3.316 \(\int \frac{(a+b \sinh ^{-1}(c x))^2}{x (d+c^2 d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=518 \[ -\frac{2 b \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^2 \sqrt{c^2 d x^2+d}}+\frac{2 b \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^2 \sqrt{c^2 d x^2+d}}+\frac{7 i b^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{3 d^2 \sqrt{c^2 d x^2+d}}-\frac{7 i b^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{3 d^2 \sqrt{c^2 d x^2+d}}+\frac{2 b^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (3,-e^{\sinh ^{-1}(c x)}\right )}{d^2 \sqrt{c^2 d x^2+d}}-\frac{2 b^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (3,e^{\sinh ^{-1}(c x)}\right )}{d^2 \sqrt{c^2 d x^2+d}}-\frac{b c x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt{c^2 x^2+1} \sqrt{c^2 d x^2+d}}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d^2 \sqrt{c^2 d x^2+d}}-\frac{14 b \sqrt{c^2 x^2+1} \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt{c^2 d x^2+d}}-\frac{2 \sqrt{c^2 x^2+1} \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{d^2 \sqrt{c^2 d x^2+d}}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d \left (c^2 d x^2+d\right )^{3/2}}-\frac{b^2}{3 d^2 \sqrt{c^2 d x^2+d}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.863378, antiderivative size = 518, normalized size of antiderivative = 1., number of steps used = 25, number of rules used = 13, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.464, Rules used = {5755, 5764, 5760, 4182, 2531, 2282, 6589, 5693, 4180, 2279, 2391, 5690, 261} \[ -\frac{2 b \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^2 \sqrt{c^2 d x^2+d}}+\frac{2 b \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^2 \sqrt{c^2 d x^2+d}}+\frac{7 i b^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{3 d^2 \sqrt{c^2 d x^2+d}}-\frac{7 i b^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{3 d^2 \sqrt{c^2 d x^2+d}}+\frac{2 b^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (3,-e^{\sinh ^{-1}(c x)}\right )}{d^2 \sqrt{c^2 d x^2+d}}-\frac{2 b^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (3,e^{\sinh ^{-1}(c x)}\right )}{d^2 \sqrt{c^2 d x^2+d}}-\frac{b c x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt{c^2 x^2+1} \sqrt{c^2 d x^2+d}}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d^2 \sqrt{c^2 d x^2+d}}-\frac{14 b \sqrt{c^2 x^2+1} \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt{c^2 d x^2+d}}-\frac{2 \sqrt{c^2 x^2+1} \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{d^2 \sqrt{c^2 d x^2+d}}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d \left (c^2 d x^2+d\right )^{3/2}}-\frac{b^2}{3 d^2 \sqrt{c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c*x])^2/(x*(d + c^2*d*x^2)^(5/2)),x]

[Out]

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

Rule 5755

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)/(2*d*f*(p + 1)), x] + (Dist[(m + 2*p + 3)/(2*d*(p
+ 1)), Int[(f*x)^m*(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)^F
racPart[p])/(2*f*(p + 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, m}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] &&  !GtQ
[m, 1] && (IntegerQ[m] || IntegerQ[p] || EqQ[n, 1])

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 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 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 5690

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(x*(d + e*x^2)^(p
+ 1)*(a + b*ArcSinh[c*x])^n)/(2*d*(p + 1)), x] + (Dist[(2*p + 3)/(2*d*(p + 1)), Int[(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])/(2*(p + 1)*(1 + c^2*x^2)^FracPar
t[p]), Int[x*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ
[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] && NeQ[p, -3/2]

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]

