3.351 \(\int \frac {(A+B x) (a+c x^2)^{5/2}}{x^8} \, dx\)

Optimal. Leaf size=115 \[ -\frac {A \left (a+c x^2\right )^{7/2}}{7 a x^7}-\frac {5 B c^3 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{16 \sqrt {a}}-\frac {5 B c^2 \sqrt {a+c x^2}}{16 x^2}-\frac {B \left (a+c x^2\right )^{5/2}}{6 x^6}-\frac {5 B c \left (a+c x^2\right )^{3/2}}{24 x^4} \]

[Out]

-5/24*B*c*(c*x^2+a)^(3/2)/x^4-1/6*B*(c*x^2+a)^(5/2)/x^6-1/7*A*(c*x^2+a)^(7/2)/a/x^7-5/16*B*c^3*arctanh((c*x^2+
a)^(1/2)/a^(1/2))/a^(1/2)-5/16*B*c^2*(c*x^2+a)^(1/2)/x^2

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Rubi [A]  time = 0.07, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {807, 266, 47, 63, 208} \[ -\frac {A \left (a+c x^2\right )^{7/2}}{7 a x^7}-\frac {5 B c^2 \sqrt {a+c x^2}}{16 x^2}-\frac {5 B c^3 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{16 \sqrt {a}}-\frac {B \left (a+c x^2\right )^{5/2}}{6 x^6}-\frac {5 B c \left (a+c x^2\right )^{3/2}}{24 x^4} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + c*x^2)^(5/2))/x^8,x]

[Out]

(-5*B*c^2*Sqrt[a + c*x^2])/(16*x^2) - (5*B*c*(a + c*x^2)^(3/2))/(24*x^4) - (B*(a + c*x^2)^(5/2))/(6*x^6) - (A*
(a + c*x^2)^(7/2))/(7*a*x^7) - (5*B*c^3*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]])/(16*Sqrt[a])

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 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 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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 807

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

Rubi steps

\begin {align*} \int \frac {(A+B x) \left (a+c x^2\right )^{5/2}}{x^8} \, dx &=-\frac {A \left (a+c x^2\right )^{7/2}}{7 a x^7}+B \int \frac {\left (a+c x^2\right )^{5/2}}{x^7} \, dx\\ &=-\frac {A \left (a+c x^2\right )^{7/2}}{7 a x^7}+\frac {1}{2} B \operatorname {Subst}\left (\int \frac {(a+c x)^{5/2}}{x^4} \, dx,x,x^2\right )\\ &=-\frac {B \left (a+c x^2\right )^{5/2}}{6 x^6}-\frac {A \left (a+c x^2\right )^{7/2}}{7 a x^7}+\frac {1}{12} (5 B c) \operatorname {Subst}\left (\int \frac {(a+c x)^{3/2}}{x^3} \, dx,x,x^2\right )\\ &=-\frac {5 B c \left (a+c x^2\right )^{3/2}}{24 x^4}-\frac {B \left (a+c x^2\right )^{5/2}}{6 x^6}-\frac {A \left (a+c x^2\right )^{7/2}}{7 a x^7}+\frac {1}{16} \left (5 B c^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+c x}}{x^2} \, dx,x,x^2\right )\\ &=-\frac {5 B c^2 \sqrt {a+c x^2}}{16 x^2}-\frac {5 B c \left (a+c x^2\right )^{3/2}}{24 x^4}-\frac {B \left (a+c x^2\right )^{5/2}}{6 x^6}-\frac {A \left (a+c x^2\right )^{7/2}}{7 a x^7}+\frac {1}{32} \left (5 B c^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+c x}} \, dx,x,x^2\right )\\ &=-\frac {5 B c^2 \sqrt {a+c x^2}}{16 x^2}-\frac {5 B c \left (a+c x^2\right )^{3/2}}{24 x^4}-\frac {B \left (a+c x^2\right )^{5/2}}{6 x^6}-\frac {A \left (a+c x^2\right )^{7/2}}{7 a x^7}+\frac {1}{16} \left (5 B c^2\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{c}+\frac {x^2}{c}} \, dx,x,\sqrt {a+c x^2}\right )\\ &=-\frac {5 B c^2 \sqrt {a+c x^2}}{16 x^2}-\frac {5 B c \left (a+c x^2\right )^{3/2}}{24 x^4}-\frac {B \left (a+c x^2\right )^{5/2}}{6 x^6}-\frac {A \left (a+c x^2\right )^{7/2}}{7 a x^7}-\frac {5 B c^3 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{16 \sqrt {a}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 123, normalized size = 1.07 \[ \frac {-\frac {\left (a+c x^2\right ) \left (8 a^3 (6 A+7 B x)+2 a^2 c x^2 (72 A+91 B x)+3 a c^2 x^4 (48 A+77 B x)+48 A c^3 x^6\right )}{a x^7}-105 B c^3 \sqrt {\frac {c x^2}{a}+1} \tanh ^{-1}\left (\sqrt {\frac {c x^2}{a}+1}\right )}{336 \sqrt {a+c x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + c*x^2)^(5/2))/x^8,x]

