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

Optimal. Leaf size=119 \[ \frac{15 \sqrt{\pi } \text{FresnelC}\left (\frac{2 \sqrt{\cos ^{-1}(a x)}}{\sqrt{\pi }}\right )}{128 a^2}-\frac{5 x \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^{3/2}}{8 a}-\frac{\cos ^{-1}(a x)^{5/2}}{4 a^2}+\frac{15 \sqrt{\cos ^{-1}(a x)}}{64 a^2}+\frac{1}{2} x^2 \cos ^{-1}(a x)^{5/2}-\frac{15}{32} x^2 \sqrt{\cos ^{-1}(a x)} \]

[Out]

(15*Sqrt[ArcCos[a*x]])/(64*a^2) - (15*x^2*Sqrt[ArcCos[a*x]])/32 - (5*x*Sqrt[1 - a^2*x^2]*ArcCos[a*x]^(3/2))/(8
*a) - ArcCos[a*x]^(5/2)/(4*a^2) + (x^2*ArcCos[a*x]^(5/2))/2 + (15*Sqrt[Pi]*FresnelC[(2*Sqrt[ArcCos[a*x]])/Sqrt
[Pi]])/(128*a^2)

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Rubi [A]  time = 0.287317, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.7, Rules used = {4630, 4708, 4642, 4724, 3312, 3304, 3352} \[ \frac{15 \sqrt{\pi } \text{FresnelC}\left (\frac{2 \sqrt{\cos ^{-1}(a x)}}{\sqrt{\pi }}\right )}{128 a^2}-\frac{5 x \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^{3/2}}{8 a}-\frac{\cos ^{-1}(a x)^{5/2}}{4 a^2}+\frac{15 \sqrt{\cos ^{-1}(a x)}}{64 a^2}+\frac{1}{2} x^2 \cos ^{-1}(a x)^{5/2}-\frac{15}{32} x^2 \sqrt{\cos ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*Sqrt[ArcCos[a*x]])/(64*a^2) - (15*x^2*Sqrt[ArcCos[a*x]])/32 - (5*x*Sqrt[1 - a^2*x^2]*ArcCos[a*x]^(3/2))/(8
*a) - ArcCos[a*x]^(5/2)/(4*a^2) + (x^2*ArcCos[a*x]^(5/2))/2 + (15*Sqrt[Pi]*FresnelC[(2*Sqrt[ArcCos[a*x]])/Sqrt
[Pi]])/(128*a^2)

Rule 4630

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

Rule 4708

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

Rule 4642

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> -Simp[(a + b*ArcCos[c*x])
^(n + 1)/(b*c*Sqrt[d]*(n + 1)), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e, 0] && GtQ[d, 0] && NeQ[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 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

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 x \cos ^{-1}(a x)^{5/2} \, dx &=\frac{1}{2} x^2 \cos ^{-1}(a x)^{5/2}+\frac{1}{4} (5 a) \int \frac{x^2 \cos ^{-1}(a x)^{3/2}}{\sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{5 x \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^{3/2}}{8 a}+\frac{1}{2} x^2 \cos ^{-1}(a x)^{5/2}-\frac{15}{16} \int x \sqrt{\cos ^{-1}(a x)} \, dx+\frac{5 \int \frac{\cos ^{-1}(a x)^{3/2}}{\sqrt{1-a^2 x^2}} \, dx}{8 a}\\ &=-\frac{15}{32} x^2 \sqrt{\cos ^{-1}(a x)}-\frac{5 x \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^{3/2}}{8 a}-\frac{\cos ^{-1}(a x)^{5/2}}{4 a^2}+\frac{1}{2} x^2 \cos ^{-1}(a x)^{5/2}-\frac{1}{64} (15 a) \int \frac{x^2}{\sqrt{1-a^2 x^2} \sqrt{\cos ^{-1}(a x)}} \, dx\\ &=-\frac{15}{32} x^2 \sqrt{\cos ^{-1}(a x)}-\frac{5 x \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^{3/2}}{8 a}-\frac{\cos ^{-1}(a x)^{5/2}}{4 a^2}+\frac{1}{2} x^2 \cos ^{-1}(a x)^{5/2}+\frac{15 \operatorname{Subst}\left (\int \frac{\cos ^2(x)}{\sqrt{x}} \, dx,x,\cos ^{-1}(a x)\right )}{64 a^2}\\ &=-\frac{15}{32} x^2 \sqrt{\cos ^{-1}(a x)}-\frac{5 x \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^{3/2}}{8 a}-\frac{\cos ^{-1}(a x)^{5/2}}{4 a^2}+\frac{1}{2} x^2 \cos ^{-1}(a x)^{5/2}+\frac{15 \operatorname{Subst}\left (\int \left (\frac{1}{2 \sqrt{x}}+\frac{\cos (2 x)}{2 \sqrt{x}}\right ) \, dx,x,\cos ^{-1}(a x)\right )}{64 a^2}\\ &=\frac{15 \sqrt{\cos ^{-1}(a x)}}{64 a^2}-\frac{15}{32} x^2 \sqrt{\cos ^{-1}(a x)}-\frac{5 x \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^{3/2}}{8 a}-\frac{\cos ^{-1}(a x)^{5/2}}{4 a^2}+\frac{1}{2} x^2 \cos ^{-1}(a x)^{5/2}+\frac{15 \operatorname{Subst}\left (\int \frac{\cos (2 x)}{\sqrt{x}} \, dx,x,\cos ^{-1}(a x)\right )}{128 a^2}\\ &=\frac{15 \sqrt{\cos ^{-1}(a x)}}{64 a^2}-\frac{15}{32} x^2 \sqrt{\cos ^{-1}(a x)}-\frac{5 x \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^{3/2}}{8 a}-\frac{\cos ^{-1}(a x)^{5/2}}{4 a^2}+\frac{1}{2} x^2 \cos ^{-1}(a x)^{5/2}+\frac{15 \operatorname{Subst}\left (\int \cos \left (2 x^2\right ) \, dx,x,\sqrt{\cos ^{-1}(a x)}\right )}{64 a^2}\\ &=\frac{15 \sqrt{\cos ^{-1}(a x)}}{64 a^2}-\frac{15}{32} x^2 \sqrt{\cos ^{-1}(a x)}-\frac{5 x \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^{3/2}}{8 a}-\frac{\cos ^{-1}(a x)^{5/2}}{4 a^2}+\frac{1}{2} x^2 \cos ^{-1}(a x)^{5/2}+\frac{15 \sqrt{\pi } C\left (\frac{2 \sqrt{\cos ^{-1}(a x)}}{\sqrt{\pi }}\right )}{128 a^2}\\ \end{align*}

