\(\int \frac {x^3 (a+b \arccos (c x))}{(d-c^2 d x^2)^2} \, dx\) [9]

Optimal result

Integrand size = 25, antiderivative size = 155 \[ \int \frac {x^3 (a+b \arccos (c x))}{\left (d-c^2 d x^2\right )^2} \, dx=\frac {b x}{2 c^3 d^2 \sqrt {1-c^2 x^2}}+\frac {x^2 (a+b \arccos (c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac {i (a+b \arccos (c x))^2}{2 b c^4 d^2}-\frac {b \arcsin (c x)}{2 c^4 d^2}+\frac {(a+b \arccos (c x)) \log \left (1-e^{2 i \arccos (c x)}\right )}{c^4 d^2}-\frac {i b \operatorname {PolyLog}\left (2,e^{2 i \arccos (c x)}\right )}{2 c^4 d^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.31 \[ \int \frac {x^3 (a+b \arccos (c x))}{\left (d-c^2 d x^2\right )^2} \, dx=\frac {\frac {b \sqrt {1-c^2 x^2}}{1-c x}-\frac {b \sqrt {1-c^2 x^2}}{1+c x}-\frac {2 a}{-1+c^2 x^2}+\frac {b \arccos (c x)}{1-c x}+\frac {b \arccos (c x)}{1+c x}-2 i b \arccos (c x)^2+4 b \arccos (c x) \log \left (1-e^{i \arccos (c x)}\right )+4 b \arccos (c x) \log \left (1+e^{i \arccos (c x)}\right )+2 a \log \left (1-c^2 x^2\right )-4 i b \operatorname {PolyLog}\left (2,-e^{i \arccos (c x)}\right )-4 i b \operatorname {PolyLog}\left (2,e^{i \arccos (c x)}\right )}{4 c^4 d^2} \] Input:

Integrate[(x^3*(a + b*ArcCos[c*x]))/(d - c^2*d*x^2)^2,x]
 

Output:

((b*Sqrt[1 - c^2*x^2])/(1 - c*x) - (b*Sqrt[1 - c^2*x^2])/(1 + c*x) - (2*a) 
/(-1 + c^2*x^2) + (b*ArcCos[c*x])/(1 - c*x) + (b*ArcCos[c*x])/(1 + c*x) - 
(2*I)*b*ArcCos[c*x]^2 + 4*b*ArcCos[c*x]*Log[1 - E^(I*ArcCos[c*x])] + 4*b*A 
rcCos[c*x]*Log[1 + E^(I*ArcCos[c*x])] + 2*a*Log[1 - c^2*x^2] - (4*I)*b*Pol 
yLog[2, -E^(I*ArcCos[c*x])] - (4*I)*b*PolyLog[2, E^(I*ArcCos[c*x])])/(4*c^ 
4*d^2)
 

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.98, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {5207, 27, 252, 223, 5181, 3042, 25, 4200, 25, 2620, 2715, 2838}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x^3 (a+b \arccos (c x))}{\left (d-c^2 d x^2\right )^2} \, dx\)

\(\Big \downarrow \) 5207

\(\displaystyle -\frac {\int \frac {x (a+b \arccos (c x))}{d \left (1-c^2 x^2\right )}dx}{c^2 d}+\frac {b \int \frac {x^2}{\left (1-c^2 x^2\right )^{3/2}}dx}{2 c d^2}+\frac {x^2 (a+b \arccos (c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {x (a+b \arccos (c x))}{1-c^2 x^2}dx}{c^2 d^2}+\frac {b \int \frac {x^2}{\left (1-c^2 x^2\right )^{3/2}}dx}{2 c d^2}+\frac {x^2 (a+b \arccos (c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}\)

\(\Big \downarrow \) 252

\(\displaystyle -\frac {\int \frac {x (a+b \arccos (c x))}{1-c^2 x^2}dx}{c^2 d^2}+\frac {b \left (\frac {x}{c^2 \sqrt {1-c^2 x^2}}-\frac {\int \frac {1}{\sqrt {1-c^2 x^2}}dx}{c^2}\right )}{2 c d^2}+\frac {x^2 (a+b \arccos (c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}\)

\(\Big \downarrow \) 223

\(\displaystyle -\frac {\int \frac {x (a+b \arccos (c x))}{1-c^2 x^2}dx}{c^2 d^2}+\frac {x^2 (a+b \arccos (c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac {b \left (\frac {x}{c^2 \sqrt {1-c^2 x^2}}-\frac {\arcsin (c x)}{c^3}\right )}{2 c d^2}\)