[Out]

(-(((a + c*x^2)*(48*A*c^3*x^6 + 8*a^3*(6*A + 7*B*x) + 3*a*c^2*x^4*(48*A + 77*B*x) + 2*a^2*c*x^2*(72*A + 91*B*x
)))/(a*x^7)) - 105*B*c^3*Sqrt[1 + (c*x^2)/a]*ArcTanh[Sqrt[1 + (c*x^2)/a]])/(336*Sqrt[a + c*x^2])

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fricas [A]  time = 1.05, size = 238, normalized size = 2.07 \[ \left [\frac {105 \, B \sqrt {a} c^{3} x^{7} \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (48 \, A c^{3} x^{6} + 231 \, B a c^{2} x^{5} + 144 \, A a c^{2} x^{4} + 182 \, B a^{2} c x^{3} + 144 \, A a^{2} c x^{2} + 56 \, B a^{3} x + 48 \, A a^{3}\right )} \sqrt {c x^{2} + a}}{672 \, a x^{7}}, \frac {105 \, B \sqrt {-a} c^{3} x^{7} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) - {\left (48 \, A c^{3} x^{6} + 231 \, B a c^{2} x^{5} + 144 \, A a c^{2} x^{4} + 182 \, B a^{2} c x^{3} + 144 \, A a^{2} c x^{2} + 56 \, B a^{3} x + 48 \, A a^{3}\right )} \sqrt {c x^{2} + a}}{336 \, a x^{7}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(5/2)/x^8,x, algorithm="fricas")

[Out]

[1/672*(105*B*sqrt(a)*c^3*x^7*log(-(c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(48*A*c^3*x^6 + 231*B*a*
c^2*x^5 + 144*A*a*c^2*x^4 + 182*B*a^2*c*x^3 + 144*A*a^2*c*x^2 + 56*B*a^3*x + 48*A*a^3)*sqrt(c*x^2 + a))/(a*x^7
), 1/336*(105*B*sqrt(-a)*c^3*x^7*arctan(sqrt(-a)/sqrt(c*x^2 + a)) - (48*A*c^3*x^6 + 231*B*a*c^2*x^5 + 144*A*a*
c^2*x^4 + 182*B*a^2*c*x^3 + 144*A*a^2*c*x^2 + 56*B*a^3*x + 48*A*a^3)*sqrt(c*x^2 + a))/(a*x^7)]

