\(\int \frac {x^2}{(a+b \arccos (c x))^2} \, dx\) [163]

Optimal result

Integrand size = 14, antiderivative size = 155 \[ \int \frac {x^2}{(a+b \arccos (c x))^2} \, dx=\frac {x^2 \sqrt {1-c^2 x^2}}{b c (a+b \arccos (c x))}-\frac {\cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arccos (c x)}{b}\right )}{4 b^2 c^3}-\frac {3 \cos \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {3 (a+b \arccos (c x))}{b}\right )}{4 b^2 c^3}-\frac {\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arccos (c x)}{b}\right )}{4 b^2 c^3}-\frac {3 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arccos (c x))}{b}\right )}{4 b^2 c^3} \]

[Out]

-1/4*Ci((a+b*arccos(c*x))/b)*cos(a/b)/b^2/c^3-3/4*Ci(3*(a+b*arccos(c*x))/b)*cos(3*a/b)/b^2/c^3-1/4*Si((a+b*arc
cos(c*x))/b)*sin(a/b)/b^2/c^3-3/4*Si(3*(a+b*arccos(c*x))/b)*sin(3*a/b)/b^2/c^3+x^2*(-c^2*x^2+1)^(1/2)/b/c/(a+b
*arccos(c*x))

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {4728, 3384, 3380, 3383} \[ \int \frac {x^2}{(a+b \arccos (c x))^2} \, dx=-\frac {\cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arccos (c x)}{b}\right )}{4 b^2 c^3}-\frac {3 \cos \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {3 (a+b \arccos (c x))}{b}\right )}{4 b^2 c^3}-\frac {\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arccos (c x)}{b}\right )}{4 b^2 c^3}-\frac {3 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arccos (c x))}{b}\right )}{4 b^2 c^3}+\frac {x^2 \sqrt {1-c^2 x^2}}{b c (a+b \arccos (c x))} \]

[In]

Int[x^2/(a + b*ArcCos[c*x])^2,x]

[Out]

(x^2*Sqrt[1 - c^2*x^2])/(b*c*(a + b*ArcCos[c*x])) - (Cos[a/b]*CosIntegral[(a + b*ArcCos[c*x])/b])/(4*b^2*c^3)
- (3*Cos[(3*a)/b]*CosIntegral[(3*(a + b*ArcCos[c*x]))/b])/(4*b^2*c^3) - (Sin[a/b]*SinIntegral[(a + b*ArcCos[c*
x])/b])/(4*b^2*c^3) - (3*Sin[(3*a)/b]*SinIntegral[(3*(a + b*ArcCos[c*x]))/b])/(4*b^2*c^3)

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 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 3384

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[c*(f/d) + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[c*(f/d) + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - 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-c^2 x^2}}{b c (a+b \arccos (c x))}+\frac {\text {Subst}\left (\int \left (-\frac {3 \cos \left (\frac {3 a}{b}-\frac {3 x}{b}\right )}{4 x}-\frac {\cos \left (\frac {a}{b}-\frac {x}{b}\right )}{4 x}\right ) \, dx,x,a+b \arccos (c x)\right )}{b^2 c^3} \\ & = \frac {x^2 \sqrt {1-c^2 x^2}}{b c (a+b \arccos (c x))}-\frac {\text {Subst}\left (\int \frac {\cos \left (\frac {a}{b}-\frac {x}{b}\right )}{x} \, dx,x,a+b \arccos (c x)\right )}{4 b^2 c^3}-\frac {3 \text {Subst}\left (\int \frac {\cos \left (\frac {3 a}{b}-\frac {3 x}{b}\right )}{x} \, dx,x,a+b \arccos (c x)\right )}{4 b^2 c^3} \\ & = \frac {x^2 \sqrt {1-c^2 x^2}}{b c (a+b \arccos (c x))}-\frac {\cos \left (\frac {a}{b}\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {x}{b}\right )}{x} \, dx,x,a+b \arccos (c x)\right )}{4 b^2 c^3}-\frac {\left (3 \cos \left (\frac {3 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {3 x}{b}\right )}{x} \, dx,x,a+b \arccos (c x)\right )}{4 b^2 c^3}-\frac {\sin \left (\frac {a}{b}\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {x}{b}\right )}{x} \, dx,x,a+b \arccos (c x)\right )}{4 b^2 c^3}-\frac {\left (3 \sin \left (\frac {3 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {3 x}{b}\right )}{x} \, dx,x,a+b \arccos (c x)\right )}{4 b^2 c^3} \\ & = \frac {x^2 \sqrt {1-c^2 x^2}}{b c (a+b \arccos (c x))}-\frac {\cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arccos (c x)}{b}\right )}{4 b^2 c^3}-\frac {3 \cos \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {3 (a+b \arccos (c x))}{b}\right )}{4 b^2 c^3}-\frac {\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arccos (c x)}{b}\right )}{4 b^2 c^3}-\frac {3 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arccos (c x))}{b}\right )}{4 b^2 c^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.80 \[ \int \frac {x^2}{(a+b \arccos (c x))^2} \, dx=-\frac {-\frac {4 b c^2 x^2 \sqrt {1-c^2 x^2}}{a+b \arccos (c x)}+\cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a}{b}+\arccos (c x)\right )+3 \cos \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (3 \left (\frac {a}{b}+\arccos (c x)\right )\right )+\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\arccos (c x)\right )+3 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\arccos (c x)\right )\right )}{4 b^2 c^3} \]