\(\Big \downarrow \) 5181

\(\displaystyle \frac {\int \frac {c x (a+b \arccos (c x))}{\sqrt {1-c^2 x^2}}d\arccos (c x)}{c^4 d^2}+\frac {x^2 (a+b \arccos (c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac {b \left (\frac {x}{c^2 \sqrt {1-c^2 x^2}}-\frac {\arcsin (c x)}{c^3}\right )}{2 c d^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int -\left ((a+b \arccos (c x)) \tan \left (\arccos (c x)+\frac {\pi }{2}\right )\right )d\arccos (c x)}{c^4 d^2}+\frac {x^2 (a+b \arccos (c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac {b \left (\frac {x}{c^2 \sqrt {1-c^2 x^2}}-\frac {\arcsin (c x)}{c^3}\right )}{2 c d^2}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\int (a+b \arccos (c x)) \tan \left (\arccos (c x)+\frac {\pi }{2}\right )d\arccos (c x)}{c^4 d^2}+\frac {x^2 (a+b \arccos (c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac {b \left (\frac {x}{c^2 \sqrt {1-c^2 x^2}}-\frac {\arcsin (c x)}{c^3}\right )}{2 c d^2}\)

\(\Big \downarrow \) 4200

\(\displaystyle \frac {2 i \int -\frac {e^{2 i \arccos (c x)} (a+b \arccos (c x))}{1-e^{2 i \arccos (c x)}}d\arccos (c x)-\frac {i (a+b \arccos (c x))^2}{2 b}}{c^4 d^2}+\frac {x^2 (a+b \arccos (c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac {b \left (\frac {x}{c^2 \sqrt {1-c^2 x^2}}-\frac {\arcsin (c x)}{c^3}\right )}{2 c d^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {-2 i \int \frac {e^{2 i \arccos (c x)} (a+b \arccos (c x))}{1-e^{2 i \arccos (c x)}}d\arccos (c x)-\frac {i (a+b \arccos (c x))^2}{2 b}}{c^4 d^2}+\frac {x^2 (a+b \arccos (c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac {b \left (\frac {x}{c^2 \sqrt {1-c^2 x^2}}-\frac {\arcsin (c x)}{c^3}\right )}{2 c d^2}\)

\(\Big \downarrow \) 2620

\(\displaystyle \frac {-2 i \left (\frac {1}{2} i \log \left (1-e^{2 i \arccos (c x)}\right ) (a+b \arccos (c x))-\frac {1}{2} i b \int \log \left (1-e^{2 i \arccos (c x)}\right )d\arccos (c x)\right )-\frac {i (a+b \arccos (c x))^2}{2 b}}{c^4 d^2}+\frac {x^2 (a+b \arccos (c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac {b \left (\frac {x}{c^2 \sqrt {1-c^2 x^2}}-\frac {\arcsin (c x)}{c^3}\right )}{2 c d^2}\)

\(\Big \downarrow \) 2715

\(\displaystyle \frac {-2 i \left (\frac {1}{2} i \log \left (1-e^{2 i \arccos (c x)}\right ) (a+b \arccos (c x))-\frac {1}{4} b \int e^{-2 i \arccos (c x)} \log \left (1-e^{2 i \arccos (c x)}\right )de^{2 i \arccos (c x)}\right )-\frac {i (a+b \arccos (c x))^2}{2 b}}{c^4 d^2}+\frac {x^2 (a+b \arccos (c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac {b \left (\frac {x}{c^2 \sqrt {1-c^2 x^2}}-\frac {\arcsin (c x)}{c^3}\right )}{2 c d^2}\)

\(\Big \downarrow \) 2838

\(\displaystyle \frac {-2 i \left (\frac {1}{2} i \log \left (1-e^{2 i \arccos (c x)}\right ) (a+b \arccos (c x))+\frac {1}{4} b \operatorname {PolyLog}\left (2,e^{2 i \arccos (c x)}\right )\right )-\frac {i (a+b \arccos (c x))^2}{2 b}}{c^4 d^2}+\frac {x^2 (a+b \arccos (c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac {b \left (\frac {x}{c^2 \sqrt {1-c^2 x^2}}-\frac {\arcsin (c x)}{c^3}\right )}{2 c d^2}\)

Input:

Int[(x^3*(a + b*ArcCos[c*x]))/(d - c^2*d*x^2)^2,x]
 

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt 
[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
 

rule 252

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x 
)^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* 
(p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c 
}, x] && LtQ[p, -1] && GtQ[m, 1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi 
alQ[a, b, c, 2, m, p, x]
 

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[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 2838

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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4200

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol 
] :> Simp[I*((c + d*x)^(m + 1)/(d*(m + 1))), x] - Simp[2*I   Int[(c + d*x)^ 
m*E^(2*I*k*Pi)*(E^(2*I*(e + f*x))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))), x] 
, x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]
 

rule 5181

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

rule 5207

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] + (-Simp[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] - Simp 
[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] && IG 
tQ[m, 1]
 

