3.90 \(\int \cos ^{-1}(a x)^{5/2} \, dx\)

Optimal. Leaf size=88 \[ -\frac{5 \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^{3/2}}{2 a}+\frac{15 \sqrt{\frac{\pi }{2}} \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{\cos ^{-1}(a x)}\right )}{4 a}+x \cos ^{-1}(a x)^{5/2}-\frac{15}{4} x \sqrt{\cos ^{-1}(a x)} \]

[Out]

(-15*x*Sqrt[ArcCos[a*x]])/4 - (5*Sqrt[1 - a^2*x^2]*ArcCos[a*x]^(3/2))/(2*a) + x*ArcCos[a*x]^(5/2) + (15*Sqrt[P
i/2]*FresnelC[Sqrt[2/Pi]*Sqrt[ArcCos[a*x]]])/(4*a)

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Rubi [A]  time = 0.160196, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {4620, 4678, 4724, 3304, 3352} \[ -\frac{5 \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^{3/2}}{2 a}+\frac{15 \sqrt{\frac{\pi }{2}} \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{\cos ^{-1}(a x)}\right )}{4 a}+x \cos ^{-1}(a x)^{5/2}-\frac{15}{4} x \sqrt{\cos ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

Int[ArcCos[a*x]^(5/2),x]

[Out]

(-15*x*Sqrt[ArcCos[a*x]])/4 - (5*Sqrt[1 - a^2*x^2]*ArcCos[a*x]^(3/2))/(2*a) + x*ArcCos[a*x]^(5/2) + (15*Sqrt[P
i/2]*FresnelC[Sqrt[2/Pi]*Sqrt[ArcCos[a*x]]])/(4*a)

Rule 4620

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*ArcCos[c*x])^n, x] + Dist[b*c*n, Int[
(x*(a + b*ArcCos[c*x])^(n - 1))/Sqrt[1 - c^2*x^2], x], x] /; FreeQ[{a, b, c}, x] && GtQ[n, 0]

Rule 4678

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)^
(p + 1)*(a + b*ArcCos[c*x])^n)/(2*e*(p + 1)), x] - Dist[(b*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p + 1
)*(1 - c^2*x^2)^FracPart[p]), Int[(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcCos[c*x])^(n - 1), x], x] /; FreeQ[{a, b,
c, d, e, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && NeQ[p, -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 \cos ^{-1}(a x)^{5/2} \, dx &=x \cos ^{-1}(a x)^{5/2}+\frac{1}{2} (5 a) \int \frac{x \cos ^{-1}(a x)^{3/2}}{\sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{5 \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^{3/2}}{2 a}+x \cos ^{-1}(a x)^{5/2}-\frac{15}{4} \int \sqrt{\cos ^{-1}(a x)} \, dx\\ &=-\frac{15}{4} x \sqrt{\cos ^{-1}(a x)}-\frac{5 \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^{3/2}}{2 a}+x \cos ^{-1}(a x)^{5/2}-\frac{1}{8} (15 a) \int \frac{x}{\sqrt{1-a^2 x^2} \sqrt{\cos ^{-1}(a x)}} \, dx\\ &=-\frac{15}{4} x \sqrt{\cos ^{-1}(a x)}-\frac{5 \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^{3/2}}{2 a}+x \cos ^{-1}(a x)^{5/2}+\frac{15 \operatorname{Subst}\left (\int \frac{\cos (x)}{\sqrt{x}} \, dx,x,\cos ^{-1}(a x)\right )}{8 a}\\ &=-\frac{15}{4} x \sqrt{\cos ^{-1}(a x)}-\frac{5 \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^{3/2}}{2 a}+x \cos ^{-1}(a x)^{5/2}+\frac{15 \operatorname{Subst}\left (\int \cos \left (x^2\right ) \, dx,x,\sqrt{\cos ^{-1}(a x)}\right )}{4 a}\\ &=-\frac{15}{4} x \sqrt{\cos ^{-1}(a x)}-\frac{5 \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^{3/2}}{2 a}+x \cos ^{-1}(a x)^{5/2}+\frac{15 \sqrt{\frac{\pi }{2}} C\left (\sqrt{\frac{2}{\pi }} \sqrt{\cos ^{-1}(a x)}\right )}{4 a}\\ \end{align*}

