3.138 \(\int (c e+d e x)^3 (a+b \sinh ^{-1}(c+d x))^3 \, dx\)

Optimal. Leaf size=279 \[ \frac{3 b^2 e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )}{32 d}-\frac{9 b^2 e^3 (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )}{32 d}+\frac{e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{4 d}-\frac{3 b e^3 \sqrt{(c+d x)^2+1} (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{16 d}+\frac{9 b e^3 \sqrt{(c+d x)^2+1} (c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{32 d}-\frac{3 e^3 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{32 d}-\frac{3 b^3 e^3 \sqrt{(c+d x)^2+1} (c+d x)^3}{128 d}+\frac{45 b^3 e^3 \sqrt{(c+d x)^2+1} (c+d x)}{256 d}-\frac{45 b^3 e^3 \sinh ^{-1}(c+d x)}{256 d} \]

[Out]

(45*b^3*e^3*(c + d*x)*Sqrt[1 + (c + d*x)^2])/(256*d) - (3*b^3*e^3*(c + d*x)^3*Sqrt[1 + (c + d*x)^2])/(128*d) -
(45*b^3*e^3*ArcSinh[c + d*x])/(256*d) - (9*b^2*e^3*(c + d*x)^2*(a + b*ArcSinh[c + d*x]))/(32*d) + (3*b^2*e^3*
(c + d*x)^4*(a + b*ArcSinh[c + d*x]))/(32*d) + (9*b*e^3*(c + d*x)*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x
])^2)/(32*d) - (3*b*e^3*(c + d*x)^3*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x])^2)/(16*d) - (3*e^3*(a + b*A
rcSinh[c + d*x])^3)/(32*d) + (e^3*(c + d*x)^4*(a + b*ArcSinh[c + d*x])^3)/(4*d)

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Rubi [A]  time = 0.385716, antiderivative size = 279, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {5865, 12, 5661, 5758, 5675, 321, 215} \[ \frac{3 b^2 e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )}{32 d}-\frac{9 b^2 e^3 (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )}{32 d}+\frac{e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{4 d}-\frac{3 b e^3 \sqrt{(c+d x)^2+1} (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{16 d}+\frac{9 b e^3 \sqrt{(c+d x)^2+1} (c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{32 d}-\frac{3 e^3 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{32 d}-\frac{3 b^3 e^3 \sqrt{(c+d x)^2+1} (c+d x)^3}{128 d}+\frac{45 b^3 e^3 \sqrt{(c+d x)^2+1} (c+d x)}{256 d}-\frac{45 b^3 e^3 \sinh ^{-1}(c+d x)}{256 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(45*b^3*e^3*(c + d*x)*Sqrt[1 + (c + d*x)^2])/(256*d) - (3*b^3*e^3*(c + d*x)^3*Sqrt[1 + (c + d*x)^2])/(128*d) -
(45*b^3*e^3*ArcSinh[c + d*x])/(256*d) - (9*b^2*e^3*(c + d*x)^2*(a + b*ArcSinh[c + d*x]))/(32*d) + (3*b^2*e^3*
(c + d*x)^4*(a + b*ArcSinh[c + d*x]))/(32*d) + (9*b*e^3*(c + d*x)*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x
])^2)/(32*d) - (3*b*e^3*(c + d*x)^3*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x])^2)/(16*d) - (3*e^3*(a + b*A
rcSinh[c + d*x])^3)/(32*d) + (e^3*(c + d*x)^4*(a + b*ArcSinh[c + d*x])^3)/(4*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 5675

