3.145 \(\int \frac{(a+b \sinh ^{-1}(c+d x))^3}{(c e+d e x)^4} \, dx\)

Optimal. Leaf size=261 \[ \frac{b^2 \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^4}-\frac{b^2 \text{PolyLog}\left (2,e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^4}-\frac{b^3 \text{PolyLog}\left (3,-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac{b^3 \text{PolyLog}\left (3,e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac{b^2 \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac{b \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac{b \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^4}-\frac{b^3 \tanh ^{-1}\left (\sqrt{(c+d x)^2+1}\right )}{d e^4} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.410011, antiderivative size = 261, normalized size of antiderivative = 1., number of steps used = 16, 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, 2282, 6589, 266, 63, 207} \[ \frac{b^2 \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^4}-\frac{b^2 \text{PolyLog}\left (2,e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^4}-\frac{b^3 \text{PolyLog}\left (3,-e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}+\frac{b^3 \text{PolyLog}\left (3,e^{\sinh ^{-1}(c+d x)}\right )}{d e^4}-\frac{b^2 \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac{b \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac{b \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^4}-\frac{b^3 \tanh ^{-1}\left (\sqrt{(c+d x)^2+1}\right )}{d e^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c + d*x])^3/(c*e + d*e*x)^4,x]

[Out]

-((b^2*(a + b*ArcSinh[c + d*x]))/(d*e^4*(c + d*x))) - (b*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x])^2)/(2*
d*e^4*(c + d*x)^2) - (a + b*ArcSinh[c + d*x])^3/(3*d*e^4*(c + d*x)^3) + (b*(a + b*ArcSinh[c + d*x])^2*ArcTanh[
E^ArcSinh[c + d*x]])/(d*e^4) - (b^3*ArcTanh[Sqrt[1 + (c + d*x)^2]])/(d*e^4) + (b^2*(a + b*ArcSinh[c + d*x])*Po
lyLog[2, -E^ArcSinh[c + d*x]])/(d*e^4) - (b^2*(a + b*ArcSinh[c + d*x])*PolyLog[2, E^ArcSinh[c + d*x]])/(d*e^4)
- (b^3*PolyLog[3, -E^ArcSinh[c + d*x]])/(d*e^4) + (b^3*PolyLog[3, 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 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 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 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [B]  time = 7.53295, size = 694, normalized size = 2.66 \[ \frac{a b^2 \left (-8 \text{PolyLog}\left (2,-e^{-\sinh ^{-1}(c+d x)}\right )-\frac{2 \left (-4 (c+d x)^3 \text{PolyLog}\left (2,e^{-\sinh ^{-1}(c+d x)}\right )+4 \sinh ^{-1}(c+d x)^2+2 \sinh ^{-1}(c+d x) \sinh \left (2 \sinh ^{-1}(c+d x)\right )-3 (c+d x) \sinh ^{-1}(c+d x) \log \left (1-e^{-\sinh ^{-1}(c+d x)}\right )+3 (c+d x) \sinh ^{-1}(c+d x) \log \left (e^{-\sinh ^{-1}(c+d x)}+1\right )+\sinh ^{-1}(c+d x) \sinh \left (3 \sinh ^{-1}(c+d x)\right ) \log \left (1-e^{-\sinh ^{-1}(c+d x)}\right )-\sinh ^{-1}(c+d x) \sinh \left (3 \sinh ^{-1}(c+d x)\right ) \log \left (e^{-\sinh ^{-1}(c+d x)}+1\right )+2 \cosh \left (2 \sinh ^{-1}(c+d x)\right )-2\right )}{(c+d x)^3}\right )}{8 d e^4}+\frac{b^3 \left (-48 \sinh ^{-1}(c+d x) \text{PolyLog}\left (2,-e^{-\sinh ^{-1}(c+d x)}\right )+48 \sinh ^{-1}(c+d x) \text{PolyLog}\left (2,e^{-\sinh ^{-1}(c+d x)}\right )-48 \text{PolyLog}\left (3,-e^{-\sinh ^{-1}(c+d x)}\right )+48 \text{PolyLog}\left (3,e^{-\sinh ^{-1}(c+d x)}\right )-\frac{16 \sinh ^{-1}(c+d x)^3 \sinh ^4\left (\frac{1}{2} \sinh ^{-1}(c+d x)\right )}{(c+d x)^3}-24 \sinh ^{-1}(c+d x)^2 \log \left (1-e^{-\sinh ^{-1}(c+d x)}\right )+24 \sinh ^{-1}(c+d x)^2 \log \left (e^{-\sinh ^{-1}(c+d x)}+1\right )-4 \sinh ^{-1}(c+d x)^3 \tanh \left (\frac{1}{2} \sinh ^{-1}(c+d x)\right )+24 \sinh ^{-1}(c+d x) \tanh \left (\frac{1}{2} \sinh ^{-1}(c+d x)\right )+4 \sinh ^{-1}(c+d x)^3 \coth \left (\frac{1}{2} \sinh ^{-1}(c+d x)\right )-24 \sinh ^{-1}(c+d x) \coth \left (\frac{1}{2} \sinh ^{-1}(c+d x)\right )-(c+d x) \sinh ^{-1}(c+d x)^3 \text{csch}^4\left (\frac{1}{2} \sinh ^{-1}(c+d x)\right )-6 \sinh ^{-1}(c+d x)^2 \text{csch}^2\left (\frac{1}{2} \sinh ^{-1}(c+d x)\right )-6 \sinh ^{-1}(c+d x)^2 \text{sech}^2\left (\frac{1}{2} \sinh ^{-1}(c+d x)\right )+48 \log \left (\tanh \left (\frac{1}{2} \sinh ^{-1}(c+d x)\right )\right )\right )}{48 d e^4}-\frac{a^2 b \sqrt{c^2+2 c d x+d^2 x^2+1}}{2 d e^4 (c+d x)^2}+\frac{a^2 b \log \left (\sqrt{c^2+2 c d x+d^2 x^2+1}+1\right )}{2 d e^4}-\frac{a^2 b \log (c+d x)}{2 d e^4}-\frac{a^2 b \sinh ^{-1}(c+d x)}{d e^4 (c+d x)^3}-\frac{a^3}{3 d e^4 (c+d x)^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSinh[c + d*x])^3/(c*e + d*e*x)^4,x]

