\(\int \frac {x^4 (a+b \arccos (c x))}{\sqrt {d-c^2 d x^2}} \, dx\) [112]

Optimal result

Integrand size = 27, antiderivative size = 200 \[ \int \frac {x^4 (a+b \arccos (c x))}{\sqrt {d-c^2 d x^2}} \, dx=\frac {3 b x^2 \sqrt {1-c^2 x^2}}{16 c^3 \sqrt {d-c^2 d x^2}}+\frac {b x^4 \sqrt {1-c^2 x^2}}{16 c \sqrt {d-c^2 d x^2}}-\frac {3 x \sqrt {d-c^2 d x^2} (a+b \arccos (c x))}{8 c^4 d}-\frac {x^3 \sqrt {d-c^2 d x^2} (a+b \arccos (c x))}{4 c^2 d}+\frac {3 \sqrt {1-c^2 x^2} (a+b \arccos (c x))^2}{16 b c^5 \sqrt {d-c^2 d x^2}} \] Output:

3/16*b*x^2*(-c^2*x^2+1)^(1/2)/c^3/(-c^2*d*x^2+d)^(1/2)+1/16*b*x^4*(-c^2*x^ 
2+1)^(1/2)/c/(-c^2*d*x^2+d)^(1/2)-3/8*x*(-c^2*d*x^2+d)^(1/2)*(a+b*arccos(c 
*x))/c^4/d-1/4*x^3*(-c^2*d*x^2+d)^(1/2)*(a+b*arccos(c*x))/c^2/d+3/16*(-c^2 
*x^2+1)^(1/2)*(a+b*arccos(c*x))^2/b/c^5/(-c^2*d*x^2+d)^(1/2)
 

Mathematica [A] (verified)

Time = 0.97 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.80 \[ \int \frac {x^4 (a+b \arccos (c x))}{\sqrt {d-c^2 d x^2}} \, dx=-\frac {\frac {16 a c x \left (3+2 c^2 x^2\right ) \sqrt {d-c^2 d x^2}}{d}+\frac {48 a \arctan \left (\frac {c x \sqrt {d-c^2 d x^2}}{\sqrt {d} \left (-1+c^2 x^2\right )}\right )}{\sqrt {d}}+\frac {b \sqrt {1-c^2 x^2} (16 \cos (2 \arccos (c x))+\cos (4 \arccos (c x))+4 \arccos (c x) (6 \arccos (c x)+8 \sin (2 \arccos (c x))+\sin (4 \arccos (c x))))}{\sqrt {d-c^2 d x^2}}}{128 c^5} \] Input:

Integrate[(x^4*(a + b*ArcCos[c*x]))/Sqrt[d - c^2*d*x^2],x]
 

Output:

-1/128*((16*a*c*x*(3 + 2*c^2*x^2)*Sqrt[d - c^2*d*x^2])/d + (48*a*ArcTan[(c 
*x*Sqrt[d - c^2*d*x^2])/(Sqrt[d]*(-1 + c^2*x^2))])/Sqrt[d] + (b*Sqrt[1 - c 
^2*x^2]*(16*Cos[2*ArcCos[c*x]] + Cos[4*ArcCos[c*x]] + 4*ArcCos[c*x]*(6*Arc 
Cos[c*x] + 8*Sin[2*ArcCos[c*x]] + Sin[4*ArcCos[c*x]])))/Sqrt[d - c^2*d*x^2 
])/c^5
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {5211, 15, 5211, 15, 5153}

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.

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

\(\Big \downarrow \) 5211

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

\(\Big \downarrow \) 15

\(\displaystyle \frac {3 \int \frac {x^2 (a+b \arccos (c x))}{\sqrt {d-c^2 d x^2}}dx}{4 c^2}-\frac {x^3 \sqrt {d-c^2 d x^2} (a+b \arccos (c x))}{4 c^2 d}-\frac {b x^4 \sqrt {1-c^2 x^2}}{16 c \sqrt {d-c^2 d x^2}}\)

\(\Big \downarrow \) 5211

\(\displaystyle \frac {3 \left (\frac {\int \frac {a+b \arccos (c x)}{\sqrt {d-c^2 d x^2}}dx}{2 c^2}-\frac {b \sqrt {1-c^2 x^2} \int xdx}{2 c \sqrt {d-c^2 d x^2}}-\frac {x \sqrt {d-c^2 d x^2} (a+b \arccos (c x))}{2 c^2 d}\right )}{4 c^2}-\frac {x^3 \sqrt {d-c^2 d x^2} (a+b \arccos (c x))}{4 c^2 d}-\frac {b x^4 \sqrt {1-c^2 x^2}}{16 c \sqrt {d-c^2 d x^2}}\)

