3.115 \(\int (c e+d e x)^2 (a+b \cosh ^{-1}(c+d x))^3 \, dx\)

Optimal. Leaf size=262 \[ \frac{2 b^2 e^2 (c+d x)^3 \left (a+b \cosh ^{-1}(c+d x)\right )}{9 d}+\frac{4}{3} a b^2 e^2 x-\frac{b e^2 \sqrt{c+d x-1} (c+d x)^2 \sqrt{c+d x+1} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{3 d}-\frac{2 b e^2 \sqrt{c+d x-1} \sqrt{c+d x+1} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{3 d}+\frac{e^2 (c+d x)^3 \left (a+b \cosh ^{-1}(c+d x)\right )^3}{3 d}-\frac{2 b^3 e^2 \sqrt{c+d x-1} (c+d x)^2 \sqrt{c+d x+1}}{27 d}-\frac{40 b^3 e^2 \sqrt{c+d x-1} \sqrt{c+d x+1}}{27 d}+\frac{4 b^3 e^2 (c+d x) \cosh ^{-1}(c+d x)}{3 d} \]

[Out]

(4*a*b^2*e^2*x)/3 - (40*b^3*e^2*Sqrt[-1 + c + d*x]*Sqrt[1 + c + d*x])/(27*d) - (2*b^3*e^2*Sqrt[-1 + c + d*x]*(
c + d*x)^2*Sqrt[1 + c + d*x])/(27*d) + (4*b^3*e^2*(c + d*x)*ArcCosh[c + d*x])/(3*d) + (2*b^2*e^2*(c + d*x)^3*(
a + b*ArcCosh[c + d*x]))/(9*d) - (2*b*e^2*Sqrt[-1 + c + d*x]*Sqrt[1 + c + d*x]*(a + b*ArcCosh[c + d*x])^2)/(3*
d) - (b*e^2*Sqrt[-1 + c + d*x]*(c + d*x)^2*Sqrt[1 + c + d*x]*(a + b*ArcCosh[c + d*x])^2)/(3*d) + (e^2*(c + d*x
)^3*(a + b*ArcCosh[c + d*x])^3)/(3*d)

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Rubi [A]  time = 0.472206, antiderivative size = 262, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {5866, 12, 5662, 5759, 5718, 5654, 74, 100} \[ \frac{2 b^2 e^2 (c+d x)^3 \left (a+b \cosh ^{-1}(c+d x)\right )}{9 d}+\frac{4}{3} a b^2 e^2 x-\frac{b e^2 \sqrt{c+d x-1} (c+d x)^2 \sqrt{c+d x+1} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{3 d}-\frac{2 b e^2 \sqrt{c+d x-1} \sqrt{c+d x+1} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{3 d}+\frac{e^2 (c+d x)^3 \left (a+b \cosh ^{-1}(c+d x)\right )^3}{3 d}-\frac{2 b^3 e^2 \sqrt{c+d x-1} (c+d x)^2 \sqrt{c+d x+1}}{27 d}-\frac{40 b^3 e^2 \sqrt{c+d x-1} \sqrt{c+d x+1}}{27 d}+\frac{4 b^3 e^2 (c+d x) \cosh ^{-1}(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^2*(a + b*ArcCosh[c + d*x])^3,x]

[Out]

(4*a*b^2*e^2*x)/3 - (40*b^3*e^2*Sqrt[-1 + c + d*x]*Sqrt[1 + c + d*x])/(27*d) - (2*b^3*e^2*Sqrt[-1 + c + d*x]*(
c + d*x)^2*Sqrt[1 + c + d*x])/(27*d) + (4*b^3*e^2*(c + d*x)*ArcCosh[c + d*x])/(3*d) + (2*b^2*e^2*(c + d*x)^3*(
a + b*ArcCosh[c + d*x]))/(9*d) - (2*b*e^2*Sqrt[-1 + c + d*x]*Sqrt[1 + c + d*x]*(a + b*ArcCosh[c + d*x])^2)/(3*
d) - (b*e^2*Sqrt[-1 + c + d*x]*(c + d*x)^2*Sqrt[1 + c + d*x]*(a + b*ArcCosh[c + d*x])^2)/(3*d) + (e^2*(c + d*x
)^3*(a + b*ArcCosh[c + d*x])^3)/(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 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 5654

