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

Optimal. Leaf size=331 \[ \frac{e^2 \sinh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )}{24 b^4 d}-\frac{9 e^2 \sinh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{8 b^4 d}-\frac{e^2 \cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )}{24 b^4 d}+\frac{9 e^2 \cosh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{8 b^4 d}-\frac{e^2 (c+d x)^3}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{3 e^2 \sqrt{(c+d x)^2+1} (c+d x)^2}{2 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{e^2 (c+d x)}{3 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{e^2 \sqrt{(c+d x)^2+1}}{3 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{e^2 \sqrt{(c+d x)^2+1} (c+d x)^2}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3} \]

[Out]

-(e^2*(c + d*x)^2*Sqrt[1 + (c + d*x)^2])/(3*b*d*(a + b*ArcSinh[c + d*x])^3) - (e^2*(c + d*x))/(3*b^2*d*(a + b*
ArcSinh[c + d*x])^2) - (e^2*(c + d*x)^3)/(2*b^2*d*(a + b*ArcSinh[c + d*x])^2) - (e^2*Sqrt[1 + (c + d*x)^2])/(3
*b^3*d*(a + b*ArcSinh[c + d*x])) - (3*e^2*(c + d*x)^2*Sqrt[1 + (c + d*x)^2])/(2*b^3*d*(a + b*ArcSinh[c + d*x])
) + (e^2*CoshIntegral[(a + b*ArcSinh[c + d*x])/b]*Sinh[a/b])/(24*b^4*d) - (9*e^2*CoshIntegral[(3*(a + b*ArcSin
h[c + d*x]))/b]*Sinh[(3*a)/b])/(8*b^4*d) - (e^2*Cosh[a/b]*SinhIntegral[(a + b*ArcSinh[c + d*x])/b])/(24*b^4*d)
+ (9*e^2*Cosh[(3*a)/b]*SinhIntegral[(3*(a + b*ArcSinh[c + d*x]))/b])/(8*b^4*d)

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Rubi [A]  time = 0.674519, antiderivative size = 327, normalized size of antiderivative = 0.99, number of steps used = 18, number of rules used = 10, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.435, Rules used = {5865, 12, 5667, 5774, 5665, 3303, 3298, 3301, 5655, 5779} \[ \frac{e^2 \sinh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )}{24 b^4 d}-\frac{9 e^2 \sinh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c+d x)\right )}{8 b^4 d}-\frac{e^2 \cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )}{24 b^4 d}+\frac{9 e^2 \cosh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c+d x)\right )}{8 b^4 d}-\frac{e^2 (c+d x)^3}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{3 e^2 \sqrt{(c+d x)^2+1} (c+d x)^2}{2 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{e^2 (c+d x)}{3 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{e^2 \sqrt{(c+d x)^2+1}}{3 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{e^2 \sqrt{(c+d x)^2+1} (c+d x)^2}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(e^2*(c + d*x)^2*Sqrt[1 + (c + d*x)^2])/(3*b*d*(a + b*ArcSinh[c + d*x])^3) - (e^2*(c + d*x))/(3*b^2*d*(a + b*
ArcSinh[c + d*x])^2) - (e^2*(c + d*x)^3)/(2*b^2*d*(a + b*ArcSinh[c + d*x])^2) - (e^2*Sqrt[1 + (c + d*x)^2])/(3
*b^3*d*(a + b*ArcSinh[c + d*x])) - (3*e^2*(c + d*x)^2*Sqrt[1 + (c + d*x)^2])/(2*b^3*d*(a + b*ArcSinh[c + d*x])
) + (e^2*CoshIntegral[a/b + ArcSinh[c + d*x]]*Sinh[a/b])/(24*b^4*d) - (9*e^2*CoshIntegral[(3*a)/b + 3*ArcSinh[
c + d*x]]*Sinh[(3*a)/b])/(8*b^4*d) - (e^2*Cosh[a/b]*SinhIntegral[a/b + ArcSinh[c + d*x]])/(24*b^4*d) + (9*e^2*
Cosh[(3*a)/b]*SinhIntegral[(3*a)/b + 3*ArcSinh[c + d*x]])/(8*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]

Rule 5655

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

Rule 5779

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c^
(m + 1), Subst[Int[(a + b*x)^n*Sinh[x]^m*Cosh[x]^(2*p + 1), x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e,
n}, x] && EqQ[e, c^2*d] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || GtQ[d, 0])

