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

Optimal. Leaf size=99 \[ \frac{x \cos \left (\frac{a}{2 b}\right ) \text{CosIntegral}\left (\frac{a+b \cos ^{-1}\left (d x^2+1\right )}{2 b}\right )}{\sqrt{2} b \sqrt{-d x^2}}+\frac{x \sin \left (\frac{a}{2 b}\right ) \text{Si}\left (\frac{a+b \cos ^{-1}\left (d x^2+1\right )}{2 b}\right )}{\sqrt{2} b \sqrt{-d x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0311988, antiderivative size = 99, 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 = {4817} \[ \frac{x \cos \left (\frac{a}{2 b}\right ) \text{CosIntegral}\left (\frac{a+b \cos ^{-1}\left (d x^2+1\right )}{2 b}\right )}{\sqrt{2} b \sqrt{-d x^2}}+\frac{x \sin \left (\frac{a}{2 b}\right ) \text{Si}\left (\frac{a+b \cos ^{-1}\left (d x^2+1\right )}{2 b}\right )}{\sqrt{2} b \sqrt{-d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCos[1 + d*x^2])^(-1),x]

[Out]

(x*Cos[a/(2*b)]*CosIntegral[(a + b*ArcCos[1 + d*x^2])/(2*b)])/(Sqrt[2]*b*Sqrt[-(d*x^2)]) + (x*Sin[a/(2*b)]*Sin
Integral[(a + b*ArcCos[1 + d*x^2])/(2*b)])/(Sqrt[2]*b*Sqrt[-(d*x^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}{a+b \cos ^{-1}\left (1+d x^2\right )} \, dx &=\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 )}{\sqrt{2} b \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 )}{\sqrt{2} b \sqrt{-d x^2}}\\ \end{align*}

Mathematica [A]  time = 0.103861, size = 85, normalized size = 0.86 \[ -\frac{\sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \left (\cos \left (\frac{a}{2 b}\right ) \text{CosIntegral}\left (\frac{a+b \cos ^{-1}\left (d x^2+1\right )}{2 b}\right )+\sin \left (\frac{a}{2 b}\right ) \text{Si}\left (\frac{a+b \cos ^{-1}\left (d x^2+1\right )}{2 b}\right )\right )}{b d x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((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)]*SinInteg
ral[(a + b*ArcCos[1 + d*x^2])/(2*b)]))/(b*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 ) ^{-1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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