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

Optimal. Leaf size=281 \[ \frac{160 b^3 e^2 \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )}{27 d}-\frac{8 b^3 e^2 (c+d x)^2 \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )}{27 d}+\frac{4 b^2 e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{9 d}-\frac{8 b^2 e^2 (c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d}+\frac{8 b e^2 \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{9 d}-\frac{4 b e^2 (c+d x)^2 \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{9 d}+\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^4}{3 d}+\frac{8 b^4 e^2 (c+d x)^3}{81 d}-\frac{160}{27} b^4 e^2 x \]

[Out]

(-160*b^4*e^2*x)/27 + (8*b^4*e^2*(c + d*x)^3)/(81*d) + (160*b^3*e^2*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d
*x]))/(27*d) - (8*b^3*e^2*(c + d*x)^2*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x]))/(27*d) - (8*b^2*e^2*(c +
d*x)*(a + b*ArcSinh[c + d*x])^2)/(3*d) + (4*b^2*e^2*(c + d*x)^3*(a + b*ArcSinh[c + d*x])^2)/(9*d) + (8*b*e^2*
Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x])^3)/(9*d) - (4*b*e^2*(c + d*x)^2*Sqrt[1 + (c + d*x)^2]*(a + b*Ar
cSinh[c + d*x])^3)/(9*d) + (e^2*(c + d*x)^3*(a + b*ArcSinh[c + d*x])^4)/(3*d)

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Rubi [A]  time = 0.484952, antiderivative size = 281, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {5865, 12, 5661, 5758, 5717, 5653, 8, 30} \[ \frac{160 b^3 e^2 \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )}{27 d}-\frac{8 b^3 e^2 (c+d x)^2 \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )}{27 d}+\frac{4 b^2 e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{9 d}-\frac{8 b^2 e^2 (c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d}+\frac{8 b e^2 \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{9 d}-\frac{4 b e^2 (c+d x)^2 \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{9 d}+\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^4}{3 d}+\frac{8 b^4 e^2 (c+d x)^3}{81 d}-\frac{160}{27} b^4 e^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-160*b^4*e^2*x)/27 + (8*b^4*e^2*(c + d*x)^3)/(81*d) + (160*b^3*e^2*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d
*x]))/(27*d) - (8*b^3*e^2*(c + d*x)^2*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x]))/(27*d) - (8*b^2*e^2*(c +
d*x)*(a + b*ArcSinh[c + d*x])^2)/(3*d) + (4*b^2*e^2*(c + d*x)^3*(a + b*ArcSinh[c + d*x])^2)/(9*d) + (8*b*e^2*
Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x])^3)/(9*d) - (4*b*e^2*(c + d*x)^2*Sqrt[1 + (c + d*x)^2]*(a + b*Ar
cSinh[c + d*x])^3)/(9*d) + (e^2*(c + d*x)^3*(a + b*ArcSinh[c + d*x])^4)/(3*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 5653