[In]

Integrate[x^2/(a + b*ArcCos[c*x])^2,x]

[Out]

-1/4*((-4*b*c^2*x^2*Sqrt[1 - c^2*x^2])/(a + b*ArcCos[c*x]) + Cos[a/b]*CosIntegral[a/b + ArcCos[c*x]] + 3*Cos[(
3*a)/b]*CosIntegral[3*(a/b + ArcCos[c*x])] + Sin[a/b]*SinIntegral[a/b + ArcCos[c*x]] + 3*Sin[(3*a)/b]*SinInteg
ral[3*(a/b + ArcCos[c*x])])/(b^2*c^3)

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(\frac {\frac {\sin \left (3 \arccos \left (c x \right )\right )}{4 \left (a +b \arccos \left (c x \right )\right ) b}-\frac {3 \left (\operatorname {Si}\left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right )+\operatorname {Ci}\left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right )\right )}{4 b^{2}}+\frac {\sqrt {-c^{2} x^{2}+1}}{4 \left (a +b \arccos \left (c x \right )\right ) b}-\frac {\operatorname {Si}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )+\operatorname {Ci}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )}{4 b^{2}}}{c^{3}}\)	\(147\)
default	\(\frac {\frac {\sin \left (3 \arccos \left (c x \right )\right )}{4 \left (a +b \arccos \left (c x \right )\right ) b}-\frac {3 \left (\operatorname {Si}\left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right )+\operatorname {Ci}\left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right )\right )}{4 b^{2}}+\frac {\sqrt {-c^{2} x^{2}+1}}{4 \left (a +b \arccos \left (c x \right )\right ) b}-\frac {\operatorname {Si}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )+\operatorname {Ci}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )}{4 b^{2}}}{c^{3}}\)	\(147\)

[In]

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

[Out]

1/c^3*(1/4*sin(3*arccos(c*x))/(a+b*arccos(c*x))/b-3/4*(Si(3*arccos(c*x)+3*a/b)*sin(3*a/b)+Ci(3*arccos(c*x)+3*a
/b)*cos(3*a/b))/b^2+1/4*(-c^2*x^2+1)^(1/2)/(a+b*arccos(c*x))/b-1/4*(Si(arccos(c*x)+a/b)*sin(a/b)+Ci(arccos(c*x
)+a/b)*cos(a/b))/b^2)

Fricas [F]

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

[In]

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

[Out]

integral(x^2/(b^2*arccos(c*x)^2 + 2*a*b*arccos(c*x) + a^2), x)

Sympy [F]

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

[In]

integrate(x**2/(a+b*acos(c*x))**2,x)

[Out]

Integral(x**2/(a + b*acos(c*x))**2, x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 615 vs. \(2 (145) = 290\).

