3.253 \(\int \frac {1}{(a+b \cosh ^{-1}(-1+d x^2))^3} \, dx\)

Optimal. Leaf size=181 \[ -\frac {x \sinh \left (\frac {a}{2 b}\right ) \text {Chi}\left (\frac {a+b \cosh ^{-1}\left (d x^2-1\right )}{2 b}\right )}{8 \sqrt {2} b^3 \sqrt {d x^2}}+\frac {x \cosh \left (\frac {a}{2 b}\right ) \text {Shi}\left (\frac {a+b \cosh ^{-1}\left (d x^2-1\right )}{2 b}\right )}{8 \sqrt {2} b^3 \sqrt {d x^2}}-\frac {x}{8 b^2 \left (a+b \cosh ^{-1}\left (d x^2-1\right )\right )}+\frac {2 x^2-d x^4}{4 b x \sqrt {d x^2} \sqrt {d x^2-2} \left (a+b \cosh ^{-1}\left (d x^2-1\right )\right )^2} \]

[Out]

-1/8*x/b^2/(a+b*arccosh(d*x^2-1))+1/16*x*cosh(1/2*a/b)*Shi(1/2*(a+b*arccosh(d*x^2-1))/b)/b^3*2^(1/2)/(d*x^2)^(
1/2)-1/16*x*Chi(1/2*(a+b*arccosh(d*x^2-1))/b)*sinh(1/2*a/b)/b^3*2^(1/2)/(d*x^2)^(1/2)+1/4*(-d*x^4+2*x^2)/b/x/(
a+b*arccosh(d*x^2-1))^2/(d*x^2)^(1/2)/(d*x^2-2)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5889, 5882} \[ -\frac {x \sinh \left (\frac {a}{2 b}\right ) \text {Chi}\left (\frac {a+b \cosh ^{-1}\left (d x^2-1\right )}{2 b}\right )}{8 \sqrt {2} b^3 \sqrt {d x^2}}+\frac {x \cosh \left (\frac {a}{2 b}\right ) \text {Shi}\left (\frac {a+b \cosh ^{-1}\left (d x^2-1\right )}{2 b}\right )}{8 \sqrt {2} b^3 \sqrt {d x^2}}-\frac {x}{8 b^2 \left (a+b \cosh ^{-1}\left (d x^2-1\right )\right )}+\frac {2 x^2-d x^4}{4 b x \sqrt {d x^2} \sqrt {d x^2-2} \left (a+b \cosh ^{-1}\left (d x^2-1\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCosh[-1 + d*x^2])^(-3),x]

[Out]

(2*x^2 - d*x^4)/(4*b*x*Sqrt[d*x^2]*Sqrt[-2 + d*x^2]*(a + b*ArcCosh[-1 + d*x^2])^2) - x/(8*b^2*(a + b*ArcCosh[-
1 + d*x^2])) - (x*CoshIntegral[(a + b*ArcCosh[-1 + d*x^2])/(2*b)]*Sinh[a/(2*b)])/(8*Sqrt[2]*b^3*Sqrt[d*x^2]) +
(x*Cosh[a/(2*b)]*SinhIntegral[(a + b*ArcCosh[-1 + d*x^2])/(2*b)])/(8*Sqrt[2]*b^3*Sqrt[d*x^2])

Rule 5882

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

Rule 5889

Int[((a_.) + ArcCosh[(c_) + (d_.)*(x_)^2]*(b_.))^(n_), x_Symbol] :> -Simp[(x*(a + b*ArcCosh[c + d*x^2])^(n + 2
))/(4*b^2*(n + 1)*(n + 2)), x] + (Dist[1/(4*b^2*(n + 1)*(n + 2)), Int[(a + b*ArcCosh[c + d*x^2])^(n + 2), x],
x] + Simp[((2*c*x^2 + d*x^4)*(a + b*ArcCosh[c + d*x^2])^(n + 1))/(2*b*(n + 1)*x*Sqrt[-1 + c + d*x^2]*Sqrt[1 +
c + d*x^2]), x]) /; FreeQ[{a, b, c, d}, x] && EqQ[c^2, 1] && LtQ[n, -1] && NeQ[n, -2]

