3.272 \(\int \frac {(d+c^2 d x^2)^{3/2} (a+b \sinh ^{-1}(c x))^2}{x^3} \, dx\)

Optimal. Leaf size=541 \[ -\frac {3 b c^2 d \sqrt {c^2 d x^2+d} \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {c^2 x^2+1}}+\frac {3 b c^2 d \sqrt {c^2 d x^2+d} \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {c^2 x^2+1}}+\frac {3}{2} c^2 d \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {b c d \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{x \sqrt {c^2 x^2+1}}-\frac {\left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 x^2}-\frac {3 c^2 d \sqrt {c^2 d x^2+d} \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt {c^2 x^2+1}}-\frac {3 a b c^3 d x \sqrt {c^2 d x^2+d}}{\sqrt {c^2 x^2+1}}+\frac {b c^3 d x \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {c^2 x^2+1}}+\frac {3 b^2 c^2 d \sqrt {c^2 d x^2+d} \text {Li}_3\left (-e^{\sinh ^{-1}(c x)}\right )}{\sqrt {c^2 x^2+1}}-\frac {3 b^2 c^2 d \sqrt {c^2 d x^2+d} \text {Li}_3\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {c^2 x^2+1}}+2 b^2 c^2 d \sqrt {c^2 d x^2+d}-\frac {b^2 c^2 d \sqrt {c^2 d x^2+d} \tanh ^{-1}\left (\sqrt {c^2 x^2+1}\right )}{\sqrt {c^2 x^2+1}}-\frac {3 b^2 c^3 d x \sqrt {c^2 d x^2+d} \sinh ^{-1}(c x)}{\sqrt {c^2 x^2+1}} \]

[Out]

-1/2*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))^2/x^2+2*b^2*c^2*d*(c^2*d*x^2+d)^(1/2)+3/2*c^2*d*(a+b*arcsinh(c*x))
^2*(c^2*d*x^2+d)^(1/2)-3*a*b*c^3*d*x*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)-3*b^2*c^3*d*x*arcsinh(c*x)*(c^2*d*x
^2+d)^(1/2)/(c^2*x^2+1)^(1/2)-b*c*d*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2)/x/(c^2*x^2+1)^(1/2)+b*c^3*d*x*(a+b*
arcsinh(c*x))*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)-3*c^2*d*(a+b*arcsinh(c*x))^2*arctanh(c*x+(c^2*x^2+1)^(1/2)
)*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)-b^2*c^2*d*arctanh((c^2*x^2+1)^(1/2))*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(
1/2)-3*b*c^2*d*(a+b*arcsinh(c*x))*polylog(2,-c*x-(c^2*x^2+1)^(1/2))*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)+3*b*
c^2*d*(a+b*arcsinh(c*x))*polylog(2,c*x+(c^2*x^2+1)^(1/2))*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)+3*b^2*c^2*d*po
lylog(3,-c*x-(c^2*x^2+1)^(1/2))*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)-3*b^2*c^2*d*polylog(3,c*x+(c^2*x^2+1)^(1
/2))*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.65, antiderivative size = 541, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 15, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.536, Rules used = {5739, 5742, 5760, 4182, 2531, 2282, 6589, 5653, 261, 14, 5730, 446, 80, 63, 208} \[ -\frac {3 b c^2 d \sqrt {c^2 d x^2+d} \text {PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {c^2 x^2+1}}+\frac {3 b c^2 d \sqrt {c^2 d x^2+d} \text {PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {c^2 x^2+1}}+\frac {3 b^2 c^2 d \sqrt {c^2 d x^2+d} \text {PolyLog}\left (3,-e^{\sinh ^{-1}(c x)}\right )}{\sqrt {c^2 x^2+1}}-\frac {3 b^2 c^2 d \sqrt {c^2 d x^2+d} \text {PolyLog}\left (3,e^{\sinh ^{-1}(c x)}\right )}{\sqrt {c^2 x^2+1}}-\frac {3 a b c^3 d x \sqrt {c^2 d x^2+d}}{\sqrt {c^2 x^2+1}}+\frac {b c^3 d x \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {c^2 x^2+1}}+\frac {3}{2} c^2 d \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {b c d \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{x \sqrt {c^2 x^2+1}}-\frac {\left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 x^2}-\frac {3 c^2 d \sqrt {c^2 d x^2+d} \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt {c^2 x^2+1}}+2 b^2 c^2 d \sqrt {c^2 d x^2+d}-\frac {3 b^2 c^3 d x \sqrt {c^2 d x^2+d} \sinh ^{-1}(c x)}{\sqrt {c^2 x^2+1}}-\frac {b^2 c^2 d \sqrt {c^2 d x^2+d} \tanh ^{-1}\left (\sqrt {c^2 x^2+1}\right )}{\sqrt {c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

