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

Optimal. Leaf size=385 \[ -\frac{4 b^3 \text{PolyLog}\left (3,-e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^4}+\frac{4 b^3 \text{PolyLog}\left (3,e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^4}+\frac{2 b^2 \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^4}-\frac{2 b^2 \text{PolyLog}\left (2,e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^4}-\frac{4 b^4 \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac{4 b^4 \text{PolyLog}\left (2,e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac{4 b^4 \text{PolyLog}\left (4,-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac{4 b^4 \text{PolyLog}\left (4,e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac{2 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^4 (c+d x)}-\frac{8 b^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^4}-\frac{2 b \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^2}-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{3 d e^4 (c+d x)^3}+\frac{4 b \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d e^4} \]

[Out]

(-2*b^2*(a + b*ArcSinh[c + d*x])^2)/(d*e^4*(c + d*x)) - (2*b*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x])^3)
/(3*d*e^4*(c + d*x)^2) - (a + b*ArcSinh[c + d*x])^4/(3*d*e^4*(c + d*x)^3) - (8*b^3*(a + b*ArcSinh[c + d*x])*Ar
cTanh[E^ArcSinh[c + d*x]])/(d*e^4) + (4*b*(a + b*ArcSinh[c + d*x])^3*ArcTanh[E^ArcSinh[c + d*x]])/(3*d*e^4) -
(4*b^4*PolyLog[2, -E^ArcSinh[c + d*x]])/(d*e^4) + (2*b^2*(a + b*ArcSinh[c + d*x])^2*PolyLog[2, -E^ArcSinh[c +
d*x]])/(d*e^4) + (4*b^4*PolyLog[2, E^ArcSinh[c + d*x]])/(d*e^4) - (2*b^2*(a + b*ArcSinh[c + d*x])^2*PolyLog[2,
E^ArcSinh[c + d*x]])/(d*e^4) - (4*b^3*(a + b*ArcSinh[c + d*x])*PolyLog[3, -E^ArcSinh[c + d*x]])/(d*e^4) + (4*
b^3*(a + b*ArcSinh[c + d*x])*PolyLog[3, E^ArcSinh[c + d*x]])/(d*e^4) + (4*b^4*PolyLog[4, -E^ArcSinh[c + d*x]])
/(d*e^4) - (4*b^4*PolyLog[4, E^ArcSinh[c + d*x]])/(d*e^4)

________________________________________________________________________________________

Rubi [A]  time = 0.564892, antiderivative size = 385, normalized size of antiderivative = 1., number of steps used = 21, number of rules used = 12, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.522, Rules used = {5865, 12, 5661, 5747, 5760, 4182, 2531, 6609, 2282, 6589, 2279, 2391} \[ -\frac{4 b^3 \text{PolyLog}\left (3,-e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^4}+\frac{4 b^3 \text{PolyLog}\left (3,e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^4}+\frac{2 b^2 \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^4}-\frac{2 b^2 \text{PolyLog}\left (2,e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^4}-\frac{4 b^4 \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac{4 b^4 \text{PolyLog}\left (2,e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac{4 b^4 \text{PolyLog}\left (4,-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac{4 b^4 \text{PolyLog}\left (4,e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac{2 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^4 (c+d x)}-\frac{8 b^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^4}-\frac{2 b \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^2}-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{3 d e^4 (c+d x)^3}+\frac{4 b \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b^2*(a + b*ArcSinh[c + d*x])^2)/(d*e^4*(c + d*x)) - (2*b*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x])^3)
/(3*d*e^4*(c + d*x)^2) - (a + b*ArcSinh[c + d*x])^4/(3*d*e^4*(c + d*x)^3) - (8*b^3*(a + b*ArcSinh[c + d*x])*Ar
cTanh[E^ArcSinh[c + d*x]])/(d*e^4) + (4*b*(a + b*ArcSinh[c + d*x])^3*ArcTanh[E^ArcSinh[c + d*x]])/(3*d*e^4) -
(4*b^4*PolyLog[2, -E^ArcSinh[c + d*x]])/(d*e^4) + (2*b^2*(a + b*ArcSinh[c + d*x])^2*PolyLog[2, -E^ArcSinh[c +
d*x]])/(d*e^4) + (4*b^4*PolyLog[2, E^ArcSinh[c + d*x]])/(d*e^4) - (2*b^2*(a + b*ArcSinh[c + d*x])^2*PolyLog[2,
E^ArcSinh[c + d*x]])/(d*e^4) - (4*b^3*(a + b*ArcSinh[c + d*x])*PolyLog[3, -E^ArcSinh[c + d*x]])/(d*e^4) + (4*
b^3*(a + b*ArcSinh[c + d*x])*PolyLog[3, E^ArcSinh[c + d*x]])/(d*e^4) + (4*b^4*PolyLog[4, -E^ArcSinh[c + d*x]])
/(d*e^4) - (4*b^4*PolyLog[4, E^ArcSinh[c + d*x]])/(d*e^4)