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

Rule 321

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

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rubi steps

\begin{align*} \int (c e+d e x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^3 \, dx &=\frac{\operatorname{Subst}\left (\int e^3 x^3 \left (a+b \sinh ^{-1}(x)\right )^3 \, dx,x,c+d x\right )}{d}\\ &=\frac{e^3 \operatorname{Subst}\left (\int x^3 \left (a+b \sinh ^{-1}(x)\right )^3 \, dx,x,c+d x\right )}{d}\\ &=\frac{e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{4 d}-\frac{\left (3 b e^3\right ) \operatorname{Subst}\left (\int \frac{x^4 \left (a+b \sinh ^{-1}(x)\right )^2}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{4 d}\\ &=-\frac{3 b e^3 (c+d x)^3 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{16 d}+\frac{e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{4 d}+\frac{\left (9 b e^3\right ) \operatorname{Subst}\left (\int \frac{x^2 \left (a+b \sinh ^{-1}(x)\right )^2}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{16 d}+\frac{\left (3 b^2 e^3\right ) \operatorname{Subst}\left (\int x^3 \left (a+b \sinh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{8 d}\\ &=\frac{3 b^2 e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )}{32 d}+\frac{9 b e^3 (c+d x) \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{32 d}-\frac{3 b e^3 (c+d x)^3 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{16 d}+\frac{e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{4 d}-\frac{\left (9 b e^3\right ) \operatorname{Subst}\left (\int \frac{\left (a+b \sinh ^{-1}(x)\right )^2}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{32 d}-\frac{\left (9 b^2 e^3\right ) \operatorname{Subst}\left (\int x \left (a+b \sinh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{16 d}-\frac{\left (3 b^3 e^3\right ) \operatorname{Subst}\left (\int \frac{x^4}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{32 d}\\ &=-\frac{3 b^3 e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{128 d}-\frac{9 b^2 e^3 (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )}{32 d}+\frac{3 b^2 e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )}{32 d}+\frac{9 b e^3 (c+d x) \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{32 d}-\frac{3 b e^3 (c+d x)^3 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{16 d}-\frac{3 e^3 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{32 d}+\frac{e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{4 d}+\frac{\left (9 b^3 e^3\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{128 d}+\frac{\left (9 b^3 e^3\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{32 d}\\ &=\frac{45 b^3 e^3 (c+d x) \sqrt{1+(c+d x)^2}}{256 d}-\frac{3 b^3 e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{128 d}-\frac{9 b^2 e^3 (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )}{32 d}+\frac{3 b^2 e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )}{32 d}+\frac{9 b e^3 (c+d x) \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{32 d}-\frac{3 b e^3 (c+d x)^3 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{16 d}-\frac{3 e^3 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{32 d}+\frac{e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{4 d}-\frac{\left (9 b^3 e^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{256 d}-\frac{\left (9 b^3 e^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{64 d}\\ &=\frac{45 b^3 e^3 (c+d x) \sqrt{1+(c+d x)^2}}{256 d}-\frac{3 b^3 e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{128 d}-\frac{45 b^3 e^3 \sinh ^{-1}(c+d x)}{256 d}-\frac{9 b^2 e^3 (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )}{32 d}+\frac{3 b^2 e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )}{32 d}+\frac{9 b e^3 (c+d x) \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{32 d}-\frac{3 b e^3 (c+d x)^3 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{16 d}-\frac{3 e^3 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{32 d}+\frac{e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.383546, size = 303, normalized size = 1.09 \[ \frac{e^3 \left (8 a \left (8 a^2+3 b^2\right ) (c+d x)^4+3 b \sqrt{(c+d x)^2+1} (c+d x) \left (3 \left (8 a^2+5 b^2\right )-2 \left (8 a^2+b^2\right ) (c+d x)^2\right )-24 b (c+d x) \sinh ^{-1}(c+d x) \left (-8 a^2 (c+d x)^3+4 a b \sqrt{(c+d x)^2+1} (c+d x)^2-6 a b \sqrt{(c+d x)^2+1}-b^2 (c+d x)^3+3 b^2 (c+d x)\right )-9 b \left (8 a^2+5 b^2\right ) \sinh ^{-1}(c+d x)-72 a b^2 (c+d x)^2+24 b^2 \sinh ^{-1}(c+d x)^2 \left (8 a (c+d x)^4-3 a-2 b \sqrt{(c+d x)^2+1} (c+d x)^3+3 b \sqrt{(c+d x)^2+1} (c+d x)\right )+8 b^3 \left (8 (c+d x)^4-3\right ) \sinh ^{-1}(c+d x)^3\right )}{256 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^3*(-72*a*b^2*(c + d*x)^2 + 8*a*(8*a^2 + 3*b^2)*(c + d*x)^4 + 3*b*(c + d*x)*Sqrt[1 + (c + d*x)^2]*(3*(8*a^2
+ 5*b^2) - 2*(8*a^2 + b^2)*(c + d*x)^2) - 9*b*(8*a^2 + 5*b^2)*ArcSinh[c + d*x] - 24*b*(c + d*x)*(3*b^2*(c + d*
x) - 8*a^2*(c + d*x)^3 - b^2*(c + d*x)^3 - 6*a*b*Sqrt[1 + (c + d*x)^2] + 4*a*b*(c + d*x)^2*Sqrt[1 + (c + d*x)^
2])*ArcSinh[c + d*x] + 24*b^2*(-3*a + 8*a*(c + d*x)^4 + 3*b*(c + d*x)*Sqrt[1 + (c + d*x)^2] - 2*b*(c + d*x)^3*
Sqrt[1 + (c + d*x)^2])*ArcSinh[c + d*x]^2 + 8*b^3*(-3 + 8*(c + d*x)^4)*ArcSinh[c + d*x]^3))/(256*d)