Rubi steps

\begin{align*} \int \frac{(c e+d e x)^2}{\left (a+b \sinh ^{-1}(c+d x)\right )^4} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{e^2 x^2}{\left (a+b \sinh ^{-1}(x)\right )^4} \, dx,x,c+d x\right )}{d}\\ &=\frac{e^2 \operatorname{Subst}\left (\int \frac{x^2}{\left (a+b \sinh ^{-1}(x)\right )^4} \, dx,x,c+d x\right )}{d}\\ &=-\frac{e^2 (c+d x)^2 \sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}+\frac{\left (2 e^2\right ) \operatorname{Subst}\left (\int \frac{x}{\sqrt{1+x^2} \left (a+b \sinh ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{3 b d}+\frac{e^2 \operatorname{Subst}\left (\int \frac{x^3}{\sqrt{1+x^2} \left (a+b \sinh ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{b d}\\ &=-\frac{e^2 (c+d x)^2 \sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}-\frac{e^2 (c+d x)}{3 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{e^2 (c+d x)^3}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}+\frac{e^2 \operatorname{Subst}\left (\int \frac{1}{\left (a+b \sinh ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{3 b^2 d}+\frac{\left (3 e^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{\left (a+b \sinh ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{2 b^2 d}\\ &=-\frac{e^2 (c+d x)^2 \sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}-\frac{e^2 (c+d x)}{3 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{e^2 (c+d x)^3}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{e^2 \sqrt{1+(c+d x)^2}}{3 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{3 e^2 (c+d x)^2 \sqrt{1+(c+d x)^2}}{2 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac{e^2 \operatorname{Subst}\left (\int \frac{x}{\sqrt{1+x^2} \left (a+b \sinh ^{-1}(x)\right )} \, dx,x,c+d x\right )}{3 b^3 d}+\frac{\left (3 e^2\right ) \operatorname{Subst}\left (\int \left (-\frac{\sinh (x)}{4 (a+b x)}+\frac{3 \sinh (3 x)}{4 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b^3 d}\\ &=-\frac{e^2 (c+d x)^2 \sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}-\frac{e^2 (c+d x)}{3 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{e^2 (c+d x)^3}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{e^2 \sqrt{1+(c+d x)^2}}{3 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{3 e^2 (c+d x)^2 \sqrt{1+(c+d x)^2}}{2 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac{e^2 \operatorname{Subst}\left (\int \frac{\sinh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{3 b^3 d}-\frac{\left (3 e^2\right ) \operatorname{Subst}\left (\int \frac{\sinh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 b^3 d}+\frac{\left (9 e^2\right ) \operatorname{Subst}\left (\int \frac{\sinh (3 x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 b^3 d}\\ &=-\frac{e^2 (c+d x)^2 \sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}-\frac{e^2 (c+d x)}{3 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{e^2 (c+d x)^3}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{e^2 \sqrt{1+(c+d x)^2}}{3 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{3 e^2 (c+d x)^2 \sqrt{1+(c+d x)^2}}{2 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac{\left (e^2 \cosh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{3 b^3 d}-\frac{\left (3 e^2 \cosh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 b^3 d}+\frac{\left (9 e^2 \cosh \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 b^3 d}-\frac{\left (e^2 \sinh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{3 b^3 d}+\frac{\left (3 e^2 \sinh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 b^3 d}-\frac{\left (9 e^2 \sinh \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 b^3 d}\\ &=-\frac{e^2 (c+d x)^2 \sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}-\frac{e^2 (c+d x)}{3 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{e^2 (c+d x)^3}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{e^2 \sqrt{1+(c+d x)^2}}{3 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{3 e^2 (c+d x)^2 \sqrt{1+(c+d x)^2}}{2 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac{e^2 \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right ) \sinh \left (\frac{a}{b}\right )}{24 b^4 d}-\frac{9 e^2 \text{Chi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c+d x)\right ) \sinh \left (\frac{3 a}{b}\right )}{8 b^4 d}-\frac{e^2 \cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )}{24 b^4 d}+\frac{9 e^2 \cosh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c+d x)\right )}{8 b^4 d}\\ \end{align*}

