\(\int \frac {x^3 (a+b \arccos (c x))}{(d-c^2 d x^2)^{3/2}} \, dx\) [123]

Optimal result

Integrand size = 27, antiderivative size = 142 \[ \int \frac {x^3 (a+b \arccos (c x))}{\left (d-c^2 d x^2\right )^{3/2}} \, dx=-\frac {b x \sqrt {d-c^2 d x^2}}{c^3 d^2 \sqrt {1-c^2 x^2}}+\frac {a+b \arccos (c x)}{c^4 d \sqrt {d-c^2 d x^2}}+\frac {\sqrt {d-c^2 d x^2} (a+b \arccos (c x))}{c^4 d^2}-\frac {b \sqrt {d-c^2 d x^2} \text {arctanh}(c x)}{c^4 d^2 \sqrt {1-c^2 x^2}} \] Output:

-b*x*(-c^2*d*x^2+d)^(1/2)/c^3/d^2/(-c^2*x^2+1)^(1/2)+(a+b*arccos(c*x))/c^4 
/d/(-c^2*d*x^2+d)^(1/2)+(-c^2*d*x^2+d)^(1/2)*(a+b*arccos(c*x))/c^4/d^2-b*( 
-c^2*d*x^2+d)^(1/2)*arctanh(c*x)/c^4/d^2/(-c^2*x^2+1)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 0.28 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.96 \[ \int \frac {x^3 (a+b \arccos (c x))}{\left (d-c^2 d x^2\right )^{3/2}} \, dx=\frac {\sqrt {d-c^2 d x^2} \left (\sqrt {-c^2} \left (-2 a+a c^2 x^2-b c x \sqrt {1-c^2 x^2}+b \left (-2+c^2 x^2\right ) \arccos (c x)\right )+i b c \sqrt {1-c^2 x^2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-c^2} x\right ),1\right )\right )}{c^4 \sqrt {-c^2} d^2 \left (-1+c^2 x^2\right )} \] Input:

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

Output:

(Sqrt[d - c^2*d*x^2]*(Sqrt[-c^2]*(-2*a + a*c^2*x^2 - b*c*x*Sqrt[1 - c^2*x^ 
2] + b*(-2 + c^2*x^2)*ArcCos[c*x]) + I*b*c*Sqrt[1 - c^2*x^2]*EllipticF[I*A 
rcSinh[Sqrt[-c^2]*x], 1]))/(c^4*Sqrt[-c^2]*d^2*(-1 + c^2*x^2))
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.76, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {5195, 27, 299, 219}

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.

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

\(\Big \downarrow \) 5195

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 299

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(a + b*ArcCos[c*x])/(c^4*d*Sqrt[d - c^2*d*x^2]) + (Sqrt[d - c^2*d*x^2]*(a 
+ b*ArcCos[c*x]))/(c^4*d^2) + (b*Sqrt[d - c^2*d*x^2]*(x + ArcTanh[c*x]/c)) 
/(c^3*d^2*Sqrt[1 - c^2*x^2])
 

Defintions of rubi rules used

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 219

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

rule 299

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x 
*((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 
*p + 3))   Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0] && NeQ[2*p + 3, 0]
 

rule 5195

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(p_) 
, x_Symbol] :> With[{u = IntHide[x^m*(d + e*x^2)^p, x]}, Simp[(a + b*ArcCos 
[c*x])   u, x] + Simp[b*c*Simp[Sqrt[d + e*x^2]/Sqrt[1 - c^2*x^2]]   Int[Sim 
plifyIntegrand[u/Sqrt[d + e*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e}, x] 
&& EqQ[c^2*d + e, 0] && IntegerQ[p - 1/2] && NeQ[p, -2^(-1)] && (IGtQ[(m + 
1)/2, 0] || ILtQ[(m + 2*p + 3)/2, 0])
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.67 (sec) , antiderivative size = 308, normalized size of antiderivative = 2.17

