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

Optimal. Leaf size=360 \[ \frac{b c^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{c^2 d x^2+d}}-\frac{b c^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{c^2 d x^2+d}}-\frac{b^2 c^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (3,-e^{\sinh ^{-1}(c x)}\right )}{\sqrt{c^2 d x^2+d}}+\frac{b^2 c^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (3,e^{\sinh ^{-1}(c x)}\right )}{\sqrt{c^2 d x^2+d}}-\frac{b c \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{x \sqrt{c^2 d x^2+d}}-\frac{\sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d x^2}+\frac{c^2 \sqrt{c^2 x^2+1} \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt{c^2 d x^2+d}}-\frac{b^2 c^2 \sqrt{c^2 x^2+1} \tanh ^{-1}\left (\sqrt{c^2 x^2+1}\right )}{\sqrt{c^2 d x^2+d}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.570366, antiderivative size = 360, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 11, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.393, Rules used = {5747, 5764, 5760, 4182, 2531, 2282, 6589, 5661, 266, 63, 208} \[ \frac{b c^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{c^2 d x^2+d}}-\frac{b c^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{c^2 d x^2+d}}-\frac{b^2 c^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (3,-e^{\sinh ^{-1}(c x)}\right )}{\sqrt{c^2 d x^2+d}}+\frac{b^2 c^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (3,e^{\sinh ^{-1}(c x)}\right )}{\sqrt{c^2 d x^2+d}}-\frac{b c \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{x \sqrt{c^2 d x^2+d}}-\frac{\sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d x^2}+\frac{c^2 \sqrt{c^2 x^2+1} \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt{c^2 d x^2+d}}-\frac{b^2 c^2 \sqrt{c^2 x^2+1} \tanh ^{-1}\left (\sqrt{c^2 x^2+1}\right )}{\sqrt{c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5747

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 + 1)*(a + b*ArcSinh[c*x])^n)/(d*f*(m + 1)), x] + (-Dist[(c^2*(m + 2*p + 3))/(f^2
*(m + 1)), Int[(f*x)^(m + 2)*(d + e*x^2)^p*(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*ArcSin
h[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[m, -1] && Int
egerQ[m]

Rule 5764

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

Rule 5661

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcS
inh[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSinh[c*x])^(n - 1))/Sqrt
[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

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 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]

Rubi steps

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

Mathematica [A]  time = 5.4416, size = 455, normalized size = 1.26 \[ \frac{\frac{2 a b c^2 d^2 \left (c^2 x^2+1\right )^{3/2} \left (-4 \text{PolyLog}\left (2,-e^{-\sinh ^{-1}(c x)}\right )+4 \text{PolyLog}\left (2,e^{-\sinh ^{-1}(c x)}\right )-4 \sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )+4 \sinh ^{-1}(c x) \log \left (e^{-\sinh ^{-1}(c x)}+1\right )+2 \tanh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )-2 \coth \left (\frac{1}{2} \sinh ^{-1}(c x)\right )-\sinh ^{-1}(c x) \text{csch}^2\left (\frac{1}{2} \sinh ^{-1}(c x)\right )-\sinh ^{-1}(c x) \text{sech}^2\left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )}{\left (c^2 d x^2+d\right )^{3/2}}+\frac{b^2 c^2 d^2 \left (c^2 x^2+1\right )^{3/2} \left (-8 \sinh ^{-1}(c x) \text{PolyLog}\left (2,-e^{-\sinh ^{-1}(c x)}\right )+8 \sinh ^{-1}(c x) \text{PolyLog}\left (2,e^{-\sinh ^{-1}(c x)}\right )-8 \text{PolyLog}\left (3,-e^{-\sinh ^{-1}(c x)}\right )+8 \text{PolyLog}\left (3,e^{-\sinh ^{-1}(c x)}\right )-4 \sinh ^{-1}(c x)^2 \log \left (1-e^{-\sinh ^{-1}(c x)}\right )+4 \sinh ^{-1}(c x)^2 \log \left (e^{-\sinh ^{-1}(c x)}+1\right )+4 \sinh ^{-1}(c x) \tanh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )-4 \sinh ^{-1}(c x) \coth \left (\frac{1}{2} \sinh ^{-1}(c x)\right )+\sinh ^{-1}(c x)^2 \left (-\text{csch}^2\left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )-\sinh ^{-1}(c x)^2 \text{sech}^2\left (\frac{1}{2} \sinh ^{-1}(c x)\right )+8 \log \left (\tanh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )\right )}{\left (c^2 d x^2+d\right )^{3/2}}-\frac{4 a^2 \sqrt{c^2 d x^2+d}}{x^2}+4 a^2 c^2 \sqrt{d} \log \left (\sqrt{d} \sqrt{c^2 d x^2+d}+d\right )-4 a^2 c^2 \sqrt{d} \log (x)}{8 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-4*a^2*Sqrt[d + c^2*d*x^2])/x^2 - 4*a^2*c^2*Sqrt[d]*Log[x] + 4*a^2*c^2*Sqrt[d]*Log[d + Sqrt[d]*Sqrt[d + c^2*
d*x^2]] + (2*a*b*c^2*d^2*(1 + c^2*x^2)^(3/2)*(-2*Coth[ArcSinh[c*x]/2] - ArcSinh[c*x]*Csch[ArcSinh[c*x]/2]^2 -
4*ArcSinh[c*x]*Log[1 - E^(-ArcSinh[c*x])] + 4*ArcSinh[c*x]*Log[1 + E^(-ArcSinh[c*x])] - 4*PolyLog[2, -E^(-ArcS
inh[c*x])] + 4*PolyLog[2, E^(-ArcSinh[c*x])] - ArcSinh[c*x]*Sech[ArcSinh[c*x]/2]^2 + 2*Tanh[ArcSinh[c*x]/2]))/
(d + c^2*d*x^2)^(3/2) + (b^2*c^2*d^2*(1 + c^2*x^2)^(3/2)*(-4*ArcSinh[c*x]*Coth[ArcSinh[c*x]/2] - ArcSinh[c*x]^
2*Csch[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*Se
ch[ArcSinh[c*x]/2]^2 + 4*ArcSinh[c*x]*Tanh[ArcSinh[c*x]/2]))/(d + c^2*d*x^2)^(3/2))/(8*d)