[Out]

-a^3/(3*d*e^4*(c + d*x)^3) - (a^2*b*Sqrt[1 + c^2 + 2*c*d*x + d^2*x^2])/(2*d*e^4*(c + d*x)^2) - (a^2*b*ArcSinh[
c + d*x])/(d*e^4*(c + d*x)^3) - (a^2*b*Log[c + d*x])/(2*d*e^4) + (a^2*b*Log[1 + Sqrt[1 + c^2 + 2*c*d*x + d^2*x
^2]])/(2*d*e^4) + (a*b^2*(-8*PolyLog[2, -E^(-ArcSinh[c + d*x])] - (2*(-2 + 4*ArcSinh[c + d*x]^2 + 2*Cosh[2*Arc
Sinh[c + d*x]] - 3*(c + d*x)*ArcSinh[c + d*x]*Log[1 - E^(-ArcSinh[c + d*x])] + 3*(c + d*x)*ArcSinh[c + d*x]*Lo
g[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*Arc
Sinh[c + d*x]] + ArcSinh[c + d*x]*Log[1 - E^(-ArcSinh[c + d*x])]*Sinh[3*ArcSinh[c + d*x]] - ArcSinh[c + d*x]*L
og[1 + E^(-ArcSinh[c + d*x])]*Sinh[3*ArcSinh[c + d*x]]))/(c + d*x)^3))/(8*d*e^4) + (b^3*(-24*ArcSinh[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^(-Arc
Sinh[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^(-ArcSinh[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[ArcS
inh[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]))/(48*d*e^4)

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Maple [B]  time = 0.094, size = 651, 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(d*x+c))^3/(d*e*x+c*e)^4,x)

[Out]