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

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)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^4 \, dx &=\frac{\operatorname{Subst}\left (\int e^2 x^2 \left (a+b \sinh ^{-1}(x)\right )^4 \, dx,x,c+d x\right )}{d}\\ &=\frac{e^2 \operatorname{Subst}\left (\int x^2 \left (a+b \sinh ^{-1}(x)\right )^4 \, dx,x,c+d x\right )}{d}\\ &=\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^4}{3 d}-\frac{\left (4 b e^2\right ) \operatorname{Subst}\left (\int \frac{x^3 \left (a+b \sinh ^{-1}(x)\right )^3}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{3 d}\\ &=-\frac{4 b e^2 (c+d x)^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{9 d}+\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^4}{3 d}+\frac{\left (8 b e^2\right ) \operatorname{Subst}\left (\int \frac{x \left (a+b \sinh ^{-1}(x)\right )^3}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{9 d}+\frac{\left (4 b^2 e^2\right ) \operatorname{Subst}\left (\int x^2 \left (a+b \sinh ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{3 d}\\ &=\frac{4 b^2 e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{9 d}+\frac{8 b e^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{9 d}-\frac{4 b e^2 (c+d x)^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{9 d}+\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^4}{3 d}-\frac{\left (8 b^2 e^2\right ) \operatorname{Subst}\left (\int \left (a+b \sinh ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{3 d}-\frac{\left (8 b^3 e^2\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 )}{9 d}\\ &=-\frac{8 b^3 e^2 (c+d x)^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )}{27 d}-\frac{8 b^2 e^2 (c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d}+\frac{4 b^2 e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{9 d}+\frac{8 b e^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{9 d}-\frac{4 b e^2 (c+d x)^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{9 d}+\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^4}{3 d}+\frac{\left (16 b^3 e^2\right ) \operatorname{Subst}\left (\int \frac{x \left (a+b \sinh ^{-1}(x)\right )}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{27 d}+\frac{\left (16 b^3 e^2\right ) \operatorname{Subst}\left (\int \frac{x \left (a+b \sinh ^{-1}(x)\right )}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{3 d}+\frac{\left (8 b^4 e^2\right ) \operatorname{Subst}\left (\int x^2 \, dx,x,c+d x\right )}{27 d}\\ &=\frac{8 b^4 e^2 (c+d x)^3}{81 d}+\frac{160 b^3 e^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )}{27 d}-\frac{8 b^3 e^2 (c+d x)^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )}{27 d}-\frac{8 b^2 e^2 (c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d}+\frac{4 b^2 e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{9 d}+\frac{8 b e^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{9 d}-\frac{4 b e^2 (c+d x)^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{9 d}+\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^4}{3 d}-\frac{\left (16 b^4 e^2\right ) \operatorname{Subst}(\int 1 \, dx,x,c+d x)}{27 d}-\frac{\left (16 b^4 e^2\right ) \operatorname{Subst}(\int 1 \, dx,x,c+d x)}{3 d}\\ &=-\frac{160}{27} b^4 e^2 x+\frac{8 b^4 e^2 (c+d x)^3}{81 d}+\frac{160 b^3 e^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )}{27 d}-\frac{8 b^3 e^2 (c+d x)^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )}{27 d}-\frac{8 b^2 e^2 (c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d}+\frac{4 b^2 e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{9 d}+\frac{8 b e^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{9 d}-\frac{4 b e^2 (c+d x)^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{9 d}+\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^4}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.441229, size = 412, normalized size = 1.47 \[ \frac{e^2 \left (\left (36 a^2 b^2+27 a^4+8 b^4\right ) (c+d x)^3-24 b^2 \left (9 a^2+20 b^2\right ) (c+d x)+12 a b \sqrt{(c+d x)^2+1} \left (-\left (3 a^2+2 b^2\right ) (c+d x)^2+6 a^2+40 b^2\right )+18 b^2 \sinh ^{-1}(c+d x)^2 \left (9 a^2 (c+d x)^3-6 a b \sqrt{(c+d x)^2+1} (c+d x)^2+12 a b \sqrt{(c+d x)^2+1}+2 b^2 (c+d x)^3-12 b^2 (c+d x)\right )+12 b \sinh ^{-1}(c+d x) \left (18 a^2 b \sqrt{(c+d x)^2+1}-9 a^2 b (c+d x)^2 \sqrt{(c+d x)^2+1}+9 a^3 (c+d x)^3+6 a b^2 (c+d x)^3-36 a b^2 (c+d x)-2 b^3 (c+d x)^2 \sqrt{(c+d x)^2+1}+40 b^3 \sqrt{(c+d x)^2+1}\right )-36 b^3 \sinh ^{-1}(c+d x)^3 \left (-3 a (c+d x)^3+b \sqrt{(c+d x)^2+1} (c+d x)^2-2 b \sqrt{(c+d x)^2+1}\right )+27 b^4 (c+d x)^3 \sinh ^{-1}(c+d x)^4\right )}{81 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^2*(-24*b^2*(9*a^2 + 20*b^2)*(c + d*x) + (27*a^4 + 36*a^2*b^2 + 8*b^4)*(c + d*x)^3 + 12*a*b*Sqrt[1 + (c + d*
x)^2]*(6*a^2 + 40*b^2 - (3*a^2 + 2*b^2)*(c + d*x)^2) + 12*b*(-36*a*b^2*(c + d*x) + 9*a^3*(c + d*x)^3 + 6*a*b^2
*(c + d*x)^3 + 18*a^2*b*Sqrt[1 + (c + d*x)^2] + 40*b^3*Sqrt[1 + (c + d*x)^2] - 9*a^2*b*(c + d*x)^2*Sqrt[1 + (c
+ d*x)^2] - 2*b^3*(c + d*x)^2*Sqrt[1 + (c + d*x)^2])*ArcSinh[c + d*x] + 18*b^2*(-12*b^2*(c + d*x) + 9*a^2*(c
+ d*x)^3 + 2*b^2*(c + d*x)^3 + 12*a*b*Sqrt[1 + (c + d*x)^2] - 6*a*b*(c + d*x)^2*Sqrt[1 + (c + d*x)^2])*ArcSinh
[c + d*x]^2 - 36*b^3*(-3*a*(c + d*x)^3 - 2*b*Sqrt[1 + (c + d*x)^2] + b*(c + d*x)^2*Sqrt[1 + (c + d*x)^2])*ArcS
inh[c + d*x]^3 + 27*b^4*(c + d*x)^3*ArcSinh[c + d*x]^4))/(81*d)