Rubi steps

\begin {align*} \int \frac {1}{\left (a+b \cosh ^{-1}\left (-1+d x^2\right )\right )^3} \, dx &=\frac {2 x^2-d x^4}{4 b x \sqrt {d x^2} \sqrt {-2+d x^2} \left (a+b \cosh ^{-1}\left (-1+d x^2\right )\right )^2}-\frac {x}{8 b^2 \left (a+b \cosh ^{-1}\left (-1+d x^2\right )\right )}+\frac {\int \frac {1}{a+b \cosh ^{-1}\left (-1+d x^2\right )} \, dx}{8 b^2}\\ &=\frac {2 x^2-d x^4}{4 b x \sqrt {d x^2} \sqrt {-2+d x^2} \left (a+b \cosh ^{-1}\left (-1+d x^2\right )\right )^2}-\frac {x}{8 b^2 \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 )}{8 \sqrt {2} b^3 \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 )}{8 \sqrt {2} b^3 \sqrt {d x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.69, size = 168, normalized size = 0.93 \[ -\frac {\frac {2 b^2 \sqrt {d x^2} \sqrt {d x^2-2}}{d \left (a+b \cosh ^{-1}\left (d x^2-1\right )\right )^2}+\frac {1}{2} x^2 \sqrt {1-\frac {2}{d x^2}} \text {csch}\left (\frac {1}{2} \cosh ^{-1}\left (d x^2-1\right )\right ) \left (\sinh \left (\frac {a}{2 b}\right ) \text {Chi}\left (\frac {a+b \cosh ^{-1}\left (d x^2-1\right )}{2 b}\right )-\cosh \left (\frac {a}{2 b}\right ) \text {Shi}\left (\frac {a+b \cosh ^{-1}\left (d x^2-1\right )}{2 b}\right )\right )+\frac {b x^2}{a+b \cosh ^{-1}\left (d x^2-1\right )}}{8 b^3 x} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcCosh[-1 + d*x^2])^(-3),x]

[Out]

-1/8*((2*b^2*Sqrt[d*x^2]*Sqrt[-2 + d*x^2])/(d*(a + b*ArcCosh[-1 + d*x^2])^2) + (b*x^2)/(a + b*ArcCosh[-1 + d*x
^2]) + (Sqrt[1 - 2/(d*x^2)]*x^2*Csch[ArcCosh[-1 + d*x^2]/2]*(CoshIntegral[(a + b*ArcCosh[-1 + d*x^2])/(2*b)]*S
inh[a/(2*b)] - Cosh[a/(2*b)]*SinhIntegral[(a + b*ArcCosh[-1 + d*x^2])/(2*b)]))/2)/(b^3*x)

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fricas [F]  time = 0.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{b^{3} \operatorname {arcosh}\left (d x^{2} - 1\right )^{3} + 3 \, a b^{2} \operatorname {arcosh}\left (d x^{2} - 1\right )^{2} + 3 \, a^{2} b \operatorname {arcosh}\left (d x^{2} - 1\right ) + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arccosh(d*x^2-1))^3,x, algorithm="fricas")

[Out]

integral(1/(b^3*arccosh(d*x^2 - 1)^3 + 3*a*b^2*arccosh(d*x^2 - 1)^2 + 3*a^2*b*arccosh(d*x^2 - 1) + a^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \operatorname {arcosh}\left (d x^{2} - 1\right ) + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arccosh(d*x^2-1))^3,x, algorithm="giac")

[Out]

integrate((b*arccosh(d*x^2 - 1) + a)^(-3), x)

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a +b \,\mathrm {arccosh}\left (d \,x^{2}-1\right )\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*arccosh(d*x^2-1))^3,x)

[Out]

int(1/(a+b*arccosh(d*x^2-1))^3,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arccosh(d*x^2-1))^3,x, algorithm="maxima")

[Out]