-1/3/d*a^3/e^4/(d*x+c)^3-1/2/d*b^3/e^4/(d*x+c)^2*arcsinh(d*x+c)^2*(1+(d*x+c)^2)^(1/2)-1/3/d*b^3/e^4/(d*x+c)^3*
arcsinh(d*x+c)^3-1/d*b^3/e^4/(d*x+c)*arcsinh(d*x+c)+1/2/d*b^3/e^4*arcsinh(d*x+c)^2*ln(1+d*x+c+(1+(d*x+c)^2)^(1
/2))+1/d*b^3/e^4*arcsinh(d*x+c)*polylog(2,-d*x-c-(1+(d*x+c)^2)^(1/2))-b^3*polylog(3,-d*x-c-(1+(d*x+c)^2)^(1/2)
)/d/e^4-1/2/d*b^3/e^4*arcsinh(d*x+c)^2*ln(1-d*x-c-(1+(d*x+c)^2)^(1/2))-1/d*b^3/e^4*arcsinh(d*x+c)*polylog(2,d*
x+c+(1+(d*x+c)^2)^(1/2))+b^3*polylog(3,d*x+c+(1+(d*x+c)^2)^(1/2))/d/e^4-2/d*b^3/e^4*arctanh(d*x+c+(1+(d*x+c)^2
)^(1/2))-1/d*a*b^2/e^4/(d*x+c)^2*arcsinh(d*x+c)*(1+(d*x+c)^2)^(1/2)-1/d*a*b^2/e^4/(d*x+c)^3*arcsinh(d*x+c)^2-1
/d*a*b^2/e^4/(d*x+c)+1/d*a*b^2/e^4*arcsinh(d*x+c)*ln(1+d*x+c+(1+(d*x+c)^2)^(1/2))+1/d*a*b^2/e^4*polylog(2,-d*x
-c-(1+(d*x+c)^2)^(1/2))-1/d*a*b^2/e^4*arcsinh(d*x+c)*ln(1-d*x-c-(1+(d*x+c)^2)^(1/2))-1/d*a*b^2/e^4*polylog(2,d
*x+c+(1+(d*x+c)^2)^(1/2))-1/d*a^2*b/e^4/(d*x+c)^3*arcsinh(d*x+c)-1/2/d*a^2*b/e^4/(d*x+c)^2*(1+(d*x+c)^2)^(1/2)
+1/2/d*a^2*b/e^4*arctanh(1/(1+(d*x+c)^2)^(1/2))

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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))^3/(d*e*x+c*e)^4,x, algorithm="maxima")

[Out]

-1/3*b^3*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))^3/(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^3/(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(((3*(c^3 + c)*
a*b^2 + (c^3 + c)*b^3 + (3*a*b^2*d^3 + b^3*d^3)*x^3 + 3*(3*a*b^2*c*d^2 + b^3*c*d^2)*x^2 + (3*(3*c^2*d + d)*a*b
^2 + (3*c^2*d + d)*b^3)*x + (b^3*c^2 + 3*(c^2 + 1)*a*b^2 + (3*a*b^2*d^2 + b^3*d^2)*x^2 + 2*(3*a*b^2*c*d + b^3*
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))^2 + 3*(a^2*b*d^3*x
^3 + 3*a^2*b*c*d^2*x^2 + (3*c^2*d + d)*a^2*b*x + (c^3 + c)*a^2*b + (a^2*b*d^2*x^2 + 2*a^2*b*c*d*x + (c^2 + 1)*
a^2*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^{3} \operatorname{arsinh}\left (d x + c\right )^{3} + 3 \, a b^{2} \operatorname{arsinh}\left (d x + c\right )^{2} + 3 \, a^{2} b \operatorname{arsinh}\left (d x + c\right ) + a^{3}}{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))^3/(d*e*x+c*e)^4,x, algorithm="fricas")

[Out]

integral((b^3*arcsinh(d*x + c)^3 + 3*a*b^2*arcsinh(d*x + c)^2 + 3*a^2*b*arcsinh(d*x + c) + a^3)/(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^{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}}\, dx + \int \frac{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{3 a 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{3 a^{2} 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))**3/(d*e*x+c*e)**4,x)

[Out]

(Integral(a**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(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(3*a*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(3*a**2*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 )}^{3}}{{\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))^3/(d*e*x+c*e)^4,x, algorithm="giac")

[Out]

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