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

Optimal. Leaf size=31 \[ \frac{\text{Chi}\left (2 \sinh ^{-1}(a+b x)\right )}{2 b}+\frac{\log \left (\sinh ^{-1}(a+b x)\right )}{2 b} \]

[Out]

CoshIntegral[2*ArcSinh[a + b*x]]/(2*b) + Log[ArcSinh[a + b*x]]/(2*b)

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Rubi [A]  time = 0.119285, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {5867, 5699, 3312, 3301} \[ \frac{\text{Chi}\left (2 \sinh ^{-1}(a+b x)\right )}{2 b}+\frac{\log \left (\sinh ^{-1}(a+b x)\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

CoshIntegral[2*ArcSinh[a + b*x]]/(2*b) + Log[ArcSinh[a + b*x]]/(2*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 5699

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c, Subst[Int[
(a + b*x)^n*Cosh[x]^(2*p + 1), x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && IG
tQ[2*p, 0] && (IntegerQ[p] || GtQ[d, 0])

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 3301

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[(c*f*fz)/d
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rubi steps

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

Mathematica [A]  time = 0.0629579, size = 24, normalized size = 0.77 \[ \frac{\text{Chi}\left (2 \sinh ^{-1}(a+b x)\right )+\log \left (\sinh ^{-1}(a+b x)\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(CoshIntegral[2*ArcSinh[a + b*x]] + Log[ArcSinh[a + b*x]])/(2*b)

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Maple [A]  time = 0.052, size = 28, normalized size = 0.9 \begin{align*}{\frac{{\it Chi} \left ( 2\,{\it Arcsinh} \left ( bx+a \right ) \right ) }{2\,b}}+{\frac{\ln \left ({\it Arcsinh} \left ( bx+a \right ) \right ) }{2\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*Chi(2*arcsinh(b*x+a))/b+1/2*ln(arcsinh(b*x+a))/b

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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