-1/8*((a*d^5 + 2*b*d^5)*sqrt(d)*x^10 - 2*(3*a*d^4 + 7*b*d^4)*sqrt(d)*x^8 + (11*a*d^3 + 36*b*d^3)*sqrt(d)*x^6 -
2*(a*d^2 + 20*b*d^2)*sqrt(d)*x^4 - 4*(3*a*d - 4*b*d)*sqrt(d)*x^2 + ((a*d^4 + 2*b*d^4)*x^7 - (3*a*d^3 + 8*b*d^
3)*x^5 + 2*(2*a*d^2 + 5*b*d^2)*x^3 - 4*(a*d + b*d)*x)*(d*x^2 - 2)^(3/2) + (3*(a*d^4 + 2*b*d^4)*sqrt(d)*x^8 - 6
*(2*a*d^3 + 5*b*d^3)*sqrt(d)*x^6 + 2*(8*a*d^2 + 25*b*d^2)*sqrt(d)*x^4 - 10*(a*d + 3*b*d)*sqrt(d)*x^2 + 4*(a +
b)*sqrt(d))*(d*x^2 - 2) + (b*d^(11/2)*x^10 - 6*b*d^(9/2)*x^8 + 11*b*d^(7/2)*x^6 - 2*b*d^(5/2)*x^4 - 12*b*d^(3/
2)*x^2 + (b*d^4*x^7 - 3*b*d^3*x^5 + 4*b*d^2*x^3 - 4*b*d*x)*(d*x^2 - 2)^(3/2) + (3*b*d^(9/2)*x^8 - 12*b*d^(7/2)
*x^6 + 16*b*d^(5/2)*x^4 - 10*b*d^(3/2)*x^2 + 4*b*sqrt(d))*(d*x^2 - 2) + (3*b*d^5*x^9 - 15*b*d^4*x^7 + 23*b*d^3
*x^5 - 7*b*d^2*x^3 - 6*b*d*x)*sqrt(d*x^2 - 2) + 8*b*sqrt(d))*log(d*x^2 + sqrt(d*x^2 - 2)*sqrt(d)*x - 1) + (3*(
a*d^5 + 2*b*d^5)*x^9 - 3*(5*a*d^4 + 12*b*d^4)*x^7 + (23*a*d^3 + 76*b*d^3)*x^5 - (7*a*d^2 + 64*b*d^2)*x^3 - 2*(
3*a*d - 8*b*d)*x)*sqrt(d*x^2 - 2) + 8*a*sqrt(d))/(a^2*b^2*d^(11/2)*x^9 - 6*a^2*b^2*d^(9/2)*x^7 + 12*a^2*b^2*d^
(7/2)*x^5 - 8*a^2*b^2*d^(5/2)*x^3 + (b^4*d^(11/2)*x^9 - 6*b^4*d^(9/2)*x^7 + 12*b^4*d^(7/2)*x^5 - 8*b^4*d^(5/2)
*x^3 + (b^4*d^4*x^6 - 3*b^4*d^3*x^4 + 3*b^4*d^2*x^2 - b^4*d)*(d*x^2 - 2)^(3/2) + 3*(b^4*d^(9/2)*x^7 - 4*b^4*d^
(7/2)*x^5 + 5*b^4*d^(5/2)*x^3 - 2*b^4*d^(3/2)*x)*(d*x^2 - 2) + 3*(b^4*d^5*x^8 - 5*b^4*d^4*x^6 + 8*b^4*d^3*x^4
- 4*b^4*d^2*x^2)*sqrt(d*x^2 - 2))*log(d*x^2 + sqrt(d*x^2 - 2)*sqrt(d)*x - 1)^2 + (a^2*b^2*d^4*x^6 - 3*a^2*b^2*
d^3*x^4 + 3*a^2*b^2*d^2*x^2 - a^2*b^2*d)*(d*x^2 - 2)^(3/2) + 3*(a^2*b^2*d^(9/2)*x^7 - 4*a^2*b^2*d^(7/2)*x^5 +
5*a^2*b^2*d^(5/2)*x^3 - 2*a^2*b^2*d^(3/2)*x)*(d*x^2 - 2) + 2*(a*b^3*d^(11/2)*x^9 - 6*a*b^3*d^(9/2)*x^7 + 12*a*
b^3*d^(7/2)*x^5 - 8*a*b^3*d^(5/2)*x^3 + (a*b^3*d^4*x^6 - 3*a*b^3*d^3*x^4 + 3*a*b^3*d^2*x^2 - a*b^3*d)*(d*x^2 -
2)^(3/2) + 3*(a*b^3*d^(9/2)*x^7 - 4*a*b^3*d^(7/2)*x^5 + 5*a*b^3*d^(5/2)*x^3 - 2*a*b^3*d^(3/2)*x)*(d*x^2 - 2)
+ 3*(a*b^3*d^5*x^8 - 5*a*b^3*d^4*x^6 + 8*a*b^3*d^3*x^4 - 4*a*b^3*d^2*x^2)*sqrt(d*x^2 - 2))*log(d*x^2 + sqrt(d*