Rubi steps

\begin{align*} \int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{x \left (d+c^2 d x^2\right )^{5/2}} \, dx &=\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d \left (d+c^2 d x^2\right )^{3/2}}+\frac{\int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{x \left (d+c^2 d x^2\right )^{3/2}} \, dx}{d}-\frac{\left (2 b c \sqrt{1+c^2 x^2}\right ) \int \frac{a+b \sinh ^{-1}(c x)}{\left (1+c^2 x^2\right )^2} \, dx}{3 d^2 \sqrt{d+c^2 d x^2}}\\ &=-\frac{b c x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt{1+c^2 x^2} \sqrt{d+c^2 d x^2}}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d \left (d+c^2 d x^2\right )^{3/2}}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d^2 \sqrt{d+c^2 d x^2}}+\frac{\int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{x \sqrt{d+c^2 d x^2}} \, dx}{d^2}-\frac{\left (b c \sqrt{1+c^2 x^2}\right ) \int \frac{a+b \sinh ^{-1}(c x)}{1+c^2 x^2} \, dx}{3 d^2 \sqrt{d+c^2 d x^2}}-\frac{\left (2 b c \sqrt{1+c^2 x^2}\right ) \int \frac{a+b \sinh ^{-1}(c x)}{1+c^2 x^2} \, dx}{d^2 \sqrt{d+c^2 d x^2}}+\frac{\left (b^2 c^2 \sqrt{1+c^2 x^2}\right ) \int \frac{x}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{3 d^2 \sqrt{d+c^2 d x^2}}\\ &=-\frac{b^2}{3 d^2 \sqrt{d+c^2 d x^2}}-\frac{b c x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt{1+c^2 x^2} \sqrt{d+c^2 d x^2}}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d \left (d+c^2 d x^2\right )^{3/2}}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d^2 \sqrt{d+c^2 d x^2}}+\frac{\sqrt{1+c^2 x^2} \int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{x \sqrt{1+c^2 x^2}} \, dx}{d^2 \sqrt{d+c^2 d x^2}}-\frac{\left (b \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int (a+b x) \text{sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{3 d^2 \sqrt{d+c^2 d x^2}}-\frac{\left (2 b \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int (a+b x) \text{sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{d^2 \sqrt{d+c^2 d x^2}}\\ &=-\frac{b^2}{3 d^2 \sqrt{d+c^2 d x^2}}-\frac{b c x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt{1+c^2 x^2} \sqrt{d+c^2 d x^2}}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d \left (d+c^2 d x^2\right )^{3/2}}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d^2 \sqrt{d+c^2 d x^2}}-\frac{14 b \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{3 d^2 \sqrt{d+c^2 d x^2}}+\frac{\sqrt{1+c^2 x^2} \operatorname{Subst}\left (\int (a+b x)^2 \text{csch}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{d^2 \sqrt{d+c^2 d x^2}}+\frac{\left (i b^2 \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{3 d^2 \sqrt{d+c^2 d x^2}}-\frac{\left (i b^2 \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{3 d^2 \sqrt{d+c^2 d x^2}}+\frac{\left (2 i b^2 \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d^2 \sqrt{d+c^2 d x^2}}-\frac{\left (2 i b^2 \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d^2 \sqrt{d+c^2 d x^2}}\\ &=-\frac{b^2}{3 d^2 \sqrt{d+c^2 d x^2}}-\frac{b c x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt{1+c^2 x^2} \sqrt{d+c^2 d x^2}}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d \left (d+c^2 d x^2\right )^{3/2}}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d^2 \sqrt{d+c^2 d x^2}}-\frac{14 b \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{3 d^2 \sqrt{d+c^2 d x^2}}-\frac{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 )}{d^2 \sqrt{d+c^2 d x^2}}-\frac{\left (2 b \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 )}{d^2 \sqrt{d+c^2 d x^2}}+\frac{\left (2 b \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 )}{d^2 \sqrt{d+c^2 d x^2}}+\frac{\left (i b^2 \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{3 d^2 \sqrt{d+c^2 d x^2}}-\frac{\left (i b^2 \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{3 d^2 \sqrt{d+c^2 d x^2}}+\frac{\left (2 i b^2 \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{d^2 \sqrt{d+c^2 d x^2}}-\frac{\left (2 i b^2 \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{d^2 \sqrt{d+c^2 d x^2}}\\ &=-\frac{b^2}{3 d^2 \sqrt{d+c^2 d x^2}}-\frac{b c x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt{1+c^2 x^2} \sqrt{d+c^2 d x^2}}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d \left (d+c^2 d x^2\right )^{3/2}}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d^2 \sqrt{d+c^2 d x^2}}-\frac{14 b \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{3 d^2 \sqrt{d+c^2 d x^2}}-\frac{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 )}{d^2 \sqrt{d+c^2 d x^2}}-\frac{2 b \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{d^2 \sqrt{d+c^2 d x^2}}+\frac{7 i b^2 \sqrt{1+c^2 x^2} \text{Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{3 d^2 \sqrt{d+c^2 d x^2}}-\frac{7 i b^2 \sqrt{1+c^2 x^2} \text{Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{3 d^2 \sqrt{d+c^2 d x^2}}+\frac{2 b \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{d^2 \sqrt{d+c^2 d x^2}}+\frac{\left (2 b^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 )}{d^2 \sqrt{d+c^2 d x^2}}-\frac{\left (2 b^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 )}{d^2 \sqrt{d+c^2 d x^2}}\\ &=-\frac{b^2}{3 d^2 \sqrt{d+c^2 d x^2}}-\frac{b c x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt{1+c^2 x^2} \sqrt{d+c^2 d x^2}}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d \left (d+c^2 d x^2\right )^{3/2}}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d^2 \sqrt{d+c^2 d x^2}}-\frac{14 b \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{3 d^2 \sqrt{d+c^2 d x^2}}-\frac{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 )}{d^2 \sqrt{d+c^2 d x^2}}-\frac{2 b \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{d^2 \sqrt{d+c^2 d x^2}}+\frac{7 i b^2 \sqrt{1+c^2 x^2} \text{Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{3 d^2 \sqrt{d+c^2 d x^2}}-\frac{7 i b^2 \sqrt{1+c^2 x^2} \text{Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{3 d^2 \sqrt{d+c^2 d x^2}}+\frac{2 b \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{d^2 \sqrt{d+c^2 d x^2}}+\frac{\left (2 b^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 )}{d^2 \sqrt{d+c^2 d x^2}}-\frac{\left (2 b^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 )}{d^2 \sqrt{d+c^2 d x^2}}\\ &=-\frac{b^2}{3 d^2 \sqrt{d+c^2 d x^2}}-\frac{b c x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt{1+c^2 x^2} \sqrt{d+c^2 d x^2}}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d \left (d+c^2 d x^2\right )^{3/2}}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d^2 \sqrt{d+c^2 d x^2}}-\frac{14 b \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{3 d^2 \sqrt{d+c^2 d x^2}}-\frac{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 )}{d^2 \sqrt{d+c^2 d x^2}}-\frac{2 b \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{d^2 \sqrt{d+c^2 d x^2}}+\frac{7 i b^2 \sqrt{1+c^2 x^2} \text{Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{3 d^2 \sqrt{d+c^2 d x^2}}-\frac{7 i b^2 \sqrt{1+c^2 x^2} \text{Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{3 d^2 \sqrt{d+c^2 d x^2}}+\frac{2 b \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{d^2 \sqrt{d+c^2 d x^2}}+\frac{2 b^2 \sqrt{1+c^2 x^2} \text{Li}_3\left (-e^{\sinh ^{-1}(c x)}\right )}{d^2 \sqrt{d+c^2 d x^2}}-\frac{2 b^2 \sqrt{1+c^2 x^2} \text{Li}_3\left (e^{\sinh ^{-1}(c x)}\right )}{d^2 \sqrt{d+c^2 d x^2}}\\ \end{align*}

