3.2.58 \(\int \frac {x^2}{a+b \text {ArcCos}(c x)} \, dx\) [158]

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

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 121, normalized size of antiderivative = 1.00, 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 = {4732, 4491, 3384, 3380, 3383} \begin {gather*} \frac {\sin \left (\frac {a}{b}\right ) \text {CosIntegral}\left (\frac {a+b \text {ArcCos}(c x)}{b}\right )}{4 b c^3}+\frac {\sin \left (\frac {3 a}{b}\right ) \text {CosIntegral}\left (\frac {3 (a+b \text {ArcCos}(c x))}{b}\right )}{4 b c^3}-\frac {\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \text {ArcCos}(c x)}{b}\right )}{4 b c^3}-\frac {\cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \text {ArcCos}(c x))}{b}\right )}{4 b c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b*ArcCos[c*x]),x]

[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)

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 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 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]

Rubi steps

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

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Mathematica [A]
time = 0.12, size = 91, normalized size = 0.75 \begin {gather*} -\frac {-\text {CosIntegral}\left (\frac {a}{b}+\text {ArcCos}(c x)\right ) \sin \left (\frac {a}{b}\right )-\text {CosIntegral}\left (3 \left (\frac {a}{b}+\text {ArcCos}(c x)\right )\right ) \sin \left (\frac {3 a}{b}\right )+\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\text {ArcCos}(c x)\right )+\cos \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\text {ArcCos}(c x)\right )\right )}{4 b c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a + b*ArcCos[c*x]),x]

[Out]

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

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Maple [A]
time = 0.08, size = 102, normalized size = 0.84

method	result	size
derivativedivides	\(\frac {-\frac {\sinIntegral \left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right )}{4 b}+\frac {\cosineIntegral \left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right )}{4 b}-\frac {\sinIntegral \left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )}{4 b}+\frac {\cosineIntegral \left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )}{4 b}}{c^{3}}\)	\(102\)
default	\(\frac {-\frac {\sinIntegral \left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right )}{4 b}+\frac {\cosineIntegral \left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right )}{4 b}-\frac {\sinIntegral \left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )}{4 b}+\frac {\cosineIntegral \left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )}{4 b}}{c^{3}}\)	\(102\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{a + b \operatorname {acos}{\left (c x \right )}}\, dx \end {gather*}

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 = 0.40, size = 172, normalized size = 1.42 \begin {gather*} \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 {gather*}

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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a + b*acos(c*x)),x)

[Out]

int(x^2/(a + b*acos(c*x)), x)

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