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

Optimal. Leaf size=340 \[ -\frac{e^3 \cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{3 b^4 d}+\frac{4 e^3 \cosh \left (\frac{4 a}{b}\right ) \text{Chi}\left (\frac{4 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{3 b^4 d}+\frac{e^3 \sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{3 b^4 d}-\frac{4 e^3 \sinh \left (\frac{4 a}{b}\right ) \text{Shi}\left (\frac{4 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{3 b^4 d}-\frac{2 e^3 (c+d x)^4}{3 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{8 e^3 \sqrt{(c+d x)^2+1} (c+d x)^3}{3 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{e^3 (c+d x)^2}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{e^3 \sqrt{(c+d x)^2+1} (c+d x)}{b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{e^3 \sqrt{(c+d x)^2+1} (c+d x)^3}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3} \]

[Out]

-(e^3*(c + d*x)^3*Sqrt[1 + (c + d*x)^2])/(3*b*d*(a + b*ArcSinh[c + d*x])^3) - (e^3*(c + d*x)^2)/(2*b^2*d*(a +
b*ArcSinh[c + d*x])^2) - (2*e^3*(c + d*x)^4)/(3*b^2*d*(a + b*ArcSinh[c + d*x])^2) - (e^3*(c + d*x)*Sqrt[1 + (c
+ d*x)^2])/(b^3*d*(a + b*ArcSinh[c + d*x])) - (8*e^3*(c + d*x)^3*Sqrt[1 + (c + d*x)^2])/(3*b^3*d*(a + b*ArcSi
nh[c + d*x])) - (e^3*Cosh[(2*a)/b]*CoshIntegral[(2*(a + b*ArcSinh[c + d*x]))/b])/(3*b^4*d) + (4*e^3*Cosh[(4*a)
/b]*CoshIntegral[(4*(a + b*ArcSinh[c + d*x]))/b])/(3*b^4*d) + (e^3*Sinh[(2*a)/b]*SinhIntegral[(2*(a + b*ArcSin
h[c + d*x]))/b])/(3*b^4*d) - (4*e^3*Sinh[(4*a)/b]*SinhIntegral[(4*(a + b*ArcSinh[c + d*x]))/b])/(3*b^4*d)

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Rubi [A]  time = 0.695904, antiderivative size = 340, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {5865, 12, 5667, 5774, 5665, 3303, 3298, 3301} \[ -\frac{e^3 \cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c+d x)\right )}{3 b^4 d}+\frac{4 e^3 \cosh \left (\frac{4 a}{b}\right ) \text{Chi}\left (\frac{4 a}{b}+4 \sinh ^{-1}(c+d x)\right )}{3 b^4 d}+\frac{e^3 \sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c+d x)\right )}{3 b^4 d}-\frac{4 e^3 \sinh \left (\frac{4 a}{b}\right ) \text{Shi}\left (\frac{4 a}{b}+4 \sinh ^{-1}(c+d x)\right )}{3 b^4 d}-\frac{2 e^3 (c+d x)^4}{3 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{8 e^3 \sqrt{(c+d x)^2+1} (c+d x)^3}{3 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{e^3 (c+d x)^2}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{e^3 \sqrt{(c+d x)^2+1} (c+d x)}{b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{e^3 \sqrt{(c+d x)^2+1} (c+d x)^3}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(e^3*(c + d*x)^3*Sqrt[1 + (c + d*x)^2])/(3*b*d*(a + b*ArcSinh[c + d*x])^3) - (e^3*(c + d*x)^2)/(2*b^2*d*(a +
b*ArcSinh[c + d*x])^2) - (2*e^3*(c + d*x)^4)/(3*b^2*d*(a + b*ArcSinh[c + d*x])^2) - (e^3*(c + d*x)*Sqrt[1 + (c
+ d*x)^2])/(b^3*d*(a + b*ArcSinh[c + d*x])) - (8*e^3*(c + d*x)^3*Sqrt[1 + (c + d*x)^2])/(3*b^3*d*(a + b*ArcSi
nh[c + d*x])) - (e^3*Cosh[(2*a)/b]*CoshIntegral[(2*a)/b + 2*ArcSinh[c + d*x]])/(3*b^4*d) + (4*e^3*Cosh[(4*a)/b
]*CoshIntegral[(4*a)/b + 4*ArcSinh[c + d*x]])/(3*b^4*d) + (e^3*Sinh[(2*a)/b]*SinhIntegral[(2*a)/b + 2*ArcSinh[
c + d*x]])/(3*b^4*d) - (4*e^3*Sinh[(4*a)/b]*SinhIntegral[(4*a)/b + 4*ArcSinh[c + d*x]])/(3*b^4*d)

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 5667

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

Rule 5774

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

Rule 5665

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[(x^m*Sqrt[1 + c^2*x^2]*(a + b*ArcSi
nh[c*x])^(n + 1))/(b*c*(n + 1)), x] - Dist[1/(b*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[(a + b*x)^(n +
1), Sinh[x]^(m - 1)*(m + (m + 1)*Sinh[x]^2), x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0]
&& GeQ[n, -2] && LtQ[n, -1]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3298