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

Rule 74

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

Rule 100

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

Rubi steps

\begin{align*} \int (c e+d e x)^2 \left (a+b \cosh ^{-1}(c+d x)\right )^3 \, dx &=\frac{\operatorname{Subst}\left (\int e^2 x^2 \left (a+b \cosh ^{-1}(x)\right )^3 \, dx,x,c+d x\right )}{d}\\ &=\frac{e^2 \operatorname{Subst}\left (\int x^2 \left (a+b \cosh ^{-1}(x)\right )^3 \, dx,x,c+d x\right )}{d}\\ &=\frac{e^2 (c+d x)^3 \left (a+b \cosh ^{-1}(c+d x)\right )^3}{3 d}-\frac{\left (b e^2\right ) \operatorname{Subst}\left (\int \frac{x^3 \left (a+b \cosh ^{-1}(x)\right )^2}{\sqrt{-1+x} \sqrt{1+x}} \, dx,x,c+d x\right )}{d}\\ &=-\frac{b e^2 \sqrt{-1+c+d x} (c+d x)^2 \sqrt{1+c+d x} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{3 d}+\frac{e^2 (c+d x)^3 \left (a+b \cosh ^{-1}(c+d x)\right )^3}{3 d}-\frac{\left (2 b e^2\right ) \operatorname{Subst}\left (\int \frac{x \left (a+b \cosh ^{-1}(x)\right )^2}{\sqrt{-1+x} \sqrt{1+x}} \, dx,x,c+d x\right )}{3 d}+\frac{\left (2 b^2 e^2\right ) \operatorname{Subst}\left (\int x^2 \left (a+b \cosh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{3 d}\\ &=\frac{2 b^2 e^2 (c+d x)^3 \left (a+b \cosh ^{-1}(c+d x)\right )}{9 d}-\frac{2 b e^2 \sqrt{-1+c+d x} \sqrt{1+c+d x} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{3 d}-\frac{b e^2 \sqrt{-1+c+d x} (c+d x)^2 \sqrt{1+c+d x} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{3 d}+\frac{e^2 (c+d x)^3 \left (a+b \cosh ^{-1}(c+d x)\right )^3}{3 d}+\frac{\left (4 b^2 e^2\right ) \operatorname{Subst}\left (\int \left (a+b \cosh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{3 d}-\frac{\left (2 b^3 e^2\right ) \operatorname{Subst}\left (\int \frac{x^3}{\sqrt{-1+x} \sqrt{1+x}} \, dx,x,c+d x\right )}{9 d}\\ &=\frac{4}{3} a b^2 e^2 x-\frac{2 b^3 e^2 \sqrt{-1+c+d x} (c+d x)^2 \sqrt{1+c+d x}}{27 d}+\frac{2 b^2 e^2 (c+d x)^3 \left (a+b \cosh ^{-1}(c+d x)\right )}{9 d}-\frac{2 b e^2 \sqrt{-1+c+d x} \sqrt{1+c+d x} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{3 d}-\frac{b e^2 \sqrt{-1+c+d x} (c+d x)^2 \sqrt{1+c+d x} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{3 d}+\frac{e^2 (c+d x)^3 \left (a+b \cosh ^{-1}(c+d x)\right )^3}{3 d}-\frac{\left (2 b^3 e^2\right ) \operatorname{Subst}\left (\int \frac{2 x}{\sqrt{-1+x} \sqrt{1+x}} \, dx,x,c+d x\right )}{27 d}+\frac{\left (4 b^3 e^2\right ) \operatorname{Subst}\left (\int \cosh ^{-1}(x) \, dx,x,c+d x\right )}{3 d}\\ &=\frac{4}{3} a b^2 e^2 x-\frac{2 b^3 e^2 \sqrt{-1+c+d x} (c+d x)^2 \sqrt{1+c+d x}}{27 d}+\frac{4 b^3 e^2 (c+d x) \cosh ^{-1}(c+d x)}{3 d}+\frac{2 b^2 e^2 (c+d x)^3 \left (a+b \cosh ^{-1}(c+d x)\right )}{9 d}-\frac{2 b e^2 \sqrt{-1+c+d x} \sqrt{1+c+d x} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{3 d}-\frac{b e^2 \sqrt{-1+c+d x} (c+d x)^2 \sqrt{1+c+d x} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{3 d}+\frac{e^2 (c+d x)^3 \left (a+b \cosh ^{-1}(c+d x)\right )^3}{3 d}-\frac{\left (4 b^3 e^2\right ) \operatorname{Subst}\left (\int \frac{x}{\sqrt{-1+x} \sqrt{1+x}} \, dx,x,c+d x\right )}{27 d}-\frac{\left (4 b^3 e^2\right ) \operatorname{Subst}\left (\int \frac{x}{\sqrt{-1+x} \sqrt{1+x}} \, dx,x,c+d x\right )}{3 d}\\ &=\frac{4}{3} a b^2 e^2 x-\frac{40 b^3 e^2 \sqrt{-1+c+d x} \sqrt{1+c+d x}}{27 d}-\frac{2 b^3 e^2 \sqrt{-1+c+d x} (c+d x)^2 \sqrt{1+c+d x}}{27 d}+\frac{4 b^3 e^2 (c+d x) \cosh ^{-1}(c+d x)}{3 d}+\frac{2 b^2 e^2 (c+d x)^3 \left (a+b \cosh ^{-1}(c+d x)\right )}{9 d}-\frac{2 b e^2 \sqrt{-1+c+d x} \sqrt{1+c+d x} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{3 d}-\frac{b e^2 \sqrt{-1+c+d x} (c+d x)^2 \sqrt{1+c+d x} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{3 d}+\frac{e^2 (c+d x)^3 \left (a+b \cosh ^{-1}(c+d x)\right )^3}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.392803, size = 296, normalized size = 1.13 \[ \frac{e^2 \left (a \left (3 a^2+2 b^2\right ) (c+d x)^3+\frac{1}{3} b \sqrt{c+d x-1} \sqrt{c+d x+1} \left (-\left (9 a^2+2 b^2\right ) (c+d x)^2-2 \left (9 a^2+20 b^2\right )\right )-b \cosh ^{-1}(c+d x) \left (-9 a^2 (c+d x)^3+6 a b \sqrt{c+d x-1} \sqrt{c+d x+1} (c+d x)^2+12 a b \sqrt{c+d x-1} \sqrt{c+d x+1}-2 b^2 (c+d x)^3-12 b^2 (c+d x)\right )+12 a b^2 (c+d x)-3 b^2 \cosh ^{-1}(c+d x)^2 \left (-3 a (c+d x)^3+b \sqrt{c+d x-1} \sqrt{c+d x+1} (c+d x)^2+2 b \sqrt{c+d x-1} \sqrt{c+d x+1}\right )+3 b^3 (c+d x)^3 \cosh ^{-1}(c+d x)^3\right )}{9 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)^2*(a + b*ArcCosh[c + d*x])^3,x]