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giac [B]  time = 0.24, size = 316, normalized size = 2.75 \[ \frac {5 \, B c^{3} \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + a}}{\sqrt {-a}}\right )}{8 \, \sqrt {-a}} + \frac {231 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{13} B c^{3} + 336 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{12} A c^{\frac {7}{2}} - 196 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{11} B a c^{3} + 595 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{9} B a^{2} c^{3} + 1680 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{8} A a^{2} c^{\frac {7}{2}} - 595 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{5} B a^{4} c^{3} + 1008 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{4} A a^{4} c^{\frac {7}{2}} + 196 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{3} B a^{5} c^{3} - 231 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} B a^{6} c^{3} + 48 \, A a^{6} c^{\frac {7}{2}}}{168 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} - a\right )}^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(5/2)/x^8,x, algorithm="giac")

[Out]

5/8*B*c^3*arctan(-(sqrt(c)*x - sqrt(c*x^2 + a))/sqrt(-a))/sqrt(-a) + 1/168*(231*(sqrt(c)*x - sqrt(c*x^2 + a))^
13*B*c^3 + 336*(sqrt(c)*x - sqrt(c*x^2 + a))^12*A*c^(7/2) - 196*(sqrt(c)*x - sqrt(c*x^2 + a))^11*B*a*c^3 + 595
*(sqrt(c)*x - sqrt(c*x^2 + a))^9*B*a^2*c^3 + 1680*(sqrt(c)*x - sqrt(c*x^2 + a))^8*A*a^2*c^(7/2) - 595*(sqrt(c)
*x - sqrt(c*x^2 + a))^5*B*a^4*c^3 + 1008*(sqrt(c)*x - sqrt(c*x^2 + a))^4*A*a^4*c^(7/2) + 196*(sqrt(c)*x - sqrt
(c*x^2 + a))^3*B*a^5*c^3 - 231*(sqrt(c)*x - sqrt(c*x^2 + a))*B*a^6*c^3 + 48*A*a^6*c^(7/2))/((sqrt(c)*x - sqrt(
c*x^2 + a))^2 - a)^7

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maple [A]  time = 0.08, size = 164, normalized size = 1.43 \[ -\frac {5 B \,c^{3} \ln \left (\frac {2 a +2 \sqrt {c \,x^{2}+a}\, \sqrt {a}}{x}\right )}{16 \sqrt {a}}+\frac {5 \sqrt {c \,x^{2}+a}\, B \,c^{3}}{16 a}+\frac {5 \left (c \,x^{2}+a \right )^{\frac {3}{2}} B \,c^{3}}{48 a^{2}}+\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} B \,c^{3}}{16 a^{3}}-\frac {\left (c \,x^{2}+a \right )^{\frac {7}{2}} B \,c^{2}}{16 a^{3} x^{2}}-\frac {\left (c \,x^{2}+a \right )^{\frac {7}{2}} B c}{24 a^{2} x^{4}}-\frac {\left (c \,x^{2}+a \right )^{\frac {7}{2}} B}{6 a \,x^{6}}-\frac {\left (c \,x^{2}+a \right )^{\frac {7}{2}} A}{7 a \,x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+a)^(5/2)/x^8,x)

[Out]

-1/7*A*(c*x^2+a)^(7/2)/a/x^7-1/6*B/a/x^6*(c*x^2+a)^(7/2)-1/24*B*c/a^2/x^4*(c*x^2+a)^(7/2)-1/16*B*c^2/a^3/x^2*(
c*x^2+a)^(7/2)+1/16*B*c^3/a^3*(c*x^2+a)^(5/2)+5/48*B*c^3/a^2*(c*x^2+a)^(3/2)-5/16*B*c^3/a^(1/2)*ln((2*a+2*(c*x
^2+a)^(1/2)*a^(1/2))/x)+5/16*B*c^3/a*(c*x^2+a)^(1/2)

