3.4.35 \(\int \frac {(A+B x) (a+c x^2)^{3/2}}{x^5} \, dx\)

Optimal. Leaf size=111 \[ -\frac {c \sqrt {a+c x^2} (3 A+8 B x)}{8 x^2}-\frac {\left (a+c x^2\right )^{3/2} (3 A+4 B x)}{12 x^4}-\frac {3 A c^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{8 \sqrt {a}}+B c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right ) \]

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Rubi [A]  time = 0.09, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {811, 844, 217, 206, 266, 63, 208} \begin {gather*} -\frac {c \sqrt {a+c x^2} (3 A+8 B x)}{8 x^2}-\frac {\left (a+c x^2\right )^{3/2} (3 A+4 B x)}{12 x^4}-\frac {3 A c^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{8 \sqrt {a}}+B c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + c*x^2)^(3/2))/x^5,x]

[Out]

-(c*(3*A + 8*B*x)*Sqrt[a + c*x^2])/(8*x^2) - ((3*A + 4*B*x)*(a + c*x^2)^(3/2))/(12*x^4) + B*c^(3/2)*ArcTanh[(S
qrt[c]*x)/Sqrt[a + c*x^2]] - (3*A*c^2*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]])/(8*Sqrt[a])

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 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 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 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]

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 811

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

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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] &&  !IGtQ[m, 0]

Rubi steps

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

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Mathematica [C]  time = 0.08, size = 114, normalized size = 1.03 \begin {gather*} -\frac {\sqrt {a+c x^2} \left (8 a^2 B x \, _2F_1\left (-\frac {3}{2},-\frac {3}{2};-\frac {1}{2};-\frac {c x^2}{a}\right )+9 A c^2 x^4 \tanh ^{-1}\left (\sqrt {\frac {c x^2}{a}+1}\right )+3 a A \left (2 a+5 c x^2\right ) \sqrt {\frac {c x^2}{a}+1}\right )}{24 a x^4 \sqrt {\frac {c x^2}{a}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + c*x^2)^(3/2))/x^5,x]

[Out]

-1/24*(Sqrt[a + c*x^2]*(3*a*A*(2*a + 5*c*x^2)*Sqrt[1 + (c*x^2)/a] + 9*A*c^2*x^4*ArcTanh[Sqrt[1 + (c*x^2)/a]] +
8*a^2*B*x*Hypergeometric2F1[-3/2, -3/2, -1/2, -((c*x^2)/a)]))/(a*x^4*Sqrt[1 + (c*x^2)/a])

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IntegrateAlgebraic [A]  time = 0.63, size = 117, normalized size = 1.05 \begin {gather*} \frac {\sqrt {a+c x^2} \left (-6 a A-8 a B x-15 A c x^2-32 B c x^3\right )}{24 x^4}+\frac {3 A c^2 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}-\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{4 \sqrt {a}}-B c^{3/2} \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(a + c*x^2)^(3/2))/x^5,x]

[Out]

(Sqrt[a + c*x^2]*(-6*a*A - 8*a*B*x - 15*A*c*x^2 - 32*B*c*x^3))/(24*x^4) + (3*A*c^2*ArcTanh[(Sqrt[c]*x)/Sqrt[a]
- Sqrt[a + c*x^2]/Sqrt[a]])/(4*Sqrt[a]) - B*c^(3/2)*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]]

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fricas [A]  time = 0.48, size = 474, normalized size = 4.27 \begin {gather*} \left [\frac {24 \, B a c^{\frac {3}{2}} x^{4} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 9 \, A \sqrt {a} c^{2} x^{4} \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (32 \, B a c x^{3} + 15 \, A a c x^{2} + 8 \, B a^{2} x + 6 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{48 \, a x^{4}}, -\frac {48 \, B a \sqrt {-c} c x^{4} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - 9 \, A \sqrt {a} c^{2} x^{4} \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (32 \, B a c x^{3} + 15 \, A a c x^{2} + 8 \, B a^{2} x + 6 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{48 \, a x^{4}}, \frac {9 \, A \sqrt {-a} c^{2} x^{4} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) + 12 \, B a c^{\frac {3}{2}} x^{4} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) - {\left (32 \, B a c x^{3} + 15 \, A a c x^{2} + 8 \, B a^{2} x + 6 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{24 \, a x^{4}}, -\frac {24 \, B a \sqrt {-c} c x^{4} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - 9 \, A \sqrt {-a} c^{2} x^{4} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) + {\left (32 \, B a c x^{3} + 15 \, A a c x^{2} + 8 \, B a^{2} x + 6 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{24 \, a x^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(3/2)/x^5,x, algorithm="fricas")

