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

Optimal. Leaf size=331 \[ -\frac{2 a^3 x}{b^3}+\frac{3 a^2 (a+b x)^2}{4 b^4}-\frac{a^4 \sinh ^{-1}(a+b x)^2}{4 b^4}+\frac{2 a^3 \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)}{b^4}+\frac{3 a^2 \sinh ^{-1}(a+b x)^2}{4 b^4}-\frac{3 a^2 (a+b x) \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)}{2 b^4}-\frac{2 a (a+b x)^3}{9 b^4}+\frac{4 a x}{3 b^3}+\frac{(a+b x)^4}{32 b^4}-\frac{3 (a+b x)^2}{32 b^4}+\frac{2 a (a+b x)^2 \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)}{3 b^4}-\frac{4 a \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)}{3 b^4}-\frac{3 \sinh ^{-1}(a+b x)^2}{32 b^4}-\frac{(a+b x)^3 \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)}{8 b^4}+\frac{3 (a+b x) \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)}{16 b^4}+\frac{1}{4} x^4 \sinh ^{-1}(a+b x)^2 \]

[Out]

(4*a*x)/(3*b^3) - (2*a^3*x)/b^3 - (3*(a + b*x)^2)/(32*b^4) + (3*a^2*(a + b*x)^2)/(4*b^4) - (2*a*(a + b*x)^3)/(
9*b^4) + (a + b*x)^4/(32*b^4) - (4*a*Sqrt[1 + (a + b*x)^2]*ArcSinh[a + b*x])/(3*b^4) + (2*a^3*Sqrt[1 + (a + b*
x)^2]*ArcSinh[a + b*x])/b^4 + (3*(a + b*x)*Sqrt[1 + (a + b*x)^2]*ArcSinh[a + b*x])/(16*b^4) - (3*a^2*(a + b*x)
*Sqrt[1 + (a + b*x)^2]*ArcSinh[a + b*x])/(2*b^4) + (2*a*(a + b*x)^2*Sqrt[1 + (a + b*x)^2]*ArcSinh[a + b*x])/(3
*b^4) - ((a + b*x)^3*Sqrt[1 + (a + b*x)^2]*ArcSinh[a + b*x])/(8*b^4) - (3*ArcSinh[a + b*x]^2)/(32*b^4) + (3*a^
2*ArcSinh[a + b*x]^2)/(4*b^4) - (a^4*ArcSinh[a + b*x]^2)/(4*b^4) + (x^4*ArcSinh[a + b*x]^2)/4

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Rubi [A]  time = 0.546772, antiderivative size = 331, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 8, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {5865, 5801, 5821, 5675, 5717, 8, 5758, 30} \[ -\frac{2 a^3 x}{b^3}+\frac{3 a^2 (a+b x)^2}{4 b^4}-\frac{a^4 \sinh ^{-1}(a+b x)^2}{4 b^4}+\frac{2 a^3 \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)}{b^4}+\frac{3 a^2 \sinh ^{-1}(a+b x)^2}{4 b^4}-\frac{3 a^2 (a+b x) \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)}{2 b^4}-\frac{2 a (a+b x)^3}{9 b^4}+\frac{4 a x}{3 b^3}+\frac{(a+b x)^4}{32 b^4}-\frac{3 (a+b x)^2}{32 b^4}+\frac{2 a (a+b x)^2 \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)}{3 b^4}-\frac{4 a \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)}{3 b^4}-\frac{3 \sinh ^{-1}(a+b x)^2}{32 b^4}-\frac{(a+b x)^3 \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)}{8 b^4}+\frac{3 (a+b x) \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)}{16 b^4}+\frac{1}{4} x^4 \sinh ^{-1}(a+b x)^2 \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcSinh[a + b*x]^2,x]

[Out]