Rule 5865

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x
]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 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 6609

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

Rubi steps

\begin{align*} \int \frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{(c e+d e x)^4} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \sinh ^{-1}(x)\right )^4}{e^4 x^4} \, dx,x,c+d x\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \sinh ^{-1}(x)\right )^4}{x^4} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{3 d e^4 (c+d x)^3}+\frac{(4 b) \operatorname{Subst}\left (\int \frac{\left (a+b \sinh ^{-1}(x)\right )^3}{x^3 \sqrt{1+x^2}} \, dx,x,c+d x\right )}{3 d e^4}\\ &=-\frac{2 b \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^2}-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{3 d e^4 (c+d x)^3}-\frac{(2 b) \operatorname{Subst}\left (\int \frac{\left (a+b \sinh ^{-1}(x)\right )^3}{x \sqrt{1+x^2}} \, dx,x,c+d x\right )}{3 d e^4}+\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{\left (a+b \sinh ^{-1}(x)\right )^2}{x^2} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac{2 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^4 (c+d x)}-\frac{2 b \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^2}-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{3 d e^4 (c+d x)^3}-\frac{(2 b) \operatorname{Subst}\left (\int (a+b x)^3 \text{csch}(x) \, dx,x,\sinh ^{-1}(c+d x)\right )}{3 d e^4}+\frac{\left (4 b^3\right ) \operatorname{Subst}\left (\int \frac{a+b \sinh ^{-1}(x)}{x \sqrt{1+x^2}} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac{2 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^4 (c+d x)}-\frac{2 b \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^2}-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{3 d e^4 (c+d x)^3}+\frac{4 b \left (a+b \sinh ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{3 d e^4}+\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int (a+b x)^2 \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^4}-\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int (a+b x)^2 \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^4}+\frac{\left (4 b^3\right ) \operatorname{Subst}\left (\int (a+b x) \text{csch}(x) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^4}\\ &=-\frac{2 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^4 (c+d x)}-\frac{2 b \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^2}-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{3 d e^4 (c+d x)^3}-\frac{8 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac{4 b \left (a+b \sinh ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{3 d e^4}+\frac{2 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text{Li}_2\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac{2 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text{Li}_2\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac{\left (4 b^3\right ) \operatorname{Subst}\left (\int (a+b x) \text{Li}_2\left (-e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^4}+\frac{\left (4 b^3\right ) \operatorname{Subst}\left (\int (a+b x) \text{Li}_2\left (e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^4}-\frac{\left (4 b^4\right ) \operatorname{Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^4}+\frac{\left (4 b^4\right ) \operatorname{Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^4}\\ &=-\frac{2 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^4 (c+d x)}-\frac{2 b \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^2}-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{3 d e^4 (c+d x)^3}-\frac{8 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac{4 b \left (a+b \sinh ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{3 d e^4}+\frac{2 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text{Li}_2\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac{2 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text{Li}_2\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac{4 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \text{Li}_3\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac{4 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \text{Li}_3\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac{\left (4 b^4\right ) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac{\left (4 b^4\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac{\left (4 b^4\right ) \operatorname{Subst}\left (\int \text{Li}_3\left (-e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^4}-\frac{\left (4 b^4\right ) \operatorname{Subst}\left (\int \text{Li}_3\left (e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^4}\\ &=-\frac{2 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^4 (c+d x)}-\frac{2 b \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^2}-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{3 d e^4 (c+d x)^3}-\frac{8 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac{4 b \left (a+b \sinh ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{3 d e^4}-\frac{4 b^4 \text{Li}_2\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac{2 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text{Li}_2\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac{4 b^4 \text{Li}_2\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac{2 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text{Li}_2\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac{4 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \text{Li}_3\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac{4 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \text{Li}_3\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac{\left (4 b^4\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-x)}{x} \, dx,x,e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac{\left (4 b^4\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(x)}{x} \, dx,x,e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}\\ &=-\frac{2 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^4 (c+d x)}-\frac{2 b \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^2}-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{3 d e^4 (c+d x)^3}-\frac{8 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac{4 b \left (a+b \sinh ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{3 d e^4}-\frac{4 b^4 \text{Li}_2\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac{2 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text{Li}_2\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac{4 b^4 \text{Li}_2\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac{2 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text{Li}_2\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac{4 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \text{Li}_3\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac{4 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \text{Li}_3\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac{4 b^4 \text{Li}_4\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac{4 b^4 \text{Li}_4\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}\\ \end{align*}