Int[((d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x])^2)/x^3,x]

[Out]

2*b^2*c^2*d*Sqrt[d + c^2*d*x^2] - (3*a*b*c^3*d*x*Sqrt[d + c^2*d*x^2])/Sqrt[1 + c^2*x^2] - (3*b^2*c^3*d*x*Sqrt[
d + c^2*d*x^2]*ArcSinh[c*x])/Sqrt[1 + c^2*x^2] - (b*c*d*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(x*Sqrt[1 +
c^2*x^2]) + (b*c^3*d*x*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/Sqrt[1 + c^2*x^2] + (3*c^2*d*Sqrt[d + c^2*d*x
^2]*(a + b*ArcSinh[c*x])^2)/2 - ((d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x])^2)/(2*x^2) - (3*c^2*d*Sqrt[d + c^2
*d*x^2]*(a + b*ArcSinh[c*x])^2*ArcTanh[E^ArcSinh[c*x]])/Sqrt[1 + c^2*x^2] - (b^2*c^2*d*Sqrt[d + c^2*d*x^2]*Arc
Tanh[Sqrt[1 + c^2*x^2]])/Sqrt[1 + c^2*x^2] - (3*b*c^2*d*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])*PolyLog[2, -E
^ArcSinh[c*x]])/Sqrt[1 + c^2*x^2] + (3*b*c^2*d*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])*PolyLog[2, E^ArcSinh[c
*x]])/Sqrt[1 + c^2*x^2] + (3*b^2*c^2*d*Sqrt[d + c^2*d*x^2]*PolyLog[3, -E^ArcSinh[c*x]])/Sqrt[1 + c^2*x^2] - (3
*b^2*c^2*d*Sqrt[d + c^2*d*x^2]*PolyLog[3, E^ArcSinh[c*x]])/Sqrt[1 + c^2*x^2]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
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 261

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

Rule 446

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

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 4182

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*Ar
cTanh[E^(-(I*e) + f*fz*x)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 - E^(-(I*e) + f*
fz*x)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e) + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, x] && IGtQ[m, 0]

Rule 5653

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*ArcSinh[c*x])^n, x] - Dist[b*c*n, In
t[(x*(a + b*ArcSinh[c*x])^(n - 1))/Sqrt[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c}, x] && GtQ[n, 0]

Rule 5730

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSinh[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/Sqrt[1
+ c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]

Rule 5739

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp
[((f*x)^(m + 1)*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n)/(f*(m + 1)), x] + (-Dist[(2*e*p)/(f^2*(m + 1)), Int[(f*x
)^(m + 2)*(d + e*x^2)^(p - 1)*(a + b*ArcSinh[c*x])^n, x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^FracPart[p
])/(f*(m + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 + c^2*x^2)^(p - 1/2)*(a + b*ArcSinh[c*x])^(n -
1), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1]

