3.75 \(\int x^2 \sinh ^{-1}(a+b x)^3 \, dx\)

Optimal. Leaf size=355 \[ -\frac{6 a^2 \sqrt{(a+b x)^2+1}}{b^3}+\frac{6 a^2 (a+b x) \sinh ^{-1}(a+b x)}{b^3}+\frac{a^3 \sinh ^{-1}(a+b x)^3}{3 b^3}-\frac{3 a^2 \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)^2}{b^3}+\frac{3 a \sqrt{(a+b x)^2+1} (a+b x)}{4 b^3}-\frac{2 \left ((a+b x)^2+1\right )^{3/2}}{27 b^3}+\frac{14 \sqrt{(a+b x)^2+1}}{9 b^3}+\frac{2 (a+b x)^3 \sinh ^{-1}(a+b x)}{9 b^3}-\frac{\sqrt{(a+b x)^2+1} (a+b x)^2 \sinh ^{-1}(a+b x)^2}{3 b^3}-\frac{3 a (a+b x)^2 \sinh ^{-1}(a+b x)}{2 b^3}+\frac{3 a \sqrt{(a+b x)^2+1} (a+b x) \sinh ^{-1}(a+b x)^2}{2 b^3}-\frac{4 (a+b x) \sinh ^{-1}(a+b x)}{3 b^3}-\frac{a \sinh ^{-1}(a+b x)^3}{2 b^3}+\frac{2 \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)^2}{3 b^3}-\frac{3 a \sinh ^{-1}(a+b x)}{4 b^3}+\frac{1}{3} x^3 \sinh ^{-1}(a+b x)^3 \]

[Out]

(14*Sqrt[1 + (a + b*x)^2])/(9*b^3) - (6*a^2*Sqrt[1 + (a + b*x)^2])/b^3 + (3*a*(a + b*x)*Sqrt[1 + (a + b*x)^2])
/(4*b^3) - (2*(1 + (a + b*x)^2)^(3/2))/(27*b^3) - (3*a*ArcSinh[a + b*x])/(4*b^3) - (4*(a + b*x)*ArcSinh[a + b*
x])/(3*b^3) + (6*a^2*(a + b*x)*ArcSinh[a + b*x])/b^3 - (3*a*(a + b*x)^2*ArcSinh[a + b*x])/(2*b^3) + (2*(a + b*
x)^3*ArcSinh[a + b*x])/(9*b^3) + (2*Sqrt[1 + (a + b*x)^2]*ArcSinh[a + b*x]^2)/(3*b^3) - (3*a^2*Sqrt[1 + (a + b
*x)^2]*ArcSinh[a + b*x]^2)/b^3 + (3*a*(a + b*x)*Sqrt[1 + (a + b*x)^2]*ArcSinh[a + b*x]^2)/(2*b^3) - ((a + b*x)
^2*Sqrt[1 + (a + b*x)^2]*ArcSinh[a + b*x]^2)/(3*b^3) - (a*ArcSinh[a + b*x]^3)/(2*b^3) + (a^3*ArcSinh[a + b*x]^
3)/(3*b^3) + (x^3*ArcSinh[a + b*x]^3)/3