Mathematica [B]  time = 8.92788, size = 1182, normalized size = 3.07 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-a^4/(3*d*e^4*(c + d*x)^3) + (a^2*b^2*(-8*PolyLog[2, -E^(-ArcSinh[c + d*x])] - (2*(-2 + 4*ArcSinh[c + d*x]^2 +
2*Cosh[2*ArcSinh[c + d*x]] - 3*(c + d*x)*ArcSinh[c + d*x]*Log[1 - E^(-ArcSinh[c + d*x])] + 3*(c + d*x)*ArcSin
h[c + d*x]*Log[1 + E^(-ArcSinh[c + d*x])] - 4*(c + d*x)^3*PolyLog[2, E^(-ArcSinh[c + d*x])] + 2*ArcSinh[c + d*
x]*Sinh[2*ArcSinh[c + d*x]] + ArcSinh[c + d*x]*Log[1 - E^(-ArcSinh[c + d*x])]*Sinh[3*ArcSinh[c + d*x]] - ArcSi
nh[c + d*x]*Log[1 + E^(-ArcSinh[c + d*x])]*Sinh[3*ArcSinh[c + d*x]]))/(c + d*x)^3))/(4*d*e^4) + (a*b^3*(-24*Ar
cSinh[c + d*x]*Coth[ArcSinh[c + d*x]/2] + 4*ArcSinh[c + d*x]^3*Coth[ArcSinh[c + d*x]/2] - 6*ArcSinh[c + d*x]^2
*Csch[ArcSinh[c + d*x]/2]^2 - (c + d*x)*ArcSinh[c + d*x]^3*Csch[ArcSinh[c + d*x]/2]^4 - 24*ArcSinh[c + d*x]^2*
Log[1 - E^(-ArcSinh[c + d*x])] + 24*ArcSinh[c + d*x]^2*Log[1 + E^(-ArcSinh[c + d*x])] + 48*Log[Tanh[ArcSinh[c
+ d*x]/2]] - 48*ArcSinh[c + d*x]*PolyLog[2, -E^(-ArcSinh[c + d*x])] + 48*ArcSinh[c + d*x]*PolyLog[2, E^(-ArcSi
nh[c + d*x])] - 48*PolyLog[3, -E^(-ArcSinh[c + d*x])] + 48*PolyLog[3, E^(-ArcSinh[c + d*x])] - 6*ArcSinh[c + d
*x]^2*Sech[ArcSinh[c + d*x]/2]^2 - (16*ArcSinh[c + d*x]^3*Sinh[ArcSinh[c + d*x]/2]^4)/(c + d*x)^3 + 24*ArcSinh
[c + d*x]*Tanh[ArcSinh[c + d*x]/2] - 4*ArcSinh[c + d*x]^3*Tanh[ArcSinh[c + d*x]/2]))/(12*d*e^4) + (b^4*(-2*Pi^