Rule 5742

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(
(f*x)^(m + 1)*Sqrt[d + e*x^2]*(a + b*ArcSinh[c*x])^n)/(f*(m + 2)), x] + (Dist[Sqrt[d + e*x^2]/((m + 2)*Sqrt[1
+ c^2*x^2]), Int[((f*x)^m*(a + b*ArcSinh[c*x])^n)/Sqrt[1 + c^2*x^2], x], x] - Dist[(b*c*n*Sqrt[d + e*x^2])/(f*
(m + 2)*Sqrt[1 + c^2*x^2]), Int[(f*x)^(m + 1)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f
, m}, x] && EqQ[e, c^2*d] && GtQ[n, 0] &&  !LtQ[m, -1] && (RationalQ[m] || EqQ[n, 1])

Rule 5760

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[1/(c^(m
+ 1)*Sqrt[d]), Subst[Int[(a + b*x)^n*Sinh[x]^m, x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[
e, c^2*d] && GtQ[d, 0] && IGtQ[n, 0] && IntegerQ[m]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int \frac {\left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{x^3} \, dx &=-\frac {\left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 x^2}+\frac {1}{2} \left (3 c^2 d\right ) \int \frac {\sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{x} \, dx+\frac {\left (b c d \sqrt {d+c^2 d x^2}\right ) \int \frac {\left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{x^2} \, dx}{\sqrt {1+c^2 x^2}}\\ &=-\frac {b c d \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{x \sqrt {1+c^2 x^2}}+\frac {b c^3 d x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}}+\frac {3}{2} c^2 d \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {\left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 x^2}+\frac {\left (3 c^2 d \sqrt {d+c^2 d x^2}\right ) \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{x \sqrt {1+c^2 x^2}} \, dx}{2 \sqrt {1+c^2 x^2}}-\frac {\left (b^2 c^2 d \sqrt {d+c^2 d x^2}\right ) \int \frac {-1+c^2 x^2}{x \sqrt {1+c^2 x^2}} \, dx}{\sqrt {1+c^2 x^2}}-\frac {\left (3 b c^3 d \sqrt {d+c^2 d x^2}\right ) \int \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{\sqrt {1+c^2 x^2}}\\ &=-\frac {3 a b c^3 d x \sqrt {d+c^2 d x^2}}{\sqrt {1+c^2 x^2}}-\frac {b c d \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{x \sqrt {1+c^2 x^2}}+\frac {b c^3 d x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}}+\frac {3}{2} c^2 d \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {\left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 x^2}+\frac {\left (3 c^2 d \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int (a+b x)^2 \text {csch}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{2 \sqrt {1+c^2 x^2}}-\frac {\left (b^2 c^2 d \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int \frac {-1+c^2 x}{x \sqrt {1+c^2 x}} \, dx,x,x^2\right )}{2 \sqrt {1+c^2 x^2}}-\frac {\left (3 b^2 c^3 d \sqrt {d+c^2 d x^2}\right ) \int \sinh ^{-1}(c x) \, dx}{\sqrt {1+c^2 x^2}}\\ &=-b^2 c^2 d \sqrt {d+c^2 d x^2}-\frac {3 a b c^3 d x \sqrt {d+c^2 d x^2}}{\sqrt {1+c^2 x^2}}-\frac {3 b^2 c^3 d x \sqrt {d+c^2 d x^2} \sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}}-\frac {b c d \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{x \sqrt {1+c^2 x^2}}+\frac {b c^3 d x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}}+\frac {3}{2} c^2 d \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {\left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 