[Out]

(e^2*(12*a*b^2*(c + d*x) + a*(3*a^2 + 2*b^2)*(c + d*x)^3 + (b*Sqrt[-1 + c + d*x]*Sqrt[1 + c + d*x]*(-2*(9*a^2
+ 20*b^2) - (9*a^2 + 2*b^2)*(c + d*x)^2))/3 - b*(-12*b^2*(c + d*x) - 9*a^2*(c + d*x)^3 - 2*b^2*(c + d*x)^3 + 1
2*a*b*Sqrt[-1 + c + d*x]*Sqrt[1 + c + d*x] + 6*a*b*Sqrt[-1 + c + d*x]*(c + d*x)^2*Sqrt[1 + c + d*x])*ArcCosh[c
+ d*x] - 3*b^2*(-3*a*(c + d*x)^3 + 2*b*Sqrt[-1 + c + d*x]*Sqrt[1 + c + d*x] + b*Sqrt[-1 + c + d*x]*(c + d*x)^
2*Sqrt[1 + c + d*x])*ArcCosh[c + d*x]^2 + 3*b^3*(c + d*x)^3*ArcCosh[c + d*x]^3))/(9*d)

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Maple [A]  time = 0.043, size = 396, normalized size = 1.5 \begin{align*}{\frac{1}{d} \left ({\frac{ \left ( dx+c \right ) ^{3}{e}^{2}{a}^{3}}{3}}+{e}^{2}{b}^{3} \left ({\frac{ \left ({\rm arccosh} \left (dx+c\right ) \right ) ^{3} \left ( dx+c-1 \right ) \left ( dx+c+1 \right ) \left ( dx+c \right ) }{3}}+{\frac{ \left ({\rm arccosh} \left (dx+c\right ) \right ) ^{3} \left ( dx+c \right ) }{3}}-{\frac{ \left ({\rm arccosh} \left (dx+c\right ) \right ) ^{2} \left ( dx+c \right ) ^{2}}{3}\sqrt{dx+c-1}\sqrt{dx+c+1}}-{\frac{2\, \left ({\rm arccosh} \left (dx+c\right ) \right ) ^{2}}{3}\sqrt{dx+c-1}\sqrt{dx+c+1}}+{\frac{2\,{\rm arccosh} \left (dx+c\right ) \left ( dx+c-1 \right ) \left ( dx+c+1 \right ) \left ( dx+c \right ) }{9}}+{\frac{ \left ( 14\,dx+14\,c \right ){\rm arccosh} \left (dx+c\right )}{9}}-{\frac{2\, \left ( dx+c \right ) ^{2}}{27}\sqrt{dx+c-1}\sqrt{dx+c+1}}-{\frac{40}{27}\sqrt{dx+c-1}\sqrt{dx+c+1}} \right ) +3\,{e}^{2}a{b}^{2} \left ( 1/3\, \left ({\rm arccosh} \left (dx+c\right ) \right ) ^{2} \left ( dx+c-1 \right ) \left ( dx+c+1 \right ) \left ( dx+c \right ) +1/3\, \left ({\rm arccosh} \left (dx+c\right ) \right ) ^{2} \left ( dx+c \right ) -2/9\,{\rm arccosh} \left (dx+c\right )\sqrt{dx+c-1}\sqrt{dx+c+1} \left ( dx+c \right ) ^{2}-4/9\,{\rm arccosh} \left (dx+c\right )\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 ) }{27}}+{\frac{14\,dx}{27}}+{\frac{14\,c}{27}} \right ) +3\,{e}^{2}{a}^{2}b \left ( 1/3\,{\rm arccosh} \left (dx+c\right ) \left ( dx+c \right ) ^{3}-1/9\,\sqrt{dx+c-1}\sqrt{dx+c+1} \left ( \left ( dx+c \right ) ^{2}+2 \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^2*(a+b*arccosh(d*x+c))^3,x)

[Out]

1/d*(1/3*(d*x+c)^3*e^2*a^3+e^2*b^3*(1/3*arccosh(d*x+c)^3*(d*x+c-1)*(d*x+c+1)*(d*x+c)+1/3*arccosh(d*x+c)^3*(d*x
+c)-1/3*arccosh(d*x+c)^2*(d*x+c)^2*(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)-2/3*arccosh(d*x+c)^2*(d*x+c-1)^(1/2)*(d*x+c
+1)^(1/2)+2/9*arccosh(d*x+c)*(d*x+c-1)*(d*x+c+1)*(d*x+c)+14/9*(d*x+c)*arccosh(d*x+c)-2/27*(d*x+c)^2*(d*x+c-1)^
(1/2)*(d*x+c+1)^(1/2)-40/27*(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2))+3*e^2*a*b^2*(1/3*arccosh(d*x+c)^2*(d*x+c-1)*(d*x+
c+1)*(d*x+c)+1/3*arccosh(d*x+c)^2*(d*x+c)-2/9*arccosh(d*x+c)*(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)*(d*x+c)^2-4/9*arc
cosh(d*x+c)*(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)+2/27*(d*x+c-1)*(d*x+c+1)*(d*x+c)+14/27*d*x+14/27*c)+3*e^2*a^2*b*(1
/3*arccosh(d*x+c)*(d*x+c)^3-1/9*(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)*((d*x+c)^2+2)))