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.34

method	result	size
derivativedivides	\(\frac {\frac {a \left (-\frac {1}{4 \left (c x -1\right )}+\frac {\ln \left (c x -1\right )}{2}+\frac {1}{4 c x +4}+\frac {\ln \left (c x +1\right )}{2}\right )}{d^{2}}+\frac {b \left (-\frac {i \arccos \left (c x \right )^{2}}{2}-\frac {i c^{2} x^{2}+c x \sqrt {-c^{2} x^{2}+1}+\arccos \left (c x \right )-i}{2 \left (c^{2} x^{2}-1\right )}+\arccos \left (c x \right ) \ln \left (1+c x +i \sqrt {-c^{2} x^{2}+1}\right )+\arccos \left (c x \right ) \ln \left (1-c x -i \sqrt {-c^{2} x^{2}+1}\right )-i \operatorname {polylog}\left (2, -c x -i \sqrt {-c^{2} x^{2}+1}\right )-i \operatorname {polylog}\left (2, c x +i \sqrt {-c^{2} x^{2}+1}\right )\right )}{d^{2}}}{c^{4}}\)	\(207\)
default	\(\frac {\frac {a \left (-\frac {1}{4 \left (c x -1\right )}+\frac {\ln \left (c x -1\right )}{2}+\frac {1}{4 c x +4}+\frac {\ln \left (c x +1\right )}{2}\right )}{d^{2}}+\frac {b \left (-\frac {i \arccos \left (c x \right )^{2}}{2}-\frac {i c^{2} x^{2}+c x \sqrt {-c^{2} x^{2}+1}+\arccos \left (c x \right )-i}{2 \left (c^{2} x^{2}-1\right )}+\arccos \left (c x \right ) \ln \left (1+c x +i \sqrt {-c^{2} x^{2}+1}\right )+\arccos \left (c x \right ) \ln \left (1-c x -i \sqrt {-c^{2} x^{2}+1}\right )-i \operatorname {polylog}\left (2, -c x -i \sqrt {-c^{2} x^{2}+1}\right )-i \operatorname {polylog}\left (2, c x +i \sqrt {-c^{2} x^{2}+1}\right )\right )}{d^{2}}}{c^{4}}\)	\(207\)
parts	\(\frac {a \left (-\frac {1}{4 \left (c x -1\right ) c^{4}}+\frac {\ln \left (c x -1\right )}{2 c^{4}}+\frac {1}{4 \left (c x +1\right ) c^{4}}+\frac {\ln \left (c x +1\right )}{2 c^{4}}\right )}{d^{2}}+\frac {b \left (-\frac {i \arccos \left (c x \right )^{2}}{2}-\frac {i c^{2} x^{2}+c x \sqrt {-c^{2} x^{2}+1}+\arccos \left (c x \right )-i}{2 \left (c^{2} x^{2}-1\right )}+\arccos \left (c x \right ) \ln \left (1+c x +i \sqrt {-c^{2} x^{2}+1}\right )+\arccos \left (c x \right ) \ln \left (1-c x -i \sqrt {-c^{2} x^{2}+1}\right )-i \operatorname {polylog}\left (2, -c x -i \sqrt {-c^{2} x^{2}+1}\right )-i \operatorname {polylog}\left (2, c x +i \sqrt {-c^{2} x^{2}+1}\right )\right )}{d^{2} c^{4}}\)	\(218\)

Input:

int(x^3*(a+b*arccos(c*x))/(-c^2*d*x^2+d)^2,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [F]

\[ \int \frac {x^3 (a+b \arccos (c x))}{\left (d-c^2 d x^2\right )^2} \, dx=\int { \frac {{\left (b \arccos \left (c x\right ) + a\right )} x^{3}}{{\left (c^{2} d x^{2} - d\right )}^{2}} \,d x } \] Input:

integrate(x^3*(a+b*arccos(c*x))/(-c^2*d*x^2+d)^2,x, algorithm="fricas")
 

Output:

integral((b*x^3*arccos(c*x) + a*x^3)/(c^4*d^2*x^4 - 2*c^2*d^2*x^2 + d^2), 
x)
 

Sympy [F]

\[ \int \frac {x^3 (a+b \arccos (c x))}{\left (d-c^2 d x^2\right )^2} \, dx=\frac {\int \frac {a x^{3}}{c^{4} x^{4} - 2 c^{2} x^{2} + 1}\, dx + \int \frac {b x^{3} \operatorname {acos}{\left (c x \right )}}{c^{4} x^{4} - 2 c^{2} x^{2} + 1}\, dx}{d^{2}} \] Input:

integrate(x**3*(a+b*acos(c*x))/(-c**2*d*x**2+d)**2,x)
                                                                                    
