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

Optimal. Leaf size=197 \[ \frac{e^4 (c+d x)^5 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{5 d}-\frac{2 b e^4 \sqrt{(c+d x)^2+1} (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )}{25 d}+\frac{8 b e^4 \sqrt{(c+d x)^2+1} (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )}{75 d}-\frac{16 b e^4 \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )}{75 d}+\frac{2 b^2 e^4 (c+d x)^5}{125 d}-\frac{8 b^2 e^4 (c+d x)^3}{225 d}+\frac{16}{75} b^2 e^4 x \]

[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)^2]*(a + b*ArcSinh[c + d*x]))/(75*d) + (8*b*e^4*(c + d*x)^2*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x
]))/(75*d) - (2*b*e^4*(c + d*x)^4*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x]))/(25*d) + (e^4*(c + d*x)^5*(a
+ b*ArcSinh[c + d*x])^2)/(5*d)

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Rubi [A]  time = 0.306813, antiderivative size = 197, 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 = {5865, 12, 5661, 5758, 5717, 8, 30} \[ \frac{e^4 (c+d x)^5 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{5 d}-\frac{2 b e^4 \sqrt{(c+d x)^2+1} (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )}{25 d}+\frac{8 b e^4 \sqrt{(c+d x)^2+1} (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )}{75 d}-\frac{16 b e^4 \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )}{75 d}+\frac{2 b^2 e^4 (c+d x)^5}{125 d}-\frac{8 b^2 e^4 (c+d x)^3}{225 d}+\frac{16}{75} b^2 e^4 x \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^4*(a + b*ArcSinh[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)^2]*(a + b*ArcSinh[c + d*x]))/(75*d) + (8*b*e^4*(c + d*x)^2*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x
]))/(75*d) - (2*b*e^4*(c + d*x)^4*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x]))/(25*d) + (e^4*(c + d*x)^5*(a
+ b*ArcSinh[c + d*x])^2)/(5*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 5661

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

Rule 5758

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

Rule 5717

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

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

Mathematica [A]  time = 0.248307, size = 192, normalized size = 0.97 \[ \frac{e^4 \left (9 \left (25 a^2+2 b^2\right ) (c+d x)^5+30 a b \sqrt{(c+d x)^2+1} \left (-3 (c+d x)^4+4 (c+d x)^2-8\right )+30 b \sinh ^{-1}(c+d x) \left (15 a (c+d x)^5-3 b \sqrt{(c+d x)^2+1} (c+d x)^4+4 b \sqrt{(c+d x)^2+1} (c+d x)^2-8 b \sqrt{(c+d x)^2+1}\right )-40 b^2 (c+d x)^3+240 b^2 (c+d x)+225 b^2 (c+d x)^5 \sinh ^{-1}(c+d x)^2\right )}{1125 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)^4*(a + b*ArcSinh[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)^2]*(
-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)^2] + 4*b*(c + d*x)^2*Sqr
t[1 + (c + d*x)^2] - 3*b*(c + d*x)^4*Sqrt[1 + (c + d*x)^2])*ArcSinh[c + d*x] + 225*b^2*(c + d*x)^5*ArcSinh[c +
d*x]^2))/(1125*d)

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Maple [A]  time = 0.039, size = 282, 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 ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{2} \left ( 1+ \left ( dx+c \right ) ^{2} \right ) }{5}}-{\frac{ \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{2} \left ( dx+c \right ) \left ( 1+ \left ( dx+c \right ) ^{2} \right ) }{5}}+{\frac{ \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{2} \left ( dx+c \right ) }{5}}-{\frac{2\,{\it Arcsinh} \left ( dx+c \right ) \left ( dx+c \right ) ^{2}}{25} \left ( 1+ \left ( dx+c \right ) ^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{14\,{\it Arcsinh} \left ( dx+c \right ) \left ( dx+c \right ) ^{2}}{75}\sqrt{1+ \left ( dx+c \right ) ^{2}}}-{\frac{16\,{\it Arcsinh} \left ( dx+c \right ) }{75}\sqrt{1+ \left ( dx+c \right ) ^{2}}}+{\frac{2\, \left ( 1+ \left ( dx+c \right ) ^{2} \right ) ^{2} \left ( dx+c \right ) }{125}}+{\frac{298\,dx}{1125}}+{\frac{298\,c}{1125}}-{\frac{ \left ( 76+76\, \left ( dx+c \right ) ^{2} \right ) \left ( dx+c \right ) }{1125}} \right ) +2\,{e}^{4}ab \left ( 1/5\, \left ( dx+c \right ) ^{5}{\it Arcsinh} \left ( dx+c \right ) -1/25\, \left ( dx+c \right ) ^{4}\sqrt{1+ \left ( dx+c \right ) ^{2}}+{\frac{4\, \left ( dx+c \right ) ^{2}\sqrt{1+ \left ( dx+c \right ) ^{2}}}{75}}-{\frac{8\,\sqrt{1+ \left ( dx+c \right ) ^{2}}}{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*arcsinh(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*arcsinh(d*x+c)^2*(1+(d*x+c)^2)-1/5*arcsinh(d*x+c)^2*(d*x+c)*
(1+(d*x+c)^2)+1/5*arcsinh(d*x+c)^2*(d*x+c)-2/25*arcsinh(d*x+c)*(d*x+c)^2*(1+(d*x+c)^2)^(3/2)+14/75*arcsinh(d*x
+c)*(d*x+c)^2*(1+(d*x+c)^2)^(1/2)-16/75*arcsinh(d*x+c)*(1+(d*x+c)^2)^(1/2)+2/125*(1+(d*x+c)^2)^2*(d*x+c)+298/1
125*d*x+298/1125*c-76/1125*(1+(d*x+c)^2)*(d*x+c))+2*e^4*a*b*(1/5*(d*x+c)^5*arcsinh(d*x+c)-1/25*(d*x+c)^4*(1+(d
*x+c)^2)^(1/2)+4/75*(d*x+c)^2*(1+(d*x+c)^2)^(1/2)-8/75*(1+(d*x+c)^2)^(1/2)))

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.69383, size = 1331, normalized size = 6.76 \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*arcsinh(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 = 11.1037, size = 1268, normalized size = 6.44 \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*asinh(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*asinh(c + d*x)/(5*d) + 2*a*b*c**4*e**4*x*asinh(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*asinh(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*asinh(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*asinh(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*asinh(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*asinh(c + d*x)**2/(5*d) + b**2*c**4*e**4*x*asinh(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)*asinh(c + d*x)/(25*d
) + 2*b**2*c**3*d*e**4*x**2*asinh(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)*asinh(c + d*x)/25 + 2*b**2*c**2*d**2*e**4*x**3*asinh(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)*asinh(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)*asinh(c + d*x)/(75*d) + b**2*c*d**3*e**4*x**4*as
inh(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)*as
inh(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)*asinh(c + d*
x)/75 + b**2*d**4*e**4*x**5*asinh(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)*asinh(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)*asinh(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)
*asinh(c + d*x)/(75*d), Ne(d, 0)), (c**4*e**4*x*(a + b*asinh(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{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)