\(\Big \downarrow \) 15

\(\displaystyle \frac {3 \left (\frac {\int \frac {a+b \arccos (c x)}{\sqrt {d-c^2 d x^2}}dx}{2 c^2}-\frac {x \sqrt {d-c^2 d x^2} (a+b \arccos (c x))}{2 c^2 d}-\frac {b x^2 \sqrt {1-c^2 x^2}}{4 c \sqrt {d-c^2 d x^2}}\right )}{4 c^2}-\frac {x^3 \sqrt {d-c^2 d x^2} (a+b \arccos (c x))}{4 c^2 d}-\frac {b x^4 \sqrt {1-c^2 x^2}}{16 c \sqrt {d-c^2 d x^2}}\)

\(\Big \downarrow \) 5153

\(\displaystyle -\frac {x^3 \sqrt {d-c^2 d x^2} (a+b \arccos (c x))}{4 c^2 d}+\frac {3 \left (-\frac {x \sqrt {d-c^2 d x^2} (a+b \arccos (c x))}{2 c^2 d}-\frac {\sqrt {1-c^2 x^2} (a+b \arccos (c x))^2}{4 b c^3 \sqrt {d-c^2 d x^2}}-\frac {b x^2 \sqrt {1-c^2 x^2}}{4 c \sqrt {d-c^2 d x^2}}\right )}{4 c^2}-\frac {b x^4 \sqrt {1-c^2 x^2}}{16 c \sqrt {d-c^2 d x^2}}\)

Input:

Int[(x^4*(a + b*ArcCos[c*x]))/Sqrt[d - c^2*d*x^2],x]
 

Output:

-1/16*(b*x^4*Sqrt[1 - c^2*x^2])/(c*Sqrt[d - c^2*d*x^2]) - (x^3*Sqrt[d - c^ 
2*d*x^2]*(a + b*ArcCos[c*x]))/(4*c^2*d) + (3*(-1/4*(b*x^2*Sqrt[1 - c^2*x^2 
])/(c*Sqrt[d - c^2*d*x^2]) - (x*Sqrt[d - c^2*d*x^2]*(a + b*ArcCos[c*x]))/( 
2*c^2*d) - (Sqrt[1 - c^2*x^2]*(a + b*ArcCos[c*x])^2)/(4*b*c^3*Sqrt[d - c^2 
*d*x^2])))/(4*c^2)
 

Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 5153

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

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(376\) vs. \(2(174)=348\).

Time = 0.51 (sec) , antiderivative size = 377, normalized size of antiderivative = 1.88

method	result	size
default	\(-\frac {a \,x^{3} \sqrt {-c^{2} d \,x^{2}+d}}{4 c^{2} d}-\frac {3 a x \sqrt {-c^{2} d \,x^{2}+d}}{8 c^{4} d}+\frac {3 a \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{8 c^{4} \sqrt {c^{2} d}}+b \left (\frac {3 \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arccos \left (c x \right )^{2}}{16 c^{5} d \left (c^{2} x^{2}-1\right )}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arccos \left (c x \right ) x}{8 c^{4} d \left (c^{2} x^{2}-1\right )}+\frac {\sqrt {-c^{2} x^{2}+1}}{16 c^{5} \sqrt {-d \left (c^{2} x^{2}-1\right )}}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arccos \left (c x \right ) \cos \left (5 \arccos \left (c x \right )\right )}{64 c^{5} d \left (c^{2} x^{2}-1\right )}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sin \left (5 \arccos \left (c x \right )\right )}{256 c^{5} d \left (c^{2} x^{2}-1\right )}-\frac {7 \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arccos \left (c x \right ) \cos \left (3 \arccos \left (c x \right )\right )}{64 c^{5} d \left (c^{2} x^{2}-1\right )}+\frac {15 \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sin \left (3 \arccos \left (c x \right )\right )}{256 c^{5} d \left (c^{2} x^{2}-1\right )}\right )\)	\(377\)
parts	\(-\frac {a \,x^{3} \sqrt {-c^{2} d \,x^{2}+d}}{4 c^{2} d}-\frac {3 a x \sqrt {-c^{2} d \,x^{2}+d}}{8 c^{4} d}+\frac {3 a \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{8 c^{4} \sqrt {c^{2} d}}+b \left (\frac {3 \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arccos \left (c x \right )^{2}}{16 c^{5} d \left (c^{2} x^{2}-1\right )}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arccos \left (c x \right ) x}{8 c^{4} d \left (c^{2} x^{2}-1\right )}+\frac {\sqrt {-c^{2} x^{2}+1}}{16 c^{5} \sqrt {-d \left (c^{2} x^{2}-1\right )}}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arccos \left (c x \right ) \cos \left (5 \arccos \left (c x \right )\right )}{64 c^{5} d \left (c^{2} x^{2}-1\right )}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sin \left (5 \arccos \left (c x \right )\right )}{256 c^{5} d \left (c^{2} x^{2}-1\right )}-\frac {7 \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arccos \left (c x \right ) \cos \left (3 \arccos \left (c x \right )\right )}{64 c^{5} d \left (c^{2} x^{2}-1\right )}+\frac {15 \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sin \left (3 \arccos \left (c x \right )\right )}{256 c^{5} d \left (c^{2} x^{2}-1\right )}\right )\)	\(377\)