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Rubi [A]  time = 0.449703, antiderivative size = 355, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 11, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.917, Rules used = {5865, 5801, 5831, 3317, 3296, 2638, 3311, 30, 2635, 8, 2633} \[ -\frac{6 a^2 \sqrt{(a+b x)^2+1}}{b^3}+\frac{6 a^2 (a+b x) \sinh ^{-1}(a+b x)}{b^3}+\frac{a^3 \sinh ^{-1}(a+b x)^3}{3 b^3}-\frac{3 a^2 \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)^2}{b^3}+\frac{3 a \sqrt{(a+b x)^2+1} (a+b x)}{4 b^3}-\frac{2 \left ((a+b x)^2+1\right )^{3/2}}{27 b^3}+\frac{14 \sqrt{(a+b x)^2+1}}{9 b^3}+\frac{2 (a+b x)^3 \sinh ^{-1}(a+b x)}{9 b^3}-\frac{\sqrt{(a+b x)^2+1} (a+b x)^2 \sinh ^{-1}(a+b x)^2}{3 b^3}-\frac{3 a (a+b x)^2 \sinh ^{-1}(a+b x)}{2 b^3}+\frac{3 a \sqrt{(a+b x)^2+1} (a+b x) \sinh ^{-1}(a+b x)^2}{2 b^3}-\frac{4 (a+b x) \sinh ^{-1}(a+b x)}{3 b^3}-\frac{a \sinh ^{-1}(a+b x)^3}{2 b^3}+\frac{2 \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)^2}{3 b^3}-\frac{3 a \sinh ^{-1}(a+b x)}{4 b^3}+\frac{1}{3} x^3 \sinh ^{-1}(a+b x)^3 \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcSinh[a + b*x]^3,x]

[Out]

(14*Sqrt[1 + (a + b*x)^2])/(9*b^3) - (6*a^2*Sqrt[1 + (a + b*x)^2])/b^3 + (3*a*(a + b*x)*Sqrt[1 + (a + b*x)^2])
/(4*b^3) - (2*(1 + (a + b*x)^2)^(3/2))/(27*b^3) - (3*a*ArcSinh[a + b*x])/(4*b^3) - (4*(a + b*x)*ArcSinh[a + b*
x])/(3*b^3) + (6*a^2*(a + b*x)*ArcSinh[a + b*x])/b^3 - (3*a*(a + b*x)^2*ArcSinh[a + b*x])/(2*b^3) + (2*(a + b*
x)^3*ArcSinh[a + b*x])/(9*b^3) + (2*Sqrt[1 + (a + b*x)^2]*ArcSinh[a + b*x]^2)/(3*b^3) - (3*a^2*Sqrt[1 + (a + b
*x)^2]*ArcSinh[a + b*x]^2)/b^3 + (3*a*(a + b*x)*Sqrt[1 + (a + b*x)^2]*ArcSinh[a + b*x]^2)/(2*b^3) - ((a + b*x)
^2*Sqrt[1 + (a + b*x)^2]*ArcSinh[a + b*x]^2)/(3*b^3) - (a*ArcSinh[a + b*x]^3)/(2*b^3) + (a^3*ArcSinh[a + b*x]^
3)/(3*b^3) + (x^3*ArcSinh[a + b*x]^3)/3

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 5801

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

Rule 5831

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

Rule 3317

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

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3311

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*m*(c + d*x)^(m - 1)*(
b*Sin[e + f*x])^n)/(f^2*n^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[(d^2*m*(m - 1))/(f^2*n^2), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[(b*(c + d*x)^m*Cos[e +
f*x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]

