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

Optimal. Leaf size=256 \[ -\frac{e^4 \sinh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )}{8 b^2 d}+\frac{9 e^4 \sinh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{16 b^2 d}-\frac{5 e^4 \sinh \left (\frac{5 a}{b}\right ) \text{Chi}\left (\frac{5 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{16 b^2 d}+\frac{e^4 \cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )}{8 b^2 d}-\frac{9 e^4 \cosh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{16 b^2 d}+\frac{5 e^4 \cosh \left (\frac{5 a}{b}\right ) \text{Shi}\left (\frac{5 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{16 b^2 d}-\frac{e^4 (c+d x)^4 \sqrt{(c+d x)^2+1}}{b d \left (a+b \sinh ^{-1}(c+d x)\right )} \]

[Out]

-((e^4*(c + d*x)^4*Sqrt[1 + (c + d*x)^2])/(b*d*(a + b*ArcSinh[c + d*x]))) - (e^4*CoshIntegral[(a + b*ArcSinh[c
+ d*x])/b]*Sinh[a/b])/(8*b^2*d) + (9*e^4*CoshIntegral[(3*(a + b*ArcSinh[c + d*x]))/b]*Sinh[(3*a)/b])/(16*b^2*
d) - (5*e^4*CoshIntegral[(5*(a + b*ArcSinh[c + d*x]))/b]*Sinh[(5*a)/b])/(16*b^2*d) + (e^4*Cosh[a/b]*SinhIntegr
al[(a + b*ArcSinh[c + d*x])/b])/(8*b^2*d) - (9*e^4*Cosh[(3*a)/b]*SinhIntegral[(3*(a + b*ArcSinh[c + d*x]))/b])
/(16*b^2*d) + (5*e^4*Cosh[(5*a)/b]*SinhIntegral[(5*(a + b*ArcSinh[c + d*x]))/b])/(16*b^2*d)

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Rubi [A]  time = 0.409048, antiderivative size = 252, normalized size of antiderivative = 0.98, number of steps used = 13, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {5865, 12, 5665, 3303, 3298, 3301} \[ -\frac{e^4 \sinh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )}{8 b^2 d}+\frac{9 e^4 \sinh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c+d x)\right )}{16 b^2 d}-\frac{5 e^4 \sinh \left (\frac{5 a}{b}\right ) \text{Chi}\left (\frac{5 a}{b}+5 \sinh ^{-1}(c+d x)\right )}{16 b^2 d}+\frac{e^4 \cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )}{8 b^2 d}-\frac{9 e^4 \cosh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c+d x)\right )}{16 b^2 d}+\frac{5 e^4 \cosh \left (\frac{5 a}{b}\right ) \text{Shi}\left (\frac{5 a}{b}+5 \sinh ^{-1}(c+d x)\right )}{16 b^2 d}-\frac{e^4 (c+d x)^4 \sqrt{(c+d x)^2+1}}{b d \left (a+b \sinh ^{-1}(c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((e^4*(c + d*x)^4*Sqrt[1 + (c + d*x)^2])/(b*d*(a + b*ArcSinh[c + d*x]))) - (e^4*CoshIntegral[a/b + ArcSinh[c
+ d*x]]*Sinh[a/b])/(8*b^2*d) + (9*e^4*CoshIntegral[(3*a)/b + 3*ArcSinh[c + d*x]]*Sinh[(3*a)/b])/(16*b^2*d) - (
5*e^4*CoshIntegral[(5*a)/b + 5*ArcSinh[c + d*x]]*Sinh[(5*a)/b])/(16*b^2*d) + (e^4*Cosh[a/b]*SinhIntegral[a/b +
ArcSinh[c + d*x]])/(8*b^2*d) - (9*e^4*Cosh[(3*a)/b]*SinhIntegral[(3*a)/b + 3*ArcSinh[c + d*x]])/(16*b^2*d) +
(5*e^4*Cosh[(5*a)/b]*SinhIntegral[(5*a)/b + 5*ArcSinh[c + d*x]])/(16*b^2*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 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)^4}{\left (a+b \sinh ^{-1}(c+d x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{e^4 x^4}{\left (a+b \sinh ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{d}\\ &=\frac{e^4 \operatorname{Subst}\left (\int \frac{x^4}{\left (a+b \sinh ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{d}\\ &=-\frac{e^4 (c+d x)^4 \sqrt{1+(c+d x)^2}}{b d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac{e^4 \operatorname{Subst}\left (\int \left (\frac{\sinh (x)}{8 (a+b x)}-\frac{9 \sinh (3 x)}{16 (a+b x)}+\frac{5 \sinh (5 x)}{16 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{b d}\\ &=-\frac{e^4 (c+d x)^4 \sqrt{1+(c+d x)^2}}{b d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac{e^4 \operatorname{Subst}\left (\int \frac{\sinh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 b d}+\frac{\left (5 e^4\right ) \operatorname{Subst}\left (\int \frac{\sinh (5 x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{16 b d}-\frac{\left (9 e^4\right ) \operatorname{Subst}\left (\int \frac{\sinh (3 x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{16 b d}\\ &=-\frac{e^4 (c+d x)^4 \sqrt{1+(c+d x)^2}}{b d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac{\left (e^4 \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 d}-\frac{\left (9 e^4 \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 )}{16 b d}+\frac{\left (5 e^4 \cosh \left (\frac{5 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{16 b d}-\frac{\left (e^4 \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 d}+\frac{\left (9 e^4 \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 )}{16 b d}-\frac{\left (5 e^4 \sinh \left (\frac{5 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{16 b d}\\ &=-\frac{e^4 (c+d x)^4 \sqrt{1+(c+d x)^2}}{b d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{e^4 \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right ) \sinh \left (\frac{a}{b}\right )}{8 b^2 d}+\frac{9 e^4 \text{Chi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c+d x)\right ) \sinh \left (\frac{3 a}{b}\right )}{16 b^2 d}-\frac{5 e^4 \text{Chi}\left (\frac{5 a}{b}+5 \sinh ^{-1}(c+d x)\right ) \sinh \left (\frac{5 a}{b}\right )}{16 b^2 d}+\frac{e^4 \cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )}{8 b^2 d}-\frac{9 e^4 \cosh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c+d x)\right )}{16 b^2 d}+\frac{5 e^4 \cosh \left (\frac{5 a}{b}\right ) \text{Shi}\left (\frac{5 a}{b}+5 \sinh ^{-1}(c+d x)\right )}{16 b^2 d}\\ \end{align*}