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Maple [B]  time = 0.342, size = 901, normalized size = 2.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))^2/x^3/(c^2*d*x^2+d)^(1/2),x)

[Out]

-1/2*a^2/d/x^2*(c^2*d*x^2+d)^(1/2)+1/2*a^2*c^2/d^(1/2)*ln((2*d+2*d^(1/2)*(c^2*d*x^2+d)^(1/2))/x)-1/2*b^2*arcsi
nh(c*x)^2*(d*(c^2*x^2+1))^(1/2)/d/(c^2*x^2+1)*c^2-b^2*arcsinh(c*x)*(d*(c^2*x^2+1))^(1/2)/x/d/(c^2*x^2+1)^(1/2)
*c-1/2*b^2*arcsinh(c*x)^2*(d*(c^2*x^2+1))^(1/2)/x^2/d/(c^2*x^2+1)+1/2*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1
/2)/d*arcsinh(c*x)^2*ln(1+c*x+(c^2*x^2+1)^(1/2))*c^2+b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d*arcsinh(c*x
)*polylog(2,-c*x-(c^2*x^2+1)^(1/2))*c^2-b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d*polylog(3,-c*x-(c^2*x^2+
1)^(1/2))*c^2-1/2*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d*arcsinh(c*x)^2*ln(1-c*x-(c^2*x^2+1)^(1/2))*c^2
-b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d*arcsinh(c*x)*polylog(2,c*x+(c^2*x^2+1)^(1/2))*c^2+b^2*(d*(c^2*x
^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d*polylog(3,c*x+(c^2*x^2+1)^(1/2))*c^2-2*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^
(1/2)/d*arctanh(c*x+(c^2*x^2+1)^(1/2))*c^2-a*b*(d*(c^2*x^2+1))^(1/2)/d/(c^2*x^2+1)*arcsinh(c*x)*c^2-a*b*(d*(c^
2*x^2+1))^(1/2)/x/d/(c^2*x^2+1)^(1/2)*c-a*b*arcsinh(c*x)*(d*(c^2*x^2+1))^(1/2)/x^2/d/(c^2*x^2+1)+a*b*(d*(c^2*x
^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d*arcsinh(c*x)*ln(1+c*x+(c^2*x^2+1)^(1/2))*c^2+a*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x
^2+1)^(1/2)/d*polylog(2,-c*x-(c^2*x^2+1)^(1/2))*c^2-a*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d*arcsinh(c*x)
*ln(1-c*x-(c^2*x^2+1)^(1/2))*c^2-a*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d*polylog(2,c*x+(c^2*x^2+1)^(1/2)
)*c^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))^2/x^3/(c^2*d*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c^{2} d x^{2} + d}{\left (b^{2} \operatorname{arsinh}\left (c x\right )^{2} + 2 \, a b \operatorname{arsinh}\left (c x\right ) + a^{2}\right )}}{c^{2} d x^{5} + d x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))^2/x^3/(c^2*d*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(c^2*d*x^2 + d)*(b^2*arcsinh(c*x)^2 + 2*a*b*arcsinh(c*x) + a^2)/(c^2*d*x^5 + d*x^3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{asinh}{\left (c x \right )}\right )^{2}}{x^{3} \sqrt{d \left (c^{2} x^{2} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}^{2}}{\sqrt{c^{2} d x^{2} + d} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))^2/x^3/(c^2*d*x^2+d)^(1/2),x, algorithm="giac")

[Out]

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