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

Optimal. Leaf size=121 \[ \frac{\sin \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a+b \cos ^{-1}(c x)}{b}\right )}{4 b c^3}+\frac{\sin \left (\frac{3 a}{b}\right ) \text{CosIntegral}\left (\frac{3 \left (a+b \cos ^{-1}(c x)\right )}{b}\right )}{4 b c^3}-\frac{\cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a+b \cos ^{-1}(c x)}{b}\right )}{4 b c^3}-\frac{\cos \left (\frac{3 a}{b}\right ) \text{Si}\left (\frac{3 \left (a+b \cos ^{-1}(c x)\right )}{b}\right )}{4 b c^3} \]

[Out]

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

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Rubi [A]  time = 0.21562, antiderivative size = 117, normalized size of antiderivative = 0.97, number of steps used = 9, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {4636, 4406, 3303, 3299, 3302} \[ \frac{\sin \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\cos ^{-1}(c x)\right )}{4 b c^3}+\frac{\sin \left (\frac{3 a}{b}\right ) \text{CosIntegral}\left (\frac{3 a}{b}+3 \cos ^{-1}(c x)\right )}{4 b c^3}-\frac{\cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\cos ^{-1}(c x)\right )}{4 b c^3}-\frac{\cos \left (\frac{3 a}{b}\right ) \text{Si}\left (\frac{3 a}{b}+3 \cos ^{-1}(c x)\right )}{4 b c^3} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b*ArcCos[c*x]),x]

[Out]

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

Rule 4636

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

Rule 4406

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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.160972, size = 91, normalized size = 0.75 \[ -\frac{\sin \left (\frac{a}{b}\right ) \left (-\text{CosIntegral}\left (\frac{a}{b}+\cos ^{-1}(c x)\right )\right )-\sin \left (\frac{3 a}{b}\right ) \text{CosIntegral}\left (3 \left (\frac{a}{b}+\cos ^{-1}(c x)\right )\right )+\cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\cos ^{-1}(c x)\right )+\cos \left (\frac{3 a}{b}\right ) \text{Si}\left (3 \left (\frac{a}{b}+\cos ^{-1}(c x)\right )\right )}{4 b c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a + b*ArcCos[c*x]),x]

[Out]

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

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Maple [A]  time = 0.05, size = 102, normalized size = 0.8 \begin{align*}{\frac{1}{{c}^{3}} \left ( -{\frac{1}{4\,b}{\it Si} \left ( 3\,\arccos \left ( cx \right ) +3\,{\frac{a}{b}} \right ) \cos \left ( 3\,{\frac{a}{b}} \right ) }+{\frac{1}{4\,b}{\it Ci} \left ( 3\,\arccos \left ( cx \right ) +3\,{\frac{a}{b}} \right ) \sin \left ( 3\,{\frac{a}{b}} \right ) }-{\frac{1}{4\,b}{\it Si} \left ( \arccos \left ( cx \right ) +{\frac{a}{b}} \right ) \cos \left ({\frac{a}{b}} \right ) }+{\frac{1}{4\,b}{\it Ci} \left ( \arccos \left ( cx \right ) +{\frac{a}{b}} \right ) \sin \left ({\frac{a}{b}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{b \arccos \left (c x\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/(b*arccos(c*x) + a), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{2}}{b \arccos \left (c x\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2/(b*arccos(c*x) + a), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{a + b \operatorname{acos}{\left (c x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2/(a + b*acos(c*x)), x)

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Giac [A]  time = 1.20722, size = 232, normalized size = 1.92 \begin{align*} \frac{\cos \left (\frac{a}{b}\right )^{2} \operatorname{Ci}\left (\frac{3 \, a}{b} + 3 \, \arccos \left (c x\right )\right ) \sin \left (\frac{a}{b}\right )}{b c^{3}} - \frac{\cos \left (\frac{a}{b}\right )^{3} \operatorname{Si}\left (\frac{3 \, a}{b} + 3 \, \arccos \left (c x\right )\right )}{b c^{3}} - \frac{\operatorname{Ci}\left (\frac{3 \, a}{b} + 3 \, \arccos \left (c x\right )\right ) \sin \left (\frac{a}{b}\right )}{4 \, b c^{3}} + \frac{\operatorname{Ci}\left (\frac{a}{b} + \arccos \left (c x\right )\right ) \sin \left (\frac{a}{b}\right )}{4 \, b c^{3}} + \frac{3 \, \cos \left (\frac{a}{b}\right ) \operatorname{Si}\left (\frac{3 \, a}{b} + 3 \, \arccos \left (c x\right )\right )}{4 \, b c^{3}} - \frac{\cos \left (\frac{a}{b}\right ) \operatorname{Si}\left (\frac{a}{b} + \arccos \left (c x\right )\right )}{4 \, b c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

cos(a/b)^2*cos_integral(3*a/b + 3*arccos(c*x))*sin(a/b)/(b*c^3) - cos(a/b)^3*sin_integral(3*a/b + 3*arccos(c*x
))/(b*c^3) - 1/4*cos_integral(3*a/b + 3*arccos(c*x))*sin(a/b)/(b*c^3) + 1/4*cos_integral(a/b + arccos(c*x))*si
n(a/b)/(b*c^3) + 3/4*cos(a/b)*sin_integral(3*a/b + 3*arccos(c*x))/(b*c^3) - 1/4*cos(a/b)*sin_integral(a/b + ar
ccos(c*x))/(b*c^3)