Rule 30

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

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

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

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps

\begin{align*} \int x^2 \sinh ^{-1}(a+b x)^3 \, dx &=\frac{\operatorname{Subst}\left (\int \left (-\frac{a}{b}+\frac{x}{b}\right )^2 \sinh ^{-1}(x)^3 \, dx,x,a+b x\right )}{b}\\ &=\frac{1}{3} x^3 \sinh ^{-1}(a+b x)^3-\operatorname{Subst}\left (\int \frac{\left (-\frac{a}{b}+\frac{x}{b}\right )^3 \sinh ^{-1}(x)^2}{\sqrt{1+x^2}} \, dx,x,a+b x\right )\\ &=\frac{1}{3} x^3 \sinh ^{-1}(a+b x)^3-\operatorname{Subst}\left (\int x^2 \left (-\frac{a}{b}+\frac{\sinh (x)}{b}\right )^3 \, dx,x,\sinh ^{-1}(a+b x)\right )\\ &=\frac{1}{3} x^3 \sinh ^{-1}(a+b x)^3-\operatorname{Subst}\left (\int \left (-\frac{a^3 x^2}{b^3}+\frac{3 a^2 x^2 \sinh (x)}{b^3}-\frac{3 a x^2 \sinh ^2(x)}{b^3}+\frac{x^2 \sinh ^3(x)}{b^3}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )\\ &=\frac{a^3 \sinh ^{-1}(a+b x)^3}{3 b^3}+\frac{1}{3} x^3 \sinh ^{-1}(a+b x)^3-\frac{\operatorname{Subst}\left (\int x^2 \sinh ^3(x) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^3}+\frac{(3 a) \operatorname{Subst}\left (\int x^2 \sinh ^2(x) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^3}-\frac{\left (3 a^2\right ) \operatorname{Subst}\left (\int x^2 \sinh (x) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^3}\\ &=-\frac{3 a (a+b x)^2 \sinh ^{-1}(a+b x)}{2 b^3}+\frac{2 (a+b x)^3 \sinh ^{-1}(a+b x)}{9 b^3}-\frac{3 a^2 \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)^2}{b^3}+\frac{3 a (a+b x) \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)^2}{2 b^3}-\frac{(a+b x)^2 \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)^2}{3 b^3}+\frac{a^3 \sinh ^{-1}(a+b x)^3}{3 b^3}+\frac{1}{3} x^3 \sinh ^{-1}(a+b x)^3-\frac{2 \operatorname{Subst}\left (\int \sinh ^3(x) \, dx,x,\sinh ^{-1}(a+b x)\right )}{9 b^3}+\frac{2 \operatorname{Subst}\left (\int x^2 \sinh (x) \, dx,x,\sinh ^{-1}(a+b x)\right )}{3 b^3}-\frac{(3 a) \operatorname{Subst}\left (\int x^2 \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b^3}+\frac{(3 a) \operatorname{Subst}\left (\int \sinh ^2(x) \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b^3}+\frac{\left (6 a^2\right ) \operatorname{Subst}\left (\int x \cosh (x) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^3}\\ &=\frac{3 a (a+b x) \sqrt{1+(a+b x)^2}}{4 b^3}+\frac{6 a^2 (a+b x) \sinh ^{-1}(a+b x)}{b^3}-\frac{3 a (a+b x)^2 \sinh ^{-1}(a+b x)}{2 b^3}+\frac{2 (a+b x)^3 \sinh ^{-1}(a+b x)}{9 b^3}+\frac{2 \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)^2}{3 b^3}-\frac{3 a^2 \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)^2}{b^3}+\frac{3 a (a+b x) \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)^2}{2 b^3}-\frac{(a+b x)^2 \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)^2}{3 b^3}-\frac{a \sinh ^{-1}(a+b x)^3}{2 b^3}+\frac{a^3 \sinh ^{-1}(a+b x)^3}{3 b^3}+\frac{1}{3} x^3 \sinh ^{-1}(a+b x)^3+\frac{2 \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\sqrt{1+(a+b x)^2}\right )}{9 b^3}-\frac{4 \operatorname{Subst}\left (\int x \cosh (x) \, dx,x,\sinh ^{-1}(a+b x)\right )}{3 b^3}-\frac{(3 a) \operatorname{Subst}\left (\int 1 \, dx,x,\sinh ^{-1}(a+b x)\right )}{4 b^3}-\frac{\left (6 a^2\right ) \operatorname{Subst}\left (\int \sinh (x) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^3}\\ &=\frac{2 \sqrt{1+(a+b x)^2}}{9 b^3}-\frac{6 a^2 \sqrt{1+(a+b x)^2}}{b^3}+\frac{3 a (a+b x) \sqrt{1+(a+b x)^2}}{4 b^3}-\frac{2 \left (1+(a+b x)^2\right )^{3/2}}{27 b^3}-\frac{3 a \sinh ^{-1}(a+b x)}{4 b^3}-\frac{4 (a+b x) \sinh ^{-1}(a+b x)}{3 b^3}+\frac{6 a^2 (a+b x) \sinh ^{-1}(a+b x)}{b^3}-\frac{3 a (a+b x)^2 \sinh ^{-1}(a+b x)}{2 b^3}+\frac{2 (a+b x)^3 \sinh ^{-1}(a+b x)}{9 b^3}+\frac{2 \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)^2}{3 b^3}-\frac{3 a^2 \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)^2}{b^3}+\frac{3 a (a+b x) \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)^2}{2 b^3}-\frac{(a+b x)^2 \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)^2}{3 b^3}-\frac{a \sinh ^{-1}(a+b x)^3}{2 b^3}+\frac{a^3 \sinh ^{-1}(a+b x)^3}{3 b^3}+\frac{1}{3} x^3 \sinh ^{-1}(a+b x)^3+\frac{4 \operatorname{Subst}\left (\int \sinh (x) \, dx,x,\sinh ^{-1}(a+b x)\right )}{3 b^3}\\ &=\frac{14 \sqrt{1+(a+b x)^2}}{9 b^3}-\frac{6 a^2 \sqrt{1+(a+b x)^2}}{b^3}+\frac{3 a (a+b x) \sqrt{1+(a+b x)^2}}{4 b^3}-\frac{2 \left (1+(a+b x)^2\right )^{3/2}}{27 b^3}-\frac{3 a \sinh ^{-1}(a+b x)}{4 b^3}-\frac{4 (a+b x) \sinh ^{-1}(a+b x)}{3 b^3}+\frac{6 a^2 (a+b x) \sinh ^{-1}(a+b x)}{b^3}-\frac{3 a (a+b x)^2 \sinh ^{-1}(a+b x)}{2 b^3}+\frac{2 (a+b x)^3 \sinh ^{-1}(a+b x)}{9 b^3}+\frac{2 \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)^2}{3 b^3}-\frac{3 a^2 \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)^2}{b^3}+\frac{3 a (a+b x) \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)^2}{2 b^3}-\frac{(a+b x)^2 \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)^2}{3 b^3}-\frac{a \sinh ^{-1}(a+b x)^3}{2 b^3}+\frac{a^3 \sinh ^{-1}(a+b x)^3}{3 b^3}+\frac{1}{3} x^3 \sinh ^{-1}(a+b x)^3\\ \end{align*}