4 + 4*ArcSinh[c + d*x]^4 - 24*ArcSinh[c + d*x]^2*Coth[ArcSinh[c + d*x]/2] + 2*ArcSinh[c + d*x]^4*Coth[ArcSinh[
c + d*x]/2] - 4*ArcSinh[c + d*x]^3*Csch[ArcSinh[c + d*x]/2]^2 - ((c + d*x)*ArcSinh[c + d*x]^4*Csch[ArcSinh[c +
d*x]/2]^4)/2 + 96*ArcSinh[c + d*x]*Log[1 - E^(-ArcSinh[c + d*x])] - 96*ArcSinh[c + d*x]*Log[1 + E^(-ArcSinh[c
+ d*x])] + 16*ArcSinh[c + d*x]^3*Log[1 + E^(-ArcSinh[c + d*x])] - 16*ArcSinh[c + d*x]^3*Log[1 - E^ArcSinh[c +
d*x]] - 48*(-2 + ArcSinh[c + d*x]^2)*PolyLog[2, -E^(-ArcSinh[c + d*x])] - 96*PolyLog[2, E^(-ArcSinh[c + d*x])
] - 48*ArcSinh[c + d*x]^2*PolyLog[2, E^ArcSinh[c + d*x]] - 96*ArcSinh[c + d*x]*PolyLog[3, -E^(-ArcSinh[c + d*x
])] + 96*ArcSinh[c + d*x]*PolyLog[3, E^ArcSinh[c + d*x]] - 96*PolyLog[4, -E^(-ArcSinh[c + d*x])] - 96*PolyLog[
4, E^ArcSinh[c + d*x]] - 4*ArcSinh[c + d*x]^3*Sech[ArcSinh[c + d*x]/2]^2 - (8*ArcSinh[c + d*x]^4*Sinh[ArcSinh[
c + d*x]/2]^4)/(c + d*x)^3 + 24*ArcSinh[c + d*x]^2*Tanh[ArcSinh[c + d*x]/2] - 2*ArcSinh[c + d*x]^4*Tanh[ArcSin
h[c + d*x]/2]))/(24*d*e^4) + (4*a^3*b*((ArcSinh[c + d*x]*Coth[ArcSinh[c + d*x]/2])/12 - Csch[ArcSinh[c + d*x]/
2]^2/24 - (ArcSinh[c + d*x]*Coth[ArcSinh[c + d*x]/2]*Csch[ArcSinh[c + d*x]/2]^2)/24 - Log[Tanh[ArcSinh[c + d*x
]/2]]/6 - Sech[ArcSinh[c + d*x]/2]^2/24 - (ArcSinh[c + d*x]*Tanh[ArcSinh[c + d*x]/2])/12 - (ArcSinh[c + d*x]*S
ech[ArcSinh[c + d*x]/2]^2*Tanh[ArcSinh[c + d*x]/2])/24))/(d*e^4)

