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]

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

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Rubi [A]  time = 0.07, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.02, 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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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{\operatorname {arsinh}\left (b x + a\right )^{2}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\operatorname {arsinh}\left (b x + a\right )^{2}}\,{d x} \]

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)

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maple [A]  time = 0.05, size = 34, normalized size = 0.89 \[ \frac {-\frac {\sqrt {1+\left (b x +a \right )^{2}}}{\arcsinh \left (b x +a \right )}+\Shi \left (\arcsinh \left (b x +a \right )\right )}{b} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ -\frac {b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} b + b\right )} x + {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}} + a}{{\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (b^{2} x + a b\right )} + b\right )} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} + \int \frac {b^{4} x^{4} + 4 \, a b^{3} x^{3} + a^{4} + 2 \, {\left (3 \, a^{2} b^{2} + b^{2}\right )} x^{2} + {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )} {\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )} + 2 \, a^{2} + 4 \, {\left (a^{3} b + a b\right )} x + {\left (2 \, b^{3} x^{3} + 6 \, a b^{2} x^{2} + 2 \, a^{3} + {\left (6 \, a^{2} b + b\right )} x + a\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} + 1}{{\left (b^{4} x^{4} + 4 \, a b^{3} x^{3} + a^{4} + 2 \, {\left (3 \, a^{2} b^{2} + b^{2}\right )} x^{2} + {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )} {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} + 2 \, a^{2} + 4 \, {\left (a^{3} b + a b\right )} x + 2 \, {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} b + b\right )} x + a\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} + 1\right )} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {1}{{\mathrm {asinh}\left (a+b\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/asinh(a + b*x)^2,x)

[Out]

int(1/asinh(a + b*x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\operatorname {asinh}^{2}{\left (a + b x \right )}}\, dx \]

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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