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

Optimal. Leaf size=115 \[ -\frac{i b \text{PolyLog}\left (2,-e^{i \cos ^{-1}(c x)}\right )}{c^3 d}+\frac{i b \text{PolyLog}\left (2,e^{i \cos ^{-1}(c x)}\right )}{c^3 d}-\frac{x \left (a+b \cos ^{-1}(c x)\right )}{c^2 d}+\frac{2 \tanh ^{-1}\left (e^{i \cos ^{-1}(c x)}\right ) \left (a+b \cos ^{-1}(c x)\right )}{c^3 d}+\frac{b \sqrt{1-c^2 x^2}}{c^3 d} \]

[Out]

(b*Sqrt[1 - c^2*x^2])/(c^3*d) - (x*(a + b*ArcCos[c*x]))/(c^2*d) + (2*(a + b*ArcCos[c*x])*ArcTanh[E^(I*ArcCos[c
*x])])/(c^3*d) - (I*b*PolyLog[2, -E^(I*ArcCos[c*x])])/(c^3*d) + (I*b*PolyLog[2, E^(I*ArcCos[c*x])])/(c^3*d)

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Rubi [A]  time = 0.134022, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {4716, 4658, 4183, 2279, 2391, 261} \[ -\frac{i b \text{PolyLog}\left (2,-e^{i \cos ^{-1}(c x)}\right )}{c^3 d}+\frac{i b \text{PolyLog}\left (2,e^{i \cos ^{-1}(c x)}\right )}{c^3 d}-\frac{x \left (a+b \cos ^{-1}(c x)\right )}{c^2 d}+\frac{2 \tanh ^{-1}\left (e^{i \cos ^{-1}(c x)}\right ) \left (a+b \cos ^{-1}(c x)\right )}{c^3 d}+\frac{b \sqrt{1-c^2 x^2}}{c^3 d} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(a + b*ArcCos[c*x]))/(d - c^2*d*x^2),x]

[Out]

(b*Sqrt[1 - c^2*x^2])/(c^3*d) - (x*(a + b*ArcCos[c*x]))/(c^2*d) + (2*(a + b*ArcCos[c*x])*ArcTanh[E^(I*ArcCos[c
*x])])/(c^3*d) - (I*b*PolyLog[2, -E^(I*ArcCos[c*x])])/(c^3*d) + (I*b*PolyLog[2, E^(I*ArcCos[c*x])])/(c^3*d)

Rule 4716

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(
f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*(a + b*ArcCos[c*x])^n)/(e*(m + 2*p + 1)), x] + (Dist[(f^2*(m - 1))/(c^2*(m
+ 2*p + 1)), Int[(f*x)^(m - 2)*(d + e*x^2)^p*(a + b*ArcCos[c*x])^n, x], x] - Dist[(b*f*n*d^IntPart[p]*(d + e*
x^2)^FracPart[p])/(c*(m + 2*p + 1)*(1 - c^2*x^2)^FracPart[p]), Int[(f*x)^(m - 1)*(1 - c^2*x^2)^(p + 1/2)*(a +
b*ArcCos[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[m,
1] && NeQ[m + 2*p + 1, 0] && IntegerQ[m]

Rule 4658

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Dist[(c*d)^(-1), Subst[Int[(
a + b*x)^n*Csc[x], x], x, ArcCos[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[n, 0]

Rule 4183

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E^(I*(e + f*
x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Dist[(d*m)/f, Int[(c +
d*x)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

\begin{align*} \int \frac{x^2 \left (a+b \cos ^{-1}(c x)\right )}{d-c^2 d x^2} \, dx &=-\frac{x \left (a+b \cos ^{-1}(c x)\right )}{c^2 d}+\frac{\int \frac{a+b \cos ^{-1}(c x)}{d-c^2 d x^2} \, dx}{c^2}-\frac{b \int \frac{x}{\sqrt{1-c^2 x^2}} \, dx}{c d}\\ &=\frac{b \sqrt{1-c^2 x^2}}{c^3 d}-\frac{x \left (a+b \cos ^{-1}(c x)\right )}{c^2 d}-\frac{\operatorname{Subst}\left (\int (a+b x) \csc (x) \, dx,x,\cos ^{-1}(c x)\right )}{c^3 d}\\ &=\frac{b \sqrt{1-c^2 x^2}}{c^3 d}-\frac{x \left (a+b \cos ^{-1}(c x)\right )}{c^2 d}+\frac{2 \left (a+b \cos ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \cos ^{-1}(c x)}\right )}{c^3 d}+\frac{b \operatorname{Subst}\left (\int \log \left (1-e^{i x}\right ) \, dx,x,\cos ^{-1}(c x)\right )}{c^3 d}-\frac{b \operatorname{Subst}\left (\int \log \left (1+e^{i x}\right ) \, dx,x,\cos ^{-1}(c x)\right )}{c^3 d}\\ &=\frac{b \sqrt{1-c^2 x^2}}{c^3 d}-\frac{x \left (a+b \cos ^{-1}(c x)\right )}{c^2 d}+\frac{2 \left (a+b \cos ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \cos ^{-1}(c x)}\right )}{c^3 d}-\frac{(i b) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{i \cos ^{-1}(c x)}\right )}{c^3 d}+\frac{(i b) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{i \cos ^{-1}(c x)}\right )}{c^3 d}\\ &=\frac{b \sqrt{1-c^2 x^2}}{c^3 d}-\frac{x \left (a+b \cos ^{-1}(c x)\right )}{c^2 d}+\frac{2 \left (a+b \cos ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \cos ^{-1}(c x)}\right )}{c^3 d}-\frac{i b \text{Li}_2\left (-e^{i \cos ^{-1}(c x)}\right )}{c^3 d}+\frac{i b \text{Li}_2\left (e^{i \cos ^{-1}(c x)}\right )}{c^3 d}\\ \end{align*}