Mathematica [A]  time = 0.1985, size = 175, normalized size = 0.49 \[ \frac{\left (-575 a^2+65 a b x-8 b^2 x^2+160\right ) \sqrt{a^2+2 a b x+b^2 x^2+1}+18 \left (2 a^3-3 a+2 b^3 x^3\right ) \sinh ^{-1}(a+b x)^3-18 \sqrt{a^2+2 a b x+b^2 x^2+1} \left (11 a^2-5 a b x+2 b^2 x^2-4\right ) \sinh ^{-1}(a+b x)^2+3 \left (132 a^2 b x+170 a^3-15 a \left (2 b^2 x^2+5\right )+8 b x \left (b^2 x^2-6\right )\right ) \sinh ^{-1}(a+b x)}{108 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcSinh[a + b*x]^3,x]

[Out]

((160 - 575*a^2 + 65*a*b*x - 8*b^2*x^2)*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2] + 3*(170*a^3 + 132*a^2*b*x + 8*b*x*(
-6 + b^2*x^2) - 15*a*(5 + 2*b^2*x^2))*ArcSinh[a + b*x] - 18*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]*(-4 + 11*a^2 - 5
*a*b*x + 2*b^2*x^2)*ArcSinh[a + b*x]^2 + 18*(-3*a + 2*a^3 + 2*b^3*x^3)*ArcSinh[a + b*x]^3)/(108*b^3)

