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

Optimal. Leaf size=218 \[ \frac{e^4 (c+d x)^5 \left (a+b \cosh ^{-1}(c+d x)\right )^2}{5 d}-\frac{2 b e^4 \sqrt{c+d x-1} \sqrt{c+d x+1} (c+d x)^4 \left (a+b \cosh ^{-1}(c+d x)\right )}{25 d}-\frac{8 b e^4 \sqrt{c+d x-1} \sqrt{c+d x+1} (c+d x)^2 \left (a+b \cosh ^{-1}(c+d x)\right )}{75 d}-\frac{16 b e^4 \sqrt{c+d x-1} \sqrt{c+d x+1} \left (a+b \cosh ^{-1}(c+d x)\right )}{75 d}+\frac{2 b^2 e^4 (c+d x)^5}{125 d}+\frac{8 b^2 e^4 (c+d x)^3}{225 d}+\frac{16}{75} b^2 e^4 x \]

[Out]

(16*b^2*e^4*x)/75 + (8*b^2*e^4*(c + d*x)^3)/(225*d) + (2*b^2*e^4*(c + d*x)^5)/(125*d) - (16*b*e^4*Sqrt[-1 + c
+ d*x]*Sqrt[1 + c + d*x]*(a + b*ArcCosh[c + d*x]))/(75*d) - (8*b*e^4*Sqrt[-1 + c + d*x]*(c + d*x)^2*Sqrt[1 + c
+ d*x]*(a + b*ArcCosh[c + d*x]))/(75*d) - (2*b*e^4*Sqrt[-1 + c + d*x]*(c + d*x)^4*Sqrt[1 + c + d*x]*(a + b*Ar
cCosh[c + d*x]))/(25*d) + (e^4*(c + d*x)^5*(a + b*ArcCosh[c + d*x])^2)/(5*d)

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Rubi [A]  time = 0.485118, antiderivative size = 218, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {5866, 12, 5662, 5759, 5718, 8, 30} \[ \frac{e^4 (c+d x)^5 \left (a+b \cosh ^{-1}(c+d x)\right )^2}{5 d}-\frac{2 b e^4 \sqrt{c+d x-1} \sqrt{c+d x+1} (c+d x)^4 \left (a+b \cosh ^{-1}(c+d x)\right )}{25 d}-\frac{8 b e^4 \sqrt{c+d x-1} \sqrt{c+d x+1} (c+d x)^2 \left (a+b \cosh ^{-1}(c+d x)\right )}{75 d}-\frac{16 b e^4 \sqrt{c+d x-1} \sqrt{c+d x+1} \left (a+b \cosh ^{-1}(c+d x)\right )}{75 d}+\frac{2 b^2 e^4 (c+d x)^5}{125 d}+\frac{8 b^2 e^4 (c+d x)^3}{225 d}+\frac{16}{75} b^2 e^4 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*b^2*e^4*x)/75 + (8*b^2*e^4*(c + d*x)^3)/(225*d) + (2*b^2*e^4*(c + d*x)^5)/(125*d) - (16*b*e^4*Sqrt[-1 + c
+ d*x]*Sqrt[1 + c + d*x]*(a + b*ArcCosh[c + d*x]))/(75*d) - (8*b*e^4*Sqrt[-1 + c + d*x]*(c + d*x)^2*Sqrt[1 + c
+ d*x]*(a + b*ArcCosh[c + d*x]))/(75*d) - (2*b*e^4*Sqrt[-1 + c + d*x]*(c + d*x)^4*Sqrt[1 + c + d*x]*(a + b*Ar
cCosh[c + d*x]))/(25*d) + (e^4*(c + d*x)^5*(a + b*ArcCosh[c + d*x])^2)/(5*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 5662

