3.233 \(\int \frac{(a+b \sinh ^{-1}(c x))^2}{x^4 (d+c^2 d x^2)} \, dx\)

Optimal. Leaf size=297 \[ -\frac{2 i b c^3 \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d}+\frac{2 i b c^3 \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d}+\frac{7 b^2 c^3 \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )}{3 d}-\frac{7 b^2 c^3 \text{PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )}{3 d}+\frac{2 i b^2 c^3 \text{PolyLog}\left (3,-i e^{\sinh ^{-1}(c x)}\right )}{d}-\frac{2 i b^2 c^3 \text{PolyLog}\left (3,i e^{\sinh ^{-1}(c x)}\right )}{d}-\frac{b c \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{3 d x^2}+\frac{c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{d x}+\frac{2 c^3 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{d}+\frac{14 b c^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 d}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d x^3}-\frac{b^2 c^2}{3 d x} \]

[Out]

-(b^2*c^2)/(3*d*x) - (b*c*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(3*d*x^2) - (a + b*ArcSinh[c*x])^2/(3*d*x^3)
+ (c^2*(a + b*ArcSinh[c*x])^2)/(d*x) + (2*c^3*(a + b*ArcSinh[c*x])^2*ArcTan[E^ArcSinh[c*x]])/d + (14*b*c^3*(a
+ b*ArcSinh[c*x])*ArcTanh[E^ArcSinh[c*x]])/(3*d) + (7*b^2*c^3*PolyLog[2, -E^ArcSinh[c*x]])/(3*d) - ((2*I)*b*c
^3*(a + b*ArcSinh[c*x])*PolyLog[2, (-I)*E^ArcSinh[c*x]])/d + ((2*I)*b*c^3*(a + b*ArcSinh[c*x])*PolyLog[2, I*E^
ArcSinh[c*x]])/d - (7*b^2*c^3*PolyLog[2, E^ArcSinh[c*x]])/(3*d) + ((2*I)*b^2*c^3*PolyLog[3, (-I)*E^ArcSinh[c*x
]])/d - ((2*I)*b^2*c^3*PolyLog[3, I*E^ArcSinh[c*x]])/d

________________________________________________________________________________________

Rubi [A]  time = 0.638874, antiderivative size = 297, normalized size of antiderivative = 1., number of steps used = 24, number of rules used = 11, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.423, Rules used = {5747, 5693, 4180, 2531, 2282, 6589, 5760, 4182, 2279, 2391, 30} \[ -\frac{2 i b c^3 \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d}+\frac{2 i b c^3 \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d}+\frac{7 b^2 c^3 \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )}{3 d}-\frac{7 b^2 c^3 \text{PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )}{3 d}+\frac{2 i b^2 c^3 \text{PolyLog}\left (3,-i e^{\sinh ^{-1}(c x)}\right )}{d}-\frac{2 i b^2 c^3 \text{PolyLog}\left (3,i e^{\sinh ^{-1}(c x)}\right )}{d}-\frac{b c \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{3 d x^2}+\frac{c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{d x}+\frac{2 c^3 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{d}+\frac{14 b c^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 d}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d x^3}-\frac{b^2 c^2}{3 d x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c*x])^2/(x^4*(d + c^2*d*x^2)),x]

[Out]

-(b^2*c^2)/(3*d*x) - (b*c*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(3*d*x^2) - (a + b*ArcSinh[c*x])^2/(3*d*x^3)
+ (c^2*(a + b*ArcSinh[c*x])^2)/(d*x) + (2*c^3*(a + b*ArcSinh[c*x])^2*ArcTan[E^ArcSinh[c*x]])/d + (14*b*c^3*(a
+ b*ArcSinh[c*x])*ArcTanh[E^ArcSinh[c*x]])/(3*d) + (7*b^2*c^3*PolyLog[2, -E^ArcSinh[c*x]])/(3*d) - ((2*I)*b*c
^3*(a + b*ArcSinh[c*x])*PolyLog[2, (-I)*E^ArcSinh[c*x]])/d + ((2*I)*b*c^3*(a + b*ArcSinh[c*x])*PolyLog[2, I*E^
ArcSinh[c*x]])/d - (7*b^2*c^3*PolyLog[2, E^ArcSinh[c*x]])/(3*d) + ((2*I)*b^2*c^3*PolyLog[3, (-I)*E^ArcSinh[c*x
]])/d - ((2*I)*b^2*c^3*PolyLog[3, I*E^ArcSinh[c*x]])/d

