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

Optimal. Leaf size=45 \[ \frac{(a+b x) \sinh ^{-1}(a+b x)^2}{b}-\frac{2 \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)}{b}+2 x \]

[Out]

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

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Rubi [A]  time = 0.0506684, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5863, 5653, 5717, 8} \[ \frac{(a+b x) \sinh ^{-1}(a+b x)^2}{b}-\frac{2 \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)}{b}+2 x \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a + b*x]^2,x]

[Out]

2*x - (2*Sqrt[1 + (a + b*x)^2]*ArcSinh[a + b*x])/b + ((a + b*x)*ArcSinh[a + b*x]^2)/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 8

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

Rubi steps

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

Mathematica [A]  time = 0.0217076, size = 47, normalized size = 1.04 \[ \frac{2 (a+b x)+(a+b x) \sinh ^{-1}(a+b x)^2-2 \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[a + b*x]^2,x]

[Out]

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

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Maple [A]  time = 0.027, size = 46, normalized size = 1. \begin{align*}{\frac{1}{b} \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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.42553, size = 217, normalized size = 4.82 \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 )^{2} + 2 \, b x - 2 \, \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 )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b*x + a)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^2 + 2*b*x - 2*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*lo
g(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)))/b

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Sympy [A]  time = 0.262953, size = 63, normalized size = 1.4 \begin{align*} \begin{cases} \frac{a \operatorname{asinh}^{2}{\left (a + b x \right )}}{b} + x \operatorname{asinh}^{2}{\left (a + b x \right )} + 2 x - \frac{2 \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname{asinh}{\left (a + b x \right )}}{b} & \text{for}\: b \neq 0 \\x \operatorname{asinh}^{2}{\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)**2,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsinh(b*x + a)^2, x)