3.246 \(\int \frac{1}{(a+b \cosh ^{-1}(1+d x^2))^3} \, dx\)
Optimal. Leaf size=180 \[ \frac{x \cosh \left (\frac{a}{2 b}\right ) \text{Chi}\left (\frac{a+b \cosh ^{-1}\left (d x^2+1\right )}{2 b}\right )}{8 \sqrt{2} b^3 \sqrt{d x^2}}-\frac{x \sinh \left (\frac{a}{2 b}\right ) \text{Shi}\left (\frac{a+b \cosh ^{-1}\left (d x^2+1\right )}{2 b}\right )}{8 \sqrt{2} b^3 \sqrt{d x^2}}-\frac{x}{8 b^2 \left (a+b \cosh ^{-1}\left (d x^2+1\right )\right )}-\frac{d x^4+2 x^2}{4 b x \sqrt{d x^2} \sqrt{d x^2+2} \left (a+b \cosh ^{-1}\left (d x^2+1\right )\right )^2} \]
[Out]
-(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*Cosh[a/(2*b)]*CoshIntegral[(a + b*ArcCosh[1 + d*x^2])/(2*b)])/(8*Sqrt[2]*b^3*Sqrt[d*x^2]) - (x
*Sinh[a/(2*b)]*SinhIntegral[(a + b*ArcCosh[1 + d*x^2])/(2*b)])/(8*Sqrt[2]*b^3*Sqrt[d*x^2])
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Rubi [A] time = 0.0380817, antiderivative size = 180, normalized size of antiderivative = 1.,
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, 5881} \[ \frac{x \cosh \left (\frac{a}{2 b}\right ) \text{Chi}\left (\frac{a+b \cosh ^{-1}\left (d x^2+1\right )}{2 b}\right )}{8 \sqrt{2} b^3 \sqrt{d x^2}}-\frac{x \sinh \left (\frac{a}{2 b}\right ) \text{Shi}\left (\frac{a+b \cosh ^{-1}\left (d x^2+1\right )}{2 b}\right )}{8 \sqrt{2} b^3 \sqrt{d x^2}}-\frac{x}{8 b^2 \left (a+b \cosh ^{-1}\left (d x^2+1\right )\right )}-\frac{d x^4+2 x^2}{4 b x \sqrt{d x^2} \sqrt{d x^2+2} \left (a+b \cosh ^{-1}\left (d x^2+1\right )\right )^2} \]
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*Cosh[a/(2*b)]*CoshIntegral[(a + b*ArcCosh[1 + d*x^2])/(2*b)])/(8*Sqrt[2]*b^3*Sqrt[d*x^2]) - (x
*Sinh[a/(2*b)]*SinhIntegral[(a + b*ArcCosh[1 + d*x^2])/(2*b)])/(8*Sqrt[2]*b^3*Sqrt[d*x^2])
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]
Rule 5881
Int[((a_.) + ArcCosh[1 + (d_.)*(x_)^2]*(b_.))^(-1), x_Symbol] :> Simp[(x*Cosh[a/(2*b)]*CoshIntegral[(a + b*Arc
Cosh[1 + d*x^2])/(2*b)])/(Sqrt[2]*b*Sqrt[d*x^2]), x] - Simp[(x*Sinh[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]
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 \cosh \left (\frac{a}{2 b}\right ) \text{Chi}\left (\frac{a+b \cosh ^{-1}\left (1+d x^2\right )}{2 b}\right )}{8 \sqrt{2} b^3 \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 )}{8 \sqrt{2} b^3 \sqrt{d x^2}}\\ \end{align*}
Mathematica [A] time = 0.457619, size = 152, normalized size = 0.84 \[ \frac{-\frac{2 b^2 \sqrt{d x^2} \sqrt{d x^2+2}}{d \left (a+b \cosh ^{-1}\left (d x^2+1\right )\right )^2}+\frac{\sinh \left (\frac{1}{2} \cosh ^{-1}\left (d x^2+1\right )\right ) \left (\cosh \left (\frac{a}{2 b}\right ) \text{Chi}\left (\frac{a+b \cosh ^{-1}\left (d x^2+1\right )}{2 b}\right )-\sinh \left (\frac{a}{2 b}\right ) \text{Shi}\left (\frac{a+b \cosh ^{-1}\left (d x^2+1\right )}{2 b}\right )\right )}{d}-\frac{b x^2}{a+b \cosh ^{-1}\left (d x^2+1\right )}}{8 b^3 x} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*ArcCosh[1 + d*x^2])^(-3),x]
[Out]
((-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]) +
(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)]*Sinh
Integral[(a + b*ArcCosh[1 + d*x^2])/(2*b)]))/d)/(8*b^3*x)
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Maple [F] time = 0.064, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b{\rm arccosh} \left (d{x}^{2}+1\right ) \right ) ^{-3}\, dx \end{align*}
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., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}
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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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}
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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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 )^{3}}\, dx \end{align*}
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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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 )}^{3}}\,{d x} \end{align*}
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)