method	result	size
default	\(a \left (-\frac {x^{2}}{c^{2} d \sqrt {-c^{2} d \,x^{2}+d}}+\frac {2}{d \,c^{4} \sqrt {-c^{2} d \,x^{2}+d}}\right )+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arccos \left (c x \right ) x^{2}}{d^{2} c^{2} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, x}{d^{2} c^{3} \left (c^{2} x^{2}-1\right )}-\frac {2 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arccos \left (c x \right )}{d^{2} c^{4} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \ln \left (1+c x +i \sqrt {-c^{2} x^{2}+1}\right )}{d^{2} c^{4} \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \ln \left (i \sqrt {-c^{2} x^{2}+1}+c x -1\right )}{d^{2} c^{4} \left (c^{2} x^{2}-1\right )}\)	\(308\)
parts	\(a \left (-\frac {x^{2}}{c^{2} d \sqrt {-c^{2} d \,x^{2}+d}}+\frac {2}{d \,c^{4} \sqrt {-c^{2} d \,x^{2}+d}}\right )+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arccos \left (c x \right ) x^{2}}{d^{2} c^{2} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, x}{d^{2} c^{3} \left (c^{2} x^{2}-1\right )}-\frac {2 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arccos \left (c x \right )}{d^{2} c^{4} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \ln \left (1+c x +i \sqrt {-c^{2} x^{2}+1}\right )}{d^{2} c^{4} \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \ln \left (i \sqrt {-c^{2} x^{2}+1}+c x -1\right )}{d^{2} c^{4} \left (c^{2} x^{2}-1\right )}\)	\(308\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 383, normalized size of antiderivative = 2.70 \[ \int \frac {x^3 (a+b \arccos (c x))}{\left (d-c^2 d x^2\right )^{3/2}} \, dx=\left [-\frac {4 \, \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} b c x - {\left (b c^{2} x^{2} - b\right )} \sqrt {d} \log \left (-\frac {c^{6} d x^{6} + 5 \, c^{4} d x^{4} - 5 \, c^{2} d x^{2} - 4 \, {\left (c^{3} x^{3} + c x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} \sqrt {d} - d}{c^{6} x^{6} - 3 \, c^{4} x^{4} + 3 \, c^{2} x^{2} - 1}\right ) - 4 \, {\left (a c^{2} x^{2} + {\left (b c^{2} x^{2} - 2 \, b\right )} \arccos \left (c x\right ) - 2 \, a\right )} \sqrt {-c^{2} d x^{2} + d}}{4 \, {\left (c^{6} d^{2} x^{2} - c^{4} d^{2}\right )}}, -\frac {2 \, \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} b c x - {\left (b c^{2} x^{2} - b\right )} \sqrt {-d} \arctan \left (\frac {2 \, \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} c \sqrt {-d} x}{c^{4} d x^{4} - d}\right ) - 2 \, {\left (a c^{2} x^{2} + {\left (b c^{2} x^{2} - 2 \, b\right )} \arccos \left (c x\right ) - 2 \, a\right )} \sqrt {-c^{2} d x^{2} + d}}{2 \, {\left (c^{6} d^{2} x^{2} - c^{4} d^{2}\right )}}\right ] \] Input:

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

Output:

[-1/4*(4*sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1)*b*c*x - (b*c^2*x^2 - b)*s 
qrt(d)*log(-(c^6*d*x^6 + 5*c^4*d*x^4 - 5*c^2*d*x^2 - 4*(c^3*x^3 + c*x)*sqr 
t(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1)*sqrt(d) - d)/(c^6*x^6 - 3*c^4*x^4 + 3 
*c^2*x^2 - 1)) - 4*(a*c^2*x^2 + (b*c^2*x^2 - 2*b)*arccos(c*x) - 2*a)*sqrt( 
-c^2*d*x^2 + d))/(c^6*d^2*x^2 - c^4*d^2), -1/2*(2*sqrt(-c^2*d*x^2 + d)*sqr 
t(-c^2*x^2 + 1)*b*c*x - (b*c^2*x^2 - b)*sqrt(-d)*arctan(2*sqrt(-c^2*d*x^2 
+ d)*sqrt(-c^2*x^2 + 1)*c*sqrt(-d)*x/(c^4*d*x^4 - d)) - 2*(a*c^2*x^2 + (b* 
c^2*x^2 - 2*b)*arccos(c*x) - 2*a)*sqrt(-c^2*d*x^2 + d))/(c^6*d^2*x^2 - c^4 
*d^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00 \[ \int \frac {x^3 (a+b \arccos (c x))}{\left (d-c^2 d x^2\right )^{3/2}} \, dx=\frac {1}{2} \, b c {\left (\frac {2 \, x}{c^{4} d^{\frac {3}{2}}} + \frac {\log \left (c x + 1\right )}{c^{5} d^{\frac {3}{2}}} - \frac {\log \left (c x - 1\right )}{c^{5} d^{\frac {3}{2}}}\right )} - b {\left (\frac {x^{2}}{\sqrt {-c^{2} d x^{2} + d} c^{2} d} - \frac {2}{\sqrt {-c^{2} d x^{2} + d} c^{4} d}\right )} \arccos \left (c x\right ) - a {\left (\frac {x^{2}}{\sqrt {-c^{2} d x^{2} + d} c^{2} d} - \frac {2}{\sqrt {-c^{2} d x^{2} + d} c^{4} d}\right )} \] Input:

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

Output:

1/2*b*c*(2*x/(c^4*d^(3/2)) + log(c*x + 1)/(c^5*d^(3/2)) - log(c*x - 1)/(c^ 
5*d^(3/2))) - b*(x^2/(sqrt(-c^2*d*x^2 + d)*c^2*d) - 2/(sqrt(-c^2*d*x^2 + d 
)*c^4*d))*arccos(c*x) - a*(x^2/(sqrt(-c^2*d*x^2 + d)*c^2*d) - 2/(sqrt(-c^2 
*d*x^2 + d)*c^4*d))
 

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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