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Maple [A]  time = 0.041, size = 433, normalized size = 1.6 \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)^3*(a+b*arcsinh(d*x+c))^3,x)

[Out]

1/d*(1/4*(d*x+c)^4*e^3*a^3+e^3*b^3*(1/4*(d*x+c)^2*arcsinh(d*x+c)^3*(1+(d*x+c)^2)-1/4*arcsinh(d*x+c)^3*(1+(d*x+
c)^2)-3/16*arcsinh(d*x+c)^2*(d*x+c)*(1+(d*x+c)^2)^(3/2)+15/32*arcsinh(d*x+c)^2*(1+(d*x+c)^2)^(1/2)*(d*x+c)+5/3
2*arcsinh(d*x+c)^3+3/32*arcsinh(d*x+c)*(d*x+c)^2*(1+(d*x+c)^2)-3/128*(1+(d*x+c)^2)^(3/2)*(d*x+c)+51/256*(d*x+c
)*(1+(d*x+c)^2)^(1/2)+51/256*arcsinh(d*x+c)-3/8*(1+(d*x+c)^2)*arcsinh(d*x+c))+3*e^3*a*b^2*(1/4*(d*x+c)^2*arcsi
nh(d*x+c)^2*(1+(d*x+c)^2)-1/4*arcsinh(d*x+c)^2*(1+(d*x+c)^2)-1/8*arcsinh(d*x+c)*(d*x+c)*(1+(d*x+c)^2)^(3/2)+5/
16*arcsinh(d*x+c)*(1+(d*x+c)^2)^(1/2)*(d*x+c)+5/32*arcsinh(d*x+c)^2+1/32*(d*x+c)^2*(1+(d*x+c)^2)-1/8*(d*x+c)^2
-1/8)+3*e^3*a^2*b*(1/4*(d*x+c)^4*arcsinh(d*x+c)-1/16*(d*x+c)^3*(1+(d*x+c)^2)^(1/2)+3/32*(d*x+c)*(1+(d*x+c)^2)^
(1/2)-3/32*arcsinh(d*x+c)))

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.05081, size = 1773, normalized size = 6.35 \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)^3*(a+b*arcsinh(d*x+c))^3,x, algorithm="fricas")

[Out]

