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

Optimal. Leaf size=171 \[ -\frac {x \sin \left (\frac {a}{2 b}\right ) \text {Ci}\left (\frac {a+b \cos ^{-1}\left (d x^2-1\right )}{2 b}\right )}{8 \sqrt {2} b^3 \sqrt {d x^2}}+\frac {x \cos \left (\frac {a}{2 b}\right ) \text {Si}\left (\frac {a+b \cos ^{-1}\left (d x^2-1\right )}{2 b}\right )}{8 \sqrt {2} b^3 \sqrt {d x^2}}+\frac {x}{8 b^2 \left (a+b \cos ^{-1}\left (d x^2-1\right )\right )}+\frac {\sqrt {2 d x^2-d^2 x^4}}{4 b d x \left (a+b \cos ^{-1}\left (d x^2-1\right )\right )^2} \]

[Out]

1/8*x/b^2/(a+b*arccos(d*x^2-1))+1/16*x*cos(1/2*a/b)*Si(1/2*(a+b*arccos(d*x^2-1))/b)/b^3*2^(1/2)/(d*x^2)^(1/2)-
1/16*x*Ci(1/2*(a+b*arccos(d*x^2-1))/b)*sin(1/2*a/b)/b^3*2^(1/2)/(d*x^2)^(1/2)+1/4*(-d^2*x^4+2*d*x^2)^(1/2)/b/d
/x/(a+b*arccos(d*x^2-1))^2

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Rubi [A]  time = 0.03, antiderivative size = 171, 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 = {4829, 4818} \[ -\frac {x \sin \left (\frac {a}{2 b}\right ) \text {CosIntegral}\left (\frac {a+b \cos ^{-1}\left (d x^2-1\right )}{2 b}\right )}{8 \sqrt {2} b^3 \sqrt {d x^2}}+\frac {x \cos \left (\frac {a}{2 b}\right ) \text {Si}\left (\frac {a+b \cos ^{-1}\left (d x^2-1\right )}{2 b}\right )}{8 \sqrt {2} b^3 \sqrt {d x^2}}+\frac {x}{8 b^2 \left (a+b \cos ^{-1}\left (d x^2-1\right )\right )}+\frac {\sqrt {2 d x^2-d^2 x^4}}{4 b d x \left (a+b \cos ^{-1}\left (d x^2-1\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[2*d*x^2 - d^2*x^4]/(4*b*d*x*(a + b*ArcCos[-1 + d*x^2])^2) + x/(8*b^2*(a + b*ArcCos[-1 + d*x^2])) - (x*Cos
Integral[(a + b*ArcCos[-1 + d*x^2])/(2*b)]*Sin[a/(2*b)])/(8*Sqrt[2]*b^3*Sqrt[d*x^2]) + (x*Cos[a/(2*b)]*SinInte
gral[(a + b*ArcCos[-1 + d*x^2])/(2*b)])/(8*Sqrt[2]*b^3*Sqrt[d*x^2])

Rule 4818

Int[((a_.) + ArcCos[-1 + (d_.)*(x_)^2]*(b_.))^(-1), x_Symbol] :> Simp[(x*Sin[a/(2*b)]*CosIntegral[(a + b*ArcCo
s[-1 + d*x^2])/(2*b)])/(Sqrt[2]*b*Sqrt[d*x^2]), x] - Simp[(x*Cos[a/(2*b)]*SinIntegral[(a + b*ArcCos[-1 + d*x^2
])/(2*b)])/(Sqrt[2]*b*Sqrt[d*x^2]), x] /; FreeQ[{a, b, d}, x]