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Maple [B]  time = 0.04, size = 567, normalized size = 2. \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)^2*(a+b*arcsinh(d*x+c))^4,x)

[Out]

1/d*(1/3*(d*x+c)^3*e^2*a^4+e^2*b^4*(1/3*(d*x+c)*arcsinh(d*x+c)^4*(1+(d*x+c)^2)-1/3*arcsinh(d*x+c)^4*(d*x+c)-4/
9*arcsinh(d*x+c)^3*(d*x+c)^2*(1+(d*x+c)^2)^(1/2)+8/9*arcsinh(d*x+c)^3*(1+(d*x+c)^2)^(1/2)+4/9*arcsinh(d*x+c)^2
*(d*x+c)*(1+(d*x+c)^2)-28/9*arcsinh(d*x+c)^2*(d*x+c)-8/27*arcsinh(d*x+c)*(d*x+c)^2*(1+(d*x+c)^2)^(1/2)+160/27*
arcsinh(d*x+c)*(1+(d*x+c)^2)^(1/2)+8/81*(1+(d*x+c)^2)*(d*x+c)-488/81*d*x-488/81*c)+4*e^2*a*b^3*(1/3*arcsinh(d*
x+c)^3*(d*x+c)*(1+(d*x+c)^2)-1/3*arcsinh(d*x+c)^3*(d*x+c)-1/3*arcsinh(d*x+c)^2*(d*x+c)^2*(1+(d*x+c)^2)^(1/2)+2
/3*arcsinh(d*x+c)^2*(1+(d*x+c)^2)^(1/2)+2/9*(d*x+c)*(1+(d*x+c)^2)*arcsinh(d*x+c)-14/9*(d*x+c)*arcsinh(d*x+c)-2
/27*(d*x+c)^2*(1+(d*x+c)^2)^(1/2)+40/27*(1+(d*x+c)^2)^(1/2))+6*e^2*a^2*b^2*(1/3*arcsinh(d*x+c)^2*(d*x+c)*(1+(d
*x+c)^2)-1/3*arcsinh(d*x+c)^2*(d*x+c)-2/9*arcsinh(d*x+c)*(d*x+c)^2*(1+(d*x+c)^2)^(1/2)+4/9*arcsinh(d*x+c)*(1+(
d*x+c)^2)^(1/2)+2/27*(1+(d*x+c)^2)*(d*x+c)-14/27*d*x-14/27*c)+4*e^2*a^3*b*(1/3*(d*x+c)^3*arcsinh(d*x+c)-1/9*(d
*x+c)^2*(1+(d*x+c)^2)^(1/2)+2/9*(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)^2*(a+b*arcsinh(d*x+c))^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.99754, size = 1894, normalized size = 6.74 \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*arcsinh(d*x+c))^4,x, algorithm="fricas")