1/256*(8*(8*a^3 + 3*a*b^2)*d^4*e^3*x^4 + 32*(8*a^3 + 3*a*b^2)*c*d^3*e^3*x^3 - 24*(3*a*b^2 - 2*(8*a^3 + 3*a*b^2
)*c^2)*d^2*e^3*x^2 - 16*(9*a*b^2*c - 2*(8*a^3 + 3*a*b^2)*c^3)*d*e^3*x + 8*(8*b^3*d^4*e^3*x^4 + 32*b^3*c*d^3*e^
3*x^3 + 48*b^3*c^2*d^2*e^3*x^2 + 32*b^3*c^3*d*e^3*x + (8*b^3*c^4 - 3*b^3)*e^3)*log(d*x + c + sqrt(d^2*x^2 + 2*
c*d*x + c^2 + 1))^3 + 24*(8*a*b^2*d^4*e^3*x^4 + 32*a*b^2*c*d^3*e^3*x^3 + 48*a*b^2*c^2*d^2*e^3*x^2 + 32*a*b^2*c
^3*d*e^3*x + (8*a*b^2*c^4 - 3*a*b^2)*e^3 - (2*b^3*d^3*e^3*x^3 + 6*b^3*c*d^2*e^3*x^2 + 3*(2*b^3*c^2 - b^3)*d*e^
3*x + (2*b^3*c^3 - 3*b^3*c)*e^3)*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*(8*(8*a^2*b + b^3)*d^4*e^3*x^4 + 32*(8*a^2*b + b^3)*c*d^3*e^3*x^3 - 24*(b^3 - 2*(8*a^2*b + b^3)*c
^2)*d^2*e^3*x^2 - 16*(3*b^3*c - 2*(8*a^2*b + b^3)*c^3)*d*e^3*x - (24*b^3*c^2 - 8*(8*a^2*b + b^3)*c^4 + 24*a^2*
b + 15*b^3)*e^3 - 16*(2*a*b^2*d^3*e^3*x^3 + 6*a*b^2*c*d^2*e^3*x^2 + 3*(2*a*b^2*c^2 - a*b^2)*d*e^3*x + (2*a*b^2
*c^3 - 3*a*b^2*c)*e^3)*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
*(2*(8*a^2*b + b^3)*d^3*e^3*x^3 + 6*(8*a^2*b + b^3)*c*d^2*e^3*x^2 - 3*(8*a^2*b + 5*b^3 - 2*(8*a^2*b + b^3)*c^2
)*d*e^3*x + (2*(8*a^2*b + b^3)*c^3 - 3*(8*a^2*b + 5*b^3)*c)*e^3)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))/d

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Sympy [A]  time = 12.6469, size = 1828, normalized size = 6.55 \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)**3*(a+b*asinh(d*x+c))**3,x)

[Out]

