\(\int \frac {x^2 (a+b \arccos (c x))}{(d-c^2 d x^2)^2} \, dx\) [10]

Optimal result

Integrand size = 25, antiderivative size = 136 \[ \int \frac {x^2 (a+b \arccos (c x))}{\left (d-c^2 d x^2\right )^2} \, dx=\frac {b}{2 c^3 d^2 \sqrt {1-c^2 x^2}}+\frac {x (a+b \arccos (c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac {(a+b \arccos (c x)) \text {arctanh}\left (e^{i \arccos (c x)}\right )}{c^3 d^2}+\frac {i b \operatorname {PolyLog}\left (2,-e^{i \arccos (c x)}\right )}{2 c^3 d^2}-\frac {i b \operatorname {PolyLog}\left (2,e^{i \arccos (c x)}\right )}{2 c^3 d^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {4792, 4750, 4268, 2317, 2438, 267} \[ \int \frac {x^2 (a+b \arccos (c x))}{\left (d-c^2 d x^2\right )^2} \, dx=-\frac {\text {arctanh}\left (e^{i \arccos (c x)}\right ) (a+b \arccos (c x))}{c^3 d^2}+\frac {x (a+b \arccos (c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac {i b \operatorname {PolyLog}\left (2,-e^{i \arccos (c x)}\right )}{2 c^3 d^2}-\frac {i b \operatorname {PolyLog}\left (2,e^{i \arccos (c x)}\right )}{2 c^3 d^2}+\frac {b}{2 c^3 d^2 \sqrt {1-c^2 x^2}} \]

[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 267

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 2317

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 2438

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 4268

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 4750

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 4792

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/(2*c*(p + 1)))*Simp[(d
+ e*x^2)^p/(1 - c^2*x^2)^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}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[p, -1] && IGtQ[m, 1]

Rubi steps \begin{align*} \text {integral}& = \frac {x (a+b \arccos (c x))}{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 \arccos (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 (a+b \arccos (c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac {\text {Subst}(\int (a+b x) \csc (x) \, dx,x,\arccos (c x))}{2 c^3 d^2} \\ & = \frac {b}{2 c^3 d^2 \sqrt {1-c^2 x^2}}+\frac {x (a+b \arccos (c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac {(a+b \arccos (c x)) \text {arctanh}\left (e^{i \arccos (c x)}\right )}{c^3 d^2}-\frac {b \text {Subst}\left (\int \log \left (1-e^{i x}\right ) \, dx,x,\arccos (c x)\right )}{2 c^3 d^2}+\frac {b \text {Subst}\left (\int \log \left (1+e^{i x}\right ) \, dx,x,\arccos (c x)\right )}{2 c^3 d^2} \\ & = \frac {b}{2 c^3 d^2 \sqrt {1-c^2 x^2}}+\frac {x (a+b \arccos (c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac {(a+b \arccos (c x)) \text {arctanh}\left (e^{i \arccos (c x)}\right )}{c^3 d^2}+\frac {(i b) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{i \arccos (c x)}\right )}{2 c^3 d^2}-\frac {(i b) \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{i \arccos (c x)}\right )}{2 c^3 d^2} \\ & = \frac {b}{2 c^3 d^2 \sqrt {1-c^2 x^2}}+\frac {x (a+b \arccos (c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac {(a+b \arccos (c x)) \text {arctanh}\left (e^{i \arccos (c x)}\right )}{c^3 d^2}+\frac {i b \operatorname {PolyLog}\left (2,-e^{i \arccos (c x)}\right )}{2 c^3 d^2}-\frac {i b \operatorname {PolyLog}\left (2,e^{i \arccos (c x)}\right )}{2 c^3 d^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.85 \[ \int \frac {x^2 (a+b \arccos (c x))}{\left (d-c^2 d x^2\right )^2} \, dx=-\frac {2 a c x+2 b \sqrt {1-c^2 x^2}+2 b c x \arccos (c x)+2 b \arccos (c x) \log \left (1-e^{i \arccos (c x)}\right )-2 b c^2 x^2 \arccos (c x) \log \left (1-e^{i \arccos (c x)}\right )-2 b \arccos (c x) \log \left (1+e^{i \arccos (c x)}\right )+2 b c^2 x^2 \arccos (c x) \log \left (1+e^{i \arccos (c x)}\right )+a \log (1-c x)-a c^2 x^2 \log (1-c x)-a \log (1+c x)+a c^2 x^2 \log (1+c x)-2 i b \left (-1+c^2 x^2\right ) \operatorname {PolyLog}\left (2,-e^{i \arccos (c x)}\right )+2 i b \left (-1+c^2 x^2\right ) \operatorname {PolyLog}\left (2,e^{i \arccos (c x)}\right )}{4 c^3 d^2 \left (-1+c^2 x^2\right )} \]

[In]

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

[Out]

-1/4*(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*Arc
Cos[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*Arc
Cos[c*x])])/(c^3*d^2*(-1 + c^2*x^2))

Maple [A] (verified)