Mathematica [A]  time = 0.790577, size = 258, normalized size = 0.78 \[ \frac{e^2 \left (-\frac{8 b^3 (c+d x)^2 \sqrt{(c+d x)^2+1}}{\left (a+b \sinh ^{-1}(c+d x)\right )^3}+\frac{4 b^2 \left (-3 (c+d x)^3-2 (c+d x)\right )}{\left (a+b \sinh ^{-1}(c+d x)\right )^2}+27 \left (3 \sinh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )-\sinh \left (\frac{3 a}{b}\right ) \text{Chi}\left (3 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )-3 \cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )+\cosh \left (\frac{3 a}{b}\right ) \text{Shi}\left (3 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )\right )-80 \sinh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )+80 \cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )-\frac{4 b \sqrt{(c+d x)^2+1} \left (9 (c+d x)^2+2\right )}{a+b \sinh ^{-1}(c+d x)}\right )}{24 b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^2*((-8*b^3*(c + d*x)^2*Sqrt[1 + (c + d*x)^2])/(a + b*ArcSinh[c + d*x])^3 + (4*b^2*(-2*(c + d*x) - 3*(c + d*
x)^3))/(a + b*ArcSinh[c + d*x])^2 - (4*b*Sqrt[1 + (c + d*x)^2]*(2 + 9*(c + d*x)^2))/(a + b*ArcSinh[c + d*x]) -
80*CoshIntegral[a/b + ArcSinh[c + d*x]]*Sinh[a/b] + 80*Cosh[a/b]*SinhIntegral[a/b + ArcSinh[c + d*x]] + 27*(3
*CoshIntegral[a/b + ArcSinh[c + d*x]]*Sinh[a/b] - CoshIntegral[3*(a/b + ArcSinh[c + d*x])]*Sinh[(3*a)/b] - 3*C
osh[a/b]*SinhIntegral[a/b + ArcSinh[c + d*x]] + Cosh[(3*a)/b]*SinhIntegral[3*(a/b + ArcSinh[c + d*x])])))/(24*
b^4*d)

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Maple [B]  time = 0.14, size = 709, normalized size = 2.1 \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)^2/(a+b*arcsinh(d*x+c))^4,x)

[Out]

1/d*(1/48*(4*(d*x+c)^3-4*(d*x+c)^2*(1+(d*x+c)^2)^(1/2)+3*d*x+3*c-(1+(d*x+c)^2)^(1/2))*e^2*(9*b^2*arcsinh(d*x+c
)^2+18*a*b*arcsinh(d*x+c)-3*arcsinh(d*x+c)*b^2+9*a^2-3*a*b+2*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)+9/16*e^2/b^4*exp(3*a/b)*Ei(1,3*arcsinh(d*x+c)+3*a/b)-1/48*(-(1+(d*x+c)^2)^(
1/2)+d*x+c)*e^2*(b^2*arcsinh(d*x+c)^2+2*a*b*arcsinh(d*x+c)-arcsinh(d*x+c)*b^2+a^2-a*b+2*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)-1/48*e^2/b^4*exp(a/b)*Ei(1,arcsinh(d*x+c)+a/b)+1
/24*e^2/b*(d*x+c+(1+(d*x+c)^2)^(1/2))/(a+b*arcsinh(d*x+c))^3+1/48*e^2/b^2*(d*x+c+(1+(d*x+c)^2)^(1/2))/(a+b*arc
sinh(d*x+c))^2+1/48*e^2/b^3*(d*x+c+(1+(d*x+c)^2)^(1/2))/(a+b*arcsinh(d*x+c))+1/48*e^2/b^4*exp(-a/b)*Ei(1,-arcs
inh(d*x+c)-a/b)-1/24*e^2/b*(4*(d*x+c)^3+3*d*x+3*c+4*(d*x+c)^2*(1+(d*x+c)^2)^(1/2)+(1+(d*x+c)^2)^(1/2))/(a+b*ar
csinh(d*x+c))^3-1/16*e^2/b^2*(4*(d*x+c)^3+3*d*x+3*c+4*(d*x+c)^2*(1+(d*x+c)^2)^(1/2)+(1+(d*x+c)^2)^(1/2))/(a+b*
arcsinh(d*x+c))^2-3/16*e^2/b^3*(4*(d*x+c)^3+3*d*x+3*c+4*(d*x+c)^2*(1+(d*x+c)^2)^(1/2)+(1+(d*x+c)^2)^(1/2))/(a+
b*arcsinh(d*x+c))-9/16*e^2/b^4*exp(-3*a/b)*Ei(1,-3*arcsinh(d*x+c)-3*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)^2/(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^{2} e^{2} x^{2} + 2 \, c d e^{2} x + c^{2} e^{2}}{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)^2/(a+b*arcsinh(d*x+c))^4,x, algorithm="fricas")

[Out]

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

[Out]

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