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

Optimal. Leaf size=211 \[ \frac{2 a^2 x}{b^2}+\frac{a^3 \sinh ^{-1}(a+b x)^2}{3 b^3}-\frac{2 a^2 \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)}{b^3}-\frac{a (a+b x)^2}{2 b^3}+\frac{2 (a+b x)^3}{27 b^3}-\frac{a \sinh ^{-1}(a+b x)^2}{2 b^3}+\frac{a (a+b x) \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)}{b^3}-\frac{2 (a+b x)^2 \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)}{9 b^3}+\frac{4 \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)}{9 b^3}+\frac{1}{3} x^3 \sinh ^{-1}(a+b x)^2-\frac{4 x}{9 b^2} \]

[Out]

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

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Rubi [A]  time = 0.374804, antiderivative size = 211, normalized size of antiderivative = 1., number of steps used = 14, 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} \[ \frac{2 a^2 x}{b^2}+\frac{a^3 \sinh ^{-1}(a+b x)^2}{3 b^3}-\frac{2 a^2 \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)}{b^3}-\frac{a (a+b x)^2}{2 b^3}+\frac{2 (a+b x)^3}{27 b^3}-\frac{a \sinh ^{-1}(a+b x)^2}{2 b^3}+\frac{a (a+b x) \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)}{b^3}-\frac{2 (a+b x)^2 \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)}{9 b^3}+\frac{4 \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)}{9 b^3}+\frac{1}{3} x^3 \sinh ^{-1}(a+b x)^2-\frac{4 x}{9 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcSinh[a + b*x]^2,x]

[Out]

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

Mathematica [A]  time = 0.137144, size = 107, normalized size = 0.51 \[ \frac{b x \left (66 a^2-15 a b x+4 b^2 x^2-24\right )+9 \left (2 a^3-3 a+2 b^3 x^3\right ) \sinh ^{-1}(a+b x)^2-6 \sqrt{a^2+2 a b x+b^2 x^2+1} \left (11 a^2-5 a b x+2 b^2 x^2-4\right ) \sinh ^{-1}(a+b x)}{54 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcSinh[a + b*x]^2,x]

[Out]

(b*x*(-24 + 66*a^2 - 15*a*b*x + 4*b^2*x^2) - 6*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] + 9*(-3*a + 2*a^3 + 2*b^3*x^3)*ArcSinh[a + b*x]^2)/(54*b^3)

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Maple [A]  time = 0.056, size = 219, normalized size = 1. \begin{align*}{\frac{1}{{b}^{3}} \left ( -{\frac{a}{2} \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{ \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2} \left ( bx+a \right ) }{3}}+{\frac{ \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2} \left ( bx+a \right ) \left ( 1+ \left ( bx+a \right ) ^{2} \right ) }{3}}+{\frac{4\,{\it Arcsinh} \left ( bx+a \right ) }{9}\sqrt{1+ \left ( bx+a \right ) ^{2}}}-{\frac{14\,bx}{27}}-{\frac{14\,a}{27}}-{\frac{2\,{\it Arcsinh} \left ( bx+a \right ) \left ( bx+a \right ) ^{2}}{9}\sqrt{1+ \left ( bx+a \right ) ^{2}}}+{\frac{ \left ( 2+2\, \left ( bx+a \right ) ^{2} \right ) \left ( bx+a \right ) }{27}}+{a}^{2} \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^2*arcsinh(b*x+a)^2,x)

[Out]

1/b^3*(-1/2*a*(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)-1/3*arcsinh(b*x+a)^2*(b*x+a)+1/3*arcsinh(b*x+a)^2*(b*x+a)*(1+(b*x+a)^2)+4/9*arcsinh(b*x+a)*(1+(b*x+a)
^2)^(1/2)-14/27*b*x-14/27*a-2/9*arcsinh(b*x+a)*(b*x+a)^2*(1+(b*x+a)^2)^(1/2)+2/27*(1+(b*x+a)^2)*(b*x+a)+a^2*(a
rcsinh(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^2*arcsinh(b*x+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.50995, size = 352, normalized size = 1.67 \begin{align*} \frac{4 \, b^{3} x^{3} - 15 \, a b^{2} x^{2} + 6 \,{\left (11 \, a^{2} - 4\right )} b x + 9 \,{\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 )^{2} - 6 \,{\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 )}{54 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/54*(4*b^3*x^3 - 15*a*b^2*x^2 + 6*(11*a^2 - 4)*b*x + 9*(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))^2 - 6*(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)))/b^3

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Sympy [A]  time = 1.73613, size = 243, normalized size = 1.15 \begin{align*} \begin{cases} \frac{a^{3} \operatorname{asinh}^{2}{\left (a + b x \right )}}{3 b^{3}} + \frac{11 a^{2} x}{9 b^{2}} - \frac{11 a^{2} \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname{asinh}{\left (a + b x \right )}}{9 b^{3}} - \frac{5 a x^{2}}{18 b} + \frac{5 a x \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname{asinh}{\left (a + b x \right )}}{9 b^{2}} - \frac{a \operatorname{asinh}^{2}{\left (a + b x \right )}}{2 b^{3}} + \frac{x^{3} \operatorname{asinh}^{2}{\left (a + b x \right )}}{3} + \frac{2 x^{3}}{27} - \frac{2 x^{2} \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname{asinh}{\left (a + b x \right )}}{9 b} - \frac{4 x}{9 b^{2}} + \frac{4 \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname{asinh}{\left (a + b x \right )}}{9 b^{3}} & \text{for}\: b \neq 0 \\\frac{x^{3} \operatorname{asinh}^{2}{\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)**2,x)

[Out]

Piecewise((a**3*asinh(a + b*x)**2/(3*b**3) + 11*a**2*x/(9*b**2) - 11*a**2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)
*asinh(a + b*x)/(9*b**3) - 5*a*x**2/(18*b) + 5*a*x*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*asinh(a + b*x)/(9*b**2
) - a*asinh(a + b*x)**2/(2*b**3) + x**3*asinh(a + b*x)**2/3 + 2*x**3/27 - 2*x**2*sqrt(a**2 + 2*a*b*x + b**2*x*
*2 + 1)*asinh(a + b*x)/(9*b) - 4*x/(9*b**2) + 4*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*asinh(a + b*x)/(9*b**3),
Ne(b, 0)), (x**3*asinh(a)**2/3, 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 )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*arcsinh(b*x + a)^2, x)