Mathematica [C]  time = 0.0374966, size = 76, normalized size = 0.86 \[ -\frac{\sqrt{\cos ^{-1}(a x)} \left (\sqrt{i \cos ^{-1}(a x)} \text{Gamma}\left (\frac{7}{2},-i \cos ^{-1}(a x)\right )+\sqrt{-i \cos ^{-1}(a x)} \text{Gamma}\left (\frac{7}{2},i \cos ^{-1}(a x)\right )\right )}{2 a \sqrt{\cos ^{-1}(a x)^2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCos[a*x]^(5/2),x]

[Out]

-(Sqrt[ArcCos[a*x]]*(Sqrt[I*ArcCos[a*x]]*Gamma[7/2, (-I)*ArcCos[a*x]] + Sqrt[(-I)*ArcCos[a*x]]*Gamma[7/2, I*Ar
cCos[a*x]]))/(2*a*Sqrt[ArcCos[a*x]^2])

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Maple [A]  time = 0.073, size = 88, normalized size = 1. \begin{align*}{\frac{\sqrt{2}}{8\,a\sqrt{\pi }} \left ( 4\, \left ( \arccos \left ( ax \right ) \right ) ^{5/2}\sqrt{2}\sqrt{\pi }xa-10\, \left ( \arccos \left ( ax \right ) \right ) ^{3/2}\sqrt{2}\sqrt{\pi }\sqrt{-{a}^{2}{x}^{2}+1}-15\,\sqrt{2}\sqrt{\pi }\sqrt{\arccos \left ( ax \right ) }xa+15\,\pi \,{\it FresnelC} \left ({\frac{\sqrt{2}\sqrt{\arccos \left ( ax \right ) }}{\sqrt{\pi }}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccos(a*x)^(5/2),x)

[Out]

1/8/a*2^(1/2)/Pi^(1/2)*(4*arccos(a*x)^(5/2)*2^(1/2)*Pi^(1/2)*x*a-10*arccos(a*x)^(3/2)*2^(1/2)*Pi^(1/2)*(-a^2*x
^2+1)^(1/2)-15*2^(1/2)*Pi^(1/2)*arccos(a*x)^(1/2)*x*a+15*Pi*FresnelC(2^(1/2)/Pi^(1/2)*arccos(a*x)^(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(arccos(a*x)^(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(arccos(a*x)^(5/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acos(a*x)**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 1.32006, size = 244, normalized size = 2.77 \begin{align*} \frac{5 \, i \arccos \left (a x\right )^{\frac{3}{2}} e^{\left (i \arccos \left (a x\right )\right )}}{4 \, a} + \frac{\arccos \left (a x\right )^{\frac{5}{2}} e^{\left (i \arccos \left (a x\right )\right )}}{2 \, a} - \frac{5 \, i \arccos \left (a x\right )^{\frac{3}{2}} e^{\left (-i \arccos \left (a x\right )\right )}}{4 \, a} + \frac{\arccos \left (a x\right )^{\frac{5}{2}} e^{\left (-i \arccos \left (a x\right )\right )}}{2 \, a} - \frac{15 \, \sqrt{2} \sqrt{\pi } i \operatorname{erf}\left (\frac{\sqrt{2} \sqrt{\arccos \left (a x\right )}}{i - 1}\right )}{16 \, a{\left (i - 1\right )}} - \frac{15 \, \sqrt{\arccos \left (a x\right )} e^{\left (i \arccos \left (a x\right )\right )}}{8 \, a} - \frac{15 \, \sqrt{\arccos \left (a x\right )} e^{\left (-i \arccos \left (a x\right )\right )}}{8 \, a} + \frac{15 \, \sqrt{2} \sqrt{\pi } \operatorname{erf}\left (-\frac{\sqrt{2} i \sqrt{\arccos \left (a x\right )}}{i - 1}\right )}{16 \, a{\left (i - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/4*i*arccos(a*x)^(3/2)*e^(i*arccos(a*x))/a + 1/2*arccos(a*x)^(5/2)*e^(i*arccos(a*x))/a - 5/4*i*arccos(a*x)^(3
/2)*e^(-i*arccos(a*x))/a + 1/2*arccos(a*x)^(5/2)*e^(-i*arccos(a*x))/a - 15/16*sqrt(2)*sqrt(pi)*i*erf(sqrt(2)*s
qrt(arccos(a*x))/(i - 1))/(a*(i - 1)) - 15/8*sqrt(arccos(a*x))*e^(i*arccos(a*x))/a - 15/8*sqrt(arccos(a*x))*e^
(-i*arccos(a*x))/a + 15/16*sqrt(2)*sqrt(pi)*erf(-sqrt(2)*i*sqrt(arccos(a*x))/(i - 1))/(a*(i - 1))