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

Optimal. Leaf size=327 \[ -\frac{e^4 \sinh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a+b \cosh ^{-1}(c+d x)}{b}\right )}{16 b^3 d}-\frac{27 e^4 \sinh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 \left (a+b \cosh ^{-1}(c+d x)\right )}{b}\right )}{32 b^3 d}-\frac{25 e^4 \sinh \left (\frac{5 a}{b}\right ) \text{Chi}\left (\frac{5 \left (a+b \cosh ^{-1}(c+d x)\right )}{b}\right )}{32 b^3 d}+\frac{e^4 \cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a+b \cosh ^{-1}(c+d x)}{b}\right )}{16 b^3 d}+\frac{27 e^4 \cosh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 \left (a+b \cosh ^{-1}(c+d x)\right )}{b}\right )}{32 b^3 d}+\frac{25 e^4 \cosh \left (\frac{5 a}{b}\right ) \text{Shi}\left (\frac{5 \left (a+b \cosh ^{-1}(c+d x)\right )}{b}\right )}{32 b^3 d}-\frac{5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \cosh ^{-1}(c+d x)\right )}+\frac{2 e^4 (c+d x)^3}{b^2 d \left (a+b \cosh ^{-1}(c+d x)\right )}-\frac{e^4 \sqrt{c+d x-1} \sqrt{c+d x+1} (c+d x)^4}{2 b d \left (a+b \cosh ^{-1}(c+d x)\right )^2} \]

[Out]

-(e^4*Sqrt[-1 + c + d*x]*(c + d*x)^4*Sqrt[1 + c + d*x])/(2*b*d*(a + b*ArcCosh[c + d*x])^2) + (2*e^4*(c + d*x)^
3)/(b^2*d*(a + b*ArcCosh[c + d*x])) - (5*e^4*(c + d*x)^5)/(2*b^2*d*(a + b*ArcCosh[c + d*x])) - (e^4*CoshIntegr
al[(a + b*ArcCosh[c + d*x])/b]*Sinh[a/b])/(16*b^3*d) - (27*e^4*CoshIntegral[(3*(a + b*ArcCosh[c + d*x]))/b]*Si
nh[(3*a)/b])/(32*b^3*d) - (25*e^4*CoshIntegral[(5*(a + b*ArcCosh[c + d*x]))/b]*Sinh[(5*a)/b])/(32*b^3*d) + (e^
4*Cosh[a/b]*SinhIntegral[(a + b*ArcCosh[c + d*x])/b])/(16*b^3*d) + (27*e^4*Cosh[(3*a)/b]*SinhIntegral[(3*(a +
b*ArcCosh[c + d*x]))/b])/(32*b^3*d) + (25*e^4*Cosh[(5*a)/b]*SinhIntegral[(5*(a + b*ArcCosh[c + d*x]))/b])/(32*
b^3*d)

________________________________________________________________________________________

Rubi [A]  time = 1.11063, antiderivative size = 323, normalized size of antiderivative = 0.99, number of steps used = 26, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {5866, 12, 5668, 5775, 5670, 5448, 3303, 3298, 3301} \[ -\frac{e^4 \sinh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\cosh ^{-1}(c+d x)\right )}{16 b^3 d}-\frac{27 e^4 \sinh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 a}{b}+3 \cosh ^{-1}(c+d x)\right )}{32 b^3 d}-\frac{25 e^4 \sinh \left (\frac{5 a}{b}\right ) \text{Chi}\left (\frac{5 a}{b}+5 \cosh ^{-1}(c+d x)\right )}{32 b^3 d}+\frac{e^4 \cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\cosh ^{-1}(c+d x)\right )}{16 b^3 d}+\frac{27 e^4 \cosh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 a}{b}+3 \cosh ^{-1}(c+d x)\right )}{32 b^3 d}+\frac{25 e^4 \cosh \left (\frac{5 a}{b}\right ) \text{Shi}\left (\frac{5 a}{b}+5 \cosh ^{-1}(c+d x)\right )}{32 b^3 d}-\frac{5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \cosh ^{-1}(c+d x)\right )}+\frac{2 e^4 (c+d x)^3}{b^2 d \left (a+b \cosh ^{-1}(c+d x)\right )}-\frac{e^4 \sqrt{c+d x-1} \sqrt{c+d x+1} (c+d x)^4}{2 b d \left (a+b \cosh ^{-1}(c+d x)\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(e^4*Sqrt[-1 + c + d*x]*(c + d*x)^4*Sqrt[1 + c + d*x])/(2*b*d*(a + b*ArcCosh[c + d*x])^2) + (2*e^4*(c + d*x)^
3)/(b^2*d*(a + b*ArcCosh[c + d*x])) - (5*e^4*(c + d*x)^5)/(2*b^2*d*(a + b*ArcCosh[c + d*x])) - (e^4*CoshIntegr
al[a/b + ArcCosh[c + d*x]]*Sinh[a/b])/(16*b^3*d) - (27*e^4*CoshIntegral[(3*a)/b + 3*ArcCosh[c + d*x]]*Sinh[(3*
a)/b])/(32*b^3*d) - (25*e^4*CoshIntegral[(5*a)/b + 5*ArcCosh[c + d*x]]*Sinh[(5*a)/b])/(32*b^3*d) + (e^4*Cosh[a
/b]*SinhIntegral[a/b + ArcCosh[c + d*x]])/(16*b^3*d) + (27*e^4*Cosh[(3*a)/b]*SinhIntegral[(3*a)/b + 3*ArcCosh[
c + d*x]])/(32*b^3*d) + (25*e^4*Cosh[(5*a)/b]*SinhIntegral[(5*a)/b + 5*ArcCosh[c + d*x]])/(32*b^3*d)