Rule 4829

Int[((a_.) + ArcCos[(c_) + (d_.)*(x_)^2]*(b_.))^(n_), x_Symbol] :> Simp[(x*(a + b*ArcCos[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*ArcCos[c + d*x^2])^(n + 2), x], x]
- Simp[(Sqrt[-2*c*d*x^2 - d^2*x^4]*(a + b*ArcCos[c + d*x^2])^(n + 1))/(2*b*d*(n + 1)*x), 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 \cos ^{-1}\left (-1+d x^2\right )\right )^3} \, dx &=\frac {\sqrt {2 d x^2-d^2 x^4}}{4 b d x \left (a+b \cos ^{-1}\left (-1+d x^2\right )\right )^2}+\frac {x}{8 b^2 \left (a+b \cos ^{-1}\left (-1+d x^2\right )\right )}-\frac {\int \frac {1}{a+b \cos ^{-1}\left (-1+d x^2\right )} \, dx}{8 b^2}\\ &=\frac {\sqrt {2 d x^2-d^2 x^4}}{4 b d x \left (a+b \cos ^{-1}\left (-1+d x^2\right )\right )^2}+\frac {x}{8 b^2 \left (a+b \cos ^{-1}\left (-1+d x^2\right )\right )}-\frac {x \text {Ci}\left (\frac {a+b \cos ^{-1}\left (-1+d x^2\right )}{2 b}\right ) \sin \left (\frac {a}{2 b}\right )}{8 \sqrt {2} b^3 \sqrt {d x^2}}+\frac {x \cos \left (\frac {a}{2 b}\right ) \text {Si}\left (\frac {a+b \cos ^{-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 = 0.22, size = 149, normalized size = 0.87 \[ \frac {\frac {2 b^2 \sqrt {-d x^2 \left (d x^2-2\right )}}{d \left (a+b \cos ^{-1}\left (d x^2-1\right )\right )^2}-\frac {\cos \left (\frac {1}{2} \cos ^{-1}\left (d x^2-1\right )\right ) \left (\sin \left (\frac {a}{2 b}\right ) \text {Ci}\left (\frac {a+b \cos ^{-1}\left (d x^2-1\right )}{2 b}\right )-\cos \left (\frac {a}{2 b}\right ) \text {Si}\left (\frac {a+b \cos ^{-1}\left (d x^2-1\right )}{2 b}\right )\right )}{d}+\frac {b x^2}{a+b \cos ^{-1}\left (d x^2-1\right )}}{8 b^3 x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((2*b^2*Sqrt[-(d*x^2*(-2 + d*x^2))])/(d*(a + b*ArcCos[-1 + d*x^2])^2) + (b*x^2)/(a + b*ArcCos[-1 + d*x^2]) - (
Cos[ArcCos[-1 + d*x^2]/2]*(CosIntegral[(a + b*ArcCos[-1 + d*x^2])/(2*b)]*Sin[a/(2*b)] - Cos[a/(2*b)]*SinIntegr
al[(a + b*ArcCos[-1 + d*x^2])/(2*b)]))/d)/(8*b^3*x)

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{b^{3} \arccos \left (d x^{2} - 1\right )^{3} + 3 \, a b^{2} \arccos \left (d x^{2} - 1\right )^{2} + 3 \, a^{2} b \arccos \left (d x^{2} - 1\right ) + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a +b \arccos \left (d \,x^{2}-1\right )\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {b d x \arctan \left (\sqrt {-d x^{2} + 2} \sqrt {d} x, d x^{2} - 1\right ) + a d x + 2 \, \sqrt {-d x^{2} + 2} b \sqrt {d} - {\left (b^{4} d \arctan \left (\sqrt {-d x^{2} + 2} \sqrt {d} x, d x^{2} - 1\right )^{2} + 2 \, a b^{3} d \arctan \left (\sqrt {-d x^{2} + 2} \sqrt {d} x, d x^{2} - 1\right ) + a^{2} b^{2} d\right )} \int \frac {1}{b^{3} \arctan \left (\sqrt {-d x^{2} + 2} \sqrt {d} x, d x^{2} - 1\right ) + a b^{2}}\,{d x}}{8 \, {\left (b^{4} d \arctan \left (\sqrt {-d x^{2} + 2} \sqrt {d} x, d x^{2} - 1\right )^{2} + 2 \, a b^{3} d \arctan \left (\sqrt {-d x^{2} + 2} \sqrt {d} x, d x^{2} - 1\right ) + a^{2} b^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(b*d*x*arctan2(sqrt(-d*x^2 + 2)*sqrt(d)*x, d*x^2 - 1) + a*d*x + 2*sqrt(-d*x^2 + 2)*b*sqrt(d) - 8*(b^4*d*ar
ctan2(sqrt(-d*x^2 + 2)*sqrt(d)*x, d*x^2 - 1)^2 + 2*a*b^3*d*arctan2(sqrt(-d*x^2 + 2)*sqrt(d)*x, d*x^2 - 1) + a^
2*b^2*d)*integrate(1/8/(b^3*arctan2(sqrt(-d*x^2 + 2)*sqrt(d)*x, d*x^2 - 1) + a*b^2), x))/(b^4*d*arctan2(sqrt(-
d*x^2 + 2)*sqrt(d)*x, d*x^2 - 1)^2 + 2*a*b^3*d*arctan2(sqrt(-d*x^2 + 2)*sqrt(d)*x, d*x^2 - 1) + a^2*b^2*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*acos(d*x^2 - 1))^3,x)

[Out]

int(1/(a + b*acos(d*x^2 - 1))^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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