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 5693

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

Rule 4180

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c
+ d*x)^m*ArcTanh[E^(-(I*e) + f*fz*x)/E^(I*k*Pi)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*
Log[1 - E^(-(I*e) + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e)
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && 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 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 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{x^4 \left (d+c^2 d x^2\right )} \, dx &=-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d x^3}-c^2 \int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{x^2 \left (d+c^2 d x^2\right )} \, dx+\frac{(2 b c) \int \frac{a+b \sinh ^{-1}(c x)}{x^3 \sqrt{1+c^2 x^2}} \, dx}{3 d}\\ &=-\frac{b c \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 d x^2}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d x^3}+\frac{c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{d x}+c^4 \int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d+c^2 d x^2} \, dx+\frac{\left (b^2 c^2\right ) \int \frac{1}{x^2} \, dx}{3 d}-\frac{\left (b c^3\right ) \int \frac{a+b \sinh ^{-1}(c x)}{x \sqrt{1+c^2 x^2}} \, dx}{3 d}-\frac{\left (2 b c^3\right ) \int \frac{a+b \sinh ^{-1}(c x)}{x \sqrt{1+c^2 x^2}} \, dx}{d}\\ &=-\frac{b^2 c^2}{3 d x}-\frac{b c \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 d x^2}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d x^3}+\frac{c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{d x}+\frac{c^3 \operatorname{Subst}\left (\int (a+b x)^2 \text{sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{d}-\frac{\left (b c^3\right ) \operatorname{Subst}\left (\int (a+b x) \text{csch}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{3 d}-\frac{\left (2 b c^3\right ) \operatorname{Subst}\left (\int (a+b x) \text{csch}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{d}\\ &=-\frac{b^2 c^2}{3 d x}-\frac{b c \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 d x^2}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d x^3}+\frac{c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{d x}+\frac{2 c^3 \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d}+\frac{14 b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{3 d}-\frac{\left (2 i b c^3\right ) \operatorname{Subst}\left (\int (a+b x) \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d}+\frac{\left (2 i b c^3\right ) \operatorname{Subst}\left (\int (a+b x) \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d}+\frac{\left (b^2 c^3\right ) \operatorname{Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{3 d}-\frac{\left (b^2 c^3\right ) \operatorname{Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{3 d}+\frac{\left (2 b^2 c^3\right ) \operatorname{Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d}-\frac{\left (2 b^2 c^3\right ) \operatorname{Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d}\\ &=-\frac{b^2 c^2}{3 d x}-\frac{b c \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 d x^2}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d x^3}+\frac{c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{d x}+\frac{2 c^3 \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d}+\frac{14 b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{3 d}-\frac{2 i b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{d}+\frac{2 i b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{d}+\frac{\left (2 i b^2 c^3\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d}-\frac{\left (2 i b^2 c^3\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d}+\frac{\left (b^2 c^3\right ) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{3 d}-\frac{\left (b^2 c^3\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{3 d}+\frac{\left (2 b^2 c^3\right ) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{d}-\frac{\left (2 b^2 c^3\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{d}\\ &=-\frac{b^2 c^2}{3 d x}-\frac{b c \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 d x^2}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d x^3}+\frac{c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{d x}+\frac{2 c^3 \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d}+\frac{14 b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{3 d}+\frac{7 b^2 c^3 \text{Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{3 d}-\frac{2 i b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{d}+\frac{2 i b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{d}-\frac{7 b^2 c^3 \text{Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{3 d}+\frac{\left (2 i b^2 c^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{d}-\frac{\left (2 i b^2 c^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{d}\\ &=-\frac{b^2 c^2}{3 d x}-\frac{b c \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 d x^2}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d x^3}+\frac{c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{d x}+\frac{2 c^3 \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d}+\frac{14 b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{3 d}+\frac{7 b^2 c^3 \text{Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{3 d}-\frac{2 i b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{d}+\frac{2 i b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{d}-\frac{7 b^2 c^3 \text{Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{3 d}+\frac{2 i b^2 c^3 \text{Li}_3\left (-i e^{\sinh ^{-1}(c x)}\right )}{d}-\frac{2 i b^2 c^3 \text{Li}_3\left (i e^{\sinh ^{-1}(c x)}\right )}{d}\\ \end{align*}