(4*a*x)/(3*b^3) - (2*a^3*x)/b^3 - (3*(a + b*x)^2)/(32*b^4) + (3*a^2*(a + b*x)^2)/(4*b^4) - (2*a*(a + b*x)^3)/(
9*b^4) + (a + b*x)^4/(32*b^4) - (4*a*Sqrt[1 + (a + b*x)^2]*ArcSinh[a + b*x])/(3*b^4) + (2*a^3*Sqrt[1 + (a + b*
x)^2]*ArcSinh[a + b*x])/b^4 + (3*(a + b*x)*Sqrt[1 + (a + b*x)^2]*ArcSinh[a + b*x])/(16*b^4) - (3*a^2*(a + b*x)
*Sqrt[1 + (a + b*x)^2]*ArcSinh[a + b*x])/(2*b^4) + (2*a*(a + b*x)^2*Sqrt[1 + (a + b*x)^2]*ArcSinh[a + b*x])/(3
*b^4) - ((a + b*x)^3*Sqrt[1 + (a + b*x)^2]*ArcSinh[a + b*x])/(8*b^4) - (3*ArcSinh[a + b*x]^2)/(32*b^4) + (3*a^
2*ArcSinh[a + b*x]^2)/(4*b^4) - (a^4*ArcSinh[a + b*x]^2)/(4*b^4) + (x^4*ArcSinh[a + b*x]^2)/4

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 5821

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

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 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 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 30

