3.245 \(\int \frac{1}{(a+b \cosh ^{-1}(1+d x^2))^2} \, dx\)

Optimal. Leaf size=150 \[ -\frac{x \sinh \left (\frac{a}{2 b}\right ) \text{Chi}\left (\frac{a+b \cosh ^{-1}\left (d x^2+1\right )}{2 b}\right )}{2 \sqrt{2} b^2 \sqrt{d x^2}}+\frac{x \cosh \left (\frac{a}{2 b}\right ) \text{Shi}\left (\frac{a+b \cosh ^{-1}\left (d x^2+1\right )}{2 b}\right )}{2 \sqrt{2} b^2 \sqrt{d x^2}}-\frac{\sqrt{d x^2} \sqrt{d x^2+2}}{2 b d x \left (a+b \cosh ^{-1}\left (d x^2+1\right )\right )} \]

[Out]

-(Sqrt[d*x^2]*Sqrt[2 + d*x^2])/(2*b*d*x*(a + b*ArcCosh[1 + d*x^2])) - (x*CoshIntegral[(a + b*ArcCosh[1 + d*x^2
])/(2*b)]*Sinh[a/(2*b)])/(2*Sqrt[2]*b^2*Sqrt[d*x^2]) + (x*Cosh[a/(2*b)]*SinhIntegral[(a + b*ArcCosh[1 + d*x^2]
)/(2*b)])/(2*Sqrt[2]*b^2*Sqrt[d*x^2])

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Rubi [A]  time = 0.0216295, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {5887} \[ -\frac{x \sinh \left (\frac{a}{2 b}\right ) \text{Chi}\left (\frac{a+b \cosh ^{-1}\left (d x^2+1\right )}{2 b}\right )}{2 \sqrt{2} b^2 \sqrt{d x^2}}+\frac{x \cosh \left (\frac{a}{2 b}\right ) \text{Shi}\left (\frac{a+b \cosh ^{-1}\left (d x^2+1\right )}{2 b}\right )}{2 \sqrt{2} b^2 \sqrt{d x^2}}-\frac{\sqrt{d x^2} \sqrt{d x^2+2}}{2 b d x \left (a+b \cosh ^{-1}\left (d x^2+1\right )\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCosh[1 + d*x^2])^(-2),x]

[Out]

-(Sqrt[d*x^2]*Sqrt[2 + d*x^2])/(2*b*d*x*(a + b*ArcCosh[1 + d*x^2])) - (x*CoshIntegral[(a + b*ArcCosh[1 + d*x^2
])/(2*b)]*Sinh[a/(2*b)])/(2*Sqrt[2]*b^2*Sqrt[d*x^2]) + (x*Cosh[a/(2*b)]*SinhIntegral[(a + b*ArcCosh[1 + d*x^2]
)/(2*b)])/(2*Sqrt[2]*b^2*Sqrt[d*x^2])

Rule 5887

Int[((a_.) + ArcCosh[1 + (d_.)*(x_)^2]*(b_.))^(-2), x_Symbol] :> -Simp[(Sqrt[d*x^2]*Sqrt[2 + d*x^2])/(2*b*d*x*
(a + b*ArcCosh[1 + d*x^2])), x] + (-Simp[(x*Sinh[a/(2*b)]*CoshIntegral[(a + b*ArcCosh[1 + d*x^2])/(2*b)])/(2*S
qrt[2]*b^2*Sqrt[d*x^2]), x] + Simp[(x*Cosh[a/(2*b)]*SinhIntegral[(a + b*ArcCosh[1 + d*x^2])/(2*b)])/(2*Sqrt[2]
*b^2*Sqrt[d*x^2]), x]) /; FreeQ[{a, b, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.880592, size = 130, normalized size = 0.87 \[ -\frac{x^2 \text{csch}\left (\frac{1}{2} \cosh ^{-1}\left (d x^2+1\right )\right ) \left (\sinh \left (\frac{a}{2 b}\right ) \text{Chi}\left (\frac{a+b \cosh ^{-1}\left (d x^2+1\right )}{2 b}\right )-\cosh \left (\frac{a}{2 b}\right ) \text{Shi}\left (\frac{a+b \cosh ^{-1}\left (d x^2+1\right )}{2 b}\right )\right )+\frac{2 b \sqrt{d x^2} \sqrt{d x^2+2}}{a d+b d \cosh ^{-1}\left (d x^2+1\right )}}{4 b^2 x} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcCosh[1 + d*x^2])^(-2),x]

[Out]

-((2*b*Sqrt[d*x^2]*Sqrt[2 + d*x^2])/(a*d + b*d*ArcCosh[1 + d*x^2]) + x^2*Csch[ArcCosh[1 + d*x^2]/2]*(CoshInteg
ral[(a + b*ArcCosh[1 + d*x^2])/(2*b)]*Sinh[a/(2*b)] - Cosh[a/(2*b)]*SinhIntegral[(a + b*ArcCosh[1 + d*x^2])/(2
*b)]))/(4*b^2*x)

