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

Optimal. Leaf size=136 \[ \frac{i b \text{PolyLog}\left (2,-e^{i \cos ^{-1}(c x)}\right )}{2 c^3 d^2}-\frac{i b \text{PolyLog}\left (2,e^{i \cos ^{-1}(c x)}\right )}{2 c^3 d^2}+\frac{x \left (a+b \cos ^{-1}(c x)\right )}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac{\tanh ^{-1}\left (e^{i \cos ^{-1}(c x)}\right ) \left (a+b \cos ^{-1}(c x)\right )}{c^3 d^2}+\frac{b}{2 c^3 d^2 \sqrt{1-c^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.129542, antiderivative size = 136, 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 = {4704, 4658, 4183, 2279, 2391, 261} \[ \frac{i b \text{PolyLog}\left (2,-e^{i \cos ^{-1}(c x)}\right )}{2 c^3 d^2}-\frac{i b \text{PolyLog}\left (2,e^{i \cos ^{-1}(c x)}\right )}{2 c^3 d^2}+\frac{x \left (a+b \cos ^{-1}(c x)\right )}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac{\tanh ^{-1}\left (e^{i \cos ^{-1}(c x)}\right ) \left (a+b \cos ^{-1}(c x)\right )}{c^3 d^2}+\frac{b}{2 c^3 d^2 \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4704

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)/(2*e*(p + 1)), x] + (-Dist[(f^2*(m - 1))/(2*e*(p +
1)), Int[(f*x)^(m - 2)*(d + e*x^2)^(p + 1)*(a + b*ArcCos[c*x])^n, x], x] - Dist[(b*f*n*d^IntPart[p]*(d + e*x^2
)^FracPart[p])/(2*c*(p + 1)*(1 - c^2*x^2)^FracPart[p]), Int[(f*x)^(m - 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcCo
s[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[p, -1] && Gt
Q[m, 1]

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

Mathematica [A]  time = 0.308591, size = 251, normalized size = 1.85 \[ -\frac{-2 i b \left (c^2 x^2-1\right ) \text{PolyLog}\left (2,-e^{i \cos ^{-1}(c x)}\right )+2 i b \left (c^2 x^2-1\right ) \text{PolyLog}\left (2,e^{i \cos ^{-1}(c x)}\right )-a c^2 x^2 \log (1-c x)+a c^2 x^2 \log (c x+1)+2 a c x+a \log (1-c x)-a \log (c x+1)+2 b \sqrt{1-c^2 x^2}-2 b c^2 x^2 \cos ^{-1}(c x) \log \left (1-e^{i \cos ^{-1}(c x)}\right )+2 b c^2 x^2 \cos ^{-1}(c x) \log \left (1+e^{i \cos ^{-1}(c x)}\right )+2 b c x \cos ^{-1}(c x)+2 b \cos ^{-1}(c x) \log \left (1-e^{i \cos ^{-1}(c x)}\right )-2 b \cos ^{-1}(c x) \log \left (1+e^{i \cos ^{-1}(c x)}\right )}{4 c^3 d^2 \left (c^2 x^2-1\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^2*(a + b*ArcCos[c*x]))/(d - c^2*d*x^2)^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*c^2
*x^2*ArcCos[c*x]*Log[1 - E^(I*ArcCos[c*x])] - 2*b*ArcCos[c*x]*Log[1 + E^(I*ArcCos[c*x])] + 2*b*c^2*x^2*ArcCos[
c*x]*Log[1 + E^(I*ArcCos[c*x])] + a*Log[1 - c*x] - a*c^2*x^2*Log[1 - c*x] - a*Log[1 + c*x] + a*c^2*x^2*Log[1 +
c*x] - (2*I)*b*(-1 + c^2*x^2)*PolyLog[2, -E^(I*ArcCos[c*x])] + (2*I)*b*(-1 + c^2*x^2)*PolyLog[2, E^(I*ArcCos[
c*x])])/(4*c^3*d^2*(-1 + c^2*x^2))

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Maple [A]  time = 0.151, size = 253, normalized size = 1.9 \begin{align*} -{\frac{a}{4\,{c}^{3}{d}^{2} \left ( cx-1 \right ) }}+{\frac{a\ln \left ( cx-1 \right ) }{4\,{c}^{3}{d}^{2}}}-{\frac{a}{4\,{c}^{3}{d}^{2} \left ( cx+1 \right ) }}-{\frac{a\ln \left ( cx+1 \right ) }{4\,{c}^{3}{d}^{2}}}-{\frac{b\arccos \left ( cx \right ) x}{2\,{c}^{2}{d}^{2} \left ({c}^{2}{x}^{2}-1 \right ) }}-{\frac{b}{2\,{c}^{3}{d}^{2} \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{b\arccos \left ( cx \right ) }{2\,{c}^{3}{d}^{2}}\ln \left ( 1+cx+i\sqrt{-{c}^{2}{x}^{2}+1} \right ) }+{\frac{{\frac{i}{2}}b}{{c}^{3}{d}^{2}}{\it polylog} \left ( 2,-cx-i\sqrt{-{c}^{2}{x}^{2}+1} \right ) }+{\frac{b\arccos \left ( cx \right ) }{2\,{c}^{3}{d}^{2}}\ln \left ( 1-cx-i\sqrt{-{c}^{2}{x}^{2}+1} \right ) }-{\frac{{\frac{i}{2}}b}{{c}^{3}{d}^{2}}{\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)^2,x)