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

Rubi steps

\begin{align*} \int x^3 \sinh ^{-1}(a+b x)^2 \, dx &=\frac{\operatorname{Subst}\left (\int \left (-\frac{a}{b}+\frac{x}{b}\right )^3 \sinh ^{-1}(x)^2 \, dx,x,a+b x\right )}{b}\\ &=\frac{1}{4} x^4 \sinh ^{-1}(a+b x)^2-\frac{1}{2} \operatorname{Subst}\left (\int \frac{\left (-\frac{a}{b}+\frac{x}{b}\right )^4 \sinh ^{-1}(x)}{\sqrt{1+x^2}} \, dx,x,a+b x\right )\\ &=\frac{1}{4} x^4 \sinh ^{-1}(a+b x)^2-\frac{1}{2} \operatorname{Subst}\left (\int \left (\frac{a^4 \sinh ^{-1}(x)}{b^4 \sqrt{1+x^2}}-\frac{4 a^3 x \sinh ^{-1}(x)}{b^4 \sqrt{1+x^2}}+\frac{6 a^2 x^2 \sinh ^{-1}(x)}{b^4 \sqrt{1+x^2}}-\frac{4 a x^3 \sinh ^{-1}(x)}{b^4 \sqrt{1+x^2}}+\frac{x^4 \sinh ^{-1}(x)}{b^4 \sqrt{1+x^2}}\right ) \, dx,x,a+b x\right )\\ &=\frac{1}{4} x^4 \sinh ^{-1}(a+b x)^2-\frac{\operatorname{Subst}\left (\int \frac{x^4 \sinh ^{-1}(x)}{\sqrt{1+x^2}} \, dx,x,a+b x\right )}{2 b^4}+\frac{(2 a) \operatorname{Subst}\left (\int \frac{x^3 \sinh ^{-1}(x)}{\sqrt{1+x^2}} \, dx,x,a+b x\right )}{b^4}-\frac{\left (3 a^2\right ) \operatorname{Subst}\left (\int \frac{x^2 \sinh ^{-1}(x)}{\sqrt{1+x^2}} \, dx,x,a+b x\right )}{b^4}+\frac{\left (2 a^3\right ) \operatorname{Subst}\left (\int \frac{x \sinh ^{-1}(x)}{\sqrt{1+x^2}} \, dx,x,a+b x\right )}{b^4}-\frac{a^4 \operatorname{Subst}\left (\int \frac{\sinh ^{-1}(x)}{\sqrt{1+x^2}} \, dx,x,a+b x\right )}{2 b^4}\\ &=\frac{2 a^3 \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)}{b^4}-\frac{3 a^2 (a+b x) \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)}{2 b^4}+\frac{2 a (a+b x)^2 \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)}{3 b^4}-\frac{(a+b x)^3 \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)}{8 b^4}-\frac{a^4 \sinh ^{-1}(a+b x)^2}{4 b^4}+\frac{1}{4} x^4 \sinh ^{-1}(a+b x)^2+\frac{\operatorname{Subst}\left (\int x^3 \, dx,x,a+b x\right )}{8 b^4}+\frac{3 \operatorname{Subst}\left (\int \frac{x^2 \sinh ^{-1}(x)}{\sqrt{1+x^2}} \, dx,x,a+b x\right )}{8 b^4}-\frac{(2 a) \operatorname{Subst}\left (\int x^2 \, dx,x,a+b x\right )}{3 b^4}-\frac{(4 a) \operatorname{Subst}\left (\int \frac{x \sinh ^{-1}(x)}{\sqrt{1+x^2}} \, dx,x,a+b x\right )}{3 b^4}+\frac{\left (3 a^2\right ) \operatorname{Subst}(\int x \, dx,x,a+b x)}{2 b^4}+\frac{\left (3 a^2\right ) \operatorname{Subst}\left (\int \frac{\sinh ^{-1}(x)}{\sqrt{1+x^2}} \, dx,x,a+b x\right )}{2 b^4}-\frac{\left (2 a^3\right ) \operatorname{Subst}(\int 1 \, dx,x,a+b x)}{b^4}\\ &=-\frac{2 a^3 x}{b^3}+\frac{3 a^2 (a+b x)^2}{4 b^4}-\frac{2 a (a+b x)^3}{9 b^4}+\frac{(a+b x)^4}{32 b^4}-\frac{4 a \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)}{3 b^4}+\frac{2 a^3 \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)}{b^4}+\frac{3 (a+b x) \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)}{16 b^4}-\frac{3 a^2 (a+b x) \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)}{2 b^4}+\frac{2 a (a+b x)^2 \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)}{3 b^4}-\frac{(a+b x)^3 \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)}{8 b^4}+\frac{3 a^2 \sinh ^{-1}(a+b x)^2}{4 b^4}-\frac{a^4 \sinh ^{-1}(a+b x)^2}{4 b^4}+\frac{1}{4} x^4 \sinh ^{-1}(a+b x)^2-\frac{3 \operatorname{Subst}(\int x \, dx,x,a+b x)}{16 b^4}-\frac{3 \operatorname{Subst}\left (\int \frac{\sinh ^{-1}(x)}{\sqrt{1+x^2}} \, dx,x,a+b x\right )}{16 b^4}+\frac{(4 a) \operatorname{Subst}(\int 1 \, dx,x,a+b x)}{3 b^4}\\ &=\frac{4 a x}{3 b^3}-\frac{2 a^3 x}{b^3}-\frac{3 (a+b x)^2}{32 b^4}+\frac{3 a^2 (a+b x)^2}{4 b^4}-\frac{2 a (a+b x)^3}{9 b^4}+\frac{(a+b x)^4}{32 b^4}-\frac{4 a \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)}{3 b^4}+\frac{2 a^3 \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)}{b^4}+\frac{3 (a+b x) \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)}{16 b^4}-\frac{3 a^2 (a+b x) \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)}{2 b^4}+\frac{2 a (a+b x)^2 \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)}{3 b^4}-\frac{(a+b x)^3 \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)}{8 b^4}-\frac{3 \sinh ^{-1}(a+b x)^2}{32 b^4}+\frac{3 a^2 \sinh ^{-1}(a+b x)^2}{4 b^4}-\frac{a^4 \sinh ^{-1}(a+b x)^2}{4 b^4}+\frac{1}{4} x^4 \sinh ^{-1}(a+b x)^2\\ \end{align*}

Mathematica [A]  time = 0.179018, size = 145, normalized size = 0.44 \[ \frac{b x \left (78 a^2 b x-300 a^3+a \left (330-28 b^2 x^2\right )+9 b x \left (b^2 x^2-3\right )\right )-9 \left (8 a^4-24 a^2-8 b^4 x^4+3\right ) \sinh ^{-1}(a+b x)^2+6 \sqrt{a^2+2 a b x+b^2 x^2+1} \left (-26 a^2 b x+50 a^3+a \left (14 b^2 x^2-55\right )-6 b^3 x^3+9 b x\right ) \sinh ^{-1}(a+b x)}{288 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcSinh[a + b*x]^2,x]

[Out]