Mathematica [A]  time = 0.0892726, size = 73, normalized size = 0.61 \[ \frac{15 \sqrt{\pi } \text{FresnelC}\left (\frac{2 \sqrt{\cos ^{-1}(a x)}}{\sqrt{\pi }}\right )-2 \sqrt{\cos ^{-1}(a x)} \left (\left (15-16 \cos ^{-1}(a x)^2\right ) \cos \left (2 \cos ^{-1}(a x)\right )+20 \cos ^{-1}(a x) \sin \left (2 \cos ^{-1}(a x)\right )\right )}{128 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*Sqrt[Pi]*FresnelC[(2*Sqrt[ArcCos[a*x]])/Sqrt[Pi]] - 2*Sqrt[ArcCos[a*x]]*((15 - 16*ArcCos[a*x]^2)*Cos[2*Arc
Cos[a*x]] + 20*ArcCos[a*x]*Sin[2*ArcCos[a*x]]))/(128*a^2)

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Maple [A]  time = 0.078, size = 79, normalized size = 0.7 \begin{align*}{\frac{1}{128\,{a}^{2}\sqrt{\pi }} \left ( 32\, \left ( \arccos \left ( ax \right ) \right ) ^{5/2}\sqrt{\pi }\cos \left ( 2\,\arccos \left ( ax \right ) \right ) -40\, \left ( \arccos \left ( ax \right ) \right ) ^{3/2}\sqrt{\pi }\sin \left ( 2\,\arccos \left ( ax \right ) \right ) +15\,\pi \,{\it FresnelC} \left ( 2\,{\frac{\sqrt{\arccos \left ( ax \right ) }}{\sqrt{\pi }}} \right ) -30\,\sqrt{\pi }\sqrt{\arccos \left ( ax \right ) }\cos \left ( 2\,\arccos \left ( ax \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128/a^2/Pi^(1/2)*(32*arccos(a*x)^(5/2)*Pi^(1/2)*cos(2*arccos(a*x))-40*arccos(a*x)^(3/2)*Pi^(1/2)*sin(2*arcco
s(a*x))+15*Pi*FresnelC(2*arccos(a*x)^(1/2)/Pi^(1/2))-30*Pi^(1/2)*arccos(a*x)^(1/2)*cos(2*arccos(a*x)))

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

[Out]

Timed out

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Giac [A]  time = 1.23437, size = 225, normalized size = 1.89 \begin{align*} \frac{5 \, i \arccos \left (a x\right )^{\frac{3}{2}} e^{\left (2 \, i \arccos \left (a x\right )\right )}}{32 \, a^{2}} + \frac{\arccos \left (a x\right )^{\frac{5}{2}} e^{\left (2 \, i \arccos \left (a x\right )\right )}}{8 \, a^{2}} - \frac{5 \, i \arccos \left (a x\right )^{\frac{3}{2}} e^{\left (-2 \, i \arccos \left (a x\right )\right )}}{32 \, a^{2}} + \frac{\arccos \left (a x\right )^{\frac{5}{2}} e^{\left (-2 \, i \arccos \left (a x\right )\right )}}{8 \, a^{2}} - \frac{15 \, \sqrt{\pi } i \operatorname{erf}\left (-{\left (i + 1\right )} \sqrt{\arccos \left (a x\right )}\right )}{256 \, a^{2}{\left (i - 1\right )}} - \frac{15 \, \sqrt{\arccos \left (a x\right )} e^{\left (2 \, i \arccos \left (a x\right )\right )}}{128 \, a^{2}} - \frac{15 \, \sqrt{\arccos \left (a x\right )} e^{\left (-2 \, i \arccos \left (a x\right )\right )}}{128 \, a^{2}} + \frac{15 \, \sqrt{\pi } \operatorname{erf}\left ({\left (i - 1\right )} \sqrt{\arccos \left (a x\right )}\right )}{256 \, a^{2}{\left (i - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/32*i*arccos(a*x)^(3/2)*e^(2*i*arccos(a*x))/a^2 + 1/8*arccos(a*x)^(5/2)*e^(2*i*arccos(a*x))/a^2 - 5/32*i*arcc
os(a*x)^(3/2)*e^(-2*i*arccos(a*x))/a^2 + 1/8*arccos(a*x)^(5/2)*e^(-2*i*arccos(a*x))/a^2 - 15/256*sqrt(pi)*i*er
f(-(i + 1)*sqrt(arccos(a*x)))/(a^2*(i - 1)) - 15/128*sqrt(arccos(a*x))*e^(2*i*arccos(a*x))/a^2 - 15/128*sqrt(a
rccos(a*x))*e^(-2*i*arccos(a*x))/a^2 + 15/256*sqrt(pi)*erf((i - 1)*sqrt(arccos(a*x)))/(a^2*(i - 1))