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

Rule 5759

Int[(((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_))/(Sqrt[(d1_) + (e1_.)*(x_)]*Sqrt[(d2_) + (e2_
.)*(x_)]), x_Symbol] :> Simp[(f*(f*x)^(m - 1)*Sqrt[d1 + e1*x]*Sqrt[d2 + e2*x]*(a + b*ArcCosh[c*x])^n)/(e1*e2*m
), x] + (Dist[(f^2*(m - 1))/(c^2*m), Int[((f*x)^(m - 2)*(a + b*ArcCosh[c*x])^n)/(Sqrt[d1 + e1*x]*Sqrt[d2 + e2*
x]), x], x] + Dist[(b*f*n*Sqrt[d1 + e1*x]*Sqrt[d2 + e2*x])/(c*d1*d2*m*Sqrt[1 + c*x]*Sqrt[-1 + c*x]), Int[(f*x)
^(m - 1)*(a + b*ArcCosh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d1, e1, d2, e2, f}, x] && EqQ[e1 - c*d1, 0]
&& EqQ[e2 + c*d2, 0] && GtQ[n, 0] && GtQ[m, 1] && IntegerQ[m]

Rule 5718

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d1_) + (e1_.)*(x_))^(p_.)*((d2_) + (e2_.)*(x_))^(p_.), x_
Symbol] :> Simp[((d1 + e1*x)^(p + 1)*(d2 + e2*x)^(p + 1)*(a + b*ArcCosh[c*x])^n)/(2*e1*e2*(p + 1)), x] - Dist[
(b*n*(-(d1*d2))^IntPart[p]*(d1 + e1*x)^FracPart[p]*(d2 + e2*x)^FracPart[p])/(2*c*(p + 1)*(1 + c*x)^FracPart[p]
*(-1 + c*x)^FracPart[p]), Int[(-1 + c^2*x^2)^(p + 1/2)*(a + b*ArcCosh[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c,
d1, e1, d2, e2, p}, x] && EqQ[e1 - c*d1, 0] && EqQ[e2 + c*d2, 0] && GtQ[n, 0] && NeQ[p, -1] && IntegerQ[p + 1
/2]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int (c e+d e x)^4 \left (a+b \cosh ^{-1}(c+d x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int e^4 x^4 \left (a+b \cosh ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac{e^4 \operatorname{Subst}\left (\int x^4 \left (a+b \cosh ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac{e^4 (c+d x)^5 \left (a+b \cosh ^{-1}(c+d x)\right )^2}{5 d}-\frac{\left (2 b e^4\right ) \operatorname{Subst}\left (\int \frac{x^5 \left (a+b \cosh ^{-1}(x)\right )}{\sqrt{-1+x} \sqrt{1+x}} \, dx,x,c+d x\right )}{5 d}\\ &=-\frac{2 b e^4 \sqrt{-1+c+d x} (c+d x)^4 \sqrt{1+c+d x} \left (a+b \cosh ^{-1}(c+d x)\right )}{25 d}+\frac{e^4 (c+d x)^5 \left (a+b \cosh ^{-1}(c+d x)\right )^2}{5 d}-\frac{\left (8 b e^4\right ) \operatorname{Subst}\left (\int \frac{x^3 \left (a+b \cosh ^{-1}(x)\right )}{\sqrt{-1+x} \sqrt{1+x}} \, dx,x,c+d x\right )}{25 d}+\frac{\left (2 b^2 e^4\right ) \operatorname{Subst}\left (\int x^4 \, dx,x,c+d x\right )}{25 d}\\ &=\frac{2 b^2 e^4 (c+d x)^5}{125 d}-\frac{8 b e^4 \sqrt{-1+c+d x} (c+d x)^2 \sqrt{1+c+d x} \left (a+b \cosh ^{-1}(c+d x)\right )}{75 d}-\frac{2 b e^4 \sqrt{-1+c+d x} (c+d x)^4 \sqrt{1+c+d x} \left (a+b \cosh ^{-1}(c+d x)\right )}{25 d}+\frac{e^4 (c+d x)^5 \left (a+b \cosh ^{-1}(c+d x)\right )^2}{5 d}-\frac{\left (16 b e^4\right ) \operatorname{Subst}\left (\int \frac{x \left (a+b \cosh ^{-1}(x)\right )}{\sqrt{-1+x} \sqrt{1+x}} \, dx,x,c+d x\right )}{75 d}+\frac{\left (8 b^2 e^4\right ) \operatorname{Subst}\left (\int x^2 \, dx,x,c+d x\right )}{75 d}\\ &=\frac{8 b^2 e^4 (c+d x)^3}{225 d}+\frac{2 b^2 e^4 (c+d x)^5}{125 d}-\frac{16 b e^4 \sqrt{-1+c+d x} \sqrt{1+c+d x} \left (a+b \cosh ^{-1}(c+d x)\right )}{75 d}-\frac{8 b e^4 \sqrt{-1+c+d x} (c+d x)^2 \sqrt{1+c+d x} \left (a+b \cosh ^{-1}(c+d x)\right )}{75 d}-\frac{2 b e^4 \sqrt{-1+c+d x} (c+d x)^4 \sqrt{1+c+d x} \left (a+b \cosh ^{-1}(c+d x)\right )}{25 d}+\frac{e^4 (c+d x)^5 \left (a+b \cosh ^{-1}(c+d x)\right )^2}{5 d}+\frac{\left (16 b^2 e^4\right ) \operatorname{Subst}(\int 1 \, dx,x,c+d x)}{75 d}\\ &=\frac{16}{75} b^2 e^4 x+\frac{8 b^2 e^4 (c+d x)^3}{225 d}+\frac{2 b^2 e^4 (c+d x)^5}{125 d}-\frac{16 b e^4 \sqrt{-1+c+d x} \sqrt{1+c+d x} \left (a+b \cosh ^{-1}(c+d x)\right )}{75 d}-\frac{8 b e^4 \sqrt{-1+c+d x} (c+d x)^2 \sqrt{1+c+d x} \left (a+b \cosh ^{-1}(c+d x)\right )}{75 d}-\frac{2 b e^4 \sqrt{-1+c+d x} (c+d x)^4 \sqrt{1+c+d x} \left (a+b \cosh ^{-1}(c+d x)\right )}{25 d}+\frac{e^4 (c+d x)^5 \left (a+b \cosh ^{-1}(c+d x)\right )^2}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.32343, size = 220, normalized size = 1.01 \[ \frac{e^4 \left (9 \left (25 a^2+2 b^2\right ) (c+d x)^5+30 a b \sqrt{c+d x-1} \sqrt{c+d x+1} \left (-3 (c+d x)^4-4 (c+d x)^2-8\right )+30 b \cosh ^{-1}(c+d x) \left (15 a (c+d x)^5-3 b \sqrt{c+d x-1} \sqrt{c+d x+1} (c+d x)^4-4 b \sqrt{c+d x-1} \sqrt{c+d x+1} (c+d x)^2-8 b \sqrt{c+d x-1} \sqrt{c+d x+1}\right )+40 b^2 (c+d x)^3+240 b^2 (c+d x)+225 b^2 (c+d x)^5 \cosh ^{-1}(c+d x)^2\right )}{1125 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^4*(240*b^2*(c + d*x) + 40*b^2*(c + d*x)^3 + 9*(25*a^2 + 2*b^2)*(c + d*x)^5 + 30*a*b*Sqrt[-1 + c + d*x]*Sqrt
[1 + c + d*x]*(-8 - 4*(c + d*x)^2 - 3*(c + d*x)^4) + 30*b*(15*a*(c + d*x)^5 - 8*b*Sqrt[-1 + c + d*x]*Sqrt[1 +
c + d*x] - 4*b*Sqrt[-1 + c + d*x]*(c + d*x)^2*Sqrt[1 + c + d*x] - 3*b*Sqrt[-1 + c + d*x]*(c + d*x)^4*Sqrt[1 +
c + d*x])*ArcCosh[c + d*x] + 225*b^2*(c + d*x)^5*ArcCosh[c + d*x]^2))/(1125*d)