Mathematica [B]  time = 8.01868, size = 602, normalized size = 2.03 \[ \frac{2 a b \left (-\frac{1}{2} i c^4 \left (\frac{2 \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{c}-\frac{\sinh ^{-1}(c x)^2}{2 c}+\frac{2 \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )}{c}\right )+\frac{1}{2} i c^4 \left (\frac{2 \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{c}-\frac{\sinh ^{-1}(c x)^2}{2 c}+\frac{2 \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )}{c}\right )-\frac{c \sqrt{c^2 x^2+1}}{6 x^2}+\frac{1}{6} c^3 \tanh ^{-1}\left (\sqrt{c^2 x^2+1}\right )-c^2 \left (-c \tanh ^{-1}\left (\sqrt{c^2 x^2+1}\right )-\frac{\sinh ^{-1}(c x)}{x}\right )-\frac{\sinh ^{-1}(c x)}{3 x^3}\right )}{d}+\frac{b^2 c^3 \left (-56 \text{PolyLog}\left (2,-e^{-\sinh ^{-1}(c x)}\right )-48 i \sinh ^{-1}(c x) \text{PolyLog}\left (2,-i e^{-\sinh ^{-1}(c x)}\right )+48 i \sinh ^{-1}(c x) \text{PolyLog}\left (2,i e^{-\sinh ^{-1}(c x)}\right )+56 \text{PolyLog}\left (2,e^{-\sinh ^{-1}(c x)}\right )-48 i \text{PolyLog}\left (3,-i e^{-\sinh ^{-1}(c x)}\right )+48 i \text{PolyLog}\left (3,i e^{-\sinh ^{-1}(c x)}\right )-\frac{8 \sinh ^{-1}(c x)^2 \sinh ^4\left (\frac{1}{2} \sinh ^{-1}(c x)\right )}{c^3 x^3}-56 \sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )-24 i \sinh ^{-1}(c x)^2 \log \left (1-i e^{-\sinh ^{-1}(c x)}\right )+24 i \sinh ^{-1}(c x)^2 \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )+56 \sinh ^{-1}(c x) \log \left (e^{-\sinh ^{-1}(c x)}+1\right )-14 \sinh ^{-1}(c x)^2 \tanh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )+4 \tanh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )+14 \sinh ^{-1}(c x)^2 \coth \left (\frac{1}{2} \sinh ^{-1}(c x)\right )-4 \coth \left (\frac{1}{2} \sinh ^{-1}(c x)\right )-\frac{1}{2} c x \sinh ^{-1}(c x)^2 \text{csch}^4\left (\frac{1}{2} \sinh ^{-1}(c x)\right )-2 \sinh ^{-1}(c x) \text{csch}^2\left (\frac{1}{2} \sinh ^{-1}(c x)\right )-2 \sinh ^{-1}(c x) \text{sech}^2\left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )}{24 d}+\frac{a^2 c^2}{d x}+\frac{a^2 c^3 \tan ^{-1}(c x)}{d}-\frac{a^2}{3 d x^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSinh[c*x])^2/(x^4*(d + c^2*d*x^2)),x]

[Out]