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Maple [A]  time = 0.06, size = 326, normalized size = 0.9 \begin{align*}{\frac{1}{{b}^{3}} \left ( -{\frac{a}{4} \left ( 4\, \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{3} \left ( bx+a \right ) ^{2}-6\, \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2}\sqrt{1+ \left ( bx+a \right ) ^{2}} \left ( bx+a \right ) +2\, \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{3}+6\,{\it Arcsinh} \left ( bx+a \right ) \left ( bx+a \right ) ^{2}-3\, \left ( bx+a \right ) \sqrt{1+ \left ( bx+a \right ) ^{2}}+3\,{\it Arcsinh} \left ( bx+a \right ) \right ) }-{\frac{ \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{3} \left ( bx+a \right ) }{3}}+{\frac{ \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{3} \left ( bx+a \right ) \left ( 1+ \left ( bx+a \right ) ^{2} \right ) }{3}}+{\frac{2\, \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2}}{3}\sqrt{1+ \left ( bx+a \right ) ^{2}}}-{\frac{ \left ( 14\,bx+14\,a \right ){\it Arcsinh} \left ( bx+a \right ) }{9}}+{\frac{40}{27}\sqrt{1+ \left ( bx+a \right ) ^{2}}}-{\frac{ \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2} \left ( bx+a \right ) ^{2}}{3}\sqrt{1+ \left ( bx+a \right ) ^{2}}}+{\frac{ \left ( 2\,bx+2\,a \right ) \left ( 1+ \left ( bx+a \right ) ^{2} \right ){\it Arcsinh} \left ( bx+a \right ) }{9}}-{\frac{2\, \left ( bx+a \right ) ^{2}}{27}\sqrt{1+ \left ( bx+a \right ) ^{2}}}+{a}^{2} \left ( \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{3} \left ( bx+a \right ) -3\, \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2}\sqrt{1+ \left ( bx+a \right ) ^{2}}+6\, \left ( bx+a \right ){\it Arcsinh} \left ( bx+a \right ) -6\,\sqrt{1+ \left ( bx+a \right ) ^{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arcsinh(b*x+a)^3,x)

[Out]

1/b^3*(-1/4*a*(4*arcsinh(b*x+a)^3*(b*x+a)^2-6*arcsinh(b*x+a)^2*(1+(b*x+a)^2)^(1/2)*(b*x+a)+2*arcsinh(b*x+a)^3+
6*arcsinh(b*x+a)*(b*x+a)^2-3*(b*x+a)*(1+(b*x+a)^2)^(1/2)+3*arcsinh(b*x+a))-1/3*arcsinh(b*x+a)^3*(b*x+a)+1/3*ar
csinh(b*x+a)^3*(b*x+a)*(1+(b*x+a)^2)+2/3*arcsinh(b*x+a)^2*(1+(b*x+a)^2)^(1/2)-14/9*(b*x+a)*arcsinh(b*x+a)+40/2
7*(1+(b*x+a)^2)^(1/2)-1/3*arcsinh(b*x+a)^2*(b*x+a)^2*(1+(b*x+a)^2)^(1/2)+2/9*(b*x+a)*(1+(b*x+a)^2)*arcsinh(b*x
+a)-2/27*(b*x+a)^2*(1+(b*x+a)^2)^(1/2)+a^2*(arcsinh(b*x+a)^3*(b*x+a)-3*arcsinh(b*x+a)^2*(1+(b*x+a)^2)^(1/2)+6*
(b*x+a)*arcsinh(b*x+a)-6*(1+(b*x+a)^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(x^2*arcsinh(b*x+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.55595, size = 556, normalized size = 1.57 \begin{align*} \frac{18 \,{\left (2 \, b^{3} x^{3} + 2 \, a^{3} - 3 \, a\right )} \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )^{3} - 18 \,{\left (2 \, b^{2} x^{2} - 5 \, a b x + 11 \, a^{2} - 4\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1} \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )^{2} + 3 \,{\left (8 \, b^{3} x^{3} - 30 \, a b^{2} x^{2} + 170 \, a^{3} + 12 \,{\left (11 \, a^{2} - 4\right )} b x - 75 \, a\right )} \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) -{\left (8 \, b^{2} x^{2} - 65 \, a b x + 575 \, a^{2} - 160\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{108 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arcsinh(b*x+a)^3,x, algorithm="fricas")