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Maple [A]  time = 0.044, size = 294, normalized size = 1.4 \begin{align*}{\frac{1}{d} \left ({\frac{ \left ( dx+c \right ) ^{5}{e}^{4}{a}^{2}}{5}}+{e}^{4}{b}^{2} \left ({\frac{ \left ( dx+c \right ) ^{3} \left ({\rm arccosh} \left (dx+c\right ) \right ) ^{2} \left ( dx+c-1 \right ) \left ( dx+c+1 \right ) }{5}}+{\frac{ \left ({\rm arccosh} \left (dx+c\right ) \right ) ^{2} \left ( dx+c-1 \right ) \left ( dx+c+1 \right ) \left ( dx+c \right ) }{5}}+{\frac{ \left ({\rm arccosh} \left (dx+c\right ) \right ) ^{2} \left ( dx+c \right ) }{5}}-{\frac{2\, \left ( dx+c \right ) ^{4}{\rm arccosh} \left (dx+c\right )}{25}\sqrt{dx+c-1}\sqrt{dx+c+1}}-{\frac{16\,{\rm arccosh} \left (dx+c\right )}{75}\sqrt{dx+c-1}\sqrt{dx+c+1}}-{\frac{8\,{\rm arccosh} \left (dx+c\right ) \left ( dx+c \right ) ^{2}}{75}\sqrt{dx+c-1}\sqrt{dx+c+1}}+{\frac{ \left ( 2\,dx+2\,c-2 \right ) \left ( dx+c+1 \right ) \left ( dx+c \right ) ^{3}}{125}}+{\frac{ \left ( 58\,dx+58\,c-58 \right ) \left ( dx+c+1 \right ) \left ( dx+c \right ) }{1125}}+{\frac{298\,dx}{1125}}+{\frac{298\,c}{1125}} \right ) +2\,{e}^{4}ab \left ( 1/5\, \left ( dx+c \right ) ^{5}{\rm arccosh} \left (dx+c\right )-{\frac{\sqrt{dx+c-1}\sqrt{dx+c+1} \left ( 3\, \left ( dx+c \right ) ^{4}+4\, \left ( dx+c \right ) ^{2}+8 \right ) }{75}} \right ) \right ) } \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))^2,x)

[Out]

1/d*(1/5*(d*x+c)^5*e^4*a^2+e^4*b^2*(1/5*(d*x+c)^3*arccosh(d*x+c)^2*(d*x+c-1)*(d*x+c+1)+1/5*arccosh(d*x+c)^2*(d
*x+c-1)*(d*x+c+1)*(d*x+c)+1/5*arccosh(d*x+c)^2*(d*x+c)-2/25*(d*x+c)^4*arccosh(d*x+c)*(d*x+c-1)^(1/2)*(d*x+c+1)
^(1/2)-16/75*arccosh(d*x+c)*(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)-8/75*arccosh(d*x+c)*(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2
)*(d*x+c)^2+2/125*(d*x+c-1)*(d*x+c+1)*(d*x+c)^3+58/1125*(d*x+c-1)*(d*x+c+1)*(d*x+c)+298/1125*d*x+298/1125*c)+2
*e^4*a*b*(1/5*(d*x+c)^5*arccosh(d*x+c)-1/75*(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)*(3*(d*x+c)^4+4*(d*x+c)^2+8)))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.63977, size = 1331, normalized size = 6.11 \begin{align*} \frac{9 \,{\left (25 \, a^{2} + 2 \, b^{2}\right )} d^{5} e^{4} x^{5} + 45 \,{\left (25 \, a^{2} + 2 \, b^{2}\right )} c d^{4} e^{4} x^{4} + 10 \,{\left (9 \,{\left (25 \, a^{2} + 2 \, b^{2}\right )} c^{2} + 4 \, b^{2}\right )} d^{3} e^{4} x^{3} + 30 \,{\left (3 \,{\left (25 \, a^{2} + 2 \, b^{2}\right )} c^{3} + 4 \, b^{2} c\right )} d^{2} e^{4} x^{2} + 15 \,{\left (3 \,{\left (25 \, a^{2} + 2 \, b^{2}\right )} c^{4} + 8 \, b^{2} c^{2} + 16 \, b^{2}\right )} d e^{4} x + 225 \,{\left (b^{2} d^{5} e^{4} x^{5} + 5 \, b^{2} c d^{4} e^{4} x^{4} + 10 \, b^{2} c^{2} d^{3} e^{4} x^{3} + 10 \, b^{2} c^{3} d^{2} e^{4} x^{2} + 5 \, b^{2} c^{4} d e^{4} x + b^{2} c^{5} e^{4}\right )} \log \left (d x + c + \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} - 1}\right )^{2} + 30 \,{\left (15 \, a b d^{5} e^{4} x^{5} + 75 \, a b c d^{4} e^{4} x^{4} + 150 \, a b c^{2} d^{3} e^{4} x^{3} + 150 \, a b c^{3} d^{2} e^{4} x^{2} + 75 \, a b c^{4} d e^{4} x + 15 \, a b c^{5} e^{4} -{\left (3 \, b^{2} d^{4} e^{4} x^{4} + 12 \, b^{2} c d^{3} e^{4} x^{3} + 2 \,{\left (9 \, b^{2} c^{2} + 2 \, b^{2}\right )} d^{2} e^{4} x^{2} + 4 \,{\left (3 \, b^{2} c^{3} + 2 \, b^{2} c\right )} d e^{4} x +{\left (3 \, b^{2} c^{4} + 4 \, b^{2} c^{2} + 8 \, b^{2}\right )} e^{4}\right )} \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} - 1}\right )} \log \left (d x + c + \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} - 1}\right ) - 30 \,{\left (3 \, a b d^{4} e^{4} x^{4} + 12 \, a b c d^{3} e^{4} x^{3} + 2 \,{\left (9 \, a b c^{2} + 2 \, a b\right )} d^{2} e^{4} x^{2} + 4 \,{\left (3 \, a b c^{3} + 2 \, a b c\right )} d e^{4} x +{\left (3 \, a b c^{4} + 4 \, a b c^{2} + 8 \, a b\right )} e^{4}\right )} \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} - 1}}{1125 \, d} \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))^2,x, algorithm="fricas")

