\(\int \frac {x}{\arccos (a x)^3} \, dx\) [63]

Optimal result

Integrand size = 8, antiderivative size = 63 \[ \int \frac {x}{\arccos (a x)^3} \, dx=\frac {x \sqrt {1-a^2 x^2}}{2 a \arccos (a x)^2}-\frac {1}{2 a^2 \arccos (a x)}+\frac {x^2}{\arccos (a x)}+\frac {\text {Si}(2 \arccos (a x))}{a^2} \]

[Out]

-1/2/a^2/arccos(a*x)+x^2/arccos(a*x)+Si(2*arccos(a*x))/a^2+1/2*x*(-a^2*x^2+1)^(1/2)/a/arccos(a*x)^2

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.875, Rules used = {4730, 4808, 4732, 4491, 12, 3380, 4738} \[ \int \frac {x}{\arccos (a x)^3} \, dx=\frac {\text {Si}(2 \arccos (a x))}{a^2}+\frac {x \sqrt {1-a^2 x^2}}{2 a \arccos (a x)^2}-\frac {1}{2 a^2 \arccos (a x)}+\frac {x^2}{\arccos (a x)} \]

[In]

Int[x/ArcCos[a*x]^3,x]

[Out]

(x*Sqrt[1 - a^2*x^2])/(2*a*ArcCos[a*x]^2) - 1/(2*a^2*ArcCos[a*x]) + x^2/ArcCos[a*x] + SinIntegral[2*ArcCos[a*x
]]/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 3380

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

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*} \text {integral}& = \frac {x \sqrt {1-a^2 x^2}}{2 a \arccos (a x)^2}-\frac {\int \frac {1}{\sqrt {1-a^2 x^2} \arccos (a x)^2} \, dx}{2 a}+a \int \frac {x^2}{\sqrt {1-a^2 x^2} \arccos (a x)^2} \, dx \\ & = \frac {x \sqrt {1-a^2 x^2}}{2 a \arccos (a x)^2}-\frac {1}{2 a^2 \arccos (a x)}+\frac {x^2}{\arccos (a x)}-2 \int \frac {x}{\arccos (a x)} \, dx \\ & = \frac {x \sqrt {1-a^2 x^2}}{2 a \arccos (a x)^2}-\frac {1}{2 a^2 \arccos (a x)}+\frac {x^2}{\arccos (a x)}+\frac {2 \text {Subst}\left (\int \frac {\cos (x) \sin (x)}{x} \, dx,x,\arccos (a x)\right )}{a^2} \\ & = \frac {x \sqrt {1-a^2 x^2}}{2 a \arccos (a x)^2}-\frac {1}{2 a^2 \arccos (a x)}+\frac {x^2}{\arccos (a x)}+\frac {2 \text {Subst}\left (\int \frac {\sin (2 x)}{2 x} \, dx,x,\arccos (a x)\right )}{a^2} \\ & = \frac {x \sqrt {1-a^2 x^2}}{2 a \arccos (a x)^2}-\frac {1}{2 a^2 \arccos (a x)}+\frac {x^2}{\arccos (a x)}+\frac {\text {Subst}\left (\int \frac {\sin (2 x)}{x} \, dx,x,\arccos (a x)\right )}{a^2} \\ & = \frac {x \sqrt {1-a^2 x^2}}{2 a \arccos (a x)^2}-\frac {1}{2 a^2 \arccos (a x)}+\frac {x^2}{\arccos (a x)}+\frac {\text {Si}(2 \arccos (a x))}{a^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00 \[ \int \frac {x}{\arccos (a x)^3} \, dx=\frac {x \sqrt {1-a^2 x^2}}{2 a \arccos (a x)^2}+\frac {-1+2 a^2 x^2}{2 a^2 \arccos (a x)}+\frac {\text {Si}(2 \arccos (a x))}{a^2} \]

[In]

Integrate[x/ArcCos[a*x]^3,x]

[Out]

(x*Sqrt[1 - a^2*x^2])/(2*a*ArcCos[a*x]^2) + (-1 + 2*a^2*x^2)/(2*a^2*ArcCos[a*x]) + SinIntegral[2*ArcCos[a*x]]/
a^2

Maple [A] (verified)

Time = 0.61 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.68

method	result	size
derivativedivides	\(\frac {\frac {\sin \left (2 \arccos \left (a x \right )\right )}{4 \arccos \left (a x \right )^{2}}+\frac {\cos \left (2 \arccos \left (a x \right )\right )}{2 \arccos \left (a x \right )}+\operatorname {Si}\left (2 \arccos \left (a x \right )\right )}{a^{2}}\)	\(43\)
default	\(\frac {\frac {\sin \left (2 \arccos \left (a x \right )\right )}{4 \arccos \left (a x \right )^{2}}+\frac {\cos \left (2 \arccos \left (a x \right )\right )}{2 \arccos \left (a x \right )}+\operatorname {Si}\left (2 \arccos \left (a x \right )\right )}{a^{2}}\)	\(43\)

[In]

int(x/arccos(a*x)^3,x,method=_RETURNVERBOSE)

[Out]

1/a^2*(1/4/arccos(a*x)^2*sin(2*arccos(a*x))+1/2/arccos(a*x)*cos(2*arccos(a*x))+Si(2*arccos(a*x)))

Fricas [F]

\[ \int \frac {x}{\arccos (a x)^3} \, dx=\int { \frac {x}{\arccos \left (a x\right )^{3}} \,d x } \]

[In]

integrate(x/arccos(a*x)^3,x, algorithm="fricas")

[Out]

integral(x/arccos(a*x)^3, x)

Sympy [F]

\[ \int \frac {x}{\arccos (a x)^3} \, dx=\int \frac {x}{\operatorname {acos}^{3}{\left (a x \right )}}\, dx \]

[In]

integrate(x/acos(a*x)**3,x)

[Out]

Integral(x/acos(a*x)**3, x)

Maxima [F]

\[ \int \frac {x}{\arccos (a x)^3} \, dx=\int { \frac {x}{\arccos \left (a x\right )^{3}} \,d x } \]

[In]

integrate(x/arccos(a*x)^3,x, algorithm="maxima")

[Out]

-1/2*(4*a^2*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^2*integrate(x/arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x
), x) - sqrt(a*x + 1)*sqrt(-a*x + 1)*a*x - (2*a^2*x^2 - 1)*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x))/(a^2*ar
ctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^2)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.90 \[ \int \frac {x}{\arccos (a x)^3} \, dx=\frac {x^{2}}{\arccos \left (a x\right )} + \frac {\operatorname {Si}\left (2 \, \arccos \left (a x\right )\right )}{a^{2}} + \frac {\sqrt {-a^{2} x^{2} + 1} x}{2 \, a \arccos \left (a x\right )^{2}} - \frac {1}{2 \, a^{2} \arccos \left (a x\right )} \]

[In]

integrate(x/arccos(a*x)^3,x, algorithm="giac")

[Out]

x^2/arccos(a*x) + sin_integral(2*arccos(a*x))/a^2 + 1/2*sqrt(-a^2*x^2 + 1)*x/(a*arccos(a*x)^2) - 1/2/(a^2*arcc
os(a*x))

Mupad [F(-1)]

Timed out. \[ \int \frac {x}{\arccos (a x)^3} \, dx=\int \frac {x}{{\mathrm {acos}\left (a\,x\right )}^3} \,d x \]

[In]

int(x/acos(a*x)^3,x)

[Out]

int(x/acos(a*x)^3, x)