Rule 5866

Int[((a_.) + ArcCosh[(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*ArcCosh[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 5668

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

Rule 5775

Int[(((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/(Sqrt[(d1_) + (e1_.)*(x_)]*Sqrt[(d2_) + (e2_
.)*(x_)]), x_Symbol] :> Simp[((f*x)^m*(a + b*ArcCosh[c*x])^(n + 1))/(b*c*Sqrt[-(d1*d2)]*(n + 1)), x] - Dist[(f
*m)/(b*c*Sqrt[-(d1*d2)]*(n + 1)), Int[(f*x)^(m - 1)*(a + b*ArcCosh[c*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d1
, e1, d2, e2, f, m}, x] && EqQ[e1 - c*d1, 0] && EqQ[e2 + c*d2, 0] && LtQ[n, -1] && GtQ[d1, 0] && LtQ[d2, 0]

Rule 5670

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[(a + b*x)^n*
Cosh[x]^m*Sinh[x], x], x, ArcCosh[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]

Rule 5448

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]

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 \cosh ^{-1}(c+d x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{e^4 x^4}{\left (a+b \cosh ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{d}\\ &=\frac{e^4 \operatorname{Subst}\left (\int \frac{x^4}{\left (a+b \cosh ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{d}\\ &=-\frac{e^4 \sqrt{-1+c+d x} (c+d x)^4 \sqrt{1+c+d x}}{2 b d \left (a+b \cosh ^{-1}(c+d x)\right )^2}-\frac{\left (2 e^4\right ) \operatorname{Subst}\left (\int \frac{x^3}{\sqrt{-1+x} \sqrt{1+x} \left (a+b \cosh ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{b d}+\frac{\left (5 e^4\right ) \operatorname{Subst}\left (\int \frac{x^5}{\sqrt{-1+x} \sqrt{1+x} \left (a+b \cosh ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{2 b d}\\ &=-\frac{e^4 \sqrt{-1+c+d x} (c+d x)^4 \sqrt{1+c+d x}}{2 b d \left (a+b \cosh ^{-1}(c+d x)\right )^2}+\frac{2 e^4 (c+d x)^3}{b^2 d \left (a+b \cosh ^{-1}(c+d x)\right )}-\frac{5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \cosh ^{-1}(c+d x)\right )}-\frac{\left (6 e^4\right ) \operatorname{Subst}\left (\int \frac{x^2}{a+b \cosh ^{-1}(x)} \, dx,x,c+d x\right )}{b^2 d}+\frac{\left (25 e^4\right ) \operatorname{Subst}\left (\int \frac{x^4}{a+b \cosh ^{-1}(x)} \, dx,x,c+d x\right )}{2 b^2 d}\\ &=-\frac{e^4 \sqrt{-1+c+d x} (c+d x)^4 \sqrt{1+c+d x}}{2 b d \left (a+b \cosh ^{-1}(c+d x)\right )^2}+\frac{2 e^4 (c+d x)^3}{b^2 d \left (a+b \cosh ^{-1}(c+d x)\right )}-\frac{5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \cosh ^{-1}(c+d x)\right )}-\frac{\left (6 e^4\right ) \operatorname{Subst}\left (\int \frac{\cosh ^2(x) \sinh (x)}{a+b x} \, dx,x,\cosh ^{-1}(c+d x)\right )}{b^2 d}+\frac{\left (25 e^4\right ) \operatorname{Subst}\left (\int \frac{\cosh ^4(x) \sinh (x)}{a+b x} \, dx,x,\cosh ^{-1}(c+d x)\right )}{2 b^2 d}\\ &=-\frac{e^4 \sqrt{-1+c+d x} (c+d x)^4 \sqrt{1+c+d x}}{2 b d \left (a+b \cosh ^{-1}(c+d x)\right )^2}+\frac{2 e^4 (c+d x)^3}{b^2 d \left (a+b \cosh ^{-1}(c+d x)\right )}-\frac{5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \cosh ^{-1}(c+d x)\right )}-\frac{\left (6 e^4\right ) \operatorname{Subst}\left (\int \left (\frac{\sinh (x)}{4 (a+b x)}+\frac{\sinh (3 x)}{4 (a+b x)}\right ) \, dx,x,\cosh ^{-1}(c+d x)\right )}{b^2 d}+\frac{\left (25 e^4\right ) \operatorname{Subst}\left (\int \left (\frac{\sinh (x)}{8 (a+b x)}+\frac{3 \sinh (3 x)}{16 (a+b