x^2}-\frac {3 c^2 d \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}-\frac {\left (3 b c^2 d \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int (a+b x) \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}}+\frac {\left (3 b c^2 d \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int (a+b x) \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}}+\frac {\left (b^2 c^2 d \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1+c^2 x}} \, dx,x,x^2\right )}{2 \sqrt {1+c^2 x^2}}+\frac {\left (3 b^2 c^4 d \sqrt {d+c^2 d x^2}\right ) \int \frac {x}{\sqrt {1+c^2 x^2}} \, dx}{\sqrt {1+c^2 x^2}}\\ &=2 b^2 c^2 d \sqrt {d+c^2 d x^2}-\frac {3 a b c^3 d x \sqrt {d+c^2 d x^2}}{\sqrt {1+c^2 x^2}}-\frac {3 b^2 c^3 d x \sqrt {d+c^2 d x^2} \sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}}-\frac {b c d \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{x \sqrt {1+c^2 x^2}}+\frac {b c^3 d x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}}+\frac {3}{2} c^2 d \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {\left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 x^2}-\frac {3 c^2 d \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}-\frac {3 b c^2 d \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}+\frac {3 b c^2 d \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}+\frac {\left (b^2 d \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {1}{c^2}+\frac {x^2}{c^2}} \, dx,x,\sqrt {1+c^2 x^2}\right )}{\sqrt {1+c^2 x^2}}+\frac {\left (3 b^2 c^2 d \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}}-\frac {\left (3 b^2 c^2 d \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}}\\ &=2 b^2 c^2 d \sqrt {d+c^2 d x^2}-\frac {3 a b c^3 d x \sqrt {d+c^2 d x^2}}{\sqrt {1+c^2 x^2}}-\frac {3 b^2 c^3 d x \sqrt {d+c^2 d x^2} \sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}}-\frac {b c d \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{x \sqrt {1+c^2 x^2}}+\frac {b c^3 d x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}}+\frac {3}{2} c^2 d \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {\left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 x^2}-\frac {3 c^2 d \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}-\frac {b^2 c^2 d \sqrt {d+c^2 d x^2} \tanh ^{-1}\left (\sqrt {1+c^2 x^2}\right )}{\sqrt {1+c^2 x^2}}-\frac {3 b c^2 d \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}+\frac {3 b c^2 d \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}+\frac {\left (3 b^2 c^2 d \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}-\frac {\left (3 b^2 c^2 d \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}\\ &=2 b^2 c^2 d \sqrt {d+c^2 d x^2}-\frac {3 a b c^3 d x \sqrt {d+c^2 d x^2}}{\sqrt {1+c^2 x^2}}-\frac {3 b^2 c^3 d x \sqrt {d+c^2 d x^2} \sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}}-\frac {b c d \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{x \sqrt {1+c^2 x^2}}+\frac {b c^3 d x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}}+\frac {3}{2} c^2 d \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {\left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 x^2}-\frac {3 c^2 d \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}-\frac {b^2 c^2 d \sqrt {d+c^2 d x^2} \tanh ^{-1}\left (\sqrt {1+c^2 x^2}\right )}{\sqrt {1+c^2 x^2}}-\frac {3 b c^2 d \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}+\frac {3 b c^2 d \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}+\frac {3 b^2 c^2 d \sqrt {d+c^2 d x^2} \text {Li}_3\left (-e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}-\frac {3 b^2 c^2 d \sqrt {d+c^2 d x^2} \text {Li}_3\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 7.88, size = 771, normalized size = 1.43 \[ -\frac {3}{2} a^2 c^2 d^{3/2} \log \left (\sqrt {d} \sqrt {d \left (c^2 x^2+1\right )}+d\right )+\frac {3}{2} a^2 c^2 d^{3/2} \log (x)+\sqrt {d \left (c^2 x^2+1\right )} \left (a^2 c^2 d-\frac {a^2 d}{2 x^2}\right )+\frac {2 a b c^2 d \sqrt {d \left (c^2 x^2+1\right )} \left (\sqrt {c^2 x^2+1} \sinh ^{-1}(c x)+\text {Li}_2\left (-e^{-\sinh ^{-1}(c x)}\right )-\text {Li}_2\left (e^{-\sinh ^{-1}(c x)}\right )-c x+\sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )-\sinh ^{-1}(c x) \log \left (e^{-\sinh ^{-1}(c x)}+1\right )\right )}{\sqrt {c^2 x^2+1}}+\frac {a b c^2 d \sqrt {d \left (c^2 x^2+1\right )} \left (4 \text {Li}_2\left (-e^{-\sinh ^{-1}(c x)}\right )-4 \text {Li}_2\left (e^{-\sinh ^{-1}(c x)}\right )+4 \sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )-4 \sinh ^{-1}(c x) \log \left (e^{-\sinh ^{-1}(c x)}+1\right )+2 \tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-2 \coth \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-\sinh ^{-1}(c x) \text {csch}^2\left (\frac {1}{2} \sinh ^{-1}(c x)\right )-\sinh ^{-1}(c x) \text {sech}^2\left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )}{4 \sqrt {c^2 x^2+1}}+b^2 c^2 d \sqrt {d \left (c^2 x^2+1\right )} \left (\frac {2 \sinh ^{-1}(c x) \left (\text {Li}_2\left (-e^{-\sinh ^{-1}(c x)}\right )-\text {Li}_2\left (e^{-\sinh ^{-1}(c x)}\right )\right )}{\sqrt {c^2 x^2+1}}+\frac {2 \left (\text {Li}_3\left (-e^{-\sinh ^{-1}(c x)}\right )-\text {Li}_3\left (e^{-\sinh ^{-1}(c x)}\right )\right )}{\sqrt {c^2 x^2+1}}-\frac {2 c x \sinh ^{-1}(c x)}{\sqrt {c^2 x^2+1}}+\frac {\sinh ^{-1}(c x)^2 \left (\log \left (1-e^{-\sinh ^{-1}(c x)}\right )-\log \left (e^{-\sinh ^{-1}(c x)}+1\right )\right )}{\sqrt {c^2 x^2+1}}+\sinh ^{-1}(c x)^2+2\right )+\frac {b^2 c^2 d \sqrt {d \left (c^2 x^2+1\right )} \left (8 \sinh ^{-1}(c x) \text {Li}_2\left (-e^{-\sinh ^{-1}(c x)}\right )-8 \sinh ^{-1}(c x) \text {Li}_2\left (e^{-\sinh ^{-1}(c x)}\right )+8 \text {Li}_3\left (-e^{-\sinh ^{-1}(c x)}\right )-8 \text {Li}_3\left (e^{-\sinh ^{-1}(c x)}\right )+4 \sinh ^{-1}(c x)^2 \log \left (1-e^{-\sinh ^{-1}(c x)}\right )-4 \sinh ^{-1}(c x)^2 \log \left (e^{-\sinh ^{-1}(c x)}+1\right )+4 \sinh ^{-1}(c x) \tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-4 \sinh ^{-1}(c x) \coth \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+\sinh ^{-1}(c x)^2 \left (-\text {csch}^2\left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )-\sinh ^{-1}(c x)^2 \text {sech}^2\left (\frac {1}{2} \sinh ^{-1}(c x)\right )+8 \log \left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )\right )}{8 \sqrt {c^2 x^2+1}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x])^2)/x^3,x]

