((
a*x+b)*x)^(1/2)*x^3*b*c+18*ln(1/2*(2*a*x+b+2*((a*x+b)*x)^(1/2)*a^(1/2))/a^(1/2))*x^2*a^3*b^2*d-45*ln(1/2*(2*a*
x+b+2*((a*x+b)*x)^(1/2)*a^(1/2))/a^(1/2))*x^2*a^2*b^3*c+12*a^(7/2)*((a*x+b)*x)^(3/2)*x*d-24*a^(5/2)*((a*x+b)*x
)^(3/2)*x*b*c-36*a^(7/2)*((a*x+b)*x)^(1/2)*x^2*b*d+90*a^(5/2)*((a*x+b)*x)^(1/2)*x^2*b^2*c+18*ln(1/2*(2*a*x+b+2
*((a*x+b)*x)^(1/2)*a^(1/2))/a^(1/2))*x*a^2*b^3*d-45*ln(1/2*(2*a*x+b+2*((a*x+b)*x)^(1/2)*a^(1/2))/a^(1/2))*x*a*
b^4*c+8*a^(5/2)*((a*x+b)*x)^(3/2)*b*d-20*a^(3/2)*((a*x+b)*x)^(3/2)*b^2*c-36*a^(5/2)*((a*x+b)*x)^(1/2)*x*b^2*d+
90*a^(3/2)*((a*x+b)*x)^(1/2)*x*b^3*c+6*ln(1/2*(2*a*x+b+2*((a*x+b)*x)^(1/2)*a^(1/2))/a^(1/2))*a*b^4*d-15*ln(1/2
*(2*a*x+b+2*((a*x+b)*x)^(1/2)*a^(1/2))/a^(1/2))*b^5*c-12*a^(3/2)*((a*x+b)*x)^(1/2)*b^3*d+30*a^(1/2)*((a*x+b)*x
)^(1/2)*b^4*c)/((a*x+b)*x)^(1/2)/b/(a*x+b)^3

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maxima [A]  time = 1.25, size = 170, normalized size = 1.65 \[ \frac {1}{6} \, c {\left (\frac {2 \, {\left (15 \, {\left (a + \frac {b}{x}\right )}^{2} b - 10 \, {\left (a + \frac {b}{x}\right )} a b - 2 \, a^{2} b\right )}}{{\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} a^{3} - {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a^{4}} + \frac {15 \, b \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right )}{a^{\frac {7}{2}}}\right )} - \frac {1}{3} \, d {\left (\frac {3 \, \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right )}{a^{\frac {5}{2}}} + \frac {2 \, {\left (4 \, a + \frac {3 \, b}{x}\right )}}{{\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*c*(2*(15*(a + b/x)^2*b - 10*(a + b/x)*a*b - 2*a^2*b)/((a + b/x)^(5/2)*a^3 - (a + b/x)^(3/2)*a^4) + 15*b*lo
g((sqrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a)))/a^(7/2)) - 1/3*d*(3*log((sqrt(a + b/x) - sqrt(a))/(sqrt
(a + b/x) + sqrt(a)))/a^(5/2) + 2*(4*a + 3*b/x)/((a + b/x)^(3/2)*a^2))

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mupad [B]  time = 2.91, size = 87, normalized size = 0.84 \[ \frac {2\,d\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{a^{5/2}}-\frac {\frac {2\,d}{3\,a}+\frac {2\,d\,\left (a+\frac {b}{x}\right )}{a^2}}{{\left (a+\frac {b}{x}\right )}^{3/2}}+\frac {2\,c\,x\,{\left (\frac {a\,x}{b}+1\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (\frac {5}{2},\frac {7}{2};\ \frac {9}{2};\ -\frac {a\,x}{b}\right )}{7\,{\left (a+\frac {b}{x}\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d/x)/(a + b/x)^(5/2),x)

[Out]

(2*d*atanh((a + b/x)^(1/2)/a^(1/2)))/a^(5/2) - ((2*d)/(3*a) + (2*d*(a + b/x))/a^2)/(a + b/x)^(3/2) + (2*c*x*((
a*x)/b + 1)^(5/2)*hypergeom([5/2, 7/2], 9/2, -(a*x)/b))/(7*(a + b/x)^(5/2))

