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

Optimal. Leaf size=151 \[ \frac{x \sin \left (\frac{a}{2 b}\right ) \text{CosIntegral}\left (\frac{a+b \cos ^{-1}\left (d x^2+1\right )}{2 b}\right )}{2 \sqrt{2} b^2 \sqrt{-d x^2}}-\frac{x \cos \left (\frac{a}{2 b}\right ) \text{Si}\left (\frac{a+b \cos ^{-1}\left (d x^2+1\right )}{2 b}\right )}{2 \sqrt{2} b^2 \sqrt{-d x^2}}+\frac{\sqrt{-d^2 x^4-2 d x^2}}{2 b d x \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )} \]

[Out]

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

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Rubi [A]  time = 0.0203403, antiderivative size = 151, normalized size of antiderivative = 1., 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 = {4826} \[ \frac{x \sin \left (\frac{a}{2 b}\right ) \text{CosIntegral}\left (\frac{a+b \cos ^{-1}\left (d x^2+1\right )}{2 b}\right )}{2 \sqrt{2} b^2 \sqrt{-d x^2}}-\frac{x \cos \left (\frac{a}{2 b}\right ) \text{Si}\left (\frac{a+b \cos ^{-1}\left (d x^2+1\right )}{2 b}\right )}{2 \sqrt{2} b^2 \sqrt{-d x^2}}+\frac{\sqrt{-d^2 x^4-2 d x^2}}{2 b d x \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )} \]

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

Rule 4826

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*Sin[a/(2*b)]*CosIntegral[(a + b*ArcCos[1 + d*x^2])/(2*b)])/(2*Sqrt[2]*b^2*S
qrt[(-d)*x^2]), x] - Simp[(x*Cos[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 \text{Ci}\left (\frac{a+b \cos ^{-1}\left (1+d x^2\right )}{2 b}\right ) \sin \left (\frac{a}{2 b}\right )}{2 \sqrt{2} b^2 \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 )}{2 \sqrt{2} b^2 \sqrt{-d x^2}}\\ \end{align*}

Mathematica [A]  time = 0.332094, size = 133, normalized size = 0.88 \[ \frac{\sqrt{-d x^2 \left (d x^2+2\right )} \left (\frac{b}{a+b \cos ^{-1}\left (d x^2+1\right )}-\frac{\cos \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \left (\sin \left (\frac{a}{2 b}\right ) \text{CosIntegral}\left (\frac{a+b \cos ^{-1}\left (d x^2+1\right )}{2 b}\right )-\cos \left (\frac{a}{2 b}\right ) \text{Si}\left (\frac{a+b \cos ^{-1}\left (d x^2+1\right )}{2 b}\right )\right )}{d x^2+2}\right )}{2 b^2 d x} \]

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]) - (Cos[ArcCos[1 + d*x^2]/2]*(CosIntegral[(a + b*ArcCo
s[1 + d*x^2])/(2*b)]*Sin[a/(2*b)] - Cos[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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Maple [F]  time = 0.066, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\arccos \left ( d{x}^{2}+1 \right ) \right ) ^{-2}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} -\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 )}} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b \operatorname{acos}{\left (d x^{2} + 1 \right )}\right )^{2}}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \arccos \left (d x^{2} + 1\right ) + a\right )}^{2}}\,{d x} \end{align*}

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)