Mathematica [A]  time = 0.162028, size = 138, normalized size = 1.2 \[ -\frac{2 i b \text{PolyLog}\left (2,-e^{i \cos ^{-1}(c x)}\right )-2 i b \text{PolyLog}\left (2,e^{i \cos ^{-1}(c x)}\right )+2 a c x+a \log (1-c x)-a \log (c x+1)-2 b \sqrt{1-c^2 x^2}+2 b c x \cos ^{-1}(c x)+2 b \cos ^{-1}(c x) \log \left (1-e^{i \cos ^{-1}(c x)}\right )-2 b \cos ^{-1}(c x) \log \left (1+e^{i \cos ^{-1}(c x)}\right )}{2 c^3 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(2*a*c*x - 2*b*Sqrt[1 - c^2*x^2] + 2*b*c*x*ArcCos[c*x] + 2*b*ArcCos[c*x]*Log[1 - E^(I*ArcCos[c*x])] - 2*b*Arc
Cos[c*x]*Log[1 + E^(I*ArcCos[c*x])] + a*Log[1 - c*x] - a*Log[1 + c*x] + (2*I)*b*PolyLog[2, -E^(I*ArcCos[c*x])]
- (2*I)*b*PolyLog[2, E^(I*ArcCos[c*x])])/(2*c^3*d)

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Maple [A]  time = 0.119, size = 207, normalized size = 1.8 \begin{align*} -{\frac{ax}{{c}^{2}d}}-{\frac{a\ln \left ( cx-1 \right ) }{2\,d{c}^{3}}}+{\frac{a\ln \left ( cx+1 \right ) }{2\,d{c}^{3}}}+{\frac{b}{d{c}^{3}}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{b\arccos \left ( cx \right ) }{d{c}^{3}}\ln \left ( 1-cx-i\sqrt{-{c}^{2}{x}^{2}+1} \right ) }+{\frac{b\arccos \left ( cx \right ) }{d{c}^{3}}\ln \left ( 1+cx+i\sqrt{-{c}^{2}{x}^{2}+1} \right ) }-{\frac{b\arccos \left ( cx \right ) x}{{c}^{2}d}}+{\frac{ib}{d{c}^{3}}{\it polylog} \left ( 2,cx+i\sqrt{-{c}^{2}{x}^{2}+1} \right ) }-{\frac{ib}{d{c}^{3}}{\it polylog} \left ( 2,-cx-i\sqrt{-{c}^{2}{x}^{2}+1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/c^2*a/d*x-1/2/c^3*a/d*ln(c*x-1)+1/2/c^3*a/d*ln(c*x+1)+b*(-c^2*x^2+1)^(1/2)/d/c^3-1/c^3*b/d*arccos(c*x)*ln(1
-c*x-I*(-c^2*x^2+1)^(1/2))+1/c^3*b/d*arccos(c*x)*ln(1+c*x+I*(-c^2*x^2+1)^(1/2))-1/c^2*b/d*arccos(c*x)*x+I*b*po
lylog(2,c*x+I*(-c^2*x^2+1)^(1/2))/d/c^3-I*b*polylog(2,-c*x-I*(-c^2*x^2+1)^(1/2))/d/c^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{2} \, a{\left (\frac{2 \, x}{c^{2} d} - \frac{\log \left (c x + 1\right )}{c^{3} d} + \frac{\log \left (c x - 1\right )}{c^{3} d}\right )} - \frac{-{\left (c^{3} d{\left (\frac{2 \, \sqrt{c x + 1} \sqrt{-c x + 1}}{c^{3} d} + \int -\frac{\sqrt{c x + 1} \sqrt{-c x + 1}{\left (\log \left (c x + 1\right ) - \log \left (-c x + 1\right )\right )}}{c^{4} d x^{2} - c^{2} d}\,{d x}\right )} -{\left (2 \, c x - \log \left (c x + 1\right ) + \log \left (-c x + 1\right )\right )} \arctan \left (\sqrt{c x + 1} \sqrt{-c x + 1}, c x\right )\right )} b}{2 \, c^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a*(2*x/(c^2*d) - log(c*x + 1)/(c^3*d) + log(c*x - 1)/(c^3*d)) - 1/2*(2*c^3*d*integrate(-1/2*(2*c*x - log(
c*x + 1) + log(-c*x + 1))*sqrt(c*x + 1)*sqrt(-c*x + 1)/(c^4*d*x^2 - c^2*d), x) + (2*c*x - log(c*x + 1) + log(-
c*x + 1))*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x))*b/(c^3*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{\int \frac{a x^{2}}{c^{2} x^{2} - 1}\, dx + \int \frac{b x^{2} \operatorname{acos}{\left (c x \right )}}{c^{2} x^{2} - 1}\, dx}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(Integral(a*x**2/(c**2*x**2 - 1), x) + Integral(b*x**2*acos(c*x)/(c**2*x**2 - 1), x))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(-(b*arccos(c*x) + a)*x^2/(c^2*d*x^2 - d), x)