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Maple [B]  time = 0.126, size = 1202, normalized size = 3.1 \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(d*x+c))^4/(d*e*x+c*e)^4,x)

[Out]

-4/3/d*a*b^3/e^4/(d*x+c)^3*arcsinh(d*x+c)^3-4/d*a*b^3/e^4/(d*x+c)*arcsinh(d*x+c)+2/d*a*b^3/e^4*arcsinh(d*x+c)^
2*ln(1+d*x+c+(1+(d*x+c)^2)^(1/2))+4/d*a*b^3/e^4*arcsinh(d*x+c)*polylog(2,-d*x-c-(1+(d*x+c)^2)^(1/2))-2/d*a*b^3
/e^4*arcsinh(d*x+c)^2*ln(1-d*x-c-(1+(d*x+c)^2)^(1/2))-4/3/d*a^3*b/e^4/(d*x+c)^3*arcsinh(d*x+c)-2/3/d*a^3*b/e^4
/(d*x+c)^2*(1+(d*x+c)^2)^(1/2)-4/d*a*b^3/e^4*arcsinh(d*x+c)*polylog(2,d*x+c+(1+(d*x+c)^2)^(1/2))-2/d*a^2*b^2/e
^4/(d*x+c)^3*arcsinh(d*x+c)^2+2/d*a^2*b^2/e^4*arcsinh(d*x+c)*ln(1+d*x+c+(1+(d*x+c)^2)^(1/2))-2/d*a^2*b^2/e^4*a
rcsinh(d*x+c)*ln(1-d*x-c-(1+(d*x+c)^2)^(1/2))-2/3/d*b^4/e^4/(d*x+c)^2*arcsinh(d*x+c)^3*(1+(d*x+c)^2)^(1/2)-4/d
*a*b^3/e^4*polylog(3,-d*x-c-(1+(d*x+c)^2)^(1/2))+4/d*a*b^3/e^4*polylog(3,d*x+c+(1+(d*x+c)^2)^(1/2))-8/d*a*b^3/
e^4*arctanh(d*x+c+(1+(d*x+c)^2)^(1/2))+2/d*a^2*b^2/e^4*polylog(2,-d*x-c-(1+(d*x+c)^2)^(1/2))-2/d*a^2*b^2/e^4*p
olylog(2,d*x+c+(1+(d*x+c)^2)^(1/2))+2/3/d*a^3*b/e^4*arctanh(1/(1+(d*x+c)^2)^(1/2))-1/3/d*b^4/e^4/(d*x+c)^3*arc
sinh(d*x+c)^4-2/d*b^4/e^4/(d*x+c)*arcsinh(d*x+c)^2+4/d*b^4/e^4*arcsinh(d*x+c)*polylog(3,d*x+c+(1+(d*x+c)^2)^(1
/2))-4/d*b^4/e^4*arcsinh(d*x+c)*ln(1+d*x+c+(1+(d*x+c)^2)^(1/2))-2/d*a^2*b^2/e^4/(d*x+c)+2/3/d*b^4/e^4*arcsinh(
d*x+c)^3*ln(1+d*x+c+(1+(d*x+c)^2)^(1/2))+2/d*b^4/e^4*arcsinh(d*x+c)^2*polylog(2,-d*x-c-(1+(d*x+c)^2)^(1/2))-4/
d*b^4/e^4*arcsinh(d*x+c)*polylog(3,-d*x-c-(1+(d*x+c)^2)^(1/2))-2/3/d*b^4/e^4*arcsinh(d*x+c)^3*ln(1-d*x-c-(1+(d
*x+c)^2)^(1/2))-2/d*b^4/e^4*arcsinh(d*x+c)^2*polylog(2,d*x+c+(1+(d*x+c)^2)^(1/2))-4*b^4*polylog(2,-d*x-c-(1+(d
*x+c)^2)^(1/2))/d/e^4+4*b^4*polylog(2,d*x+c+(1+(d*x+c)^2)^(1/2))/d/e^4+4*b^4*polylog(4,-d*x-c-(1+(d*x+c)^2)^(1
/2))/d/e^4-4*b^4*polylog(4,d*x+c+(1+(d*x+c)^2)^(1/2))/d/e^4+4/d*b^4/e^4*arcsinh(d*x+c)*ln(1-d*x-c-(1+(d*x+c)^2
)^(1/2))-2/d*a^2*b^2/e^4/(d*x+c)^2*arcsinh(d*x+c)*(1+(d*x+c)^2)^(1/2)-2/d*a*b^3/e^4/(d*x+c)^2*arcsinh(d*x+c)^2
*(1+(d*x+c)^2)^(1/2)-1/3/d*a^4/e^4/(d*x+c)^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*b^4*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))^4/(d^4*e^4*x^3 + 3*c*d^3*e^4*x^2 + 3*c^2*d^2*e^4*x +
c^3*d*e^4) - 1/3*a^4/(d^4*e^4*x^3 + 3*c*d^3*e^4*x^2 + 3*c^2*d^2*e^4*x + c^3*d*e^4) + integrate(2/3*(2*(3*(c^3
+ c)*a*b^3 + (c^3 + c)*b^4 + (3*a*b^3*d^3 + b^4*d^3)*x^3 + 3*(3*a*b^3*c*d^2 + b^4*c*d^2)*x^2 + (3*(3*c^2*d +
d)*a*b^3 + (3*c^2*d + d)*b^4)*x + (b^4*c^2 + 3*(c^2 + 1)*a*b^3 + (3*a*b^3*d^2 + b^4*d^2)*x^2 + 2*(3*a*b^3*c*d