[Out]

1/81*((27*a^4 + 36*a^2*b^2 + 8*b^4)*d^3*e^2*x^3 + 3*(27*a^4 + 36*a^2*b^2 + 8*b^4)*c*d^2*e^2*x^2 - 3*(72*a^2*b^
2 + 160*b^4 - (27*a^4 + 36*a^2*b^2 + 8*b^4)*c^2)*d*e^2*x + 27*(b^4*d^3*e^2*x^3 + 3*b^4*c*d^2*e^2*x^2 + 3*b^4*c
^2*d*e^2*x + b^4*c^3*e^2)*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))^4 + 36*(3*a*b^3*d^3*e^2*x^3 + 9*a*b
^3*c*d^2*e^2*x^2 + 9*a*b^3*c^2*d*e^2*x + 3*a*b^3*c^3*e^2 - (b^4*d^2*e^2*x^2 + 2*b^4*c*d*e^2*x + (b^4*c^2 - 2*b
^4)*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))^3 + 18*((9*a^2*b^
2 + 2*b^4)*d^3*e^2*x^3 + 3*(9*a^2*b^2 + 2*b^4)*c*d^2*e^2*x^2 - 3*(4*b^4 - (9*a^2*b^2 + 2*b^4)*c^2)*d*e^2*x - (
12*b^4*c - (9*a^2*b^2 + 2*b^4)*c^3)*e^2 - 6*(a*b^3*d^2*e^2*x^2 + 2*a*b^3*c*d*e^2*x + (a*b^3*c^2 - 2*a*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 + 12*(3*(3*a^3*b + 2*a
*b^3)*d^3*e^2*x^3 + 9*(3*a^3*b + 2*a*b^3)*c*d^2*e^2*x^2 - 9*(4*a*b^3 - (3*a^3*b + 2*a*b^3)*c^2)*d*e^2*x - 3*(1
2*a*b^3*c - (3*a^3*b + 2*a*b^3)*c^3)*e^2 - ((9*a^2*b^2 + 2*b^4)*d^2*e^2*x^2 + 2*(9*a^2*b^2 + 2*b^4)*c*d*e^2*x
- (18*a^2*b^2 + 40*b^4 - (9*a^2*b^2 + 2*b^4)*c^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)) - 12*((3*a^3*b + 2*a*b^3)*d^2*e^2*x^2 + 2*(3*a^3*b + 2*a*b^3)*c*d*e^2*x - (6*a^3*
b + 40*a*b^3 - (3*a^3*b + 2*a*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 = 13.9048, size = 1889, normalized size = 6.72 \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*asinh(d*x+c))**4,x)

[Out]