Time = 1.98 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.39

method	result	size
derivativedivides	\(\frac {\frac {a \left (-\frac {1}{4 \left (c x -1\right )}+\frac {\ln \left (c x -1\right )}{4}-\frac {1}{4 \left (c x +1\right )}-\frac {\ln \left (c x +1\right )}{4}\right )}{d^{2}}+\frac {b \left (-\frac {c x \arccos \left (c x \right )+\sqrt {-c^{2} x^{2}+1}}{2 \left (c^{2} x^{2}-1\right )}+\frac {\arccos \left (c x \right ) \ln \left (1-c x -i \sqrt {-c^{2} x^{2}+1}\right )}{2}-\frac {i \operatorname {polylog}\left (2, c x +i \sqrt {-c^{2} x^{2}+1}\right )}{2}-\frac {\arccos \left (c x \right ) \ln \left (1+c x +i \sqrt {-c^{2} x^{2}+1}\right )}{2}+\frac {i \operatorname {polylog}\left (2, -c x -i \sqrt {-c^{2} x^{2}+1}\right )}{2}\right )}{d^{2}}}{c^{3}}\)	\(189\)
default	\(\frac {\frac {a \left (-\frac {1}{4 \left (c x -1\right )}+\frac {\ln \left (c x -1\right )}{4}-\frac {1}{4 \left (c x +1\right )}-\frac {\ln \left (c x +1\right )}{4}\right )}{d^{2}}+\frac {b \left (-\frac {c x \arccos \left (c x \right )+\sqrt {-c^{2} x^{2}+1}}{2 \left (c^{2} x^{2}-1\right )}+\frac {\arccos \left (c x \right ) \ln \left (1-c x -i \sqrt {-c^{2} x^{2}+1}\right )}{2}-\frac {i \operatorname {polylog}\left (2, c x +i \sqrt {-c^{2} x^{2}+1}\right )}{2}-\frac {\arccos \left (c x \right ) \ln \left (1+c x +i \sqrt {-c^{2} x^{2}+1}\right )}{2}+\frac {i \operatorname {polylog}\left (2, -c x -i \sqrt {-c^{2} x^{2}+1}\right )}{2}\right )}{d^{2}}}{c^{3}}\)	\(189\)
parts	\(\frac {a \left (-\frac {1}{4 c^{3} \left (c x -1\right )}+\frac {\ln \left (c x -1\right )}{4 c^{3}}-\frac {1}{4 c^{3} \left (c x +1\right )}-\frac {\ln \left (c x +1\right )}{4 c^{3}}\right )}{d^{2}}+\frac {b \left (-\frac {c x \arccos \left (c x \right )+\sqrt {-c^{2} x^{2}+1}}{2 \left (c^{2} x^{2}-1\right )}+\frac {\arccos \left (c x \right ) \ln \left (1-c x -i \sqrt {-c^{2} x^{2}+1}\right )}{2}-\frac {i \operatorname {polylog}\left (2, c x +i \sqrt {-c^{2} x^{2}+1}\right )}{2}-\frac {\arccos \left (c x \right ) \ln \left (1+c x +i \sqrt {-c^{2} x^{2}+1}\right )}{2}+\frac {i \operatorname {polylog}\left (2, -c x -i \sqrt {-c^{2} x^{2}+1}\right )}{2}\right )}{d^{2} c^{3}}\)	\(200\)

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {x^2 (a+b \arccos (c x))}{\left (d-c^2 d x^2\right )^2} \, dx=\int { \frac {{\left (b \arccos \left (c x\right ) + a\right )} x^{2}}{{\left (c^{2} d x^{2} - d\right )}^{2}} \,d x } \]

[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)

Sympy [F]

\[ \int \frac {x^2 (a+b \arccos (c x))}{\left (d-c^2 d x^2\right )^2} \, dx=\frac {\int \frac {a x^{2}}{c^{4} x^{4} - 2 c^{2} x^{2} + 1}\, dx + \int \frac {b x^{2} \operatorname {acos}{\left (c x \right )}}{c^{4} x^{4} - 2 c^{2} x^{2} + 1}\, dx}{d^{2}} \]

[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

Maxima [F]

\[ \int \frac {x^2 (a+b \arccos (c x))}{\left (d-c^2 d x^2\right )^2} \, dx=\int { \frac {{\left (b \arccos \left (c x\right ) + a\right )} x^{2}}{{\left (c^{2} d x^{2} - d\right )}^{2}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {x^2 (a+b \arccos (c x))}{\left (d-c^2 d x^2\right )^2} \, dx=\int { \frac {{\left (b \arccos \left (c x\right ) + a\right )} x^{2}}{{\left (c^{2} d x^{2} - d\right )}^{2}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2 (a+b \arccos (c x))}{\left (d-c^2 d x^2\right )^2} \, dx=\int \frac {x^2\,\left (a+b\,\mathrm {acos}\left (c\,x\right )\right )}{{\left (d-c^2\,d\,x^2\right )}^2} \,d x \]

[In]

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

[Out]

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