+ b^4*c*d)*x)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))^3 + 9*(a^2*b
^2*d^3*x^3 + 3*a^2*b^2*c*d^2*x^2 + (3*c^2*d + d)*a^2*b^2*x + (c^3 + c)*a^2*b^2 + (a^2*b^2*d^2*x^2 + 2*a^2*b^2*
c*d*x + (c^2 + 1)*a^2*b^2)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))
^2 + 6*(a^3*b*d^3*x^3 + 3*a^3*b*c*d^2*x^2 + (3*c^2*d + d)*a^3*b*x + (c^3 + c)*a^3*b + (a^3*b*d^2*x^2 + 2*a^3*b
*c*d*x + (c^2 + 1)*a^3*b)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)))
/(d^7*e^4*x^7 + 7*c*d^6*e^4*x^6 + c^7*e^4 + c^5*e^4 + (21*c^2*d^5*e^4 + d^5*e^4)*x^5 + 5*(7*c^3*d^4*e^4 + c*d^
4*e^4)*x^4 + 5*(7*c^4*d^3*e^4 + 2*c^2*d^3*e^4)*x^3 + (21*c^5*d^2*e^4 + 10*c^3*d^2*e^4)*x^2 + (7*c^6*d*e^4 + 5*
c^4*d*e^4)*x + (d^6*e^4*x^6 + 6*c*d^5*e^4*x^5 + c^6*e^4 + c^4*e^4 + (15*c^2*d^4*e^4 + d^4*e^4)*x^4 + 4*(5*c^3*
d^3*e^4 + c*d^3*e^4)*x^3 + 3*(5*c^4*d^2*e^4 + 2*c^2*d^2*e^4)*x^2 + 2*(3*c^5*d*e^4 + 2*c^3*d*e^4)*x)*sqrt(d^2*x
^2 + 2*c*d*x + c^2 + 1)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{4} \operatorname{arsinh}\left (d x + c\right )^{4} + 4 \, a b^{3} \operatorname{arsinh}\left (d x + c\right )^{3} + 6 \, a^{2} b^{2} \operatorname{arsinh}\left (d x + c\right )^{2} + 4 \, a^{3} b \operatorname{arsinh}\left (d x + c\right ) + a^{4}}{d^{4} e^{4} x^{4} + 4 \, c d^{3} e^{4} x^{3} + 6 \, c^{2} d^{2} e^{4} x^{2} + 4 \, c^{3} d e^{4} x + c^{4} e^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^4*arcsinh(d*x + c)^4 + 4*a*b^3*arcsinh(d*x + c)^3 + 6*a^2*b^2*arcsinh(d*x + c)^2 + 4*a^3*b*arcsinh
(d*x + c) + a^4)/(d^4*e^4*x^4 + 4*c*d^3*e^4*x^3 + 6*c^2*d^2*e^4*x^2 + 4*c^3*d*e^4*x + c^4*e^4), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a^{4}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac{b^{4} \operatorname{asinh}^{4}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac{4 a b^{3} \operatorname{asinh}^{3}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac{6 a^{2} b^{2} \operatorname{asinh}^{2}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac{4 a^{3} b \operatorname{asinh}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx}{e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a**4/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x) + Integral(b**4*asinh(c
+ d*x)**4/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x) + Integral(4*a*b**3*asinh(c +
d*x)**3/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x) + Integral(6*a**2*b**2*asinh(c
+ d*x)**2/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x) + Integral(4*a**3*b*asinh(c
+ d*x)/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x))/e**4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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