-a^2/(3*d*x^3) + (a^2*c^2)/(d*x) + (a^2*c^3*ArcTan[c*x])/d + (2*a*b*(-(c*Sqrt[1 + c^2*x^2])/(6*x^2) - ArcSinh[
c*x]/(3*x^3) + (c^3*ArcTanh[Sqrt[1 + c^2*x^2]])/6 - c^2*(-(ArcSinh[c*x]/x) - c*ArcTanh[Sqrt[1 + c^2*x^2]]) - (
I/2)*c^4*(-ArcSinh[c*x]^2/(2*c) + (2*ArcSinh[c*x]*Log[1 + I*E^ArcSinh[c*x]])/c + (2*PolyLog[2, (-I)*E^ArcSinh[
c*x]])/c) + (I/2)*c^4*(-ArcSinh[c*x]^2/(2*c) + (2*ArcSinh[c*x]*Log[1 - I*E^ArcSinh[c*x]])/c + (2*PolyLog[2, I*
E^ArcSinh[c*x]])/c)))/d + (b^2*c^3*(-4*Coth[ArcSinh[c*x]/2] + 14*ArcSinh[c*x]^2*Coth[ArcSinh[c*x]/2] - 2*ArcSi
nh[c*x]*Csch[ArcSinh[c*x]/2]^2 - (c*x*ArcSinh[c*x]^2*Csch[ArcSinh[c*x]/2]^4)/2 - 56*ArcSinh[c*x]*Log[1 - E^(-A
rcSinh[c*x])] - (24*I)*ArcSinh[c*x]^2*Log[1 - I/E^ArcSinh[c*x]] + (24*I)*ArcSinh[c*x]^2*Log[1 + I/E^ArcSinh[c*
x]] + 56*ArcSinh[c*x]*Log[1 + E^(-ArcSinh[c*x])] - 56*PolyLog[2, -E^(-ArcSinh[c*x])] - (48*I)*ArcSinh[c*x]*Pol
yLog[2, (-I)/E^ArcSinh[c*x]] + (48*I)*ArcSinh[c*x]*PolyLog[2, I/E^ArcSinh[c*x]] + 56*PolyLog[2, E^(-ArcSinh[c*
x])] - (48*I)*PolyLog[3, (-I)/E^ArcSinh[c*x]] + (48*I)*PolyLog[3, I/E^ArcSinh[c*x]] - 2*ArcSinh[c*x]*Sech[ArcS
inh[c*x]/2]^2 - (8*ArcSinh[c*x]^2*Sinh[ArcSinh[c*x]/2]^4)/(c^3*x^3) + 4*Tanh[ArcSinh[c*x]/2] - 14*ArcSinh[c*x]
^2*Tanh[ArcSinh[c*x]/2]))/(24*d)

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Maple [F]  time = 0.228, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{{x}^{4} \left ({c}^{2}d{x}^{2}+d \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))^2/x^4/(c^2*d*x^2+d),x)

[Out]

int((a+b*arcsinh(c*x))^2/x^4/(c^2*d*x^2+d),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{3} \,{\left (\frac{3 \, c^{3} \arctan \left (c x\right )}{d} + \frac{3 \, c^{2} x^{2} - 1}{d x^{3}}\right )} a^{2} + \int \frac{b^{2} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )^{2}}{c^{2} d x^{6} + d x^{4}} + \frac{2 \, a b \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )}{c^{2} d x^{6} + d x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))^2/x^4/(c^2*d*x^2+d),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{2} \operatorname{arsinh}\left (c x\right )^{2} + 2 \, a b \operatorname{arsinh}\left (c x\right ) + a^{2}}{c^{2} d x^{6} + d x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))^2/x^4/(c^2*d*x^2+d),x, algorithm="fricas")

[Out]

integral((b^2*arcsinh(c*x)^2 + 2*a*b*arcsinh(c*x) + a^2)/(c^2*d*x^6 + d*x^4), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a^{2}}{c^{2} x^{6} + x^{4}}\, dx + \int \frac{b^{2} \operatorname{asinh}^{2}{\left (c x \right )}}{c^{2} x^{6} + x^{4}}\, dx + \int \frac{2 a b \operatorname{asinh}{\left (c x \right )}}{c^{2} x^{6} + x^{4}}\, dx}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(c*x))**2/x**4/(c**2*d*x**2+d),x)

[Out]

(Integral(a**2/(c**2*x**6 + x**4), x) + Integral(b**2*asinh(c*x)**2/(c**2*x**6 + x**4), x) + Integral(2*a*b*as
inh(c*x)/(c**2*x**6 + x**4), x))/d

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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}}{{\left (c^{2} d x^{2} + d\right )} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))^2/x^4/(c^2*d*x^2+d),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)^2/((c^2*d*x^2 + d)*x^4), x)