x)}+\frac{\sinh (5 x)}{16 (a+b x)}\right ) \, dx,x,\cosh ^{-1}(c+d x)\right )}{2 b^2 d}\\ &=-\frac{e^4 \sqrt{-1+c+d x} (c+d x)^4 \sqrt{1+c+d x}}{2 b d \left (a+b \cosh ^{-1}(c+d x)\right )^2}+\frac{2 e^4 (c+d x)^3}{b^2 d \left (a+b \cosh ^{-1}(c+d x)\right )}-\frac{5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \cosh ^{-1}(c+d x)\right )}+\frac{\left (25 e^4\right ) \operatorname{Subst}\left (\int \frac{\sinh (5 x)}{a+b x} \, dx,x,\cosh ^{-1}(c+d x)\right )}{32 b^2 d}-\frac{\left (3 e^4\right ) \operatorname{Subst}\left (\int \frac{\sinh (x)}{a+b x} \, dx,x,\cosh ^{-1}(c+d x)\right )}{2 b^2 d}-\frac{\left (3 e^4\right ) \operatorname{Subst}\left (\int \frac{\sinh (3 x)}{a+b x} \, dx,x,\cosh ^{-1}(c+d x)\right )}{2 b^2 d}+\frac{\left (25 e^4\right ) \operatorname{Subst}\left (\int \frac{\sinh (x)}{a+b x} \, dx,x,\cosh ^{-1}(c+d x)\right )}{16 b^2 d}+\frac{\left (75 e^4\right ) \operatorname{Subst}\left (\int \frac{\sinh (3 x)}{a+b x} \, dx,x,\cosh ^{-1}(c+d x)\right )}{32 b^2 d}\\ &=-\frac{e^4 \sqrt{-1+c+d x} (c+d x)^4 \sqrt{1+c+d x}}{2 b d \left (a+b \cosh ^{-1}(c+d x)\right )^2}+\frac{2 e^4 (c+d x)^3}{b^2 d \left (a+b \cosh ^{-1}(c+d x)\right )}-\frac{5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \cosh ^{-1}(c+d x)\right )}-\frac{\left (3 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,\cosh ^{-1}(c+d x)\right )}{2 b^2 d}+\frac{\left (25 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,\cosh ^{-1}(c+d x)\right )}{16 b^2 d}-\frac{\left (3 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,\cosh ^{-1}(c+d x)\right )}{2 b^2 d}+\frac{\left (75 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,\cosh ^{-1}(c+d x)\right )}{32 b^2 d}+\frac{\left (25 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,\cosh ^{-1}(c+d x)\right )}{32 b^2 d}+\frac{\left (3 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,\cosh ^{-1}(c+d x)\right )}{2 b^2 d}-\frac{\left (25 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,\cosh ^{-1}(c+d x)\right )}{16 b^2 d}+\frac{\left (3 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,\cosh ^{-1}(c+d x)\right )}{2 b^2 d}-\frac{\left (75 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,\cosh ^{-1}(c+d x)\right )}{32 b^2 d}-\frac{\left (25 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,\cosh ^{-1}(c+d x)\right )}{32 b^2 d}\\ &=-\frac{e^4 \sqrt{-1+c+d x} (c+d x)^4 \sqrt{1+c+d x}}{2 b d \left (a+b \cosh ^{-1}(c+d x)\right )^2}+\frac{2 e^4 (c+d x)^3}{b^2 d \left (a+b \cosh ^{-1}(c+d x)\right )}-\frac{5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \cosh ^{-1}(c+d x)\right )}-\frac{e^4 \text{Chi}\left (\frac{a}{b}+\cosh ^{-1}(c+d x)\right ) \sinh \left (\frac{a}{b}\right )}{16 b^3 d}-\frac{27 e^4 \text{Chi}\left (\frac{3 a}{b}+3 \cosh ^{-1}(c+d x)\right ) \sinh \left (\frac{3 a}{b}\right )}{32 b^3 d}-\frac{25 e^4 \text{Chi}\left (\frac{5 a}{b}+5 \cosh ^{-1}(c+d x)\right ) \sinh \left (\frac{5 a}{b}\right )}{32 b^3 d}+\frac{e^4 \cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\cosh ^{-1}(c+d x)\right )}{16 b^3 d}+\frac{27 e^4 \cosh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 a}{b}+3 \cosh ^{-1}(c+d x)\right )}{32 b^3 d}+\frac{25 e^4 \cosh \left (\frac{5 a}{b}\right ) \text{Shi}\left (\frac{5 a}{b}+5 \cosh ^{-1}(c+d x)\right )}{32 b^3 d}\\ \end{align*}