[Out]

(a^2*c^2*d - (a^2*d)/(2*x^2))*Sqrt[d*(1 + c^2*x^2)] + (3*a^2*c^2*d^(3/2)*Log[x])/2 - (3*a^2*c^2*d^(3/2)*Log[d
+ Sqrt[d]*Sqrt[d*(1 + c^2*x^2)]])/2 + (2*a*b*c^2*d*Sqrt[d*(1 + c^2*x^2)]*(-(c*x) + Sqrt[1 + c^2*x^2]*ArcSinh[c
*x] + ArcSinh[c*x]*Log[1 - E^(-ArcSinh[c*x])] - ArcSinh[c*x]*Log[1 + E^(-ArcSinh[c*x])] + PolyLog[2, -E^(-ArcS
inh[c*x])] - PolyLog[2, E^(-ArcSinh[c*x])]))/Sqrt[1 + c^2*x^2] + b^2*c^2*d*Sqrt[d*(1 + c^2*x^2)]*(2 - (2*c*x*A
rcSinh[c*x])/Sqrt[1 + c^2*x^2] + ArcSinh[c*x]^2 + (ArcSinh[c*x]^2*(Log[1 - E^(-ArcSinh[c*x])] - Log[1 + E^(-Ar
cSinh[c*x])]))/Sqrt[1 + c^2*x^2] + (2*ArcSinh[c*x]*(PolyLog[2, -E^(-ArcSinh[c*x])] - PolyLog[2, E^(-ArcSinh[c*
x])]))/Sqrt[1 + c^2*x^2] + (2*(PolyLog[3, -E^(-ArcSinh[c*x])] - PolyLog[3, E^(-ArcSinh[c*x])]))/Sqrt[1 + c^2*x
^2]) + (a*b*c^2*d*Sqrt[d*(1 + c^2*x^2)]*(-2*Coth[ArcSinh[c*x]/2] - ArcSinh[c*x]*Csch[ArcSinh[c*x]/2]^2 + 4*Arc
Sinh[c*x]*Log[1 - E^(-ArcSinh[c*x])] - 4*ArcSinh[c*x]*Log[1 + E^(-ArcSinh[c*x])] + 4*PolyLog[2, -E^(-ArcSinh[c
*x])] - 4*PolyLog[2, E^(-ArcSinh[c*x])] - ArcSinh[c*x]*Sech[ArcSinh[c*x]/2]^2 + 2*Tanh[ArcSinh[c*x]/2]))/(4*Sq
rt[1 + c^2*x^2]) + (b^2*c^2*d*Sqrt[d*(1 + c^2*x^2)]*(-4*ArcSinh[c*x]*Coth[ArcSinh[c*x]/2] - ArcSinh[c*x]^2*Csc
h[ArcSinh[c*x]/2]^2 + 4*ArcSinh[c*x]^2*Log[1 - E^(-ArcSinh[c*x])] - 4*ArcSinh[c*x]^2*Log[1 + E^(-ArcSinh[c*x])
] + 8*Log[Tanh[ArcSinh[c*x]/2]] + 8*ArcSinh[c*x]*PolyLog[2, -E^(-ArcSinh[c*x])] - 8*ArcSinh[c*x]*PolyLog[2, E^
(-ArcSinh[c*x])] + 8*PolyLog[3, -E^(-ArcSinh[c*x])] - 8*PolyLog[3, E^(-ArcSinh[c*x])] - ArcSinh[c*x]^2*Sech[Ar
cSinh[c*x]/2]^2 + 4*ArcSinh[c*x]*Tanh[ArcSinh[c*x]/2]))/(8*Sqrt[1 + c^2*x^2])

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fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a^{2} c^{2} d x^{2} + a^{2} d + {\left (b^{2} c^{2} d x^{2} + b^{2} d\right )} \operatorname {arsinh}\left (c x\right )^{2} + 2 \, {\left (a b c^{2} d x^{2} + a b d\right )} \operatorname {arsinh}\left (c x\right )\right )} \sqrt {c^{2} d x^{2} + d}}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a^2*c^2*d*x^2 + a^2*d + (b^2*c^2*d*x^2 + b^2*d)*arcsinh(c*x)^2 + 2*(a*b*c^2*d*x^2 + a*b*d)*arcsinh(c
*x))*sqrt(c^2*d*x^2 + d)/x^3, x)