Mathematica [A]  time = 4.12413, size = 547, normalized size = 1.06 \[ \frac{\frac{a b d^2 \left (c^2 x^2+1\right )^{3/2} \left (6 \text{PolyLog}\left (2,-e^{-\sinh ^{-1}(c x)}\right )-6 \text{PolyLog}\left (2,e^{-\sinh ^{-1}(c x)}\right )-\frac{c x}{c^2 x^2+1}+\frac{6 \sinh ^{-1}(c x)}{\sqrt{c^2 x^2+1}}+\frac{2 \sinh ^{-1}(c x)}{\left (c^2 x^2+1\right )^{3/2}}+6 \sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )-6 \sinh ^{-1}(c x) \log \left (e^{-\sinh ^{-1}(c x)}+1\right )-14 \tan ^{-1}\left (\tanh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )\right )}{\left (c^2 d x^2+d\right )^{3/2}}+\frac{b^2 d^2 \left (c^2 x^2+1\right )^{3/2} \left (6 \sinh ^{-1}(c x) \text{PolyLog}\left (2,-e^{-\sinh ^{-1}(c x)}\right )-6 \sinh ^{-1}(c x) \text{PolyLog}\left (2,e^{-\sinh ^{-1}(c x)}\right )+7 i \text{PolyLog}\left (2,-i e^{-\sinh ^{-1}(c x)}\right )-7 i \text{PolyLog}\left (2,i e^{-\sinh ^{-1}(c x)}\right )+6 \text{PolyLog}\left (3,-e^{-\sinh ^{-1}(c x)}\right )-6 \text{PolyLog}\left (3,e^{-\sinh ^{-1}(c x)}\right )-\frac{1}{\sqrt{c^2 x^2+1}}+\frac{3 \sinh ^{-1}(c x)^2}{\sqrt{c^2 x^2+1}}+\frac{\sinh ^{-1}(c x)^2}{\left (c^2 x^2+1\right )^{3/2}}-\frac{c x \sinh ^{-1}(c x)}{c^2 x^2+1}+3 \sinh ^{-1}(c x)^2 \log \left (1-e^{-\sinh ^{-1}(c x)}\right )-3 \sinh ^{-1}(c x)^2 \log \left (e^{-\sinh ^{-1}(c x)}+1\right )+7 i \sinh ^{-1}(c x) \log \left (1-i e^{-\sinh ^{-1}(c x)}\right )-7 i \sinh ^{-1}(c x) \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )\right )}{\left (c^2 d x^2+d\right )^{3/2}}+\frac{a^2 \left (3 c^2 x^2+4\right ) \sqrt{c^2 d x^2+d}}{\left (c^2 x^2+1\right )^2}-3 a^2 \sqrt{d} \log \left (\sqrt{d} \sqrt{c^2 d x^2+d}+d\right )+3 a^2 \sqrt{d} \log (c x)}{3 d^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSinh[c*x])^2/(x*(d + c^2*d*x^2)^(5/2)),x]