Mathematica [A]  time = 1.27099, size = 323, normalized size = 0.99 \[ \frac{e^4 \left (-\frac{16 b^2 \sqrt{c+d x-1} \sqrt{c+d x+1} (c+d x)^4}{\left (a+b \cosh ^{-1}(c+d x)\right )^2}+48 \left (\sinh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\cosh ^{-1}(c+d x)\right )+\sinh \left (\frac{3 a}{b}\right ) \text{Chi}\left (3 \left (\frac{a}{b}+\cosh ^{-1}(c+d x)\right )\right )-\cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\cosh ^{-1}(c+d x)\right )-\cosh \left (\frac{3 a}{b}\right ) \text{Shi}\left (3 \left (\frac{a}{b}+\cosh ^{-1}(c+d x)\right )\right )\right )+25 \left (-2 \sinh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\cosh ^{-1}(c+d x)\right )-3 \sinh \left (\frac{3 a}{b}\right ) \text{Chi}\left (3 \left (\frac{a}{b}+\cosh ^{-1}(c+d x)\right )\right )-\sinh \left (\frac{5 a}{b}\right ) \text{Chi}\left (5 \left (\frac{a}{b}+\cosh ^{-1}(c+d x)\right )\right )+2 \cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\cosh ^{-1}(c+d x)\right )+3 \cosh \left (\frac{3 a}{b}\right ) \text{Shi}\left (3 \left (\frac{a}{b}+\cosh ^{-1}(c+d x)\right )\right )+\cosh \left (\frac{5 a}{b}\right ) \text{Shi}\left (5 \left (\frac{a}{b}+\cosh ^{-1}(c+d x)\right )\right )\right )+\frac{16 b \left (4 (c+d x)^3-5 (c+d x)^5\right )}{a+b \cosh ^{-1}(c+d x)}\right )}{32 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^4*((-16*b^2*Sqrt[-1 + c + d*x]*(c + d*x)^4*Sqrt[1 + c + d*x])/(a + b*ArcCosh[c + d*x])^2 + (16*b*(4*(c + d*
x)^3 - 5*(c + d*x)^5))/(a + b*ArcCosh[c + d*x]) + 48*(CoshIntegral[a/b + ArcCosh[c + d*x]]*Sinh[a/b] + CoshInt
egral[3*(a/b + ArcCosh[c + d*x])]*Sinh[(3*a)/b] - Cosh[a/b]*SinhIntegral[a/b + ArcCosh[c + d*x]] - Cosh[(3*a)/
b]*SinhIntegral[3*(a/b + ArcCosh[c + d*x])]) + 25*(-2*CoshIntegral[a/b + ArcCosh[c + d*x]]*Sinh[a/b] - 3*CoshI
ntegral[3*(a/b + ArcCosh[c + d*x])]*Sinh[(3*a)/b] - CoshIntegral[5*(a/b + ArcCosh[c + d*x])]*Sinh[(5*a)/b] + 2
*Cosh[a/b]*SinhIntegral[a/b + ArcCosh[c + d*x]] + 3*Cosh[(3*a)/b]*SinhIntegral[3*(a/b + ArcCosh[c + d*x])] + C
osh[(5*a)/b]*SinhIntegral[5*(a/b + ArcCosh[c + d*x])])))/(32*b^3*d)

