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

Optimal. Leaf size=61 \[ -\frac{(a+b x)^2}{4 b}+\frac{\sqrt{(a+b x)^2+1} (a+b x) \sinh ^{-1}(a+b x)}{2 b}+\frac{\sinh ^{-1}(a+b x)^2}{4 b} \]

[Out]

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

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Rubi [A]  time = 0.0683636, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {5867, 5682, 5675, 30} \[ -\frac{(a+b x)^2}{4 b}+\frac{\sqrt{(a+b x)^2+1} (a+b x) \sinh ^{-1}(a+b x)}{2 b}+\frac{\sinh ^{-1}(a+b x)^2}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a + b*x)^2/(4*b) + ((a + b*x)*Sqrt[1 + (a + b*x)^2]*ArcSinh[a + b*x])/(2*b) + ArcSinh[a + b*x]^2/(4*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 5682

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

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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0553745, size = 61, normalized size = 1. \[ \frac{2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2+1} \sinh ^{-1}(a+b x)-b x (2 a+b x)+\sinh ^{-1}(a+b x)^2}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.047, size = 91, normalized size = 1.5 \begin{align*}{\frac{1}{4\,b} \left ( 2\,{\it Arcsinh} \left ( bx+a \right ) \sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}xb-{b}^{2}{x}^{2}+2\,{\it Arcsinh} \left ( bx+a \right ) \sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}a-2\,xab+ \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2}-{a}^{2}-1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.60704, size = 240, normalized size = 3.93 \begin{align*} -\frac{b^{2} x^{2} + 2 \, a b x - 2 \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{\left (b x + a\right )} \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )^{2}}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(b^2*x^2 + 2*a*b*x - 2*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(b*x + a)*log(b*x + a + 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)/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*asinh(a + b*x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \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(arcsinh(b*x+a)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2),x, algorithm="giac")

[Out]

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