3.1.68 \(\int x^2 \sinh ^{-1}(a+b x)^2 \, dx\) [68]

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

[Out]

-4/9*x/b^2+2*a^2*x/b^2-1/2*a*(b*x+a)^2/b^3+2/27*(b*x+a)^3/b^3-1/2*a*arcsinh(b*x+a)^2/b^3+1/3*a^3*arcsinh(b*x+a
)^2/b^3+1/3*x^3*arcsinh(b*x+a)^2+4/9*arcsinh(b*x+a)*(1+(b*x+a)^2)^(1/2)/b^3-2*a^2*arcsinh(b*x+a)*(1+(b*x+a)^2)
^(1/2)/b^3+a*(b*x+a)*arcsinh(b*x+a)*(1+(b*x+a)^2)^(1/2)/b^3-2/9*(b*x+a)^2*arcsinh(b*x+a)*(1+(b*x+a)^2)^(1/2)/b
^3

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Rubi [A]
time = 0.25, antiderivative size = 211, normalized size of antiderivative = 1.00, 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 = {5859, 5828, 5838, 5783, 5798, 8, 5812, 30} \begin {gather*} \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 {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 {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} \end {gather*}

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 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]

Rule 5783

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

Rule 5798

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/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)
^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 5812

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

Rule 5828

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 5838

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 5859

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
]

Rubi steps

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

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Mathematica [A]
time = 0.09, size = 107, normalized size = 0.51 \begin {gather*} \frac {b x \left (-24+66 a^2-15 a b x+4 b^2 x^2\right )-6 \sqrt {1+a^2+2 a b x+b^2 x^2} \left (-4+11 a^2-5 a b x+2 b^2 x^2\right ) \sinh ^{-1}(a+b x)+9 \left (-3 a+2 a^3+2 b^3 x^3\right ) \sinh ^{-1}(a+b x)^2}{54 b^3} \end {gather*}

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 [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int x^{2} \arcsinh \left (b x +a \right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^2 - integrate(2/3*(b^3*x^5 + 2*a*b^2*x^4 + (a^2*b + b
)*x^3 + (b^2*x^4 + a*b*x^3)*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*x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2*b + b)*x + (b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2) + a), x)

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Fricas [A]
time = 0.35, size = 146, normalized size = 0.69 \begin {gather*} \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 {gather*}

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 = 0.28, size = 243, normalized size = 1.15 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^2\,{\mathrm {asinh}\left (a+b\,x\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*asinh(a + b*x)^2,x)

[Out]

int(x^2*asinh(a + b*x)^2, x)

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