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

Optimal. Leaf size=11 \[ \frac{\log \left (\sinh ^{-1}(a+b x)\right )}{b} \]

[Out]

Log[ArcSinh[a + b*x]]/b

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Rubi [A]  time = 0.0751373, antiderivative size = 11, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {5867, 5673} \[ \frac{\log \left (\sinh ^{-1}(a+b x)\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[ArcSinh[a + b*x]]/b

Rule 5867

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)^(p_.), x_Symbol] :> D
ist[1/d, Subst[Int[(C/d^2 + (C*x^2)/d^2)^p*(a + b*ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, A,
B, C, n, p}, x] && EqQ[B*(1 + c^2) - 2*A*c*d, 0] && EqQ[2*c*C - B*d, 0]

Rule 5673

Int[1/(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*Sqrt[(d_) + (e_.)*(x_)^2]), x_Symbol] :> Simp[Log[a + b*ArcSinh[c*x
]]/(b*c*Sqrt[d]), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[d, 0]

Rubi steps

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

Mathematica [A]  time = 0.0304706, size = 11, normalized size = 1. \[ \frac{\log \left (\sinh ^{-1}(a+b x)\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[ArcSinh[a + b*x]]/b

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Maple [A]  time = 0.043, size = 12, normalized size = 1.1 \begin{align*}{\frac{\ln \left ({\it Arcsinh} \left ( bx+a \right ) \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/arcsinh(b*x+a)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2),x)

[Out]

ln(arcsinh(b*x+a))/b

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1} \operatorname{arsinh}\left (b x + a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*arcsinh(b*x + a)), x)

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Fricas [B]  time = 2.63541, size = 77, normalized size = 7. \begin{align*} \frac{\log \left (\log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.17822, size = 22, normalized size = 2. \begin{align*} \begin{cases} \frac{\log{\left (\operatorname{asinh}{\left (a + b x \right )} \right )}}{b} & \text{for}\: b \neq 0 \\\frac{x}{\sqrt{a^{2} + 1} \operatorname{asinh}{\left (a \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/asinh(b*x+a)/(b**2*x**2+2*a*b*x+a**2+1)**(1/2),x)

[Out]

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

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Giac [B]  time = 1.28502, size = 42, normalized size = 3.82 \begin{align*} \frac{\log \left ({\left | \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) \right |}\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(abs(log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))))/b