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maxima [A]  time = 0.63, size = 152, normalized size = 1.32 \[ -\frac {5 \, B c^{3} \operatorname {arsinh}\left (\frac {a}{\sqrt {a c} {\left | x \right |}}\right )}{16 \, \sqrt {a}} + \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} B c^{3}}{16 \, a^{3}} + \frac {5 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} B c^{3}}{48 \, a^{2}} + \frac {5 \, \sqrt {c x^{2} + a} B c^{3}}{16 \, a} - \frac {{\left (c x^{2} + a\right )}^{\frac {7}{2}} B c^{2}}{16 \, a^{3} x^{2}} - \frac {{\left (c x^{2} + a\right )}^{\frac {7}{2}} B c}{24 \, a^{2} x^{4}} - \frac {{\left (c x^{2} + a\right )}^{\frac {7}{2}} B}{6 \, a x^{6}} - \frac {{\left (c x^{2} + a\right )}^{\frac {7}{2}} A}{7 \, a x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(5/2)/x^8,x, algorithm="maxima")

[Out]

-5/16*B*c^3*arcsinh(a/(sqrt(a*c)*abs(x)))/sqrt(a) + 1/16*(c*x^2 + a)^(5/2)*B*c^3/a^3 + 5/48*(c*x^2 + a)^(3/2)*
B*c^3/a^2 + 5/16*sqrt(c*x^2 + a)*B*c^3/a - 1/16*(c*x^2 + a)^(7/2)*B*c^2/(a^3*x^2) - 1/24*(c*x^2 + a)^(7/2)*B*c
/(a^2*x^4) - 1/6*(c*x^2 + a)^(7/2)*B/(a*x^6) - 1/7*(c*x^2 + a)^(7/2)*A/(a*x^7)

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mupad [B]  time = 4.39, size = 150, normalized size = 1.30 \[ \frac {5\,B\,a\,{\left (c\,x^2+a\right )}^{3/2}}{6\,x^6}-\frac {11\,B\,{\left (c\,x^2+a\right )}^{5/2}}{16\,x^6}-\frac {A\,a^2\,\sqrt {c\,x^2+a}}{7\,x^7}-\frac {5\,B\,a^2\,\sqrt {c\,x^2+a}}{16\,x^6}-\frac {3\,A\,c^2\,\sqrt {c\,x^2+a}}{7\,x^3}-\frac {A\,c^3\,\sqrt {c\,x^2+a}}{7\,a\,x}-\frac {3\,A\,a\,c\,\sqrt {c\,x^2+a}}{7\,x^5}+\frac {B\,c^3\,\mathrm {atan}\left (\frac {\sqrt {c\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{16\,\sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + c*x^2)^(5/2)*(A + B*x))/x^8,x)

[Out]

(B*c^3*atan(((a + c*x^2)^(1/2)*1i)/a^(1/2))*5i)/(16*a^(1/2)) - (11*B*(a + c*x^2)^(5/2))/(16*x^6) + (5*B*a*(a +
c*x^2)^(3/2))/(6*x^6) - (A*a^2*(a + c*x^2)^(1/2))/(7*x^7) - (5*B*a^2*(a + c*x^2)^(1/2))/(16*x^6) - (3*A*c^2*(
a + c*x^2)^(1/2))/(7*x^3) - (A*c^3*(a + c*x^2)^(1/2))/(7*a*x) - (3*A*a*c*(a + c*x^2)^(1/2))/(7*x^5)