[Out]

1/108*(18*(2*b^3*x^3 + 2*a^3 - 3*a)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^3 - 18*(2*b^2*x^2 - 5*a*b
*x + 11*a^2 - 4)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^2 + 3*(8*b
^3*x^3 - 30*a*b^2*x^2 + 170*a^3 + 12*(11*a^2 - 4)*b*x - 75*a)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))
- (8*b^2*x^2 - 65*a*b*x + 575*a^2 - 160)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))/b^3

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Sympy [A]  time = 3.91407, size = 432, normalized size = 1.22 \begin{align*} \begin{cases} \frac{a^{3} \operatorname{asinh}^{3}{\left (a + b x \right )}}{3 b^{3}} + \frac{85 a^{3} \operatorname{asinh}{\left (a + b x \right )}}{18 b^{3}} + \frac{11 a^{2} x \operatorname{asinh}{\left (a + b x \right )}}{3 b^{2}} - \frac{11 a^{2} \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname{asinh}^{2}{\left (a + b x \right )}}{6 b^{3}} - \frac{575 a^{2} \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1}}{108 b^{3}} - \frac{5 a x^{2} \operatorname{asinh}{\left (a + b x \right )}}{6 b} + \frac{5 a x \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname{asinh}^{2}{\left (a + b x \right )}}{6 b^{2}} + \frac{65 a x \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1}}{108 b^{2}} - \frac{a \operatorname{asinh}^{3}{\left (a + b x \right )}}{2 b^{3}} - \frac{25 a \operatorname{asinh}{\left (a + b x \right )}}{12 b^{3}} + \frac{x^{3} \operatorname{asinh}^{3}{\left (a + b x \right )}}{3} + \frac{2 x^{3} \operatorname{asinh}{\left (a + b x \right )}}{9} - \frac{x^{2} \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname{asinh}^{2}{\left (a + b x \right )}}{3 b} - \frac{2 x^{2} \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1}}{27 b} - \frac{4 x \operatorname{asinh}{\left (a + b x \right )}}{3 b^{2}} + \frac{2 \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname{asinh}^{2}{\left (a + b x \right )}}{3 b^{3}} + \frac{40 \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1}}{27 b^{3}} & \text{for}\: b \neq 0 \\\frac{x^{3} \operatorname{asinh}^{3}{\left (a \right )}}{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*asinh(b*x+a)**3,x)

[Out]

Piecewise((a**3*asinh(a + b*x)**3/(3*b**3) + 85*a**3*asinh(a + b*x)/(18*b**3) + 11*a**2*x*asinh(a + b*x)/(3*b*
*2) - 11*a**2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*asinh(a + b*x)**2/(6*b**3) - 575*a**2*sqrt(a**2 + 2*a*b*x +
b**2*x**2 + 1)/(108*b**3) - 5*a*x**2*asinh(a + b*x)/(6*b) + 5*a*x*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*asinh(
a + b*x)**2/(6*b**2) + 65*a*x*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(108*b**2) - a*asinh(a + b*x)**3/(2*b**3) -
25*a*asinh(a + b*x)/(12*b**3) + x**3*asinh(a + b*x)**3/3 + 2*x**3*asinh(a + b*x)/9 - x**2*sqrt(a**2 + 2*a*b*x
+ b**2*x**2 + 1)*asinh(a + b*x)**2/(3*b) - 2*x**2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(27*b) - 4*x*asinh(a +
b*x)/(3*b**2) + 2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*asinh(a + b*x)**2/(3*b**3) + 40*sqrt(a**2 + 2*a*b*x +
b**2*x**2 + 1)/(27*b**3), Ne(b, 0)), (x**3*asinh(a)**3/3, True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \operatorname{arsinh}\left (b x + a\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arcsinh(b*x+a)^3,x, algorithm="giac")

[Out]

integrate(x^2*arcsinh(b*x + a)^3, x)