[Out]

((a^2*(4 + 3*c^2*x^2)*Sqrt[d + c^2*d*x^2])/(1 + c^2*x^2)^2 + 3*a^2*Sqrt[d]*Log[c*x] - 3*a^2*Sqrt[d]*Log[d + Sq
rt[d]*Sqrt[d + c^2*d*x^2]] + (a*b*d^2*(1 + c^2*x^2)^(3/2)*(-((c*x)/(1 + c^2*x^2)) + (2*ArcSinh[c*x])/(1 + c^2*
x^2)^(3/2) + (6*ArcSinh[c*x])/Sqrt[1 + c^2*x^2] - 14*ArcTan[Tanh[ArcSinh[c*x]/2]] + 6*ArcSinh[c*x]*Log[1 - E^(
-ArcSinh[c*x])] - 6*ArcSinh[c*x]*Log[1 + E^(-ArcSinh[c*x])] + 6*PolyLog[2, -E^(-ArcSinh[c*x])] - 6*PolyLog[2,
E^(-ArcSinh[c*x])]))/(d + c^2*d*x^2)^(3/2) + (b^2*d^2*(1 + c^2*x^2)^(3/2)*(-(1/Sqrt[1 + c^2*x^2]) - (c*x*ArcSi
nh[c*x])/(1 + c^2*x^2) + ArcSinh[c*x]^2/(1 + c^2*x^2)^(3/2) + (3*ArcSinh[c*x]^2)/Sqrt[1 + c^2*x^2] + 3*ArcSinh
[c*x]^2*Log[1 - E^(-ArcSinh[c*x])] + (7*I)*ArcSinh[c*x]*Log[1 - I/E^ArcSinh[c*x]] - (7*I)*ArcSinh[c*x]*Log[1 +
I/E^ArcSinh[c*x]] - 3*ArcSinh[c*x]^2*Log[1 + E^(-ArcSinh[c*x])] + 6*ArcSinh[c*x]*PolyLog[2, -E^(-ArcSinh[c*x]
)] + (7*I)*PolyLog[2, (-I)/E^ArcSinh[c*x]] - (7*I)*PolyLog[2, I/E^ArcSinh[c*x]] - 6*ArcSinh[c*x]*PolyLog[2, E^
(-ArcSinh[c*x])] + 6*PolyLog[3, -E^(-ArcSinh[c*x])] - 6*PolyLog[3, E^(-ArcSinh[c*x])]))/(d + c^2*d*x^2)^(3/2))
/(3*d^3)

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Maple [F]  time = 0.257, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{x} \left ({c}^{2}d{x}^{2}+d \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))^2/x/(c^2*d*x^2+d)^(5/2),x)

[Out]

int((a+b*arcsinh(c*x))^2/x/(c^2*d*x^2+d)^(5/2),x)

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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/(c^2*d*x^2+d)^(5/2),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^{6} d^{3} x^{7} + 3 \, c^{4} d^{3} x^{5} + 3 \, c^{2} d^{3} x^{3} + d^{3} x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))^2/x/(c^2*d*x^2+d)^(5/2),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^6*d^3*x^7 + 3*c^4*d^3*x^5 + 3*
c^2*d^3*x^3 + d^3*x), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(c*x))**2/x/(c**2*d*x**2+d)**(5/2),x)

[Out]

Timed out

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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 )}^{\frac{5}{2}} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))^2/x/(c^2*d*x^2+d)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)^2/((c^2*d*x^2 + d)^(5/2)*x), x)