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Maple [F]  time = 0.066, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b{\rm arccosh} \left (d{x}^{2}+1\right ) \right ) ^{-2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{d^{2} x^{4} + 3 \, d x^{2} +{\left (d^{\frac{3}{2}} x^{3} + 2 \, \sqrt{d} x\right )} \sqrt{d x^{2} + 2} + 2}{2 \,{\left (a b d^{2} x^{3} + 2 \, a b d x +{\left (b^{2} d^{2} x^{3} + 2 \, b^{2} d x +{\left (b^{2} d^{\frac{3}{2}} x^{2} + b^{2} \sqrt{d}\right )} \sqrt{d x^{2} + 2}\right )} \log \left (d x^{2} + \sqrt{d x^{2} + 2} \sqrt{d} x + 1\right ) +{\left (a b d^{\frac{3}{2}} x^{2} + a b \sqrt{d}\right )} \sqrt{d x^{2} + 2}\right )}} + \int \frac{d^{3} x^{6} + 3 \, d^{2} x^{4} +{\left (d^{2} x^{4} + d x^{2} + 2\right )}{\left (d x^{2} + 2\right )} +{\left (2 \, d^{\frac{5}{2}} x^{5} + 4 \, d^{\frac{3}{2}} x^{3} + \sqrt{d} x\right )} \sqrt{d x^{2} + 2} - 4}{2 \,{\left (a b d^{3} x^{6} + 4 \, a b d^{2} x^{4} + 4 \, a b d x^{2} +{\left (a b d^{2} x^{4} + 2 \, a b d x^{2} + a b\right )}{\left (d x^{2} + 2\right )} +{\left (b^{2} d^{3} x^{6} + 4 \, b^{2} d^{2} x^{4} + 4 \, b^{2} d x^{2} +{\left (b^{2} d^{2} x^{4} + 2 \, b^{2} d x^{2} + b^{2}\right )}{\left (d x^{2} + 2\right )} + 2 \,{\left (b^{2} d^{\frac{5}{2}} x^{5} + 3 \, b^{2} d^{\frac{3}{2}} x^{3} + 2 \, b^{2} \sqrt{d} x\right )} \sqrt{d x^{2} + 2}\right )} \log \left (d x^{2} + \sqrt{d x^{2} + 2} \sqrt{d} x + 1\right ) + 2 \,{\left (a b d^{\frac{5}{2}} x^{5} + 3 \, a b d^{\frac{3}{2}} x^{3} + 2 \, a b \sqrt{d} x\right )} \sqrt{d x^{2} + 2}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(d^2*x^4 + 3*d*x^2 + (d^(3/2)*x^3 + 2*sqrt(d)*x)*sqrt(d*x^2 + 2) + 2)/(a*b*d^2*x^3 + 2*a*b*d*x + (b^2*d^2
*x^3 + 2*b^2*d*x + (b^2*d^(3/2)*x^2 + b^2*sqrt(d))*sqrt(d*x^2 + 2))*log(d*x^2 + sqrt(d*x^2 + 2)*sqrt(d)*x + 1)
+ (a*b*d^(3/2)*x^2 + a*b*sqrt(d))*sqrt(d*x^2 + 2)) + integrate(1/2*(d^3*x^6 + 3*d^2*x^4 + (d^2*x^4 + d*x^2 +
2)*(d*x^2 + 2) + (2*d^(5/2)*x^5 + 4*d^(3/2)*x^3 + sqrt(d)*x)*sqrt(d*x^2 + 2) - 4)/(a*b*d^3*x^6 + 4*a*b*d^2*x^4
+ 4*a*b*d*x^2 + (a*b*d^2*x^4 + 2*a*b*d*x^2 + a*b)*(d*x^2 + 2) + (b^2*d^3*x^6 + 4*b^2*d^2*x^4 + 4*b^2*d*x^2 +
(b^2*d^2*x^4 + 2*b^2*d*x^2 + b^2)*(d*x^2 + 2) + 2*(b^2*d^(5/2)*x^5 + 3*b^2*d^(3/2)*x^3 + 2*b^2*sqrt(d)*x)*sqrt
(d*x^2 + 2))*log(d*x^2 + sqrt(d*x^2 + 2)*sqrt(d)*x + 1) + 2*(a*b*d^(5/2)*x^5 + 3*a*b*d^(3/2)*x^3 + 2*a*b*sqrt(
d)*x)*sqrt(d*x^2 + 2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(1/(b^2*arccosh(d*x^2 + 1)^2 + 2*a*b*arccosh(d*x^2 + 1) + a^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*acosh(d*x**2 + 1))**(-2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arccosh(d*x^2 + 1) + a)^(-2), x)