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

Optimal. Leaf size=173 \[ -\frac{x \cos \left (\frac{a}{2 b}\right ) \text{CosIntegral}\left (\frac{a+b \cos ^{-1}\left (d x^2+1\right )}{2 b}\right )}{8 \sqrt{2} b^3 \sqrt{-d x^2}}-\frac{x \sin \left (\frac{a}{2 b}\right ) \text{Si}\left (\frac{a+b \cos ^{-1}\left (d x^2+1\right )}{2 b}\right )}{8 \sqrt{2} b^3 \sqrt{-d x^2}}+\frac{x}{8 b^2 \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )}+\frac{\sqrt{-d^2 x^4-2 d x^2}}{4 b d x \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )^2} \]

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

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Rubi [A]  time = 0.0362387, antiderivative size = 173, 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 = {4829, 4817} \[ -\frac{x \cos \left (\frac{a}{2 b}\right ) \text{CosIntegral}\left (\frac{a+b \cos ^{-1}\left (d x^2+1\right )}{2 b}\right )}{8 \sqrt{2} b^3 \sqrt{-d x^2}}-\frac{x \sin \left (\frac{a}{2 b}\right ) \text{Si}\left (\frac{a+b \cos ^{-1}\left (d x^2+1\right )}{2 b}\right )}{8 \sqrt{2} b^3 \sqrt{-d x^2}}+\frac{x}{8 b^2 \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )}+\frac{\sqrt{-d^2 x^4-2 d x^2}}{4 b d x \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )^2} \]

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

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]

Rule 4817

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

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 \cos \left (\frac{a}{2 b}\right ) \text{Ci}\left (\frac{a+b \cos ^{-1}\left (1+d x^2\right )}{2 b}\right )}{8 \sqrt{2} b^3 \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 )}{8 \sqrt{2} b^3 \sqrt{-d x^2}}\\ \end{align*}

Mathematica [A]  time = 0.246495, size = 147, normalized size = 0.85 \[ \frac{\frac{2 b^2 \sqrt{-d x^2 \left (d x^2+2\right )}}{d \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )^2}+\frac{\sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \left (\cos \left (\frac{a}{2 b}\right ) \text{CosIntegral}\left (\frac{a+b \cos ^{-1}\left (d x^2+1\right )}{2 b}\right )+\sin \left (\frac{a}{2 b}\right ) \text{Si}\left (\frac{a+b \cos ^{-1}\left (d x^2+1\right )}{2 b}\right )\right )}{d}+\frac{b x^2}{a+b \cos ^{-1}\left (d x^2+1\right )}}{8 b^3 x} \]

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

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Maple [F]  time = 0.067, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\arccos \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*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., size = 0, normalized size = 0. \begin{align*} \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 )}} \end{align*}

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

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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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 )^{3}}\, dx \end{align*}

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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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 )}^{3}}\,{d x} \end{align*}

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)