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

Optimal. Leaf size=78 \[ -\frac{6 \sqrt{(a+b x)^2+1}}{b}+\frac{(a+b x) \sinh ^{-1}(a+b x)^3}{b}-\frac{3 \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)^2}{b}+\frac{6 (a+b x) \sinh ^{-1}(a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.0738137, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5863, 5653, 5717, 261} \[ -\frac{6 \sqrt{(a+b x)^2+1}}{b}+\frac{(a+b x) \sinh ^{-1}(a+b x)^3}{b}-\frac{3 \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)^2}{b}+\frac{6 (a+b x) \sinh ^{-1}(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a + b*x]^3,x]

[Out]

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

Rule 5863

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcSinh[x])^n, x
], x, c + d*x], x] /; FreeQ[{a, b, c, d, n}, x]

Rule 5653

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

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 261

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

Rubi steps

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

Mathematica [A]  time = 0.0288498, size = 70, normalized size = 0.9 \[ \frac{-6 \sqrt{(a+b x)^2+1}+(a+b x) \sinh ^{-1}(a+b x)^3-3 \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)^2+6 (a+b x) \sinh ^{-1}(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[a + b*x]^3,x]

[Out]

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

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Maple [A]  time = 0.028, size = 67, normalized size = 0.9 \begin{align*}{\frac{1}{b} \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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.38706, size = 346, normalized size = 4.44 \begin{align*} \frac{{\left (b x + a\right )} \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )^{3} - 3 \, \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} + 6 \,{\left (b x + a\right )} \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - 6 \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b*x + a)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^3 - 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))^2 + 6*(b*x + a)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)) - 6*sq
rt(b^2*x^2 + 2*a*b*x + a^2 + 1))/b

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Sympy [A]  time = 0.727157, size = 109, normalized size = 1.4 \begin{align*} \begin{cases} \frac{a \operatorname{asinh}^{3}{\left (a + b x \right )}}{b} + \frac{6 a \operatorname{asinh}{\left (a + b x \right )}}{b} + x \operatorname{asinh}^{3}{\left (a + b x \right )} + 6 x \operatorname{asinh}{\left (a + b x \right )} - \frac{3 \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname{asinh}^{2}{\left (a + b x \right )}}{b} - \frac{6 \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1}}{b} & \text{for}\: b \neq 0 \\x \operatorname{asinh}^{3}{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*asinh(a + b*x)**3/b + 6*a*asinh(a + b*x)/b + x*asinh(a + b*x)**3 + 6*x*asinh(a + b*x) - 3*sqrt(a*
*2 + 2*a*b*x + b**2*x**2 + 1)*asinh(a + b*x)**2/b - 6*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/b, Ne(b, 0)), (x*as
inh(a)**3, True))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsinh(b*x + a)^3, x)