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

Optimal. Leaf size=145 \[ -\frac{2 \sqrt{\pi } \sqrt{\frac{1}{b}} \cos \left (\frac{a}{2 b}\right ) \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )}{d x}-\frac{2 \sqrt{\pi } \sqrt{\frac{1}{b}} \sin \left (\frac{a}{2 b}\right ) \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) S\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )}{d x} \]

[Out]

(-2*Sqrt[b^(-1)]*Sqrt[Pi]*Cos[a/(2*b)]*FresnelC[(Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[1 + d*x^2]])/Sqrt[Pi]]*Sin[Arc
Cos[1 + d*x^2]/2])/(d*x) - (2*Sqrt[b^(-1)]*Sqrt[Pi]*FresnelS[(Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[1 + d*x^2]])/Sqrt
[Pi]]*Sin[a/(2*b)]*Sin[ArcCos[1 + d*x^2]/2])/(d*x)

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Rubi [A]  time = 0.0175331, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.062, Rules used = {4820} \[ -\frac{2 \sqrt{\pi } \sqrt{\frac{1}{b}} \cos \left (\frac{a}{2 b}\right ) \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )}{d x}-\frac{2 \sqrt{\pi } \sqrt{\frac{1}{b}} \sin \left (\frac{a}{2 b}\right ) \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) S\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )}{d x} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a + b*ArcCos[1 + d*x^2]],x]

[Out]

(-2*Sqrt[b^(-1)]*Sqrt[Pi]*Cos[a/(2*b)]*FresnelC[(Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[1 + d*x^2]])/Sqrt[Pi]]*Sin[Arc
Cos[1 + d*x^2]/2])/(d*x) - (2*Sqrt[b^(-1)]*Sqrt[Pi]*FresnelS[(Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[1 + d*x^2]])/Sqrt
[Pi]]*Sin[a/(2*b)]*Sin[ArcCos[1 + d*x^2]/2])/(d*x)

Rule 4820

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

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{a+b \cos ^{-1}\left (1+d x^2\right )}} \, dx &=-\frac{2 \sqrt{\frac{1}{b}} \sqrt{\pi } \cos \left (\frac{a}{2 b}\right ) C\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (1+d x^2\right )}}{\sqrt{\pi }}\right ) \sin \left (\frac{1}{2} \cos ^{-1}\left (1+d x^2\right )\right )}{d x}-\frac{2 \sqrt{\frac{1}{b}} \sqrt{\pi } S\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (1+d x^2\right )}}{\sqrt{\pi }}\right ) \sin \left (\frac{a}{2 b}\right ) \sin \left (\frac{1}{2} \cos ^{-1}\left (1+d x^2\right )\right )}{d x}\\ \end{align*}

Mathematica [A]  time = 0.161629, size = 114, normalized size = 0.79 \[ -\frac{2 \sqrt{\pi } \sqrt{\frac{1}{b}} \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \left (\cos \left (\frac{a}{2 b}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )+\sin \left (\frac{a}{2 b}\right ) S\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )\right )}{d x} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[a + b*ArcCos[1 + d*x^2]],x]

[Out]

(-2*Sqrt[b^(-1)]*Sqrt[Pi]*(Cos[a/(2*b)]*FresnelC[(Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[1 + d*x^2]])/Sqrt[Pi]] + Fres
nelS[(Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[1 + d*x^2]])/Sqrt[Pi]]*Sin[a/(2*b)])*Sin[ArcCos[1 + d*x^2]/2])/(d*x)

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Maple [F]  time = 0.069, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{\sqrt{a+b\arccos \left ( d{x}^{2}+1 \right ) }}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{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))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: UnboundLocalError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{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))**(1/2),x)

[Out]

Integral(1/sqrt(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}{\sqrt{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))^(1/2),x, algorithm="giac")

[Out]

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