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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)^2*(a+b*arccosh(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.54766, size = 1285, normalized size = 4.9 \begin{align*} \frac{3 \,{\left (3 \, a^{3} + 2 \, a b^{2}\right )} d^{3} e^{2} x^{3} + 9 \,{\left (3 \, a^{3} + 2 \, a b^{2}\right )} c d^{2} e^{2} x^{2} + 9 \,{\left (4 \, a b^{2} +{\left (3 \, a^{3} + 2 \, a b^{2}\right )} c^{2}\right )} d e^{2} x + 9 \,{\left (b^{3} d^{3} e^{2} x^{3} + 3 \, b^{3} c d^{2} e^{2} x^{2} + 3 \, b^{3} c^{2} d e^{2} x + b^{3} c^{3} e^{2}\right )} \log \left (d x + c + \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} - 1}\right )^{3} + 9 \,{\left (3 \, a b^{2} d^{3} e^{2} x^{3} + 9 \, a b^{2} c d^{2} e^{2} x^{2} + 9 \, a b^{2} c^{2} d e^{2} x + 3 \, a b^{2} c^{3} e^{2} -{\left (b^{3} d^{2} e^{2} x^{2} + 2 \, b^{3} c d e^{2} x +{\left (b^{3} c^{2} + 2 \, b^{3}\right )} e^{2}\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 )^{2} + 3 \,{\left ({\left (9 \, a^{2} b + 2 \, b^{3}\right )} d^{3} e^{2} x^{3} + 3 \,{\left (9 \, a^{2} b + 2 \, b^{3}\right )} c d^{2} e^{2} x^{2} + 3 \,{\left (4 \, b^{3} +{\left (9 \, a^{2} b + 2 \, b^{3}\right )} c^{2}\right )} d e^{2} x +{\left (12 \, b^{3} c +{\left (9 \, a^{2} b + 2 \, b^{3}\right )} c^{3}\right )} e^{2} - 6 \,{\left (a b^{2} d^{2} e^{2} x^{2} + 2 \, a b^{2} c d e^{2} x +{\left (a b^{2} c^{2} + 2 \, a b^{2}\right )} e^{2}\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 ) -{\left ({\left (9 \, a^{2} b + 2 \, b^{3}\right )} d^{2} e^{2} x^{2} + 2 \,{\left (9 \, a^{2} b + 2 \, b^{3}\right )} c d e^{2} x +{\left (18 \, a^{2} b + 40 \, b^{3} +{\left (9 \, a^{2} b + 2 \, b^{3}\right )} c^{2}\right )} e^{2}\right )} \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} - 1}}{27 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^2*(a+b*arccosh(d*x+c))^3,x, algorithm="fricas")

[Out]

1/27*(3*(3*a^3 + 2*a*b^2)*d^3*e^2*x^3 + 9*(3*a^3 + 2*a*b^2)*c*d^2*e^2*x^2 + 9*(4*a*b^2 + (3*a^3 + 2*a*b^2)*c^2
)*d*e^2*x + 9*(b^3*d^3*e^2*x^3 + 3*b^3*c*d^2*e^2*x^2 + 3*b^3*c^2*d*e^2*x + b^3*c^3*e^2)*log(d*x + c + sqrt(d^2
*x^2 + 2*c*d*x + c^2 - 1))^3 + 9*(3*a*b^2*d^3*e^2*x^3 + 9*a*b^2*c*d^2*e^2*x^2 + 9*a*b^2*c^2*d*e^2*x + 3*a*b^2*
c^3*e^2 - (b^3*d^2*e^2*x^2 + 2*b^3*c*d*e^2*x + (b^3*c^2 + 2*b^3)*e^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 + 3*((9*a^2*b + 2*b^3)*d^3*e^2*x^3 + 3*(9*a^2*b + 2*b^3)*c*d^2*e
^2*x^2 + 3*(4*b^3 + (9*a^2*b + 2*b^3)*c^2)*d*e^2*x + (12*b^3*c + (9*a^2*b + 2*b^3)*c^3)*e^2 - 6*(a*b^2*d^2*e^2
*x^2 + 2*a*b^2*c*d*e^2*x + (a*b^2*c^2 + 2*a*b^2)*e^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)) - ((9*a^2*b + 2*b^3)*d^2*e^2*x^2 + 2*(9*a^2*b + 2*b^3)*c*d*e^2*x + (18*a^2*b + 40*
b^3 + (9*a^2*b + 2*b^3)*c^2)*e^2)*sqrt(d^2*x^2 + 2*c*d*x + c^2 - 1))/d