[Out]

[1/48*(24*B*a*c^(3/2)*x^4*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 9*A*sqrt(a)*c^2*x^4*log(-(c*x^2 -
2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(32*B*a*c*x^3 + 15*A*a*c*x^2 + 8*B*a^2*x + 6*A*a^2)*sqrt(c*x^2 + a))
/(a*x^4), -1/48*(48*B*a*sqrt(-c)*c*x^4*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - 9*A*sqrt(a)*c^2*x^4*log(-(c*x^2 -
2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(32*B*a*c*x^3 + 15*A*a*c*x^2 + 8*B*a^2*x + 6*A*a^2)*sqrt(c*x^2 + a))
/(a*x^4), 1/24*(9*A*sqrt(-a)*c^2*x^4*arctan(sqrt(-a)/sqrt(c*x^2 + a)) + 12*B*a*c^(3/2)*x^4*log(-2*c*x^2 - 2*sq
rt(c*x^2 + a)*sqrt(c)*x - a) - (32*B*a*c*x^3 + 15*A*a*c*x^2 + 8*B*a^2*x + 6*A*a^2)*sqrt(c*x^2 + a))/(a*x^4), -
1/24*(24*B*a*sqrt(-c)*c*x^4*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - 9*A*sqrt(-a)*c^2*x^4*arctan(sqrt(-a)/sqrt(c*x
^2 + a)) + (32*B*a*c*x^3 + 15*A*a*c*x^2 + 8*B*a^2*x + 6*A*a^2)*sqrt(c*x^2 + a))/(a*x^4)]

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giac [B]  time = 0.22, size = 285, normalized size = 2.57 \begin {gather*} \frac {3 \, A c^{2} \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a}} - B c^{\frac {3}{2}} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right ) + \frac {15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{7} A c^{2} + 48 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{6} B a c^{\frac {3}{2}} + 9 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{5} A a c^{2} - 96 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{4} B a^{2} c^{\frac {3}{2}} + 9 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{3} A a^{2} c^{2} + 80 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} B a^{3} c^{\frac {3}{2}} + 15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} A a^{3} c^{2} - 32 \, B a^{4} c^{\frac {3}{2}}}{12 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} - a\right )}^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(3/2)/x^5,x, algorithm="giac")

[Out]

3/4*A*c^2*arctan(-(sqrt(c)*x - sqrt(c*x^2 + a))/sqrt(-a))/sqrt(-a) - B*c^(3/2)*log(abs(-sqrt(c)*x + sqrt(c*x^2
+ a))) + 1/12*(15*(sqrt(c)*x - sqrt(c*x^2 + a))^7*A*c^2 + 48*(sqrt(c)*x - sqrt(c*x^2 + a))^6*B*a*c^(3/2) + 9*
(sqrt(c)*x - sqrt(c*x^2 + a))^5*A*a*c^2 - 96*(sqrt(c)*x - sqrt(c*x^2 + a))^4*B*a^2*c^(3/2) + 9*(sqrt(c)*x - sq
rt(c*x^2 + a))^3*A*a^2*c^2 + 80*(sqrt(c)*x - sqrt(c*x^2 + a))^2*B*a^3*c^(3/2) + 15*(sqrt(c)*x - sqrt(c*x^2 + a
))*A*a^3*c^2 - 32*B*a^4*c^(3/2))/((sqrt(c)*x - sqrt(c*x^2 + a))^2 - a)^4

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maple [B]  time = 0.05, size = 202, normalized size = 1.82 \begin {gather*} -\frac {3 A \,c^{2} \ln \left (\frac {2 a +2 \sqrt {c \,x^{2}+a}\, \sqrt {a}}{x}\right )}{8 \sqrt {a}}+B \,c^{\frac {3}{2}} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )+\frac {\sqrt {c \,x^{2}+a}\, B \,c^{2} x}{a}+\frac {3 \sqrt {c \,x^{2}+a}\, A \,c^{2}}{8 a}+\frac {2 \left (c \,x^{2}+a \right )^{\frac {3}{2}} B \,c^{2} x}{3 a^{2}}+\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} A \,c^{2}}{8 a^{2}}-\frac {2 \left (c \,x^{2}+a \right )^{\frac {5}{2}} B c}{3 a^{2} x}-\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} A c}{8 a^{2} x^{2}}-\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} B}{3 a \,x^{3}}-\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} A}{4 a \,x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+a)^(3/2)/x^5,x)

[Out]

