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

Optimal. Leaf size=90 \[ \frac{4 \sqrt{2 \pi } S\left (\sqrt{\frac{2}{\pi }} \sqrt{\cos ^{-1}(a+b x)}\right )}{3 b}+\frac{4 (a+b x)}{3 b \sqrt{\cos ^{-1}(a+b x)}}+\frac{2 \sqrt{1-(a+b x)^2}}{3 b \cos ^{-1}(a+b x)^{3/2}} \]

[Out]

(2*Sqrt[1 - (a + b*x)^2])/(3*b*ArcCos[a + b*x]^(3/2)) + (4*(a + b*x))/(3*b*Sqrt[ArcCos[a + b*x]]) + (4*Sqrt[2*
Pi]*FresnelS[Sqrt[2/Pi]*Sqrt[ArcCos[a + b*x]]])/(3*b)

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Rubi [A]  time = 0.0920133, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {4804, 4622, 4720, 4624, 3305, 3351} \[ \frac{4 \sqrt{2 \pi } S\left (\sqrt{\frac{2}{\pi }} \sqrt{\cos ^{-1}(a+b x)}\right )}{3 b}+\frac{4 (a+b x)}{3 b \sqrt{\cos ^{-1}(a+b x)}}+\frac{2 \sqrt{1-(a+b x)^2}}{3 b \cos ^{-1}(a+b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[ArcCos[a + b*x]^(-5/2),x]

[Out]

(2*Sqrt[1 - (a + b*x)^2])/(3*b*ArcCos[a + b*x]^(3/2)) + (4*(a + b*x))/(3*b*Sqrt[ArcCos[a + b*x]]) + (4*Sqrt[2*
Pi]*FresnelS[Sqrt[2/Pi]*Sqrt[ArcCos[a + b*x]]])/(3*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 4720

Int[(((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> -Simp
[((f*x)^m*(a + b*ArcCos[c*x])^(n + 1))/(b*c*Sqrt[d]*(n + 1)), x] + Dist[(f*m)/(b*c*Sqrt[d]*(n + 1)), Int[(f*x)
^(m - 1)*(a + b*ArcCos[c*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && LtQ[n,
-1] && GtQ[d, 0]

Rule 4624

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

Rule 3305

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Sin[(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 3351

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelS[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)^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\cos ^{-1}(x)^{5/2}} \, dx,x,a+b x\right )}{b}\\ &=\frac{2 \sqrt{1-(a+b x)^2}}{3 b \cos ^{-1}(a+b x)^{3/2}}+\frac{2 \operatorname{Subst}\left (\int \frac{x}{\sqrt{1-x^2} \cos ^{-1}(x)^{3/2}} \, dx,x,a+b x\right )}{3 b}\\ &=\frac{2 \sqrt{1-(a+b x)^2}}{3 b \cos ^{-1}(a+b x)^{3/2}}+\frac{4 (a+b x)}{3 b \sqrt{\cos ^{-1}(a+b x)}}-\frac{4 \operatorname{Subst}\left (\int \frac{1}{\sqrt{\cos ^{-1}(x)}} \, dx,x,a+b x\right )}{3 b}\\ &=\frac{2 \sqrt{1-(a+b x)^2}}{3 b \cos ^{-1}(a+b x)^{3/2}}+\frac{4 (a+b x)}{3 b \sqrt{\cos ^{-1}(a+b x)}}+\frac{4 \operatorname{Subst}\left (\int \frac{\sin (x)}{\sqrt{x}} \, dx,x,\cos ^{-1}(a+b x)\right )}{3 b}\\ &=\frac{2 \sqrt{1-(a+b x)^2}}{3 b \cos ^{-1}(a+b x)^{3/2}}+\frac{4 (a+b x)}{3 b \sqrt{\cos ^{-1}(a+b x)}}+\frac{8 \operatorname{Subst}\left (\int \sin \left (x^2\right ) \, dx,x,\sqrt{\cos ^{-1}(a+b x)}\right )}{3 b}\\ &=\frac{2 \sqrt{1-(a+b x)^2}}{3 b \cos ^{-1}(a+b x)^{3/2}}+\frac{4 (a+b x)}{3 b \sqrt{\cos ^{-1}(a+b x)}}+\frac{4 \sqrt{2 \pi } S\left (\sqrt{\frac{2}{\pi }} \sqrt{\cos ^{-1}(a+b x)}\right )}{3 b}\\ \end{align*}

Mathematica [C]  time = 0.291454, size = 139, normalized size = 1.54 \[ -\frac{2 \left (i \left (-i \cos ^{-1}(a+b x)\right )^{3/2} \text{Gamma}\left (\frac{1}{2},-i \cos ^{-1}(a+b x)\right )-i \left (i \cos ^{-1}(a+b x)\right )^{3/2} \text{Gamma}\left (\frac{1}{2},i \cos ^{-1}(a+b x)\right )-\sqrt{1-(a+b x)^2}-e^{-i \cos ^{-1}(a+b x)} \cos ^{-1}(a+b x)-e^{i \cos ^{-1}(a+b x)} \cos ^{-1}(a+b x)\right )}{3 b \cos ^{-1}(a+b x)^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCos[a + b*x]^(-5/2),x]

[Out]

(-2*(-Sqrt[1 - (a + b*x)^2] - ArcCos[a + b*x]/E^(I*ArcCos[a + b*x]) - E^(I*ArcCos[a + b*x])*ArcCos[a + b*x] +
I*((-I)*ArcCos[a + b*x])^(3/2)*Gamma[1/2, (-I)*ArcCos[a + b*x]] - I*(I*ArcCos[a + b*x])^(3/2)*Gamma[1/2, I*Arc
Cos[a + b*x]]))/(3*b*ArcCos[a + b*x]^(3/2))

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Maple [A]  time = 0.079, size = 120, normalized size = 1.3 \begin{align*}{\frac{\sqrt{2}}{3\,b\sqrt{\pi } \left ( \arccos \left ( bx+a \right ) \right ) ^{2}} \left ( 4\,\pi \, \left ( \arccos \left ( bx+a \right ) \right ) ^{2}{\it FresnelS} \left ({\frac{\sqrt{2}\sqrt{\arccos \left ( bx+a \right ) }}{\sqrt{\pi }}} \right ) +2\, \left ( \arccos \left ( bx+a \right ) \right ) ^{3/2}\sqrt{2}\sqrt{\pi }xb+2\, \left ( \arccos \left ( bx+a \right ) \right ) ^{3/2}\sqrt{2}\sqrt{\pi }a+\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)^(5/2),x)

[Out]

1/3/b*2^(1/2)/Pi^(1/2)*(4*Pi*arccos(b*x+a)^2*FresnelS(2^(1/2)/Pi^(1/2)*arccos(b*x+a)^(1/2))+2*arccos(b*x+a)^(3
/2)*2^(1/2)*Pi^(1/2)*x*b+2*arccos(b*x+a)^(3/2)*2^(1/2)*Pi^(1/2)*a+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))/arccos(b*x+a)^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)^(5/2),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)^(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}{\operatorname{acos}^{\frac{5}{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)**(5/2),x)

[Out]

Integral(acos(a + b*x)**(-5/2), 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{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arccos(b*x+a)^(5/2),x, algorithm="giac")

[Out]

integrate(arccos(b*x + a)^(-5/2), x)