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Sympy [A]  time = 6.59907, size = 1173, normalized size = 4.48 \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)**2*(a+b*acosh(d*x+c))**3,x)

[Out]

Piecewise((a**3*c**2*e**2*x + a**3*c*d*e**2*x**2 + a**3*d**2*e**2*x**3/3 + a**2*b*c**3*e**2*acosh(c + d*x)/d +
3*a**2*b*c**2*e**2*x*acosh(c + d*x) - a**2*b*c**2*e**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 - 1)/(3*d) + 3*a**2*b*
c*d*e**2*x**2*acosh(c + d*x) - 2*a**2*b*c*e**2*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 - 1)/3 + a**2*b*d**2*e**2*x**
3*acosh(c + d*x) - a**2*b*d*e**2*x**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 - 1)/3 - 2*a**2*b*e**2*sqrt(c**2 + 2*c*d
*x + d**2*x**2 - 1)/(3*d) + a*b**2*c**3*e**2*acosh(c + d*x)**2/d + 3*a*b**2*c**2*e**2*x*acosh(c + d*x)**2 + 2*
a*b**2*c**2*e**2*x/3 - 2*a*b**2*c**2*e**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 - 1)*acosh(c + d*x)/(3*d) + 3*a*b**2
*c*d*e**2*x**2*acosh(c + d*x)**2 + 2*a*b**2*c*d*e**2*x**2/3 - 4*a*b**2*c*e**2*x*sqrt(c**2 + 2*c*d*x + d**2*x**
2 - 1)*acosh(c + d*x)/3 + a*b**2*d**2*e**2*x**3*acosh(c + d*x)**2 + 2*a*b**2*d**2*e**2*x**3/9 - 2*a*b**2*d*e**
2*x**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 - 1)*acosh(c + d*x)/3 + 4*a*b**2*e**2*x/3 - 4*a*b**2*e**2*sqrt(c**2 + 2
*c*d*x + d**2*x**2 - 1)*acosh(c + d*x)/(3*d) + b**3*c**3*e**2*acosh(c + d*x)**3/(3*d) + 2*b**3*c**3*e**2*acosh
(c + d*x)/(9*d) + b**3*c**2*e**2*x*acosh(c + d*x)**3 + 2*b**3*c**2*e**2*x*acosh(c + d*x)/3 - b**3*c**2*e**2*sq
rt(c**2 + 2*c*d*x + d**2*x**2 - 1)*acosh(c + d*x)**2/(3*d) - 2*b**3*c**2*e**2*sqrt(c**2 + 2*c*d*x + d**2*x**2
- 1)/(27*d) + b**3*c*d*e**2*x**2*acosh(c + d*x)**3 + 2*b**3*c*d*e**2*x**2*acosh(c + d*x)/3 - 2*b**3*c*e**2*x*s
qrt(c**2 + 2*c*d*x + d**2*x**2 - 1)*acosh(c + d*x)**2/3 - 4*b**3*c*e**2*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 - 1)
/27 + 4*b**3*c*e**2*acosh(c + d*x)/(3*d) + b**3*d**2*e**2*x**3*acosh(c + d*x)**3/3 + 2*b**3*d**2*e**2*x**3*aco
sh(c + d*x)/9 - b**3*d*e**2*x**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 - 1)*acosh(c + d*x)**2/3 - 2*b**3*d*e**2*x**2
*sqrt(c**2 + 2*c*d*x + d**2*x**2 - 1)/27 + 4*b**3*e**2*x*acosh(c + d*x)/3 - 2*b**3*e**2*sqrt(c**2 + 2*c*d*x +
d**2*x**2 - 1)*acosh(c + d*x)**2/(3*d) - 40*b**3*e**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 - 1)/(27*d), Ne(d, 0)),
(c**2*e**2*x*(a + b*acosh(c))**3, True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d e x + c e\right )}^{2}{\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)^2*(a+b*arccosh(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((d*e*x + c*e)^2*(b*arccosh(d*x + c) + a)^3, x)