-1/4*A/a/x^4*(c*x^2+a)^(5/2)-1/8*A*c/a^2/x^2*(c*x^2+a)^(5/2)+1/8*A*c^2/a^2*(c*x^2+a)^(3/2)-3/8*A*c^2/a^(1/2)*l
n((2*a+2*(c*x^2+a)^(1/2)*a^(1/2))/x)+3/8*A*c^2/a*(c*x^2+a)^(1/2)-1/3*B/a/x^3*(c*x^2+a)^(5/2)-2/3*B*c/a^2/x*(c*
x^2+a)^(5/2)+2/3*B*c^2/a^2*x*(c*x^2+a)^(3/2)+B*c^2/a*x*(c*x^2+a)^(1/2)+B*c^(3/2)*ln(c^(1/2)*x+(c*x^2+a)^(1/2))

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maxima [A]  time = 0.56, size = 164, normalized size = 1.48 \begin {gather*} \frac {\sqrt {c x^{2} + a} B c^{2} x}{a} + B c^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right ) - \frac {3 \, A c^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a c} {\left | x \right |}}\right )}{8 \, \sqrt {a}} + \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} A c^{2}}{8 \, a^{2}} + \frac {3 \, \sqrt {c x^{2} + a} A c^{2}}{8 \, a} - \frac {2 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} B c}{3 \, a x} - \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} A c}{8 \, a^{2} x^{2}} - \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} B}{3 \, a x^{3}} - \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} A}{4 \, a x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(3/2)/x^5,x, algorithm="maxima")

[Out]

sqrt(c*x^2 + a)*B*c^2*x/a + B*c^(3/2)*arcsinh(c*x/sqrt(a*c)) - 3/8*A*c^2*arcsinh(a/(sqrt(a*c)*abs(x)))/sqrt(a)
+ 1/8*(c*x^2 + a)^(3/2)*A*c^2/a^2 + 3/8*sqrt(c*x^2 + a)*A*c^2/a - 2/3*(c*x^2 + a)^(3/2)*B*c/(a*x) - 1/8*(c*x^
2 + a)^(5/2)*A*c/(a^2*x^2) - 1/3*(c*x^2 + a)^(5/2)*B/(a*x^3) - 1/4*(c*x^2 + a)^(5/2)*A/(a*x^4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,x^2+a\right )}^{3/2}\,\left (A+B\,x\right )}{x^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + c*x^2)^(3/2)*(A + B*x))/x^5,x)

[Out]

int(((a + c*x^2)^(3/2)*(A + B*x))/x^5, x)

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sympy [B]  time = 9.34, size = 236, normalized size = 2.13 \begin {gather*} - \frac {A a^{2}}{4 \sqrt {c} x^{5} \sqrt {\frac {a}{c x^{2}} + 1}} - \frac {3 A a \sqrt {c}}{8 x^{3} \sqrt {\frac {a}{c x^{2}} + 1}} - \frac {A c^{\frac {3}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{2 x} - \frac {A c^{\frac {3}{2}}}{8 x \sqrt {\frac {a}{c x^{2}} + 1}} - \frac {3 A c^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {c} x} \right )}}{8 \sqrt {a}} - \frac {B \sqrt {a} c}{x \sqrt {1 + \frac {c x^{2}}{a}}} - \frac {B a \sqrt {c} \sqrt {\frac {a}{c x^{2}} + 1}}{3 x^{2}} - \frac {B c^{\frac {3}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{3} + B c^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )} - \frac {B c^{2} x}{\sqrt {a} \sqrt {1 + \frac {c x^{2}}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+a)**(3/2)/x**5,x)

[Out]

-A*a**2/(4*sqrt(c)*x**5*sqrt(a/(c*x**2) + 1)) - 3*A*a*sqrt(c)/(8*x**3*sqrt(a/(c*x**2) + 1)) - A*c**(3/2)*sqrt(
a/(c*x**2) + 1)/(2*x) - A*c**(3/2)/(8*x*sqrt(a/(c*x**2) + 1)) - 3*A*c**2*asinh(sqrt(a)/(sqrt(c)*x))/(8*sqrt(a)
) - B*sqrt(a)*c/(x*sqrt(1 + c*x**2/a)) - B*a*sqrt(c)*sqrt(a/(c*x**2) + 1)/(3*x**2) - B*c**(3/2)*sqrt(a/(c*x**2
) + 1)/3 + B*c**(3/2)*asinh(sqrt(c)*x/sqrt(a)) - B*c**2*x/(sqrt(a)*sqrt(1 + c*x**2/a))

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