x^2 - 2)*sqrt(d)*x - 1) + 3*(a^2*b^2*d^5*x^8 - 5*a^2*b^2*d^4*x^6 + 8*a^2*b^2*d^3*x^4 - 4*a^2*b^2*d^2*x^2)*sqrt
(d*x^2 - 2)) + integrate(1/8*(d^6*x^12 - 8*d^5*x^10 + 27*d^4*x^8 - 56*d^3*x^6 + 88*d^2*x^4 + (d^4*x^8 - 4*d^3*
x^6 + 3*d^2*x^4 + 8*d*x^2 + 4)*(d*x^2 - 2)^2 - 96*d*x^2 + 2*(2*d^(9/2)*x^9 - 10*d^(7/2)*x^7 + 15*d^(5/2)*x^5 +
d^(3/2)*x^3 - 11*sqrt(d)*x)*(d*x^2 - 2)^(3/2) + 3*(2*d^5*x^10 - 12*d^4*x^8 + 26*d^3*x^6 - 24*d^2*x^4 + 3*d*x^
2 + 10)*(d*x^2 - 2) + 2*(2*d^(11/2)*x^11 - 14*d^(9/2)*x^9 + 39*d^(7/2)*x^7 - 61*d^(5/2)*x^5 + 61*d^(3/2)*x^3 -
30*sqrt(d)*x)*sqrt(d*x^2 - 2) + 48)/(a*b^2*d^6*x^12 - 8*a*b^2*d^5*x^10 + 24*a*b^2*d^4*x^8 - 32*a*b^2*d^3*x^6
+ 16*a*b^2*d^2*x^4 + (a*b^2*d^4*x^8 - 4*a*b^2*d^3*x^6 + 6*a*b^2*d^2*x^4 - 4*a*b^2*d*x^2 + a*b^2)*(d*x^2 - 2)^2
+ 4*(a*b^2*d^(9/2)*x^9 - 5*a*b^2*d^(7/2)*x^7 + 9*a*b^2*d^(5/2)*x^5 - 7*a*b^2*d^(3/2)*x^3 + 2*a*b^2*sqrt(d)*x)
*(d*x^2 - 2)^(3/2) + 6*(a*b^2*d^5*x^10 - 6*a*b^2*d^4*x^8 + 13*a*b^2*d^3*x^6 - 12*a*b^2*d^2*x^4 + 4*a*b^2*d*x^2
)*(d*x^2 - 2) + (b^3*d^6*x^12 - 8*b^3*d^5*x^10 + 24*b^3*d^4*x^8 - 32*b^3*d^3*x^6 + 16*b^3*d^2*x^4 + (b^3*d^4*x
^8 - 4*b^3*d^3*x^6 + 6*b^3*d^2*x^4 - 4*b^3*d*x^2 + b^3)*(d*x^2 - 2)^2 + 4*(b^3*d^(9/2)*x^9 - 5*b^3*d^(7/2)*x^7
+ 9*b^3*d^(5/2)*x^5 - 7*b^3*d^(3/2)*x^3 + 2*b^3*sqrt(d)*x)*(d*x^2 - 2)^(3/2) + 6*(b^3*d^5*x^10 - 6*b^3*d^4*x^
8 + 13*b^3*d^3*x^6 - 12*b^3*d^2*x^4 + 4*b^3*d*x^2)*(d*x^2 - 2) + 4*(b^3*d^(11/2)*x^11 - 7*b^3*d^(9/2)*x^9 + 18
*b^3*d^(7/2)*x^7 - 20*b^3*d^(5/2)*x^5 + 8*b^3*d^(3/2)*x^3)*sqrt(d*x^2 - 2))*log(d*x^2 + sqrt(d*x^2 - 2)*sqrt(d
)*x - 1) + 4*(a*b^2*d^(11/2)*x^11 - 7*a*b^2*d^(9/2)*x^9 + 18*a*b^2*d^(7/2)*x^7 - 20*a*b^2*d^(5/2)*x^5 + 8*a*b^
2*d^(3/2)*x^3)*sqrt(d*x^2 - 2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a+b\,\mathrm {acosh}\left (d\,x^2-1\right )\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*acosh(d*x^2 - 1))^3,x)

[Out]

int(1/(a + b*acosh(d*x^2 - 1))^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b \operatorname {acosh}{\left (d x^{2} - 1 \right )}\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*acosh(d*x**2-1))**3,x)

[Out]

Integral((a + b*acosh(d*x**2 - 1))**(-3), x)

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