________________________________________________________________________________________

Maple [B]  time = 0.237, size = 993, normalized size = 3. \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*arccosh(d*x+c))^3,x)

[Out]

1/d*(-1/64*(-16*(d*x+c)^4*(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)+12*(d*x+c)^2*(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)-(d*x+c-
1)^(1/2)*(d*x+c+1)^(1/2)+16*(d*x+c)^5-20*(d*x+c)^3+5*d*x+5*c)*e^4*(5*b*arccosh(d*x+c)+5*a-b)/b^2/(b^2*arccosh(
d*x+c)^2+2*a*b*arccosh(d*x+c)+a^2)+25/64*e^4/b^3*exp(5*a/b)*Ei(1,5*arccosh(d*x+c)+5*a/b)-3/64*(-4*(d*x+c)^2*(d
*x+c-1)^(1/2)*(d*x+c+1)^(1/2)+(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)+4*(d*x+c)^3-3*d*x-3*c)*e^4*(3*b*arccosh(d*x+c)+3
*a-b)/b^2/(b^2*arccosh(d*x+c)^2+2*a*b*arccosh(d*x+c)+a^2)+27/64*e^4/b^3*exp(3*a/b)*Ei(1,3*arccosh(d*x+c)+3*a/b
)-1/32*(-(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)+d*x+c)*e^4*(b*arccosh(d*x+c)+a-b)/b^2/(b^2*arccosh(d*x+c)^2+2*a*b*arc
cosh(d*x+c)+a^2)+1/32*e^4/b^3*exp(a/b)*Ei(1,arccosh(d*x+c)+a/b)-1/32*e^4/b*(d*x+c+(d*x+c-1)^(1/2)*(d*x+c+1)^(1
/2))/(a+b*arccosh(d*x+c))^2-1/32*e^4/b^2*(d*x+c+(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2))/(a+b*arccosh(d*x+c))-1/32*e^4
/b^3*exp(-a/b)*Ei(1,-arccosh(d*x+c)-a/b)-3/64*e^4/b*(4*(d*x+c)^3-3*d*x-3*c+4*(d*x+c)^2*(d*x+c-1)^(1/2)*(d*x+c+
1)^(1/2)-(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2))/(a+b*arccosh(d*x+c))^2-9/64*e^4/b^2*(4*(d*x+c)^3-3*d*x-3*c+4*(d*x+c)
^2*(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)-(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2))/(a+b*arccosh(d*x+c))-27/64*e^4/b^3*exp(-3*
a/b)*Ei(1,-3*arccosh(d*x+c)-3*a/b)-1/64*e^4/b*(16*(d*x+c)^5-20*(d*x+c)^3+16*(d*x+c)^4*(d*x+c-1)^(1/2)*(d*x+c+1
)^(1/2)+5*d*x+5*c-12*(d*x+c)^2*(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)+(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2))/(a+b*arccosh(d
*x+c))^2-5/64*e^4/b^2*(16*(d*x+c)^5-20*(d*x+c)^3+16*(d*x+c)^4*(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)+5*d*x+5*c-12*(d*
x+c)^2*(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)+(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2))/(a+b*arccosh(d*x+c))-25/64*e^4/b^3*exp
(-5*a/b)*Ei(1,-5*arccosh(d*x+c)-5*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)^4/(a+b*arccosh(d*x+c))^3,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^{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^{3} \operatorname{arcosh}\left (d x + c\right )^{3} + 3 \, a b^{2} \operatorname{arcosh}\left (d x + c\right )^{2} + 3 \, a^{2} b \operatorname{arcosh}\left (d x + c\right ) + a^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4/(a+b*arccosh(d*x+c))^3,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^3*arccosh(d*x + c)^3
+ 3*a*b^2*arccosh(d*x + c)^2 + 3*a^2*b*arccosh(d*x + c) + a^3), 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*acosh(d*x+c))**3,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{arcosh}\left (d x + c\right ) + a\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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