Time = 0.32 (sec) , antiderivative size = 615, normalized size of antiderivative = 3.97 \[ \int \frac {x^2}{(a+b \arccos (c x))^2} \, dx=-\frac {3 \, b \arccos \left (c x\right ) \cos \left (\frac {a}{b}\right )^{3} \operatorname {Ci}\left (\frac {3 \, a}{b} + 3 \, \arccos \left (c x\right )\right )}{b^{3} c^{3} \arccos \left (c x\right ) + a b^{2} c^{3}} - \frac {3 \, b \arccos \left (c x\right ) \cos \left (\frac {a}{b}\right )^{2} \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {3 \, a}{b} + 3 \, \arccos \left (c x\right )\right )}{b^{3} c^{3} \arccos \left (c x\right ) + a b^{2} c^{3}} + \frac {\sqrt {-c^{2} x^{2} + 1} b c^{2} x^{2}}{b^{3} c^{3} \arccos \left (c x\right ) + a b^{2} c^{3}} - \frac {3 \, a \cos \left (\frac {a}{b}\right )^{3} \operatorname {Ci}\left (\frac {3 \, a}{b} + 3 \, \arccos \left (c x\right )\right )}{b^{3} c^{3} \arccos \left (c x\right ) + a b^{2} c^{3}} - \frac {3 \, a \cos \left (\frac {a}{b}\right )^{2} \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {3 \, a}{b} + 3 \, \arccos \left (c x\right )\right )}{b^{3} c^{3} \arccos \left (c x\right ) + a b^{2} c^{3}} + \frac {9 \, b \arccos \left (c x\right ) \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {3 \, a}{b} + 3 \, \arccos \left (c x\right )\right )}{4 \, {\left (b^{3} c^{3} \arccos \left (c x\right ) + a b^{2} c^{3}\right )}} - \frac {b \arccos \left (c x\right ) \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {a}{b} + \arccos \left (c x\right )\right )}{4 \, {\left (b^{3} c^{3} \arccos \left (c x\right ) + a b^{2} c^{3}\right )}} + \frac {3 \, b \arccos \left (c x\right ) \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {3 \, a}{b} + 3 \, \arccos \left (c x\right )\right )}{4 \, {\left (b^{3} c^{3} \arccos \left (c x\right ) + a b^{2} c^{3}\right )}} - \frac {b \arccos \left (c x\right ) \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arccos \left (c x\right )\right )}{4 \, {\left (b^{3} c^{3} \arccos \left (c x\right ) + a b^{2} c^{3}\right )}} + \frac {9 \, a \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {3 \, a}{b} + 3 \, \arccos \left (c x\right )\right )}{4 \, {\left (b^{3} c^{3} \arccos \left (c x\right ) + a b^{2} c^{3}\right )}} - \frac {a \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {a}{b} + \arccos \left (c x\right )\right )}{4 \, {\left (b^{3} c^{3} \arccos \left (c x\right ) + a b^{2} c^{3}\right )}} + \frac {3 \, a \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {3 \, a}{b} + 3 \, \arccos \left (c x\right )\right )}{4 \, {\left (b^{3} c^{3} \arccos \left (c x\right ) + a b^{2} c^{3}\right )}} - \frac {a \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arccos \left (c x\right )\right )}{4 \, {\left (b^{3} c^{3} \arccos \left (c x\right ) + a b^{2} c^{3}\right )}} \]

[In]

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

[Out]

-3*b*arccos(c*x)*cos(a/b)^3*cos_integral(3*a/b + 3*arccos(c*x))/(b^3*c^3*arccos(c*x) + a*b^2*c^3) - 3*b*arccos
(c*x)*cos(a/b)^2*sin(a/b)*sin_integral(3*a/b + 3*arccos(c*x))/(b^3*c^3*arccos(c*x) + a*b^2*c^3) + sqrt(-c^2*x^
2 + 1)*b*c^2*x^2/(b^3*c^3*arccos(c*x) + a*b^2*c^3) - 3*a*cos(a/b)^3*cos_integral(3*a/b + 3*arccos(c*x))/(b^3*c
^3*arccos(c*x) + a*b^2*c^3) - 3*a*cos(a/b)^2*sin(a/b)*sin_integral(3*a/b + 3*arccos(c*x))/(b^3*c^3*arccos(c*x)
+ a*b^2*c^3) + 9/4*b*arccos(c*x)*cos(a/b)*cos_integral(3*a/b + 3*arccos(c*x))/(b^3*c^3*arccos(c*x) + a*b^2*c^
3) - 1/4*b*arccos(c*x)*cos(a/b)*cos_integral(a/b + arccos(c*x))/(b^3*c^3*arccos(c*x) + a*b^2*c^3) + 3/4*b*arcc
os(c*x)*sin(a/b)*sin_integral(3*a/b + 3*arccos(c*x))/(b^3*c^3*arccos(c*x) + a*b^2*c^3) - 1/4*b*arccos(c*x)*sin
(a/b)*sin_integral(a/b + arccos(c*x))/(b^3*c^3*arccos(c*x) + a*b^2*c^3) + 9/4*a*cos(a/b)*cos_integral(3*a/b +
3*arccos(c*x))/(b^3*c^3*arccos(c*x) + a*b^2*c^3) - 1/4*a*cos(a/b)*cos_integral(a/b + arccos(c*x))/(b^3*c^3*arc
cos(c*x) + a*b^2*c^3) + 3/4*a*sin(a/b)*sin_integral(3*a/b + 3*arccos(c*x))/(b^3*c^3*arccos(c*x) + a*b^2*c^3) -
1/4*a*sin(a/b)*sin_integral(a/b + arccos(c*x))/(b^3*c^3*arccos(c*x) + a*b^2*c^3)

Mupad [F(-1)]

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

[In]

int(x^2/(a + b*acos(c*x))^2,x)

[Out]

int(x^2/(a + b*acos(c*x))^2, x)