(b*x*(-300*a^3 + 78*a^2*b*x + a*(330 - 28*b^2*x^2) + 9*b*x*(-3 + b^2*x^2)) + 6*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^
2]*(50*a^3 + 9*b*x - 26*a^2*b*x - 6*b^3*x^3 + a*(-55 + 14*b^2*x^2))*ArcSinh[a + b*x] - 9*(3 - 24*a^2 + 8*a^4 -
8*b^4*x^4)*ArcSinh[a + b*x]^2)/(288*b^4)

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Maple [A]  time = 0.072, size = 387, normalized size = 1.2 \begin{align*}{\frac{1}{{b}^{4}} \left ( -{a}^{3} \left ( \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2} \left ( bx+a \right ) -2\,{\it Arcsinh} \left ( bx+a \right ) \sqrt{1+ \left ( bx+a \right ) ^{2}}+2\,bx+2\,a \right ) +{\frac{3\,{a}^{2}}{4} \left ( 2\, \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2} \left ( bx+a \right ) ^{2}-2\,{\it Arcsinh} \left ( bx+a \right ) \sqrt{1+ \left ( bx+a \right ) ^{2}} \left ( bx+a \right ) + \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2}+ \left ( bx+a \right ) ^{2}+1 \right ) }-{\frac{a}{9} \left ( 9\, \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2} \left ( bx+a \right ) ^{3}-6\,{\it Arcsinh} \left ( bx+a \right ) \left ( bx+a \right ) ^{2}\sqrt{1+ \left ( bx+a \right ) ^{2}}+27\, \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2} \left ( bx+a \right ) +2\, \left ( bx+a \right ) ^{3}-42\,{\it Arcsinh} \left ( bx+a \right ) \sqrt{1+ \left ( bx+a \right ) ^{2}}+42\,bx+42\,a \right ) }+{\frac{ \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2} \left ( bx+a \right ) ^{2} \left ( 1+ \left ( bx+a \right ) ^{2} \right ) }{4}}-{\frac{ \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2} \left ( 1+ \left ( bx+a \right ) ^{2} \right ) }{4}}-{\frac{{\it Arcsinh} \left ( bx+a \right ) \left ( bx+a \right ) }{8} \left ( 1+ \left ( bx+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{5\,{\it Arcsinh} \left ( bx+a \right ) \left ( bx+a \right ) }{16}\sqrt{1+ \left ( bx+a \right ) ^{2}}}+{\frac{5\, \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2}}{32}}+{\frac{ \left ( bx+a \right ) ^{2} \left ( 1+ \left ( bx+a \right ) ^{2} \right ) }{32}}-{\frac{ \left ( bx+a \right ) ^{2}}{8}}-{\frac{1}{8}}+3\,a \left ( \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2} \left ( bx+a \right ) -2\,{\it Arcsinh} \left ( bx+a \right ) \sqrt{1+ \left ( bx+a \right ) ^{2}}+2\,bx+2\,a \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^4*(-a^3*(arcsinh(b*x+a)^2*(b*x+a)-2*arcsinh(b*x+a)*(1+(b*x+a)^2)^(1/2)+2*b*x+2*a)+3/4*a^2*(2*arcsinh(b*x+a
)^2*(b*x+a)^2-2*arcsinh(b*x+a)*(1+(b*x+a)^2)^(1/2)*(b*x+a)+arcsinh(b*x+a)^2+(b*x+a)^2+1)-1/9*a*(9*arcsinh(b*x+
a)^2*(b*x+a)^3-6*arcsinh(b*x+a)*(b*x+a)^2*(1+(b*x+a)^2)^(1/2)+27*arcsinh(b*x+a)^2*(b*x+a)+2*(b*x+a)^3-42*arcsi
nh(b*x+a)*(1+(b*x+a)^2)^(1/2)+42*b*x+42*a)+1/4*arcsinh(b*x+a)^2*(b*x+a)^2*(1+(b*x+a)^2)-1/4*arcsinh(b*x+a)^2*(
1+(b*x+a)^2)-1/8*arcsinh(b*x+a)*(b*x+a)*(1+(b*x+a)^2)^(3/2)+5/16*arcsinh(b*x+a)*(1+(b*x+a)^2)^(1/2)*(b*x+a)+5/
32*arcsinh(b*x+a)^2+1/32*(b*x+a)^2*(1+(b*x+a)^2)-1/8*(b*x+a)^2-1/8+3*a*(arcsinh(b*x+a)^2*(b*x+a)-2*arcsinh(b*x
+a)*(1+(b*x+a)^2)^(1/2)+2*b*x+2*a))