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(I*SinhIntegral[(c*f*fz)
/d + f*fz*x])/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rule 3301

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[(c*f*fz)/d
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rubi steps

\begin{align*} \int \frac{(c e+d e x)^3}{\left (a+b \sinh ^{-1}(c+d x)\right )^4} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{e^3 x^3}{\left (a+b \sinh ^{-1}(x)\right )^4} \, dx,x,c+d x\right )}{d}\\ &=\frac{e^3 \operatorname{Subst}\left (\int \frac{x^3}{\left (a+b \sinh ^{-1}(x)\right )^4} \, dx,x,c+d x\right )}{d}\\ &=-\frac{e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}+\frac{e^3 \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1+x^2} \left (a+b \sinh ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{b d}+\frac{\left (4 e^3\right ) \operatorname{Subst}\left (\int \frac{x^4}{\sqrt{1+x^2} \left (a+b \sinh ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{3 b d}\\ &=-\frac{e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}-\frac{e^3 (c+d x)^2}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{2 e^3 (c+d x)^4}{3 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}+\frac{e^3 \operatorname{Subst}\left (\int \frac{x}{\left (a+b \sinh ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{b^2 d}+\frac{\left (8 e^3\right ) \operatorname{Subst}\left (\int \frac{x^3}{\left (a+b \sinh ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{3 b^2 d}\\ &=-\frac{e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}-\frac{e^3 (c+d x)^2}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{2 e^3 (c+d x)^4}{3 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{e^3 (c+d x) \sqrt{1+(c+d x)^2}}{b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{8 e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{3 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac{e^3 \operatorname{Subst}\left (\int \frac{\cosh (2 x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^3 d}+\frac{\left (8 e^3\right ) \operatorname{Subst}\left (\int \left (-\frac{\cosh (2 x)}{2 (a+b x)}+\frac{\cosh (4 x)}{2 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{3 b^3 d}\\ &=-\frac{e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}-\frac{e^3 (c+d x)^2}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{2 e^3 (c+d x)^4}{3 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{e^3 (c+d x) \sqrt{1+(c+d x)^2}}{b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{8 e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{3 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{\left (4 e^3\right ) \operatorname{Subst}\left (\int \frac{\cosh (2 x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{3 b^3 d}+\frac{\left (4 e^3\right ) \operatorname{Subst}\left (\int \frac{\cosh (4 x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{3 b^3 d}+\frac{\left (e^3 \cosh \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^3 d}-\frac{\left (e^3 \sinh \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^3 d}\\ &=-\frac{e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}-\frac{e^3 (c+d x)^2}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{2 e^3 (c+d x)^4}{3 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{e^3 (c+d x) \sqrt{1+(c+d x)^2}}{b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{8 e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{3 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac{e^3 \cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c+d x)\right )}{b^4 d}-\frac{e^3 \sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c+d x)\right )}{b^4 d}-\frac{\left (4 e^3 \cosh \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{3 b^3 d}+\frac{\left (4 e^3 \cosh \left (\frac{4 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{3 b^3 d}+\frac{\left (4 e^3 \sinh \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{3 b^3 d}-\frac{\left (4 e^3 \sinh \left (\frac{4 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{3 b^3 d}\\ &=-\frac{e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}-\frac{e^3 (c+d x)^2}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{2 e^3 (c+d x)^4}{3 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{e^3 (c+d x) \sqrt{1+(c+d x)^2}}{b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{8 e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{3 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{e^3 \cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c+d x)\right )}{3 b^4 d}+\frac{4 e^3 \cosh \left (\frac{4 a}{b}\right ) \text{Chi}\left (\frac{4 a}{b}+4 \sinh ^{-1}(c+d x)\right )}{3 b^4 d}+\frac{e^3 \sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c+d x)\right )}{3 b^4 d}-\frac{4 e^3 \sinh \left (\frac{4 a}{b}\right ) \text{Shi}\left (\frac{4 a}{b}+4 \sinh ^{-1}(c+d x)\right )}{3 b^4 d}\\ \end{align*}