Piecewise((a**3*c**3*e**3*x + 3*a**3*c**2*d*e**3*x**2/2 + a**3*c*d**2*e**3*x**3 + a**3*d**3*e**3*x**4/4 + 3*a*
*2*b*c**4*e**3*asinh(c + d*x)/(4*d) + 3*a**2*b*c**3*e**3*x*asinh(c + d*x) - 3*a**2*b*c**3*e**3*sqrt(c**2 + 2*c
*d*x + d**2*x**2 + 1)/(16*d) + 9*a**2*b*c**2*d*e**3*x**2*asinh(c + d*x)/2 - 9*a**2*b*c**2*e**3*x*sqrt(c**2 + 2
*c*d*x + d**2*x**2 + 1)/16 + 3*a**2*b*c*d**2*e**3*x**3*asinh(c + d*x) - 9*a**2*b*c*d*e**3*x**2*sqrt(c**2 + 2*c
*d*x + d**2*x**2 + 1)/16 + 9*a**2*b*c*e**3*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/(32*d) + 3*a**2*b*d**3*e**3*x*
*4*asinh(c + d*x)/4 - 3*a**2*b*d**2*e**3*x**3*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/16 + 9*a**2*b*e**3*x*sqrt(c
**2 + 2*c*d*x + d**2*x**2 + 1)/32 - 9*a**2*b*e**3*asinh(c + d*x)/(32*d) + 3*a*b**2*c**4*e**3*asinh(c + d*x)**2
/(4*d) + 3*a*b**2*c**3*e**3*x*asinh(c + d*x)**2 + 3*a*b**2*c**3*e**3*x/8 - 3*a*b**2*c**3*e**3*sqrt(c**2 + 2*c*
d*x + d**2*x**2 + 1)*asinh(c + d*x)/(8*d) + 9*a*b**2*c**2*d*e**3*x**2*asinh(c + d*x)**2/2 + 9*a*b**2*c**2*d*e*
*3*x**2/16 - 9*a*b**2*c**2*e**3*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)/8 + 3*a*b**2*c*d**2*e**3
*x**3*asinh(c + d*x)**2 + 3*a*b**2*c*d**2*e**3*x**3/8 - 9*a*b**2*c*d*e**3*x**2*sqrt(c**2 + 2*c*d*x + d**2*x**2
+ 1)*asinh(c + d*x)/8 - 9*a*b**2*c*e**3*x/16 + 9*a*b**2*c*e**3*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c +
d*x)/(16*d) + 3*a*b**2*d**3*e**3*x**4*asinh(c + d*x)**2/4 + 3*a*b**2*d**3*e**3*x**4/32 - 3*a*b**2*d**2*e**3*x
**3*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)/8 - 9*a*b**2*d*e**3*x**2/32 + 9*a*b**2*e**3*x*sqrt(c**
2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)/16 - 9*a*b**2*e**3*asinh(c + d*x)**2/(32*d) + b**3*c**4*e**3*asinh
(c + d*x)**3/(4*d) + 3*b**3*c**4*e**3*asinh(c + d*x)/(32*d) + b**3*c**3*e**3*x*asinh(c + d*x)**3 + 3*b**3*c**3
*e**3*x*asinh(c + d*x)/8 - 3*b**3*c**3*e**3*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)**2/(16*d) - 3*
b**3*c**3*e**3*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/(128*d) + 3*b**3*c**2*d*e**3*x**2*asinh(c + d*x)**3/2 + 9*
b**3*c**2*d*e**3*x**2*asinh(c + d*x)/16 - 9*b**3*c**2*e**3*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*
x)**2/16 - 9*b**3*c**2*e**3*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/128 - 9*b**3*c**2*e**3*asinh(c + d*x)/(32*d
) + b**3*c*d**2*e**3*x**3*asinh(c + d*x)**3 + 3*b**3*c*d**2*e**3*x**3*asinh(c + d*x)/8 - 9*b**3*c*d*e**3*x**2*
sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)**2/16 - 9*b**3*c*d*e**3*x**2*sqrt(c**2 + 2*c*d*x + d**2*x*
*2 + 1)/128 - 9*b**3*c*e**3*x*asinh(c + d*x)/16 + 9*b**3*c*e**3*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c +
d*x)**2/(32*d) + 45*b**3*c*e**3*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/(256*d) + b**3*d**3*e**3*x**4*asinh(c +
d*x)**3/4 + 3*b**3*d**3*e**3*x**4*asinh(c + d*x)/32 - 3*b**3*d**2*e**3*x**3*sqrt(c**2 + 2*c*d*x + d**2*x**2 +
1)*asinh(c + d*x)**2/16 - 3*b**3*d**2*e**3*x**3*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/128 - 9*b**3*d*e**3*x**2*
asinh(c + d*x)/32 + 9*b**3*e**3*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)**2/32 + 45*b**3*e**3*x*s
qrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/256 - 3*b**3*e**3*asinh(c + d*x)**3/(32*d) - 45*b**3*e**3*asinh(c + d*x)/(
256*d), Ne(d, 0)), (c**3*e**3*x*(a + b*asinh(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 )}^{3}{\left (b \operatorname{arsinh}\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)^3*(a+b*arcsinh(d*x+c))^3,x, algorithm="giac")

[Out]

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