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

Optimal. Leaf size=64 \[ \frac{2 \sqrt{1-(a+b x)^2}}{b \sqrt{\cos ^{-1}(a+b x)}}-\frac{2 \sqrt{2 \pi } \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{\cos ^{-1}(a+b x)}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.0860013, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4804, 4622, 4724, 3304, 3352} \[ \frac{2 \sqrt{1-(a+b x)^2}}{b \sqrt{\cos ^{-1}(a+b x)}}-\frac{2 \sqrt{2 \pi } \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{\cos ^{-1}(a+b x)}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[ArcCos[a + b*x]^(-3/2),x]

[Out]

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

Rule 4804

Int[((a_.) + ArcCos[(c_) + (d_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcCos[x])^n, x],
x, c + d*x], x] /; FreeQ[{a, b, c, d, n}, x]

Rule 4622

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(Sqrt[1 - c^2*x^2]*(a + b*ArcCos[c*x])^(n + 1)
)/(b*c*(n + 1)), x] - Dist[c/(b*(n + 1)), Int[(x*(a + b*ArcCos[c*x])^(n + 1))/Sqrt[1 - c^2*x^2], x], x] /; Fre
eQ[{a, b, c}, x] && LtQ[n, -1]

Rule 4724

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> -Dist[d^p/c^
(m + 1), Subst[Int[(a + b*x)^n*Cos[x]^m*Sin[x]^(2*p + 1), x], x, ArcCos[c*x]], x] /; FreeQ[{a, b, c, d, e, n},
x] && EqQ[c^2*d + e, 0] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || GtQ[d, 0])

Rule 3304

Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Cos[(f*x^2)/d],
x], x, Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]

Rule 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rubi steps

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

Mathematica [C]  time = 0.0535818, size = 97, normalized size = 1.52 \[ -\frac{-i \sqrt{-i \cos ^{-1}(a+b x)} \text{Gamma}\left (\frac{1}{2},-i \cos ^{-1}(a+b x)\right )+i \sqrt{i \cos ^{-1}(a+b x)} \text{Gamma}\left (\frac{1}{2},i \cos ^{-1}(a+b x)\right )-2 \sqrt{1-(a+b x)^2}}{b \sqrt{\cos ^{-1}(a+b x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCos[a + b*x]^(-3/2),x]

[Out]

-((-2*Sqrt[1 - (a + b*x)^2] - I*Sqrt[(-I)*ArcCos[a + b*x]]*Gamma[1/2, (-I)*ArcCos[a + b*x]] + I*Sqrt[I*ArcCos[
a + b*x]]*Gamma[1/2, I*ArcCos[a + b*x]])/(b*Sqrt[ArcCos[a + b*x]]))

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Maple [A]  time = 0.075, size = 84, normalized size = 1.3 \begin{align*} -{\frac{\sqrt{2}}{b\sqrt{\pi }\arccos \left ( bx+a \right ) } \left ( 2\,\arccos \left ( bx+a \right ) \pi \,{\it FresnelC} \left ({\frac{\sqrt{2}\sqrt{\arccos \left ( bx+a \right ) }}{\sqrt{\pi }}} \right ) -\sqrt{2}\sqrt{\pi }\sqrt{\arccos \left ( bx+a \right ) }\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/arccos(b*x+a)^(3/2),x)

[Out]

-1/b*2^(1/2)/Pi^(1/2)/arccos(b*x+a)*(2*arccos(b*x+a)*Pi*FresnelC(2^(1/2)/Pi^(1/2)*arccos(b*x+a)^(1/2))-2^(1/2)
*Pi^(1/2)*arccos(b*x+a)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arccos(b*x+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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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/arccos(b*x+a)^(3/2),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}{\operatorname{acos}^{\frac{3}{2}}{\left (a + b x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/acos(b*x+a)**(3/2),x)

[Out]

Integral(acos(a + b*x)**(-3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\arccos \left (b x + a\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arccos(b*x+a)^(3/2),x, algorithm="giac")

[Out]

integrate(arccos(b*x + a)^(-3/2), x)