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sympy [B]  time = 16.82, size = 605, normalized size = 5.26 \[ - \frac {15 A a^{7} c^{\frac {9}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{105 a^{5} c^{4} x^{6} + 210 a^{4} c^{5} x^{8} + 105 a^{3} c^{6} x^{10}} - \frac {33 A a^{6} c^{\frac {11}{2}} x^{2} \sqrt {\frac {a}{c x^{2}} + 1}}{105 a^{5} c^{4} x^{6} + 210 a^{4} c^{5} x^{8} + 105 a^{3} c^{6} x^{10}} - \frac {17 A a^{5} c^{\frac {13}{2}} x^{4} \sqrt {\frac {a}{c x^{2}} + 1}}{105 a^{5} c^{4} x^{6} + 210 a^{4} c^{5} x^{8} + 105 a^{3} c^{6} x^{10}} - \frac {3 A a^{4} c^{\frac {15}{2}} x^{6} \sqrt {\frac {a}{c x^{2}} + 1}}{105 a^{5} c^{4} x^{6} + 210 a^{4} c^{5} x^{8} + 105 a^{3} c^{6} x^{10}} - \frac {12 A a^{3} c^{\frac {17}{2}} x^{8} \sqrt {\frac {a}{c x^{2}} + 1}}{105 a^{5} c^{4} x^{6} + 210 a^{4} c^{5} x^{8} + 105 a^{3} c^{6} x^{10}} - \frac {8 A a^{2} c^{\frac {19}{2}} x^{10} \sqrt {\frac {a}{c x^{2}} + 1}}{105 a^{5} c^{4} x^{6} + 210 a^{4} c^{5} x^{8} + 105 a^{3} c^{6} x^{10}} - \frac {2 A a c^{\frac {3}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{5 x^{4}} - \frac {7 A c^{\frac {5}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{15 x^{2}} - \frac {A c^{\frac {7}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{15 a} - \frac {B a^{3}}{6 \sqrt {c} x^{7} \sqrt {\frac {a}{c x^{2}} + 1}} - \frac {17 B a^{2} \sqrt {c}}{24 x^{5} \sqrt {\frac {a}{c x^{2}} + 1}} - \frac {35 B a c^{\frac {3}{2}}}{48 x^{3} \sqrt {\frac {a}{c x^{2}} + 1}} - \frac {B c^{\frac {5}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{2 x} - \frac {3 B c^{\frac {5}{2}}}{16 x \sqrt {\frac {a}{c x^{2}} + 1}} - \frac {5 B c^{3} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {c} x} \right )}}{16 \sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+a)**(5/2)/x**8,x)

[Out]

-15*A*a**7*c**(9/2)*sqrt(a/(c*x**2) + 1)/(105*a**5*c**4*x**6 + 210*a**4*c**5*x**8 + 105*a**3*c**6*x**10) - 33*
A*a**6*c**(11/2)*x**2*sqrt(a/(c*x**2) + 1)/(105*a**5*c**4*x**6 + 210*a**4*c**5*x**8 + 105*a**3*c**6*x**10) - 1
7*A*a**5*c**(13/2)*x**4*sqrt(a/(c*x**2) + 1)/(105*a**5*c**4*x**6 + 210*a**4*c**5*x**8 + 105*a**3*c**6*x**10) -
3*A*a**4*c**(15/2)*x**6*sqrt(a/(c*x**2) + 1)/(105*a**5*c**4*x**6 + 210*a**4*c**5*x**8 + 105*a**3*c**6*x**10)
- 12*A*a**3*c**(17/2)*x**8*sqrt(a/(c*x**2) + 1)/(105*a**5*c**4*x**6 + 210*a**4*c**5*x**8 + 105*a**3*c**6*x**10
) - 8*A*a**2*c**(19/2)*x**10*sqrt(a/(c*x**2) + 1)/(105*a**5*c**4*x**6 + 210*a**4*c**5*x**8 + 105*a**3*c**6*x**
10) - 2*A*a*c**(3/2)*sqrt(a/(c*x**2) + 1)/(5*x**4) - 7*A*c**(5/2)*sqrt(a/(c*x**2) + 1)/(15*x**2) - A*c**(7/2)*
sqrt(a/(c*x**2) + 1)/(15*a) - B*a**3/(6*sqrt(c)*x**7*sqrt(a/(c*x**2) + 1)) - 17*B*a**2*sqrt(c)/(24*x**5*sqrt(a
/(c*x**2) + 1)) - 35*B*a*c**(3/2)/(48*x**3*sqrt(a/(c*x**2) + 1)) - B*c**(5/2)*sqrt(a/(c*x**2) + 1)/(2*x) - 3*B
*c**(5/2)/(16*x*sqrt(a/(c*x**2) + 1)) - 5*B*c**3*asinh(sqrt(a)/(sqrt(c)*x))/(16*sqrt(a))

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