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

Optimal. Leaf size=184 \[ -\frac{2 \sqrt{\pi } \sin \left (\frac{a}{2 b}\right ) \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )}{\sqrt{\frac{1}{b}} d x}+\frac{2 \sqrt{\pi } \cos \left (\frac{a}{2 b}\right ) \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) S\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )}{\sqrt{\frac{1}{b}} d x}-\frac{2 \sin ^2\left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{d x} \]

[Out]

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

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Rubi [A]  time = 0.0215428, antiderivative size = 184, 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 = {4812} \[ -\frac{2 \sqrt{\pi } \sin \left (\frac{a}{2 b}\right ) \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )}{\sqrt{\frac{1}{b}} d x}+\frac{2 \sqrt{\pi } \cos \left (\frac{a}{2 b}\right ) \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) S\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )}{\sqrt{\frac{1}{b}} d x}-\frac{2 \sin ^2\left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{d x} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*ArcCos[1 + d*x^2]],x]

[Out]

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

Rule 4812

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

Rubi steps

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

Mathematica [A]  time = 0.0848811, size = 157, normalized size = 0.85 \[ -\frac{2 \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \left (\sqrt{\pi } \sin \left (\frac{a}{2 b}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )-\sqrt{\pi } \cos \left (\frac{a}{2 b}\right ) S\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )+\sqrt{\frac{1}{b}} \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}\right )}{\sqrt{\frac{1}{b}} d x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.063, size = 0, normalized size = 0. \begin{align*} \int \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((a+b*arccos(d*x^2+1))^(1/2),x)

[Out]

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

[Out]

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

[Out]

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

[Out]

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