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.49499, size = 439, normalized size = 1.33 \begin{align*} \frac{9 \, b^{4} x^{4} - 28 \, a b^{3} x^{3} + 3 \,{\left (26 \, a^{2} - 9\right )} b^{2} x^{2} - 30 \,{\left (10 \, a^{3} - 11 \, a\right )} b x + 9 \,{\left (8 \, b^{4} x^{4} - 8 \, a^{4} + 24 \, a^{2} - 3\right )} \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )^{2} - 6 \,{\left (6 \, b^{3} x^{3} - 14 \, a b^{2} x^{2} - 50 \, a^{3} +{\left (26 \, a^{2} - 9\right )} b x + 55 \, a\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 )}{288 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/288*(9*b^4*x^4 - 28*a*b^3*x^3 + 3*(26*a^2 - 9)*b^2*x^2 - 30*(10*a^3 - 11*a)*b*x + 9*(8*b^4*x^4 - 8*a^4 + 24*
a^2 - 3)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^2 - 6*(6*b^3*x^3 - 14*a*b^2*x^2 - 50*a^3 + (26*a^2 -
9)*b*x + 55*a)*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)))/b^4

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Sympy [A]  time = 4.16391, size = 366, normalized size = 1.11 \begin{align*} \begin{cases} - \frac{a^{4} \operatorname{asinh}^{2}{\left (a + b x \right )}}{4 b^{4}} - \frac{25 a^{3} x}{24 b^{3}} + \frac{25 a^{3} \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname{asinh}{\left (a + b x \right )}}{24 b^{4}} + \frac{13 a^{2} x^{2}}{48 b^{2}} - \frac{13 a^{2} x \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname{asinh}{\left (a + b x \right )}}{24 b^{3}} + \frac{3 a^{2} \operatorname{asinh}^{2}{\left (a + b x \right )}}{4 b^{4}} - \frac{7 a x^{3}}{72 b} + \frac{7 a x^{2} \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname{asinh}{\left (a + b x \right )}}{24 b^{2}} + \frac{55 a x}{48 b^{3}} - \frac{55 a \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname{asinh}{\left (a + b x \right )}}{48 b^{4}} + \frac{x^{4} \operatorname{asinh}^{2}{\left (a + b x \right )}}{4} + \frac{x^{4}}{32} - \frac{x^{3} \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname{asinh}{\left (a + b x \right )}}{8 b} - \frac{3 x^{2}}{32 b^{2}} + \frac{3 x \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname{asinh}{\left (a + b x \right )}}{16 b^{3}} - \frac{3 \operatorname{asinh}^{2}{\left (a + b x \right )}}{32 b^{4}} & \text{for}\: b \neq 0 \\\frac{x^{4} \operatorname{asinh}^{2}{\left (a \right )}}{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**4*asinh(a + b*x)**2/(4*b**4) - 25*a**3*x/(24*b**3) + 25*a**3*sqrt(a**2 + 2*a*b*x + b**2*x**2 +
1)*asinh(a + b*x)/(24*b**4) + 13*a**2*x**2/(48*b**2) - 13*a**2*x*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*asinh(a
+ b*x)/(24*b**3) + 3*a**2*asinh(a + b*x)**2/(4*b**4) - 7*a*x**3/(72*b) + 7*a*x**2*sqrt(a**2 + 2*a*b*x + b**2*x
**2 + 1)*asinh(a + b*x)/(24*b**2) + 55*a*x/(48*b**3) - 55*a*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*asinh(a + b*x
)/(48*b**4) + x**4*asinh(a + b*x)**2/4 + x**4/32 - x**3*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*asinh(a + b*x)/(8
*b) - 3*x**2/(32*b**2) + 3*x*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*asinh(a + b*x)/(16*b**3) - 3*asinh(a + b*x)*
*2/(32*b**4), Ne(b, 0)), (x**4*asinh(a)**2/4, True))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3*arcsinh(b*x + a)^2, x)