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sympy [B]  time = 155.82, size = 1479, normalized size = 14.36 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*(6*a**17*x**4*sqrt(1 + b/(a*x))/(6*a**(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a**(33/2)*
b**3) + 46*a**16*b*x**3*sqrt(1 + b/(a*x))/(6*a**(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a*
*(33/2)*b**3) + 15*a**16*b*x**3*log(b/(a*x))/(6*a**(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6
*a**(33/2)*b**3) - 30*a**16*b*x**3*log(sqrt(1 + b/(a*x)) + 1)/(6*a**(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**
(35/2)*b**2*x + 6*a**(33/2)*b**3) + 70*a**15*b**2*x**2*sqrt(1 + b/(a*x))/(6*a**(39/2)*x**3 + 18*a**(37/2)*b*x*
*2 + 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**3) + 45*a**15*b**2*x**2*log(b/(a*x))/(6*a**(39/2)*x**3 + 18*a**(37/2
)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**3) - 90*a**15*b**2*x**2*log(sqrt(1 + b/(a*x)) + 1)/(6*a**(39/2
)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**3) + 30*a**14*b**3*x*sqrt(1 + b/(a*x))/(6*
a**(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**3) + 45*a**14*b**3*x*log(b/(a*x))/
(6*a**(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**3) - 90*a**14*b**3*x*log(sqrt(1
+ b/(a*x)) + 1)/(6*a**(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**3) + 15*a**13*
b**4*log(b/(a*x))/(6*a**(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**3) - 30*a**13
*b**4*log(sqrt(1 + b/(a*x)) + 1)/(6*a**(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a**(33/2)*b
**3)) + d*(-8*a**7*x**3*sqrt(1 + b/(a*x))/(3*a**(19/2)*x**3 + 9*a**(17/2)*b*x**2 + 9*a**(15/2)*b**2*x + 3*a**(
13/2)*b**3) - 3*a**7*x**3*log(b/(a*x))/(3*a**(19/2)*x**3 + 9*a**(17/2)*b*x**2 + 9*a**(15/2)*b**2*x + 3*a**(13/
2)*b**3) + 6*a**7*x**3*log(sqrt(1 + b/(a*x)) + 1)/(3*a**(19/2)*x**3 + 9*a**(17/2)*b*x**2 + 9*a**(15/2)*b**2*x
+ 3*a**(13/2)*b**3) - 14*a**6*b*x**2*sqrt(1 + b/(a*x))/(3*a**(19/2)*x**3 + 9*a**(17/2)*b*x**2 + 9*a**(15/2)*b*
*2*x + 3*a**(13/2)*b**3) - 9*a**6*b*x**2*log(b/(a*x))/(3*a**(19/2)*x**3 + 9*a**(17/2)*b*x**2 + 9*a**(15/2)*b**
2*x + 3*a**(13/2)*b**3) + 18*a**6*b*x**2*log(sqrt(1 + b/(a*x)) + 1)/(3*a**(19/2)*x**3 + 9*a**(17/2)*b*x**2 + 9
*a**(15/2)*b**2*x + 3*a**(13/2)*b**3) - 6*a**5*b**2*x*sqrt(1 + b/(a*x))/(3*a**(19/2)*x**3 + 9*a**(17/2)*b*x**2
+ 9*a**(15/2)*b**2*x + 3*a**(13/2)*b**3) - 9*a**5*b**2*x*log(b/(a*x))/(3*a**(19/2)*x**3 + 9*a**(17/2)*b*x**2
+ 9*a**(15/2)*b**2*x + 3*a**(13/2)*b**3) + 18*a**5*b**2*x*log(sqrt(1 + b/(a*x)) + 1)/(3*a**(19/2)*x**3 + 9*a**
(17/2)*b*x**2 + 9*a**(15/2)*b**2*x + 3*a**(13/2)*b**3) - 3*a**4*b**3*log(b/(a*x))/(3*a**(19/2)*x**3 + 9*a**(17
/2)*b*x**2 + 9*a**(15/2)*b**2*x + 3*a**(13/2)*b**3) + 6*a**4*b**3*log(sqrt(1 + b/(a*x)) + 1)/(3*a**(19/2)*x**3
+ 9*a**(17/2)*b*x**2 + 9*a**(15/2)*b**2*x + 3*a**(13/2)*b**3))

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