[Out]

1/1125*(9*(25*a^2 + 2*b^2)*d^5*e^4*x^5 + 45*(25*a^2 + 2*b^2)*c*d^4*e^4*x^4 + 10*(9*(25*a^2 + 2*b^2)*c^2 + 4*b^
2)*d^3*e^4*x^3 + 30*(3*(25*a^2 + 2*b^2)*c^3 + 4*b^2*c)*d^2*e^4*x^2 + 15*(3*(25*a^2 + 2*b^2)*c^4 + 8*b^2*c^2 +
16*b^2)*d*e^4*x + 225*(b^2*d^5*e^4*x^5 + 5*b^2*c*d^4*e^4*x^4 + 10*b^2*c^2*d^3*e^4*x^3 + 10*b^2*c^3*d^2*e^4*x^2
+ 5*b^2*c^4*d*e^4*x + b^2*c^5*e^4)*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 - 1))^2 + 30*(15*a*b*d^5*e^4*x^
5 + 75*a*b*c*d^4*e^4*x^4 + 150*a*b*c^2*d^3*e^4*x^3 + 150*a*b*c^3*d^2*e^4*x^2 + 75*a*b*c^4*d*e^4*x + 15*a*b*c^5
*e^4 - (3*b^2*d^4*e^4*x^4 + 12*b^2*c*d^3*e^4*x^3 + 2*(9*b^2*c^2 + 2*b^2)*d^2*e^4*x^2 + 4*(3*b^2*c^3 + 2*b^2*c)
*d*e^4*x + (3*b^2*c^4 + 4*b^2*c^2 + 8*b^2)*e^4)*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)) - 30*(3*a*b*d^4*e^4*x^4 + 12*a*b*c*d^3*e^4*x^3 + 2*(9*a*b*c^2 + 2*a*b)*d^2*e^4*x^2 + 4*(
3*a*b*c^3 + 2*a*b*c)*d*e^4*x + (3*a*b*c^4 + 4*a*b*c^2 + 8*a*b)*e^4)*sqrt(d^2*x^2 + 2*c*d*x + c^2 - 1))/d

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Sympy [A]  time = 10.2041, size = 1268, normalized size = 5.82 \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*acosh(d*x+c))**2,x)

[Out]

