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

Optimal. Leaf size=115 \[ \frac {2 \tanh ^{-1}\left (e^{i \cos ^{-1}(c x)}\right ) \left (a+b \cos ^{-1}(c x)\right )}{c^3 d}-\frac {x \left (a+b \cos ^{-1}(c x)\right )}{c^2 d}-\frac {i b \text {Li}_2\left (-e^{i \cos ^{-1}(c x)}\right )}{c^3 d}+\frac {i b \text {Li}_2\left (e^{i \cos ^{-1}(c x)}\right )}{c^3 d}+\frac {b \sqrt {1-c^2 x^2}}{c^3 d} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, 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 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]

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 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 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 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]

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*}

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Mathematica [A]  time = 0.18, size = 138, normalized size = 1.20 \[ -\frac {2 a c x+a \log (1-c x)-a \log (c x+1)-2 b \sqrt {1-c^2 x^2}+2 i b \text {Li}_2\left (-e^{i \cos ^{-1}(c x)}\right )-2 i b \text {Li}_2\left (e^{i \cos ^{-1}(c x)}\right )+2 b c x \cos ^{-1}(c x)+2 b \cos ^{-1}(c x) \log \left (1-e^{i \cos ^{-1}(c x)}\right )-2 b \cos ^{-1}(c x) \log \left (1+e^{i \cos ^{-1}(c x)}\right )}{2 c^3 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/2*(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
*ArcCos[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])])/(c^3*d)

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b x^{2} \arccos \left (c x\right ) + a x^{2}}{c^{2} d x^{2} - d}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {{\left (b \arccos \left (c x\right ) + a\right )} x^{2}}{c^{2} d x^{2} - d}\,{d x} \]

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)

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maple [A]  time = 0.44, size = 207, normalized size = 1.80 \[ -\frac {a x}{c^{2} d}-\frac {a \ln \left (c x -1\right )}{2 c^{3} d}+\frac {a \ln \left (c x +1\right )}{2 c^{3} d}+\frac {b \arccos \left (c x \right ) \ln \left (1+c x +i \sqrt {-c^{2} x^{2}+1}\right )}{c^{3} d}-\frac {b \arccos \left (c x \right ) \ln \left (1-c x -i \sqrt {-c^{2} x^{2}+1}\right )}{c^{3} d}-\frac {b \arccos \left (c x \right ) x}{c^{2} d}+\frac {i b \dilog \left (1-c x -i \sqrt {-c^{2} x^{2}+1}\right )}{c^{3} d}-\frac {i b \dilog \left (1+c x +i \sqrt {-c^{2} x^{2}+1}\right )}{c^{3} d}+\frac {b \sqrt {-c^{2} x^{2}+1}}{c^{3} d} \]

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)+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/c^3*b/d*dilog(1-c*x-I*(-c^2*x^2
+1)^(1/2))-I/c^3*b/d*dilog(1+c*x+I*(-c^2*x^2+1)^(1/2))+b*(-c^2*x^2+1)^(1/2)/c^3/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{2} \, a {\left (\frac {2 \, x}{c^{2} d} - \frac {\log \left (c x + 1\right )}{c^{3} d} + \frac {\log \left (c x - 1\right )}{c^{3} d}\right )} - \frac {-{\left (c^{3} d {\left (\frac {2 \, \sqrt {c x + 1} \sqrt {-c x + 1}}{c^{3} d} + \int -\frac {\sqrt {c x + 1} \sqrt {-c x + 1} {\left (\log \left (c x + 1\right ) - \log \left (-c x + 1\right )\right )}}{c^{4} d x^{2} - c^{2} d}\,{d x}\right )} - {\left (2 \, c x - \log \left (c x + 1\right ) + \log \left (-c x + 1\right )\right )} \arctan \left (\sqrt {c x + 1} \sqrt {-c x + 1}, c x\right )\right )} b}{2 \, c^{3} d} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {a x^{2}}{c^{2} x^{2} - 1}\, dx + \int \frac {b x^{2} \operatorname {acos}{\left (c x \right )}}{c^{2} x^{2} - 1}\, dx}{d} \]

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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