3.92 \(\int \frac{1}{(a+b \cos ^{-1}(1+d x^2))^{5/2}} \, dx\)

Optimal. Leaf size=221 \[ \frac{x}{3 b^2 \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}+\frac{\sqrt{-d^2 x^4-2 d x^2}}{3 b d x \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )^{3/2}}+\frac{2 \sqrt{\pi } \left (\frac{1}{b}\right )^{5/2} \cos \left (\frac{a}{2 b}\right ) \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )}{3 d x}+\frac{2 \sqrt{\pi } \left (\frac{1}{b}\right )^{5/2} \sin \left (\frac{a}{2 b}\right ) \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) S\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )}{3 d x} \]

[Out]

Sqrt[-2*d*x^2 - d^2*x^4]/(3*b*d*x*(a + b*ArcCos[1 + d*x^2])^(3/2)) + x/(3*b^2*Sqrt[a + b*ArcCos[1 + d*x^2]]) +
(2*(b^(-1))^(5/2)*Sqrt[Pi]*Cos[a/(2*b)]*FresnelC[(Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[1 + d*x^2]])/Sqrt[Pi]]*Sin[A
rcCos[1 + d*x^2]/2])/(3*d*x) + (2*(b^(-1))^(5/2)*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])/(3*d*x)

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Rubi [A]  time = 0.0500054, antiderivative size = 221, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {4829, 4820} \[ \frac{x}{3 b^2 \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}+\frac{\sqrt{-d^2 x^4-2 d x^2}}{3 b d x \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )^{3/2}}+\frac{2 \sqrt{\pi } \left (\frac{1}{b}\right )^{5/2} \cos \left (\frac{a}{2 b}\right ) \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )}{3 d x}+\frac{2 \sqrt{\pi } \left (\frac{1}{b}\right )^{5/2} \sin \left (\frac{a}{2 b}\right ) \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) S\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )}{3 d x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCos[1 + d*x^2])^(-5/2),x]

[Out]

Sqrt[-2*d*x^2 - d^2*x^4]/(3*b*d*x*(a + b*ArcCos[1 + d*x^2])^(3/2)) + x/(3*b^2*Sqrt[a + b*ArcCos[1 + d*x^2]]) +
(2*(b^(-1))^(5/2)*Sqrt[Pi]*Cos[a/(2*b)]*FresnelC[(Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[1 + d*x^2]])/Sqrt[Pi]]*Sin[A
rcCos[1 + d*x^2]/2])/(3*d*x) + (2*(b^(-1))^(5/2)*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])/(3*d*x)

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 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}{\left (a+b \cos ^{-1}\left (1+d x^2\right )\right )^{5/2}} \, dx &=\frac{\sqrt{-2 d x^2-d^2 x^4}}{3 b d x \left (a+b \cos ^{-1}\left (1+d x^2\right )\right )^{3/2}}+\frac{x}{3 b^2 \sqrt{a+b \cos ^{-1}\left (1+d x^2\right )}}-\frac{\int \frac{1}{\sqrt{a+b \cos ^{-1}\left (1+d x^2\right )}} \, dx}{3 b^2}\\ &=\frac{\sqrt{-2 d x^2-d^2 x^4}}{3 b d x \left (a+b \cos ^{-1}\left (1+d x^2\right )\right )^{3/2}}+\frac{x}{3 b^2 \sqrt{a+b \cos ^{-1}\left (1+d x^2\right )}}+\frac{2 \left (\frac{1}{b}\right )^{5/2} \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 )}{3 d x}+\frac{2 \left (\frac{1}{b}\right )^{5/2} \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 )}{3 d x}\\ \end{align*}

Mathematica [A]  time = 0.72083, size = 234, normalized size = 1.06 \[ \frac{2 \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \left (\sqrt{\pi } \sqrt{\frac{1}{b}} \cos \left (\frac{a}{2 b}\right ) \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )^{3/2} \text{FresnelC}\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )+\sqrt{\pi } \sqrt{\frac{1}{b}} \sin \left (\frac{a}{2 b}\right ) \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )^{3/2} S\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )-a \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right )+b \cos \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right )-b \cos ^{-1}\left (d x^2+1\right ) \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right )\right )}{3 b^2 d x \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcCos[1 + d*x^2])^(-5/2),x]

[Out]

(2*Sin[ArcCos[1 + d*x^2]/2]*(b*Cos[ArcCos[1 + d*x^2]/2] + Sqrt[b^(-1)]*Sqrt[Pi]*(a + b*ArcCos[1 + d*x^2])^(3/2
)*Cos[a/(2*b)]*FresnelC[(Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[1 + d*x^2]])/Sqrt[Pi]] + Sqrt[b^(-1)]*Sqrt[Pi]*(a + b*
ArcCos[1 + d*x^2])^(3/2)*FresnelS[(Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[1 + d*x^2]])/Sqrt[Pi]]*Sin[a/(2*b)] - a*Sin[
ArcCos[1 + d*x^2]/2] - b*ArcCos[1 + d*x^2]*Sin[ArcCos[1 + d*x^2]/2]))/(3*b^2*d*x*(a + b*ArcCos[1 + d*x^2])^(3/
2))

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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 ) ^{-{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*arccos(d*x^2+1))^(5/2),x)

[Out]

int(1/(a+b*arccos(d*x^2+1))^(5/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \arccos \left (d x^{2} + 1\right ) + a\right )}^{\frac{5}{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))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*arccos(d*x^2 + 1) + a)^(-5/2), 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))^(5/2),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}{\left (a + b \operatorname{acos}{\left (d x^{2} + 1 \right )}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*acos(d*x**2+1))**(5/2),x)

[Out]

Integral((a + b*acos(d*x**2 + 1))**(-5/2), 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 )}^{\frac{5}{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))^(5/2),x, algorithm="giac")

[Out]

integrate((b*arccos(d*x^2 + 1) + a)^(-5/2), x)