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

Optimal. Leaf size=149 \[ -\frac {x \cos \left (\frac {a}{2 b}\right ) \text {Ci}\left (\frac {a+b \cos ^{-1}\left (d x^2-1\right )}{2 b}\right )}{2 \sqrt {2} b^2 \sqrt {d x^2}}-\frac {x \sin \left (\frac {a}{2 b}\right ) \text {Si}\left (\frac {a+b \cos ^{-1}\left (d x^2-1\right )}{2 b}\right )}{2 \sqrt {2} b^2 \sqrt {d x^2}}+\frac {\sqrt {2 d x^2-d^2 x^4}}{2 b d x \left (a+b \cos ^{-1}\left (d x^2-1\right )\right )} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 149, 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 = {4827} \[ -\frac {x \cos \left (\frac {a}{2 b}\right ) \text {CosIntegral}\left (\frac {a+b \cos ^{-1}\left (d x^2-1\right )}{2 b}\right )}{2 \sqrt {2} b^2 \sqrt {d x^2}}-\frac {x \sin \left (\frac {a}{2 b}\right ) \text {Si}\left (\frac {a+b \cos ^{-1}\left (d x^2-1\right )}{2 b}\right )}{2 \sqrt {2} b^2 \sqrt {d x^2}}+\frac {\sqrt {2 d x^2-d^2 x^4}}{2 b d x \left (a+b \cos ^{-1}\left (d x^2-1\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[2*d*x^2 - d^2*x^4]/(2*b*d*x*(a + b*ArcCos[-1 + d*x^2])) - (x*Cos[a/(2*b)]*CosIntegral[(a + b*ArcCos[-1 +
d*x^2])/(2*b)])/(2*Sqrt[2]*b^2*Sqrt[d*x^2]) - (x*Sin[a/(2*b)]*SinIntegral[(a + b*ArcCos[-1 + d*x^2])/(2*b)])/(
2*Sqrt[2]*b^2*Sqrt[d*x^2])

Rule 4827

Int[((a_.) + ArcCos[-1 + (d_.)*(x_)^2]*(b_.))^(-2), x_Symbol] :> Simp[Sqrt[2*d*x^2 - d^2*x^4]/(2*b*d*x*(a + b*
ArcCos[-1 + d*x^2])), x] + (-Simp[(x*Cos[a/(2*b)]*CosIntegral[(a + b*ArcCos[-1 + d*x^2])/(2*b)])/(2*Sqrt[2]*b^
2*Sqrt[d*x^2]), x] - Simp[(x*Sin[a/(2*b)]*SinIntegral[(a + b*ArcCos[-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 \cos ^{-1}\left (-1+d x^2\right )\right )^2} \, dx &=\frac {\sqrt {2 d x^2-d^2 x^4}}{2 b d x \left (a+b \cos ^{-1}\left (-1+d x^2\right )\right )}-\frac {x \cos \left (\frac {a}{2 b}\right ) \text {Ci}\left (\frac {a+b \cos ^{-1}\left (-1+d x^2\right )}{2 b}\right )}{2 \sqrt {2} b^2 \sqrt {d x^2}}-\frac {x \sin \left (\frac {a}{2 b}\right ) \text {Si}\left (\frac {a+b \cos ^{-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 = 0.40, size = 131, normalized size = 0.88 \[ \frac {\sqrt {-d x^2 \left (d x^2-2\right )} \left (\frac {\sin \left (\frac {1}{2} \cos ^{-1}\left (d x^2-1\right )\right ) \left (\cos \left (\frac {a}{2 b}\right ) \text {Ci}\left (\frac {a+b \cos ^{-1}\left (d x^2-1\right )}{2 b}\right )+\sin \left (\frac {a}{2 b}\right ) \text {Si}\left (\frac {a+b \cos ^{-1}\left (d x^2-1\right )}{2 b}\right )\right )}{d x^2-2}+\frac {b}{a+b \cos ^{-1}\left (d x^2-1\right )}\right )}{2 b^2 d x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[-(d*x^2*(-2 + d*x^2))]*(b/(a + b*ArcCos[-1 + d*x^2]) + (Sin[ArcCos[-1 + d*x^2]/2]*(Cos[a/(2*b)]*CosInteg
ral[(a + b*ArcCos[-1 + d*x^2])/(2*b)] + Sin[a/(2*b)]*SinIntegral[(a + b*ArcCos[-1 + d*x^2])/(2*b)]))/(-2 + d*x
^2)))/(2*b^2*d*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(b^2*d*arctan2(sqrt(-d*x^2 + 2)*sqrt(d)*x, d*x^2 - 1) + a*b*d)*sqrt(d)*integrate(1/2*sqrt(-d*x^2 + 2)*
x/(a*b*d*x^2 - 2*a*b + (b^2*d*x^2 - 2*b^2)*arctan2(sqrt(-d*x^2 + 2)*sqrt(d)*x, d*x^2 - 1)), x) - sqrt(-d*x^2 +
2)*sqrt(d))/(b^2*d*arctan2(sqrt(-d*x^2 + 2)*sqrt(d)*x, d*x^2 - 1) + a*b*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*acos(d*x^2 - 1))^2,x)

[Out]

int(1/(a + b*acos(d*x^2 - 1))^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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