3.163 \(\int \frac {x^2}{(a+b \cos ^{-1}(c x))^2} \, dx\)

Optimal. Leaf size=155 \[ -\frac {\cos \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a+b \cos ^{-1}(c x)}{b}\right )}{4 b^2 c^3}-\frac {3 \cos \left (\frac {3 a}{b}\right ) \text {Ci}\left (\frac {3 \left (a+b \cos ^{-1}(c x)\right )}{b}\right )}{4 b^2 c^3}-\frac {\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \cos ^{-1}(c x)}{b}\right )}{4 b^2 c^3}-\frac {3 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 \left (a+b \cos ^{-1}(c x)\right )}{b}\right )}{4 b^2 c^3}+\frac {x^2 \sqrt {1-c^2 x^2}}{b c \left (a+b \cos ^{-1}(c x)\right )} \]

[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))

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Rubi [A]  time = 0.18, antiderivative size = 151, normalized size of antiderivative = 0.97, number of steps used = 8, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {4632, 3303, 3299, 3302} \[ -\frac {\cos \left (\frac {a}{b}\right ) \text {CosIntegral}\left (\frac {a}{b}+\cos ^{-1}(c x)\right )}{4 b^2 c^3}-\frac {3 \cos \left (\frac {3 a}{b}\right ) \text {CosIntegral}\left (\frac {3 a}{b}+3 \cos ^{-1}(c x)\right )}{4 b^2 c^3}-\frac {\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\cos ^{-1}(c x)\right )}{4 b^2 c^3}-\frac {3 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 a}{b}+3 \cos ^{-1}(c x)\right )}{4 b^2 c^3}+\frac {x^2 \sqrt {1-c^2 x^2}}{b c \left (a+b \cos ^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

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

Rule 3299

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 3302

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 3303

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 4632

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

Rubi steps

\begin {align*} \int \frac {x^2}{\left (a+b \cos ^{-1}(c x)\right )^2} \, dx &=\frac {x^2 \sqrt {1-c^2 x^2}}{b c \left (a+b \cos ^{-1}(c x)\right )}+\frac {\operatorname {Subst}\left (\int \left (-\frac {\cos (x)}{4 (a+b x)}-\frac {3 \cos (3 x)}{4 (a+b x)}\right ) \, dx,x,\cos ^{-1}(c x)\right )}{b c^3}\\ &=\frac {x^2 \sqrt {1-c^2 x^2}}{b c \left (a+b \cos ^{-1}(c x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {\cos (x)}{a+b x} \, dx,x,\cos ^{-1}(c x)\right )}{4 b c^3}-\frac {3 \operatorname {Subst}\left (\int \frac {\cos (3 x)}{a+b x} \, dx,x,\cos ^{-1}(c x)\right )}{4 b c^3}\\ &=\frac {x^2 \sqrt {1-c^2 x^2}}{b c \left (a+b \cos ^{-1}(c x)\right )}-\frac {\cos \left (\frac {a}{b}\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\cos ^{-1}(c x)\right )}{4 b c^3}-\frac {\left (3 \cos \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\cos ^{-1}(c x)\right )}{4 b c^3}-\frac {\sin \left (\frac {a}{b}\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\cos ^{-1}(c x)\right )}{4 b c^3}-\frac {\left (3 \sin \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\cos ^{-1}(c x)\right )}{4 b c^3}\\ &=\frac {x^2 \sqrt {1-c^2 x^2}}{b c \left (a+b \cos ^{-1}(c x)\right )}-\frac {\cos \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a}{b}+\cos ^{-1}(c x)\right )}{4 b^2 c^3}-\frac {3 \cos \left (\frac {3 a}{b}\right ) \text {Ci}\left (\frac {3 a}{b}+3 \cos ^{-1}(c x)\right )}{4 b^2 c^3}-\frac {\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\cos ^{-1}(c x)\right )}{4 b^2 c^3}-\frac {3 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 a}{b}+3 \cos ^{-1}(c x)\right )}{4 b^2 c^3}\\ \end {align*}

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Mathematica [A]  time = 0.61, size = 124, normalized size = 0.80 \[ -\frac {-\frac {4 b c^2 x^2 \sqrt {1-c^2 x^2}}{a+b \cos ^{-1}(c x)}+\cos \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a}{b}+\cos ^{-1}(c x)\right )+3 \cos \left (\frac {3 a}{b}\right ) \text {Ci}\left (3 \left (\frac {a}{b}+\cos ^{-1}(c x)\right )\right )+\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\cos ^{-1}(c x)\right )+3 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\cos ^{-1}(c x)\right )\right )}{4 b^2 c^3} \]

Antiderivative was successfully verified.

[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)

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{2}}{b^{2} \arccos \left (c x\right )^{2} + 2 \, a b \arccos \left (c x\right ) + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [B]  time = 0.23, size = 615, normalized size = 3.97 \[ -\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 )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.05, size = 147, normalized size = 0.95 \[ \frac {\frac {\sin \left (3 \arccos \left (c x \right )\right )}{4 \left (a +b \arccos \left (c x \right )\right ) b}-\frac {3 \left (\Si \left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right )+\Ci \left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right )\right )}{4 b^{2}}+\frac {\sqrt {-c^{2} x^{2}+1}}{4 \left (a +b \arccos \left (c x \right )\right ) b}-\frac {\Si \left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )+\Ci \left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )}{4 b^{2}}}{c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2}{{\left (a+b\,\mathrm {acos}\left (c\,x\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a + b*acos(c*x))^2,x)

[Out]

int(x^2/(a + b*acos(c*x))^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\left (a + b \operatorname {acos}{\left (c x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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