                                                                                    
 

Output:

(Integral(a*x**3/(c**4*x**4 - 2*c**2*x**2 + 1), x) + Integral(b*x**3*acos( 
c*x)/(c**4*x**4 - 2*c**2*x**2 + 1), x))/d**2
 

Maxima [F]

\[ \int \frac {x^3 (a+b \arccos (c x))}{\left (d-c^2 d x^2\right )^2} \, dx=\int { \frac {{\left (b \arccos \left (c x\right ) + a\right )} x^{3}}{{\left (c^{2} d x^{2} - d\right )}^{2}} \,d x } \] Input:

integrate(x^3*(a+b*arccos(c*x))/(-c^2*d*x^2+d)^2,x, algorithm="maxima")
 

Output:

-1/2*a*(1/(c^6*d^2*x^2 - c^4*d^2) - log(c^2*x^2 - 1)/(c^4*d^2)) + 1/2*(((c 
^2*x^2 - 1)*log(c*x + 1) + (c^2*x^2 - 1)*log(-c*x + 1) - 1)*arctan2(sqrt(c 
*x + 1)*sqrt(-c*x + 1), c*x) - 2*(c^6*d^2*x^2 - c^4*d^2)*integrate(1/2*((c 
^2*x^2 - 1)*e^(1/2*log(c*x + 1) + 1/2*log(-c*x + 1))*log(c*x + 1) + (c^2*x 
^2 - 1)*e^(1/2*log(c*x + 1) + 1/2*log(-c*x + 1))*log(-c*x + 1) - e^(1/2*lo 
g(c*x + 1) + 1/2*log(-c*x + 1)))/(c^9*d^2*x^6 - 2*c^7*d^2*x^4 + c^5*d^2*x^ 
2 + (c^7*d^2*x^4 - 2*c^5*d^2*x^2 + c^3*d^2)*e^(log(c*x + 1) + log(-c*x + 1 
))), x))*b/(c^6*d^2*x^2 - c^4*d^2)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {x^3 (a+b \arccos (c x))}{\left (d-c^2 d x^2\right )^2} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(x^3*(a+b*arccos(c*x))/(-c^2*d*x^2+d)^2,x, algorithm="giac")
 

Output:

Exception raised: RuntimeError >> an error occurred running a Giac command 
:INPUT:sage2OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const ve 
cteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 (a+b \arccos (c x))}{\left (d-c^2 d x^2\right )^2} \, dx=\int \frac {x^3\,\left (a+b\,\mathrm {acos}\left (c\,x\right )\right )}{{\left (d-c^2\,d\,x^2\right )}^2} \,d x \] Input:

int((x^3*(a + b*acos(c*x)))/(d - c^2*d*x^2)^2,x)
 

Output:

int((x^3*(a + b*acos(c*x)))/(d - c^2*d*x^2)^2, x)
 

Reduce [F]

\[ \int \frac {x^3 (a+b \arccos (c x))}{\left (d-c^2 d x^2\right )^2} \, dx=\frac {2 \left (\int \frac {\mathit {acos} \left (c x \right ) x^{3}}{c^{4} x^{4}-2 c^{2} x^{2}+1}d x \right ) b \,c^{6} x^{2}-2 \left (\int \frac {\mathit {acos} \left (c x \right ) x^{3}}{c^{4} x^{4}-2 c^{2} x^{2}+1}d x \right ) b \,c^{4}+\mathrm {log}\left (c^{2} x -c \right ) a \,c^{2} x^{2}-\mathrm {log}\left (c^{2} x -c \right ) a +\mathrm {log}\left (c^{2} x +c \right ) a \,c^{2} x^{2}-\mathrm {log}\left (c^{2} x +c \right ) a -a \,c^{2} x^{2}}{2 c^{4} d^{2} \left (c^{2} x^{2}-1\right )} \] Input:

int(x^3*(a+b*acos(c*x))/(-c^2*d*x^2+d)^2,x)
 

Output:

(2*int((acos(c*x)*x**3)/(c**4*x**4 - 2*c**2*x**2 + 1),x)*b*c**6*x**2 - 2*i 
nt((acos(c*x)*x**3)/(c**4*x**4 - 2*c**2*x**2 + 1),x)*b*c**4 + log(c**2*x - 
 c)*a*c**2*x**2 - log(c**2*x - c)*a + log(c**2*x + c)*a*c**2*x**2 - log(c* 
*2*x + c)*a - a*c**2*x**2)/(2*c**4*d**2*(c**2*x**2 - 1))