Piecewise((a**4*c**2*e**2*x + a**4*c*d*e**2*x**2 + a**4*d**2*e**2*x**3/3 + 4*a**3*b*c**3*e**2*asinh(c + d*x)/(
3*d) + 4*a**3*b*c**2*e**2*x*asinh(c + d*x) - 4*a**3*b*c**2*e**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/(9*d) + 4
*a**3*b*c*d*e**2*x**2*asinh(c + d*x) - 8*a**3*b*c*e**2*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/9 + 4*a**3*b*d**
2*e**2*x**3*asinh(c + d*x)/3 - 4*a**3*b*d*e**2*x**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/9 + 8*a**3*b*e**2*sqr
t(c**2 + 2*c*d*x + d**2*x**2 + 1)/(9*d) + 2*a**2*b**2*c**3*e**2*asinh(c + d*x)**2/d + 6*a**2*b**2*c**2*e**2*x*
asinh(c + d*x)**2 + 4*a**2*b**2*c**2*e**2*x/3 - 4*a**2*b**2*c**2*e**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asi
nh(c + d*x)/(3*d) + 6*a**2*b**2*c*d*e**2*x**2*asinh(c + d*x)**2 + 4*a**2*b**2*c*d*e**2*x**2/3 - 8*a**2*b**2*c*
e**2*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)/3 + 2*a**2*b**2*d**2*e**2*x**3*asinh(c + d*x)**2 +
4*a**2*b**2*d**2*e**2*x**3/9 - 4*a**2*b**2*d*e**2*x**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)/3 -
8*a**2*b**2*e**2*x/3 + 8*a**2*b**2*e**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)/(3*d) + 4*a*b**3*
c**3*e**2*asinh(c + d*x)**3/(3*d) + 8*a*b**3*c**3*e**2*asinh(c + d*x)/(9*d) + 4*a*b**3*c**2*e**2*x*asinh(c + d
*x)**3 + 8*a*b**3*c**2*e**2*x*asinh(c + d*x)/3 - 4*a*b**3*c**2*e**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh
(c + d*x)**2/(3*d) - 8*a*b**3*c**2*e**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/(27*d) + 4*a*b**3*c*d*e**2*x**2*a
sinh(c + d*x)**3 + 8*a*b**3*c*d*e**2*x**2*asinh(c + d*x)/3 - 8*a*b**3*c*e**2*x*sqrt(c**2 + 2*c*d*x + d**2*x**2
+ 1)*asinh(c + d*x)**2/3 - 16*a*b**3*c*e**2*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/27 - 16*a*b**3*c*e**2*asin
h(c + d*x)/(3*d) + 4*a*b**3*d**2*e**2*x**3*asinh(c + d*x)**3/3 + 8*a*b**3*d**2*e**2*x**3*asinh(c + d*x)/9 - 4*
a*b**3*d*e**2*x**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)**2/3 - 8*a*b**3*d*e**2*x**2*sqrt(c**2 +
2*c*d*x + d**2*x**2 + 1)/27 - 16*a*b**3*e**2*x*asinh(c + d*x)/3 + 8*a*b**3*e**2*sqrt(c**2 + 2*c*d*x + d**2*x*
*2 + 1)*asinh(c + d*x)**2/(3*d) + 160*a*b**3*e**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/(27*d) + b**4*c**3*e**2
*asinh(c + d*x)**4/(3*d) + 4*b**4*c**3*e**2*asinh(c + d*x)**2/(9*d) + b**4*c**2*e**2*x*asinh(c + d*x)**4 + 4*b
**4*c**2*e**2*x*asinh(c + d*x)**2/3 + 8*b**4*c**2*e**2*x/27 - 4*b**4*c**2*e**2*sqrt(c**2 + 2*c*d*x + d**2*x**2
+ 1)*asinh(c + d*x)**3/(9*d) - 8*b**4*c**2*e**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)/(27*d) +
b**4*c*d*e**2*x**2*asinh(c + d*x)**4 + 4*b**4*c*d*e**2*x**2*asinh(c + d*x)**2/3 + 8*b**4*c*d*e**2*x**2/27 - 8*
b**4*c*e**2*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)**3/9 - 16*b**4*c*e**2*x*sqrt(c**2 + 2*c*d*x
+ d**2*x**2 + 1)*asinh(c + d*x)/27 - 8*b**4*c*e**2*asinh(c + d*x)**2/(3*d) + b**4*d**2*e**2*x**3*asinh(c + d*x
)**4/3 + 4*b**4*d**2*e**2*x**3*asinh(c + d*x)**2/9 + 8*b**4*d**2*e**2*x**3/81 - 4*b**4*d*e**2*x**2*sqrt(c**2 +
2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)**3/9 - 8*b**4*d*e**2*x**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(
c + d*x)/27 - 8*b**4*e**2*x*asinh(c + d*x)**2/3 - 160*b**4*e**2*x/27 + 8*b**4*e**2*sqrt(c**2 + 2*c*d*x + d**2*
x**2 + 1)*asinh(c + d*x)**3/(9*d) + 160*b**4*e**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)/(27*d),
Ne(d, 0)), (c**2*e**2*x*(a + b*asinh(c))**4, 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{arsinh}\left (d x + c\right ) + a\right )}^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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