Mathematica [A]  time = 1.11992, size = 318, normalized size = 0.94 \[ \frac{e^3 \left (-\frac{2 b^3 \sqrt{(c+d x)^2+1} (c+d x)^3}{\left (a+b \sinh ^{-1}(c+d x)\right )^3}+\frac{b^2 \left (-4 (c+d x)^4-3 (c+d x)^2\right )}{\left (a+b \sinh ^{-1}(c+d x)\right )^2}+30 \left (\cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (2 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )-\sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (2 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )-\log \left (a+b \sinh ^{-1}(c+d x)\right )\right )+8 \left (-4 \cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (2 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )+\cosh \left (\frac{4 a}{b}\right ) \text{Chi}\left (4 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )+4 \sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (2 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )-\sinh \left (\frac{4 a}{b}\right ) \text{Shi}\left (4 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )+3 \log \left (a+b \sinh ^{-1}(c+d x)\right )\right )-\frac{2 b \sqrt{(c+d x)^2+1} \left (8 (c+d x)^3+3 (c+d x)\right )}{a+b \sinh ^{-1}(c+d x)}+6 \log \left (a+b \sinh ^{-1}(c+d x)\right )\right )}{6 b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^3*((-2*b^3*(c + d*x)^3*Sqrt[1 + (c + d*x)^2])/(a + b*ArcSinh[c + d*x])^3 + (b^2*(-3*(c + d*x)^2 - 4*(c + d*
x)^4))/(a + b*ArcSinh[c + d*x])^2 - (2*b*Sqrt[1 + (c + d*x)^2]*(3*(c + d*x) + 8*(c + d*x)^3))/(a + b*ArcSinh[c
+ d*x]) + 6*Log[a + b*ArcSinh[c + d*x]] + 30*(Cosh[(2*a)/b]*CoshIntegral[2*(a/b + ArcSinh[c + d*x])] - Log[a
+ b*ArcSinh[c + d*x]] - Sinh[(2*a)/b]*SinhIntegral[2*(a/b + ArcSinh[c + d*x])]) + 8*(-4*Cosh[(2*a)/b]*CoshInte
gral[2*(a/b + ArcSinh[c + d*x])] + Cosh[(4*a)/b]*CoshIntegral[4*(a/b + ArcSinh[c + d*x])] + 3*Log[a + b*ArcSin
h[c + d*x]] + 4*Sinh[(2*a)/b]*SinhIntegral[2*(a/b + ArcSinh[c + d*x])] - Sinh[(4*a)/b]*SinhIntegral[4*(a/b + A
rcSinh[c + d*x])])))/(6*b^4*d)

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Maple [B]  time = 0.151, size = 800, normalized size = 2.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(1/48*(8*(d*x+c)^4-8*(d*x+c)^3*(1+(d*x+c)^2)^(1/2)+8*(d*x+c)^2-4*(d*x+c)*(1+(d*x+c)^2)^(1/2)+1)*e^3*(8*b^2
*arcsinh(d*x+c)^2+16*a*b*arcsinh(d*x+c)-2*arcsinh(d*x+c)*b^2+8*a^2-2*a*b+b^2)/b^3/(b^3*arcsinh(d*x+c)^3+3*arcs
inh(d*x+c)^2*a*b^2+3*a^2*b*arcsinh(d*x+c)+a^3)-2/3*e^3/b^4*exp(4*a/b)*Ei(1,4*arcsinh(d*x+c)+4*a/b)-1/24*(2*(d*
x+c)^2-2*(d*x+c)*(1+(d*x+c)^2)^(1/2)+1)*e^3*(2*b^2*arcsinh(d*x+c)^2+4*a*b*arcsinh(d*x+c)-arcsinh(d*x+c)*b^2+2*
a^2-a*b+b^2)/b^3/(b^3*arcsinh(d*x+c)^3+3*arcsinh(d*x+c)^2*a*b^2+3*a^2*b*arcsinh(d*x+c)+a^3)+1/6*e^3/b^4*exp(2*
a/b)*Ei(1,2*arcsinh(d*x+c)+2*a/b)+1/24*e^3/b*(2*(d*x+c)^2+1+2*(d*x+c)*(1+(d*x+c)^2)^(1/2))/(a+b*arcsinh(d*x+c)
)^3+1/24*e^3/b^2*(2*(d*x+c)^2+1+2*(d*x+c)*(1+(d*x+c)^2)^(1/2))/(a+b*arcsinh(d*x+c))^2+1/12*e^3/b^3*(2*(d*x+c)^
2+1+2*(d*x+c)*(1+(d*x+c)^2)^(1/2))/(a+b*arcsinh(d*x+c))+1/6*e^3/b^4*exp(-2*a/b)*Ei(1,-2*arcsinh(d*x+c)-2*a/b)-
1/48*e^3/b*(8*(d*x+c)^4+8*(d*x+c)^2+8*(d*x+c)^3*(1+(d*x+c)^2)^(1/2)+4*(d*x+c)*(1+(d*x+c)^2)^(1/2)+1)/(a+b*arcs
inh(d*x+c))^3-1/24*e^3/b^2*(8*(d*x+c)^4+8*(d*x+c)^2+8*(d*x+c)^3*(1+(d*x+c)^2)^(1/2)+4*(d*x+c)*(1+(d*x+c)^2)^(1
/2)+1)/(a+b*arcsinh(d*x+c))^2-1/6*e^3/b^3*(8*(d*x+c)^4+8*(d*x+c)^2+8*(d*x+c)^3*(1+(d*x+c)^2)^(1/2)+4*(d*x+c)*(
1+(d*x+c)^2)^(1/2)+1)/(a+b*arcsinh(d*x+c))-2/3*e^3/b^4*exp(-4*a/b)*Ei(1,-4*arcsinh(d*x+c)-4*a/b))

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Maxima [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((d*e*x+c*e)^3/(a+b*arcsinh(d*x+c))^4,x, algorithm="maxima")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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