Mathematica [A]  time = 1.08561, size = 281, normalized size = 1.1 \[ \frac{e^4 \left (16 \left (3 \sinh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )-\sinh \left (\frac{3 a}{b}\right ) \text{Chi}\left (3 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )-3 \cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )+\cosh \left (\frac{3 a}{b}\right ) \text{Shi}\left (3 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )\right )-5 \left (10 \sinh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )-5 \sinh \left (\frac{3 a}{b}\right ) \text{Chi}\left (3 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )+\sinh \left (\frac{5 a}{b}\right ) \text{Chi}\left (5 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )-10 \cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )+5 \cosh \left (\frac{3 a}{b}\right ) \text{Shi}\left (3 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )-\cosh \left (\frac{5 a}{b}\right ) \text{Shi}\left (5 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )\right )-\frac{16 b \sqrt{(c+d x)^2+1} (c+d x)^4}{a+b \sinh ^{-1}(c+d x)}\right )}{16 b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^4*((-16*b*(c + d*x)^4*Sqrt[1 + (c + d*x)^2])/(a + b*ArcSinh[c + d*x]) + 16*(3*CoshIntegral[a/b + ArcSinh[c
+ d*x]]*Sinh[a/b] - CoshIntegral[3*(a/b + ArcSinh[c + d*x])]*Sinh[(3*a)/b] - 3*Cosh[a/b]*SinhIntegral[a/b + Ar
cSinh[c + d*x]] + Cosh[(3*a)/b]*SinhIntegral[3*(a/b + ArcSinh[c + d*x])]) - 5*(10*CoshIntegral[a/b + ArcSinh[c
+ d*x]]*Sinh[a/b] - 5*CoshIntegral[3*(a/b + ArcSinh[c + d*x])]*Sinh[(3*a)/b] + CoshIntegral[5*(a/b + ArcSinh[
c + d*x])]*Sinh[(5*a)/b] - 10*Cosh[a/b]*SinhIntegral[a/b + ArcSinh[c + d*x]] + 5*Cosh[(3*a)/b]*SinhIntegral[3*
(a/b + ArcSinh[c + d*x])] - Cosh[(5*a)/b]*SinhIntegral[5*(a/b + ArcSinh[c + d*x])])))/(16*b^2*d)

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Maple [B]  time = 0.207, size = 602, 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)^4/(a+b*arcsinh(d*x+c))^2,x)

[Out]

1/d*(1/32*(16*(d*x+c)^5-16*(d*x+c)^4*(1+(d*x+c)^2)^(1/2)+20*(d*x+c)^3-12*(d*x+c)^2*(1+(d*x+c)^2)^(1/2)+5*d*x+5
*c-(1+(d*x+c)^2)^(1/2))*e^4/b/(a+b*arcsinh(d*x+c))+5/32*e^4/b^2*exp(5*a/b)*Ei(1,5*arcsinh(d*x+c)+5*a/b)-3/32*(
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^4/b/(a+b*arcsinh(d*x+c))-9/32*e^4
/b^2*exp(3*a/b)*Ei(1,3*arcsinh(d*x+c)+3*a/b)+1/16*(-(1+(d*x+c)^2)^(1/2)+d*x+c)*e^4/b/(a+b*arcsinh(d*x+c))+1/16
*e^4/b^2*exp(a/b)*Ei(1,arcsinh(d*x+c)+a/b)-1/16*e^4/b*(d*x+c+(1+(d*x+c)^2)^(1/2))/(a+b*arcsinh(d*x+c))-1/16*e^
4/b^2*exp(-a/b)*Ei(1,-arcsinh(d*x+c)-a/b)+3/32*e^4/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*arcsinh(d*x+c))+9/32*e^4/b^2*exp(-3*a/b)*Ei(1,-3*arcsinh(d*x+c)-3*a/b)-1/32*e^4/b*(16*
(d*x+c)^5+20*(d*x+c)^3+16*(d*x+c)^4*(1+(d*x+c)^2)^(1/2)+5*d*x+5*c+12*(d*x+c)^2*(1+(d*x+c)^2)^(1/2)+(1+(d*x+c)^
2)^(1/2))/(a+b*arcsinh(d*x+c))-5/32*e^4/b^2*exp(-5*a/b)*Ei(1,-5*arcsinh(d*x+c)-5*a/b))