[Out]

-1/4/c^3*a/d^2/(c*x-1)+1/4/c^3*a/d^2*ln(c*x-1)-1/4/c^3*a/d^2/(c*x+1)-1/4/c^3*a/d^2*ln(c*x+1)-1/2/c^2*b/d^2/(c^
2*x^2-1)*arccos(c*x)*x-1/2/c^3*b/d^2/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)-1/2/c^3*b/d^2*arccos(c*x)*ln(1+c*x+I*(-c^2
*x^2+1)^(1/2))+1/2*I*b*polylog(2,-c*x-I*(-c^2*x^2+1)^(1/2))/c^3/d^2+1/2/c^3*b/d^2*arccos(c*x)*ln(1-c*x-I*(-c^2
*x^2+1)^(1/2))-1/2*I*b*polylog(2,c*x+I*(-c^2*x^2+1)^(1/2))/c^3/d^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{4} \, a{\left (\frac{2 \, x}{c^{4} d^{2} x^{2} - c^{2} d^{2}} + \frac{\log \left (c x + 1\right )}{c^{3} d^{2}} - \frac{\log \left (c x - 1\right )}{c^{3} d^{2}}\right )} - \frac{{\left ({\left (2 \, c x +{\left (c^{2} x^{2} - 1\right )} \log \left (c x + 1\right ) -{\left (c^{2} x^{2} - 1\right )} \log \left (-c x + 1\right )\right )} \arctan \left (\sqrt{c x + 1} \sqrt{-c x + 1}, c x\right ) -{\left (c^{5} d^{2} x^{2} - c^{3} d^{2}\right )} \int \frac{{\left (2 \, c x +{\left (c^{2} x^{2} - 1\right )} \log \left (c x + 1\right ) -{\left (c^{2} x^{2} - 1\right )} \log \left (-c x + 1\right )\right )} \sqrt{c x + 1} \sqrt{-c x + 1}}{c^{6} d^{2} x^{4} - 2 \, c^{4} d^{2} x^{2} + c^{2} d^{2}}\,{d x}\right )} b}{4 \,{\left (c^{5} d^{2} x^{2} - c^{3} d^{2}\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)^2,x, algorithm="maxima")

[Out]

-1/4*a*(2*x/(c^4*d^2*x^2 - c^2*d^2) + log(c*x + 1)/(c^3*d^2) - log(c*x - 1)/(c^3*d^2)) - 1/4*((2*c*x + (c^2*x^
2 - 1)*log(c*x + 1) - (c^2*x^2 - 1)*log(-c*x + 1))*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x) - 4*(c^5*d^2*x^2
- c^3*d^2)*integrate(1/4*(2*c*x + (c^2*x^2 - 1)*log(c*x + 1) - (c^2*x^2 - 1)*log(-c*x + 1))*sqrt(c*x + 1)*sqr
t(-c*x + 1)/(c^6*d^2*x^4 - 2*c^4*d^2*x^2 + c^2*d^2), x))*b/(c^5*d^2*x^2 - c^3*d^2)

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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^{4} d^{2} x^{4} - 2 \, c^{2} d^{2} x^{2} + d^{2}}, 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)^2,x, algorithm="fricas")

[Out]

integral((b*x^2*arccos(c*x) + a*x^2)/(c^4*d^2*x^4 - 2*c^2*d^2*x^2 + d^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a x^{2}}{c^{4} x^{4} - 2 c^{2} x^{2} + 1}\, dx + \int \frac{b x^{2} \operatorname{acos}{\left (c x \right )}}{c^{4} x^{4} - 2 c^{2} x^{2} + 1}\, dx}{d^{2}} \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)**2,x)

[Out]

(Integral(a*x**2/(c**4*x**4 - 2*c**2*x**2 + 1), x) + Integral(b*x**2*acos(c*x)/(c**4*x**4 - 2*c**2*x**2 + 1),
x))/d**2

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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}}{{\left (c^{2} d x^{2} - d\right )}^{2}}\,{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)^2,x, algorithm="giac")

[Out]

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