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

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

[Out]

1/4*x*Chi(1/2*(a+b*arccosh(d*x^2-1))/b)*cosh(1/2*a/b)/b^2*2^(1/2)/(d*x^2)^(1/2)-1/4*x*Shi(1/2*(a+b*arccosh(d*x
^2-1))/b)*sinh(1/2*a/b)/b^2*2^(1/2)/(d*x^2)^(1/2)-1/2*(d*x^2)^(1/2)*(d*x^2-2)^(1/2)/b/d/x/(a+b*arccosh(d*x^2-1
))

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Rubi [A]  time = 0.02, antiderivative size = 150, normalized size of antiderivative = 1.00, 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 = {5888} \[ \frac {x \cosh \left (\frac {a}{2 b}\right ) \text {Chi}\left (\frac {a+b \cosh ^{-1}\left (d x^2-1\right )}{2 b}\right )}{2 \sqrt {2} b^2 \sqrt {d x^2}}-\frac {x \sinh \left (\frac {a}{2 b}\right ) \text {Shi}\left (\frac {a+b \cosh ^{-1}\left (d x^2-1\right )}{2 b}\right )}{2 \sqrt {2} b^2 \sqrt {d x^2}}-\frac {\sqrt {d x^2} \sqrt {d x^2-2}}{2 b d x \left (a+b \cosh ^{-1}\left (d x^2-1\right )\right )} \]

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

Rule 5888

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*Cosh[a/(2*b)]*CoshIntegral[(a + b*ArcCosh[-1 + d*x^2])/(2*b)])/(
2*Sqrt[2]*b^2*Sqrt[d*x^2]), x] - Simp[(x*Sinh[a/(2*b)]*SinhIntegral[(a + b*ArcCosh[-1 + d*x^2])/(2*b)])/(2*Sqr
t[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 \cosh \left (\frac {a}{2 b}\right ) \text {Chi}\left (\frac {a+b \cosh ^{-1}\left (-1+d x^2\right )}{2 b}\right )}{2 \sqrt {2} b^2 \sqrt {d x^2}}-\frac {x \sinh \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*}

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Mathematica [A]  time = 0.75, size = 141, normalized size = 0.94 \[ \frac {\frac {\sinh \left (\frac {1}{2} \cosh ^{-1}\left (d x^2-1\right )\right ) \left (\cosh \left (\frac {a}{2 b}\right ) \text {Chi}\left (\frac {a+b \cosh ^{-1}\left (d x^2-1\right )}{2 b}\right )-\sinh \left (\frac {a}{2 b}\right ) \text {Shi}\left (\frac {a+b \cosh ^{-1}\left (d x^2-1\right )}{2 b}\right )\right )}{\sqrt {1-\frac {2}{d x^2}}}-\frac {b \sqrt {d x^2} \sqrt {d x^2-2}}{a+b \cosh ^{-1}\left (d x^2-1\right )}}{2 b^2 d x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{b^{2} \operatorname {arcosh}\left (d x^{2} - 1\right )^{2} + 2 \, a b \operatorname {arcosh}\left (d x^{2} - 1\right ) + a^{2}}, x\right ) \]

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

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)

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a +b \,\mathrm {arccosh}\left (d \,x^{2}-1\right )\right )^{2}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ -\frac {d^{2} x^{4} - 3 \, d x^{2} + {\left (d^{\frac {3}{2}} x^{3} - 2 \, \sqrt {d} x\right )} \sqrt {d x^{2} - 2} + 2}{2 \, {\left (a b d^{2} x^{3} - 2 \, a b d x + {\left (b^{2} d^{2} x^{3} - 2 \, b^{2} d x + {\left (b^{2} d^{\frac {3}{2}} x^{2} - b^{2} \sqrt {d}\right )} \sqrt {d x^{2} - 2}\right )} \log \left (d x^{2} + \sqrt {d x^{2} - 2} \sqrt {d} x - 1\right ) + {\left (a b d^{\frac {3}{2}} x^{2} - a b \sqrt {d}\right )} \sqrt {d x^{2} - 2}\right )}} + \int \frac {d^{3} x^{6} - 3 \, d^{2} x^{4} + {\left (d^{2} x^{4} - d x^{2} + 2\right )} {\left (d x^{2} - 2\right )} + {\left (2 \, d^{\frac {5}{2}} x^{5} - 4 \, d^{\frac {3}{2}} x^{3} + \sqrt {d} x\right )} \sqrt {d x^{2} - 2} + 4}{2 \, {\left (a b d^{3} x^{6} - 4 \, a b d^{2} x^{4} + 4 \, a b d x^{2} + {\left (a b d^{2} x^{4} - 2 \, a b d x^{2} + a b\right )} {\left (d x^{2} - 2\right )} + {\left (b^{2} d^{3} x^{6} - 4 \, b^{2} d^{2} x^{4} + 4 \, b^{2} d x^{2} + {\left (b^{2} d^{2} x^{4} - 2 \, b^{2} d x^{2} + b^{2}\right )} {\left (d x^{2} - 2\right )} + 2 \, {\left (b^{2} d^{\frac {5}{2}} x^{5} - 3 \, b^{2} d^{\frac {3}{2}} x^{3} + 2 \, b^{2} \sqrt {d} x\right )} \sqrt {d x^{2} - 2}\right )} \log \left (d x^{2} + \sqrt {d x^{2} - 2} \sqrt {d} x - 1\right ) + 2 \, {\left (a b d^{\frac {5}{2}} x^{5} - 3 \, a b d^{\frac {3}{2}} x^{3} + 2 \, a b \sqrt {d} x\right )} \sqrt {d x^{2} - 2}\right )}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a+b\,\mathrm {acosh}\left (d\,x^2-1\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*acosh(d*x^2 - 1))^2,x)

[Out]

int(1/(a + b*acosh(d*x^2 - 1))^2, x)

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

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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