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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((d*e*x+c*e)^4/(a+b*arcsinh(d*x+c))^2,x, algorithm="maxima")

[Out]

-(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))/(a*b*d^3*x^2 + 2*a*b*c*d^2*x + (c^2*d + d)*a*b + (b^2*d^3*x^2 + 2*b^2*c*d^2*x + (c^2*
d + d)*b^2 + (b^2*d^2*x + b^2*c*d)*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)) + (a*b*d^2*x + a*b*c*d)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)) + integrate((5*d^8*e^4*x^8 + 40*c*d^7*e^4*
x^7 + 5*c^8*e^4 + 10*c^6*e^4 + 5*c^4*e^4 + 10*(14*c^2*d^6*e^4 + d^6*e^4)*x^6 + 20*(14*c^3*d^5*e^4 + 3*c*d^5*e^
4)*x^5 + 5*(70*c^4*d^4*e^4 + 30*c^2*d^4*e^4 + d^4*e^4)*x^4 + 20*(14*c^5*d^3*e^4 + 10*c^3*d^3*e^4 + c*d^3*e^4)*
x^3 + 10*(14*c^6*d^2*e^4 + 15*c^4*d^2*e^4 + 3*c^2*d^2*e^4)*x^2 + (5*d^6*e^4*x^6 + 30*c*d^5*e^4*x^5 + 5*c^6*e^4
+ 3*c^4*e^4 + 3*(25*c^2*d^4*e^4 + d^4*e^4)*x^4 + 4*(25*c^3*d^3*e^4 + 3*c*d^3*e^4)*x^3 + 3*(25*c^4*d^2*e^4 + 6
*c^2*d^2*e^4)*x^2 + 6*(5*c^5*d*e^4 + 2*c^3*d*e^4)*x)*(d^2*x^2 + 2*c*d*x + c^2 + 1) + 20*(2*c^7*d*e^4 + 3*c^5*d
*e^4 + c^3*d*e^4)*x + (10*d^7*e^4*x^7 + 70*c*d^6*e^4*x^6 + 10*c^7*e^4 + 13*c^5*e^4 + 4*c^3*e^4 + (210*c^2*d^5*
e^4 + 13*d^5*e^4)*x^5 + 5*(70*c^3*d^4*e^4 + 13*c*d^4*e^4)*x^4 + 2*(175*c^4*d^3*e^4 + 65*c^2*d^3*e^4 + 2*d^3*e^
4)*x^3 + 2*(105*c^5*d^2*e^4 + 65*c^3*d^2*e^4 + 6*c*d^2*e^4)*x^2 + (70*c^6*d*e^4 + 65*c^4*d*e^4 + 12*c^2*d*e^4)
*x)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))/(a*b*d^4*x^4 + 4*a*b*c*d^3*x^3 + 2*(3*c^2*d^2 + d^2)*a*b*x^2 + 4*(c^3*d
+ c*d)*a*b*x + (c^4 + 2*c^2 + 1)*a*b + (a*b*d^2*x^2 + 2*a*b*c*d*x + a*b*c^2)*(d^2*x^2 + 2*c*d*x + c^2 + 1) +
(b^2*d^4*x^4 + 4*b^2*c*d^3*x^3 + 2*(3*c^2*d^2 + d^2)*b^2*x^2 + 4*(c^3*d + c*d)*b^2*x + (c^4 + 2*c^2 + 1)*b^2 +
(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*(d^2*x^2 + 2*c*d*x + c^2 + 1) + 2*(b^2*d^3*x^3 + 3*b^2*c*d^2*x^2 + (3*c
^2*d + d)*b^2*x + (c^3 + c)*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*(a*b*d^3*x^3 + 3*a*b*c*d^2*x^2 + (3*c^2*d + d)*a*b*x + (c^3 + c)*a*b)*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{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}}{b^{2} \operatorname{arsinh}\left (d x + c\right )^{2} + 2 \, a b \operatorname{arsinh}\left (d x + c\right ) + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((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)/(b^2*arcsinh(d*x + c)^2
+ 2*a*b*arcsinh(d*x + c) + a^2), 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)**4/(a+b*asinh(d*x+c))**2,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 )}^{4}}{{\left (b \operatorname{arsinh}\left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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