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [B]  time = 0.48, size = 1131, normalized size = 2.09 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^2*c^2*(c^2*d*x^2+d)^(3/2)+3/2*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)*arcsinh(c*x)^2*ln(1-c*x-(c^2*x
^2+1)^(1/2))*c^2*d-3/2*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)*arcsinh(c*x)^2*ln(1+c*x+(c^2*x^2+1)^(1/2))*
c^2*d-3*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)*arcsinh(c*x)*polylog(2,-c*x-(c^2*x^2+1)^(1/2))*c^2*d+a*b*(
d*(c^2*x^2+1))^(1/2)*c^2*d/(c^2*x^2+1)*arcsinh(c*x)-a*b*arcsinh(c*x)*(d*(c^2*x^2+1))^(1/2)*d/x^2/(c^2*x^2+1)+3
*a*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)*polylog(2,c*x+(c^2*x^2+1)^(1/2))*c^2*d-3*a*b*(d*(c^2*x^2+1))^(1/2
)/(c^2*x^2+1)^(1/2)*polylog(2,-c*x-(c^2*x^2+1)^(1/2))*c^2*d+2*b^2*(d*(c^2*x^2+1))^(1/2)*c^2*d/(c^2*x^2+1)-1/2*
a^2/d/x^2*(c^2*d*x^2+d)^(5/2)-3/2*a^2*c^2*d^(3/2)*ln((2*d+2*d^(1/2)*(c^2*d*x^2+d)^(1/2))/x)+3/2*a^2*c^2*(c^2*d
*x^2+d)^(1/2)*d+2*a*b*(d*(c^2*x^2+1))^(1/2)*c^4*d/(c^2*x^2+1)*arcsinh(c*x)*x^2+3*a*b*(d*(c^2*x^2+1))^(1/2)/(c^
2*x^2+1)^(1/2)*arcsinh(c*x)*ln(1-c*x-(c^2*x^2+1)^(1/2))*c^2*d-3*a*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)*ar
csinh(c*x)*ln(1+c*x+(c^2*x^2+1)^(1/2))*c^2*d-2*a*b*(d*(c^2*x^2+1))^(1/2)*c^3*d/(c^2*x^2+1)^(1/2)*x-a*b*(d*(c^2
*x^2+1))^(1/2)*d/x/(c^2*x^2+1)^(1/2)*c-2*b^2*(d*(c^2*x^2+1))^(1/2)*c^3*d/(c^2*x^2+1)^(1/2)*arcsinh(c*x)*x-b^2*
arcsinh(c*x)*(d*(c^2*x^2+1))^(1/2)*d/x/(c^2*x^2+1)^(1/2)*c+3*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)*arcsi
nh(c*x)*polylog(2,c*x+(c^2*x^2+1)^(1/2))*c^2*d+b^2*(d*(c^2*x^2+1))^(1/2)*c^4*d/(c^2*x^2+1)*arcsinh(c*x)^2*x^2+
2*b^2*(d*(c^2*x^2+1))^(1/2)*c^4*d/(c^2*x^2+1)*x^2+1/2*b^2*(d*(c^2*x^2+1))^(1/2)*c^2*d/(c^2*x^2+1)*arcsinh(c*x)
^2+3*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)*polylog(3,-c*x-(c^2*x^2+1)^(1/2))*c^2*d-2*b^2*(d*(c^2*x^2+1))
^(1/2)/(c^2*x^2+1)^(1/2)*arctanh(c*x+(c^2*x^2+1)^(1/2))*c^2*d-3*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)*po
lylog(3,c*x+(c^2*x^2+1)^(1/2))*c^2*d-1/2*b^2*arcsinh(c*x)^2*(d*(c^2*x^2+1))^(1/2)*d/x^2/(c^2*x^2+1)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{2} \, {\left (3 \, c^{2} d^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {1}{c {\left | x \right |}}\right ) - {\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2} - 3 \, \sqrt {c^{2} d x^{2} + d} c^{2} d + \frac {{\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}}}{d x^{2}}\right )} a^{2} + \int \frac {{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} b^{2} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2}}{x^{3}} + \frac {2 \, {\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} a b \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(3*c^2*d^(3/2)*arcsinh(1/(c*abs(x))) - (c^2*d*x^2 + d)^(3/2)*c^2 - 3*sqrt(c^2*d*x^2 + d)*c^2*d + (c^2*d*x
^2 + d)^(5/2)/(d*x^2))*a^2 + integrate((c^2*d*x^2 + d)^(3/2)*b^2*log(c*x + sqrt(c^2*x^2 + 1))^2/x^3 + 2*(c^2*d
*x^2 + d)^(3/2)*a*b*log(c*x + sqrt(c^2*x^2 + 1))/x^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2\,{\left (d\,c^2\,x^2+d\right )}^{3/2}}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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