3.2.10 \(\int \frac {x}{\text {ArcCos}(a x)^{5/2}} \, dx\) [110]

Optimal. Leaf size=89 \[ \frac {2 x \sqrt {1-a^2 x^2}}{3 a \text {ArcCos}(a x)^{3/2}}-\frac {4}{3 a^2 \sqrt {\text {ArcCos}(a x)}}+\frac {8 x^2}{3 \sqrt {\text {ArcCos}(a x)}}+\frac {8 \sqrt {\pi } S\left (\frac {2 \sqrt {\text {ArcCos}(a x)}}{\sqrt {\pi }}\right )}{3 a^2} \]

[Out]

8/3*FresnelS(2*arccos(a*x)^(1/2)/Pi^(1/2))*Pi^(1/2)/a^2+2/3*x*(-a^2*x^2+1)^(1/2)/a/arccos(a*x)^(3/2)-4/3/a^2/a
rccos(a*x)^(1/2)+8/3*x^2/arccos(a*x)^(1/2)

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Rubi [A]
time = 0.12, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {4730, 4808, 4732, 4491, 12, 3386, 3432, 4738} \begin {gather*} \frac {8 \sqrt {\pi } S\left (\frac {2 \sqrt {\text {ArcCos}(a x)}}{\sqrt {\pi }}\right )}{3 a^2}+\frac {2 x \sqrt {1-a^2 x^2}}{3 a \text {ArcCos}(a x)^{3/2}}-\frac {4}{3 a^2 \sqrt {\text {ArcCos}(a x)}}+\frac {8 x^2}{3 \sqrt {\text {ArcCos}(a x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*Sqrt[1 - a^2*x^2])/(3*a*ArcCos[a*x]^(3/2)) - 4/(3*a^2*Sqrt[ArcCos[a*x]]) + (8*x^2)/(3*Sqrt[ArcCos[a*x]])
+ (8*Sqrt[Pi]*FresnelS[(2*Sqrt[ArcCos[a*x]])/Sqrt[Pi]])/(3*a^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3386

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 3432

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

Rule 4491

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rule 4730

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

Rule 4732

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[-(b*c^(m + 1))^(-1), Subst[Int[x^n*C
os[-a/b + x/b]^m*Sin[-a/b + x/b], x], x, a + b*ArcCos[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]

Rule 4738

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

Rule 4808

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

Rubi steps

\begin {align*} \int \frac {x}{\cos ^{-1}(a x)^{5/2}} \, dx &=\frac {2 x \sqrt {1-a^2 x^2}}{3 a \cos ^{-1}(a x)^{3/2}}-\frac {2 \int \frac {1}{\sqrt {1-a^2 x^2} \cos ^{-1}(a x)^{3/2}} \, dx}{3 a}+\frac {1}{3} (4 a) \int \frac {x^2}{\sqrt {1-a^2 x^2} \cos ^{-1}(a x)^{3/2}} \, dx\\ &=\frac {2 x \sqrt {1-a^2 x^2}}{3 a \cos ^{-1}(a x)^{3/2}}-\frac {4}{3 a^2 \sqrt {\cos ^{-1}(a x)}}+\frac {8 x^2}{3 \sqrt {\cos ^{-1}(a x)}}-\frac {16}{3} \int \frac {x}{\sqrt {\cos ^{-1}(a x)}} \, dx\\ &=\frac {2 x \sqrt {1-a^2 x^2}}{3 a \cos ^{-1}(a x)^{3/2}}-\frac {4}{3 a^2 \sqrt {\cos ^{-1}(a x)}}+\frac {8 x^2}{3 \sqrt {\cos ^{-1}(a x)}}+\frac {16 \text {Subst}\left (\int \frac {\cos (x) \sin (x)}{\sqrt {x}} \, dx,x,\cos ^{-1}(a x)\right )}{3 a^2}\\ &=\frac {2 x \sqrt {1-a^2 x^2}}{3 a \cos ^{-1}(a x)^{3/2}}-\frac {4}{3 a^2 \sqrt {\cos ^{-1}(a x)}}+\frac {8 x^2}{3 \sqrt {\cos ^{-1}(a x)}}+\frac {16 \text {Subst}\left (\int \frac {\sin (2 x)}{2 \sqrt {x}} \, dx,x,\cos ^{-1}(a x)\right )}{3 a^2}\\ &=\frac {2 x \sqrt {1-a^2 x^2}}{3 a \cos ^{-1}(a x)^{3/2}}-\frac {4}{3 a^2 \sqrt {\cos ^{-1}(a x)}}+\frac {8 x^2}{3 \sqrt {\cos ^{-1}(a x)}}+\frac {8 \text {Subst}\left (\int \frac {\sin (2 x)}{\sqrt {x}} \, dx,x,\cos ^{-1}(a x)\right )}{3 a^2}\\ &=\frac {2 x \sqrt {1-a^2 x^2}}{3 a \cos ^{-1}(a x)^{3/2}}-\frac {4}{3 a^2 \sqrt {\cos ^{-1}(a x)}}+\frac {8 x^2}{3 \sqrt {\cos ^{-1}(a x)}}+\frac {16 \text {Subst}\left (\int \sin \left (2 x^2\right ) \, dx,x,\sqrt {\cos ^{-1}(a x)}\right )}{3 a^2}\\ &=\frac {2 x \sqrt {1-a^2 x^2}}{3 a \cos ^{-1}(a x)^{3/2}}-\frac {4}{3 a^2 \sqrt {\cos ^{-1}(a x)}}+\frac {8 x^2}{3 \sqrt {\cos ^{-1}(a x)}}+\frac {8 \sqrt {\pi } S\left (\frac {2 \sqrt {\cos ^{-1}(a x)}}{\sqrt {\pi }}\right )}{3 a^2}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 61, normalized size = 0.69 \begin {gather*} \frac {8 \sqrt {\pi } S\left (\frac {2 \sqrt {\text {ArcCos}(a x)}}{\sqrt {\pi }}\right )+\frac {4 \text {ArcCos}(a x) \cos (2 \text {ArcCos}(a x))+\sin (2 \text {ArcCos}(a x))}{\text {ArcCos}(a x)^{3/2}}}{3 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(8*Sqrt[Pi]*FresnelS[(2*Sqrt[ArcCos[a*x]])/Sqrt[Pi]] + (4*ArcCos[a*x]*Cos[2*ArcCos[a*x]] + Sin[2*ArcCos[a*x]])
/ArcCos[a*x]^(3/2))/(3*a^2)

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Maple [A]
time = 0.17, size = 56, normalized size = 0.63

method	result	size
default	\(\frac {8 \sqrt {\pi }\, \mathrm {S}\left (\frac {2 \sqrt {\arccos \left (a x \right )}}{\sqrt {\pi }}\right ) \arccos \left (a x \right )^{\frac {3}{2}}+4 \arccos \left (a x \right ) \cos \left (2 \arccos \left (a x \right )\right )+\sin \left (2 \arccos \left (a x \right )\right )}{3 a^{2} \arccos \left (a x \right )^{\frac {3}{2}}}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/a^2*(8*Pi^(1/2)*FresnelS(2*arccos(a*x)^(1/2)/Pi^(1/2))*arccos(a*x)^(3/2)+4*arccos(a*x)*cos(2*arccos(a*x))+
sin(2*arccos(a*x)))/arccos(a*x)^(3/2)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

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 >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\operatorname {acos}^{\frac {5}{2}}{\left (a x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/acos(a*x)**(5/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/arccos(a*x)^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x}{{\mathrm {acos}\left (a\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x/acos(a*x)^(5/2), x)

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