Piecewise((a**2*c**4*e**4*x + 2*a**2*c**3*d*e**4*x**2 + 2*a**2*c**2*d**2*e**4*x**3 + a**2*c*d**3*e**4*x**4 + a
**2*d**4*e**4*x**5/5 + 2*a*b*c**5*e**4*acosh(c + d*x)/(5*d) + 2*a*b*c**4*e**4*x*acosh(c + d*x) - 2*a*b*c**4*e*
*4*sqrt(c**2 + 2*c*d*x + d**2*x**2 - 1)/(25*d) + 4*a*b*c**3*d*e**4*x**2*acosh(c + d*x) - 8*a*b*c**3*e**4*x*sqr
t(c**2 + 2*c*d*x + d**2*x**2 - 1)/25 + 4*a*b*c**2*d**2*e**4*x**3*acosh(c + d*x) - 12*a*b*c**2*d*e**4*x**2*sqrt
(c**2 + 2*c*d*x + d**2*x**2 - 1)/25 - 8*a*b*c**2*e**4*sqrt(c**2 + 2*c*d*x + d**2*x**2 - 1)/(75*d) + 2*a*b*c*d*
*3*e**4*x**4*acosh(c + d*x) - 8*a*b*c*d**2*e**4*x**3*sqrt(c**2 + 2*c*d*x + d**2*x**2 - 1)/25 - 16*a*b*c*e**4*x
*sqrt(c**2 + 2*c*d*x + d**2*x**2 - 1)/75 + 2*a*b*d**4*e**4*x**5*acosh(c + d*x)/5 - 2*a*b*d**3*e**4*x**4*sqrt(c
**2 + 2*c*d*x + d**2*x**2 - 1)/25 - 8*a*b*d*e**4*x**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 - 1)/75 - 16*a*b*e**4*sq
rt(c**2 + 2*c*d*x + d**2*x**2 - 1)/(75*d) + b**2*c**5*e**4*acosh(c + d*x)**2/(5*d) + b**2*c**4*e**4*x*acosh(c
+ d*x)**2 + 2*b**2*c**4*e**4*x/25 - 2*b**2*c**4*e**4*sqrt(c**2 + 2*c*d*x + d**2*x**2 - 1)*acosh(c + d*x)/(25*d
) + 2*b**2*c**3*d*e**4*x**2*acosh(c + d*x)**2 + 4*b**2*c**3*d*e**4*x**2/25 - 8*b**2*c**3*e**4*x*sqrt(c**2 + 2*
c*d*x + d**2*x**2 - 1)*acosh(c + d*x)/25 + 2*b**2*c**2*d**2*e**4*x**3*acosh(c + d*x)**2 + 4*b**2*c**2*d**2*e**
4*x**3/25 - 12*b**2*c**2*d*e**4*x**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 - 1)*acosh(c + d*x)/25 + 8*b**2*c**2*e**4
*x/75 - 8*b**2*c**2*e**4*sqrt(c**2 + 2*c*d*x + d**2*x**2 - 1)*acosh(c + d*x)/(75*d) + b**2*c*d**3*e**4*x**4*ac
osh(c + d*x)**2 + 2*b**2*c*d**3*e**4*x**4/25 - 8*b**2*c*d**2*e**4*x**3*sqrt(c**2 + 2*c*d*x + d**2*x**2 - 1)*ac
osh(c + d*x)/25 + 8*b**2*c*d*e**4*x**2/75 - 16*b**2*c*e**4*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 - 1)*acosh(c + d*
x)/75 + b**2*d**4*e**4*x**5*acosh(c + d*x)**2/5 + 2*b**2*d**4*e**4*x**5/125 - 2*b**2*d**3*e**4*x**4*sqrt(c**2
+ 2*c*d*x + d**2*x**2 - 1)*acosh(c + d*x)/25 + 8*b**2*d**2*e**4*x**3/225 - 8*b**2*d*e**4*x**2*sqrt(c**2 + 2*c*
d*x + d**2*x**2 - 1)*acosh(c + d*x)/75 + 16*b**2*e**4*x/75 - 16*b**2*e**4*sqrt(c**2 + 2*c*d*x + d**2*x**2 - 1)
*acosh(c + d*x)/(75*d), Ne(d, 0)), (c**4*e**4*x*(a + b*acosh(c))**2, True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d e x + c e\right )}^{4}{\left (b \operatorname{arcosh}\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*arccosh(d*x+c))^2,x, algorithm="giac")

[Out]

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