\(\int \frac {x^2}{\arccos (a x)^2} \, dx\) [55]

Optimal result

Integrand size = 10, antiderivative size = 54 \[ \int \frac {x^2}{\arccos (a x)^2} \, dx=\frac {x^2 \sqrt {1-a^2 x^2}}{a \arccos (a x)}-\frac {\operatorname {CosIntegral}(\arccos (a x))}{4 a^3}-\frac {3 \operatorname {CosIntegral}(3 \arccos (a x))}{4 a^3} \]

[Out]

-1/4*Ci(arccos(a*x))/a^3-3/4*Ci(3*arccos(a*x))/a^3+x^2*(-a^2*x^2+1)^(1/2)/a/arccos(a*x)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4728, 3383} \[ \int \frac {x^2}{\arccos (a x)^2} \, dx=-\frac {\operatorname {CosIntegral}(\arccos (a x))}{4 a^3}-\frac {3 \operatorname {CosIntegral}(3 \arccos (a x))}{4 a^3}+\frac {x^2 \sqrt {1-a^2 x^2}}{a \arccos (a x)} \]

[In]

Int[x^2/ArcCos[a*x]^2,x]

[Out]

(x^2*Sqrt[1 - a^2*x^2])/(a*ArcCos[a*x]) - CosIntegral[ArcCos[a*x]]/(4*a^3) - (3*CosIntegral[3*ArcCos[a*x]])/(4
*a^3)

Rule 3383

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 4728

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[1/(b^2*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[x^(n + 1), C
os[-a/b + x/b]^(m - 1)*(m - (m + 1)*Cos[-a/b + x/b]^2), x], x], x, a + b*ArcCos[c*x]], x] /; FreeQ[{a, b, c},
x] && IGtQ[m, 0] && GeQ[n, -2] && LtQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {x^2 \sqrt {1-a^2 x^2}}{a \arccos (a x)}+\frac {\text {Subst}\left (\int \left (-\frac {\cos (x)}{4 x}-\frac {3 \cos (3 x)}{4 x}\right ) \, dx,x,\arccos (a x)\right )}{a^3} \\ & = \frac {x^2 \sqrt {1-a^2 x^2}}{a \arccos (a x)}-\frac {\text {Subst}\left (\int \frac {\cos (x)}{x} \, dx,x,\arccos (a x)\right )}{4 a^3}-\frac {3 \text {Subst}\left (\int \frac {\cos (3 x)}{x} \, dx,x,\arccos (a x)\right )}{4 a^3} \\ & = \frac {x^2 \sqrt {1-a^2 x^2}}{a \arccos (a x)}-\frac {\operatorname {CosIntegral}(\arccos (a x))}{4 a^3}-\frac {3 \operatorname {CosIntegral}(3 \arccos (a x))}{4 a^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.93 \[ \int \frac {x^2}{\arccos (a x)^2} \, dx=-\frac {-\frac {4 a^2 x^2 \sqrt {1-a^2 x^2}}{\arccos (a x)}+\operatorname {CosIntegral}(\arccos (a x))+3 \operatorname {CosIntegral}(3 \arccos (a x))}{4 a^3} \]

[In]

Integrate[x^2/ArcCos[a*x]^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.06

method	result	size
derivativedivides	\(\frac {\frac {\sin \left (3 \arccos \left (a x \right )\right )}{4 \arccos \left (a x \right )}-\frac {3 \,\operatorname {Ci}\left (3 \arccos \left (a x \right )\right )}{4}+\frac {\sqrt {-a^{2} x^{2}+1}}{4 \arccos \left (a x \right )}-\frac {\operatorname {Ci}\left (\arccos \left (a x \right )\right )}{4}}{a^{3}}\)	\(57\)
default	\(\frac {\frac {\sin \left (3 \arccos \left (a x \right )\right )}{4 \arccos \left (a x \right )}-\frac {3 \,\operatorname {Ci}\left (3 \arccos \left (a x \right )\right )}{4}+\frac {\sqrt {-a^{2} x^{2}+1}}{4 \arccos \left (a x \right )}-\frac {\operatorname {Ci}\left (\arccos \left (a x \right )\right )}{4}}{a^{3}}\)	\(57\)

[In]

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

[Out]

1/a^3*(1/4/arccos(a*x)*sin(3*arccos(a*x))-3/4*Ci(3*arccos(a*x))+1/4*(-a^2*x^2+1)^(1/2)/arccos(a*x)-1/4*Ci(arcc
os(a*x)))

Fricas [F]

\[ \int \frac {x^2}{\arccos (a x)^2} \, dx=\int { \frac {x^{2}}{\arccos \left (a x\right )^{2}} \,d x } \]

[In]

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

[Out]

integral(x^2/arccos(a*x)^2, x)

Sympy [F]

\[ \int \frac {x^2}{\arccos (a x)^2} \, dx=\int \frac {x^{2}}{\operatorname {acos}^{2}{\left (a x \right )}}\, dx \]

[In]

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

[Out]

Integral(x**2/acos(a*x)**2, x)

Maxima [F]

\[ \int \frac {x^2}{\arccos (a x)^2} \, dx=\int { \frac {x^{2}}{\arccos \left (a x\right )^{2}} \,d x } \]

[In]

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

[Out]

(sqrt(a*x + 1)*sqrt(-a*x + 1)*x^2 - a*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)*integrate((3*a^2*x^3 - 2*x)*s
qrt(a*x + 1)*sqrt(-a*x + 1)/((a^3*x^2 - a)*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)), x))/(a*arctan2(sqrt(a*
x + 1)*sqrt(-a*x + 1), a*x))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.89 \[ \int \frac {x^2}{\arccos (a x)^2} \, dx=\frac {\sqrt {-a^{2} x^{2} + 1} x^{2}}{a \arccos \left (a x\right )} - \frac {3 \, \operatorname {Ci}\left (3 \, \arccos \left (a x\right )\right )}{4 \, a^{3}} - \frac {\operatorname {Ci}\left (\arccos \left (a x\right )\right )}{4 \, a^{3}} \]

[In]

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

[Out]

sqrt(-a^2*x^2 + 1)*x^2/(a*arccos(a*x)) - 3/4*cos_integral(3*arccos(a*x))/a^3 - 1/4*cos_integral(arccos(a*x))/a
^3

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2}{\arccos (a x)^2} \, dx=\int \frac {x^2}{{\mathrm {acos}\left (a\,x\right )}^2} \,d x \]

[In]

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

[Out]

int(x^2/acos(a*x)^2, x)