3.1.1 \(\int \frac {x^3 (a+b \arccos (c x))}{d-c^2 d x^2} \, dx\) [1]

3.1.1.1 Optimal result

Integrand size = 25, antiderivative size = 144 \[ \int \frac {x^3 (a+b \arccos (c x))}{d-c^2 d x^2} \, dx=\frac {b x \sqrt {1-c^2 x^2}}{4 c^3 d}-\frac {x^2 (a+b \arccos (c x))}{2 c^2 d}+\frac {i (a+b \arccos (c x))^2}{2 b c^4 d}-\frac {b \arcsin (c x)}{4 c^4 d}-\frac {(a+b \arccos (c x)) \log \left (1-e^{2 i \arccos (c x)}\right )}{c^4 d}+\frac {i b \operatorname {PolyLog}\left (2,e^{2 i \arccos (c x)}\right )}{2 c^4 d} \]

output

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

3.1.1.2 Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.25 \[ \int \frac {x^3 (a+b \arccos (c x))}{d-c^2 d x^2} \, dx=-\frac {2 a c^2 x^2-b c x \sqrt {1-c^2 x^2}+2 b c^2 x^2 \arccos (c x)-2 i b \arccos (c x)^2+2 b \arctan \left (\frac {c x}{-1+\sqrt {1-c^2 x^2}}\right )+4 b \arccos (c x) \log \left (1-e^{i \arccos (c x)}\right )+4 b \arccos (c x) \log \left (1+e^{i \arccos (c x)}\right )+2 a \log \left (1-c^2 x^2\right )-4 i b \operatorname {PolyLog}\left (2,-e^{i \arccos (c x)}\right )-4 i b \operatorname {PolyLog}\left (2,e^{i \arccos (c x)}\right )}{4 c^4 d} \]

input

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

output

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

3.1.1.3 Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.01, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {5211, 27, 262, 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.

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

\(\Big \downarrow \) 5211

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 262

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

\(\Big \downarrow \) 223

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

\(\Big \downarrow \) 5181

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4200

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

input

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

output

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

3.1.1.3.1 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 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[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 5211

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] + (Simp[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] - S 
imp[b*f*(n/(c*(m + 2*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, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && IGtQ[m 
, 1] && NeQ[m + 2*p + 1, 0]
 

3.1.1.4 Maple [A] (verified)

Time = 1.67 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.22

method	result	size
parts	\(-\frac {a \,x^{2}}{2 d \,c^{2}}-\frac {a \ln \left (c^{2} x^{2}-1\right )}{2 d \,c^{4}}-\frac {b \left (-\frac {i \arccos \left (c x \right )^{2}}{2}+\arccos \left (c x \right ) \ln \left (1-c x -i \sqrt {-c^{2} x^{2}+1}\right )-i \operatorname {polylog}\left (2, c x +i \sqrt {-c^{2} x^{2}+1}\right )+\arccos \left (c x \right ) \ln \left (1+c x +i \sqrt {-c^{2} x^{2}+1}\right )-i \operatorname {polylog}\left (2, -c x -i \sqrt {-c^{2} x^{2}+1}\right )+\frac {\arccos \left (c x \right ) \cos \left (2 \arccos \left (c x \right )\right )}{4}-\frac {\sin \left (2 \arccos \left (c x \right )\right )}{8}\right )}{d \,c^{4}}\)	\(176\)
derivativedivides	\(\frac {-\frac {a \left (\frac {c^{2} x^{2}}{2}+\frac {\ln \left (c x -1\right )}{2}+\frac {\ln \left (c x +1\right )}{2}\right )}{d}-\frac {b \left (-\frac {i \arccos \left (c x \right )^{2}}{2}+\arccos \left (c x \right ) \ln \left (1-c x -i \sqrt {-c^{2} x^{2}+1}\right )-i \operatorname {polylog}\left (2, c x +i \sqrt {-c^{2} x^{2}+1}\right )+\arccos \left (c x \right ) \ln \left (1+c x +i \sqrt {-c^{2} x^{2}+1}\right )-i \operatorname {polylog}\left (2, -c x -i \sqrt {-c^{2} x^{2}+1}\right )+\frac {\arccos \left (c x \right ) \cos \left (2 \arccos \left (c x \right )\right )}{4}-\frac {\sin \left (2 \arccos \left (c x \right )\right )}{8}\right )}{d}}{c^{4}}\)	\(177\)
default	\(\frac {-\frac {a \left (\frac {c^{2} x^{2}}{2}+\frac {\ln \left (c x -1\right )}{2}+\frac {\ln \left (c x +1\right )}{2}\right )}{d}-\frac {b \left (-\frac {i \arccos \left (c x \right )^{2}}{2}+\arccos \left (c x \right ) \ln \left (1-c x -i \sqrt {-c^{2} x^{2}+1}\right )-i \operatorname {polylog}\left (2, c x +i \sqrt {-c^{2} x^{2}+1}\right )+\arccos \left (c x \right ) \ln \left (1+c x +i \sqrt {-c^{2} x^{2}+1}\right )-i \operatorname {polylog}\left (2, -c x -i \sqrt {-c^{2} x^{2}+1}\right )+\frac {\arccos \left (c x \right ) \cos \left (2 \arccos \left (c x \right )\right )}{4}-\frac {\sin \left (2 \arccos \left (c x \right )\right )}{8}\right )}{d}}{c^{4}}\)	\(177\)

input

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

output

-1/2*a/d/c^2*x^2-1/2*a/d/c^4*ln(c^2*x^2-1)-b/d/c^4*(-1/2*I*arccos(c*x)^2+a 
rccos(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))+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))+1/4*arccos(c*x)*cos(2*arccos(c*x))-1/8*sin(2*arccos(c*x)))
 

3.1.1.5 Fricas [F]

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

input

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

output

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

3.1.1.6 Sympy [F]

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

input

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

output

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

3.1.1.7 Maxima [F]

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

input

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

output

-1/2*a*(x^2/(c^2*d) + log(c^2*x^2 - 1)/(c^4*d)) + 1/2*(2*c^4*d*integrate(1 
/2*(c^2*x^2*e^(1/2*log(c*x + 1) + 1/2*log(-c*x + 1)) + e^(1/2*log(c*x + 1) 
 + 1/2*log(-c*x + 1))*log(c*x + 1) + e^(1/2*log(c*x + 1) + 1/2*log(-c*x + 
1))*log(-c*x + 1))/(c^7*d*x^4 - c^5*d*x^2 + (c^5*d*x^2 - c^3*d)*e^(log(c*x 
 + 1) + log(-c*x + 1))), x) - (c^2*x^2 + log(c*x + 1) + log(-c*x + 1))*arc 
tan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x))*b/(c^4*d)
 

3.1.1.8 Giac [F]

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

input

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

output

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

3.1.1.9 Mupad [F(-1)]

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

input

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

output

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