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

Optimal. Leaf size=38 \[ \frac{\text{Shi}\left (\sinh ^{-1}(a+b x)\right )}{b}-\frac{\sqrt{(a+b x)^2+1}}{b \sinh ^{-1}(a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.07259, antiderivative size = 38, 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, 5655, 5779, 3298} \[ \frac{\text{Shi}\left (\sinh ^{-1}(a+b x)\right )}{b}-\frac{\sqrt{(a+b x)^2+1}}{b \sinh ^{-1}(a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a + b*x]^(-2),x]

[Out]

-(Sqrt[1 + (a + b*x)^2]/(b*ArcSinh[a + b*x])) + SinhIntegral[ArcSinh[a + b*x]]/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 5655

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

Rule 5779

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

Rule 3298

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

Rubi steps

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

Mathematica [A]  time = 0.0219833, size = 35, normalized size = 0.92 \[ \frac{\text{Shi}\left (\sinh ^{-1}(a+b x)\right )-\frac{\sqrt{(a+b x)^2+1}}{\sinh ^{-1}(a+b x)}}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[a + b*x]^(-2),x]

[Out]

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

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Maple [A]  time = 0.027, size = 34, normalized size = 0.9 \begin{align*}{\frac{1}{b} \left ( -{\frac{1}{{\it Arcsinh} \left ( bx+a \right ) }\sqrt{1+ \left ( bx+a \right ) ^{2}}}+{\it Shi} \left ({\it Arcsinh} \left ( bx+a \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (3 \, a^{2} b + b\right )} x +{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac{3}{2}} + a}{{\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{\left (b^{2} x + a b\right )} + b\right )} \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} + \int \frac{b^{4} x^{4} + 4 \, a b^{3} x^{3} + a^{4} + 2 \,{\left (3 \, a^{2} b^{2} + b^{2}\right )} x^{2} +{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )} + 2 \, a^{2} + 4 \,{\left (a^{3} b + a b\right )} x +{\left (2 \, b^{3} x^{3} + 6 \, a b^{2} x^{2} + 2 \, a^{3} +{\left (6 \, a^{2} b + b\right )} x + a\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1} + 1}{{\left (b^{4} x^{4} + 4 \, a b^{3} x^{3} + a^{4} + 2 \,{\left (3 \, a^{2} b^{2} + b^{2}\right )} x^{2} +{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} + 2 \, a^{2} + 4 \,{\left (a^{3} b + a b\right )} x + 2 \,{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (3 \, a^{2} b + b\right )} x + a\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1} + 1\right )} \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2*b + b)*x + (b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2) + a)/((b^3*x^2 + 2*a*b^2
*x + a^2*b + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(b^2*x + a*b) + b)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 +
1))) + integrate((b^4*x^4 + 4*a*b^3*x^3 + a^4 + 2*(3*a^2*b^2 + b^2)*x^2 + (b^2*x^2 + 2*a*b*x + a^2 + 1)*(b^2*
x^2 + 2*a*b*x + a^2 - 1) + 2*a^2 + 4*(a^3*b + a*b)*x + (2*b^3*x^3 + 6*a*b^2*x^2 + 2*a^3 + (6*a^2*b + b)*x + a)
*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) + 1)/((b^4*x^4 + 4*a*b^3*x^3 + a^4 + 2*(3*a^2*b^2 + b^2)*x^2 + (b^2*x^2 + 2
*a*b*x + a^2 + 1)*(b^2*x^2 + 2*a*b*x + a^2) + 2*a^2 + 4*(a^3*b + a*b)*x + 2*(b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*
a^2*b + b)*x + a)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) + 1)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsinh(b*x + a)^(-2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asinh(a + b*x)**(-2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsinh(b*x + a)^(-2), x)