Input:

int(x^4*(a+b*arccos(c*x))/(-c^2*d*x^2+d)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/4*a*x^3/c^2/d*(-c^2*d*x^2+d)^(1/2)-3/8*a/c^4*x/d*(-c^2*d*x^2+d)^(1/2)+3 
/8*a/c^4/(c^2*d)^(1/2)*arctan((c^2*d)^(1/2)*x/(-c^2*d*x^2+d)^(1/2))+b*(3/1 
6*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c^5/d/(c^2*x^2-1)*arccos(c*x)^ 
2+1/8*(-d*(c^2*x^2-1))^(1/2)/c^4/d/(c^2*x^2-1)*arccos(c*x)*x+1/16/c^5/(-d* 
(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)-1/64*(-d*(c^2*x^2-1))^(1/2)/c^5/d/(c 
^2*x^2-1)*arccos(c*x)*cos(5*arccos(c*x))+1/256*(-d*(c^2*x^2-1))^(1/2)/c^5/ 
d/(c^2*x^2-1)*sin(5*arccos(c*x))-7/64*(-d*(c^2*x^2-1))^(1/2)/c^5/d/(c^2*x^ 
2-1)*arccos(c*x)*cos(3*arccos(c*x))+15/256*(-d*(c^2*x^2-1))^(1/2)/c^5/d/(c 
^2*x^2-1)*sin(3*arccos(c*x)))
 

Fricas [F]

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

integrate(x^4*(a+b*arccos(c*x))/(-c^2*d*x^2+d)^(1/2),x, algorithm="fricas" 
)
 

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(x^4*(a+b*arccos(c*x))/(-c^2*d*x^2+d)^(1/2),x, algorithm="maxima" 
)
 

Output:

-1/8*a*(2*sqrt(-c^2*d*x^2 + d)*x^3/(c^2*d) + 3*sqrt(-c^2*d*x^2 + d)*x/(c^4 
*d) - 3*arcsin(c*x)/(c^5*sqrt(d))) + b*integrate(x^4*arctan2(sqrt(c*x + 1) 
*sqrt(-c*x + 1), c*x)/(sqrt(c*x + 1)*sqrt(-c*x + 1)), x)/sqrt(d)
 

Giac [A] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.68 \[ \int \frac {x^4 (a+b \arccos (c x))}{\sqrt {d-c^2 d x^2}} \, dx=-\frac {8 \, b c^{3} x^{4} + 32 \, \sqrt {-c^{2} x^{2} + 1} b c^{2} x^{3} \arccos \left (c x\right ) + 32 \, \sqrt {-c^{2} x^{2} + 1} a c^{2} x^{3} + 24 \, b c x^{2} + 48 \, \sqrt {-c^{2} x^{2} + 1} b x \arccos \left (c x\right ) + 48 \, \sqrt {-c^{2} x^{2} + 1} a x + \frac {24 \, b \arccos \left (c x\right )^{2}}{c} + \frac {48 \, a \arccos \left (c x\right )}{c} - \frac {15 \, b}{c}}{128 \, c^{4} \sqrt {d}} \] Input:

integrate(x^4*(a+b*arccos(c*x))/(-c^2*d*x^2+d)^(1/2),x, algorithm="giac")
 

Output:

-1/128*(8*b*c^3*x^4 + 32*sqrt(-c^2*x^2 + 1)*b*c^2*x^3*arccos(c*x) + 32*sqr 
t(-c^2*x^2 + 1)*a*c^2*x^3 + 24*b*c*x^2 + 48*sqrt(-c^2*x^2 + 1)*b*x*arccos( 
c*x) + 48*sqrt(-c^2*x^2 + 1)*a*x + 24*b*arccos(c*x)^2/c + 48*a*arccos(c*x) 
/c - 15*b/c)/(c^4*sqrt(d))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4 (a+b \arccos (c x))}{\sqrt {d-c^2 d x^2}} \, dx=\int \frac {x^4\,\left (a+b\,\mathrm {acos}\left (c\,x\right )\right )}{\sqrt {d-c^2\,d\,x^2}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {x^4 (a+b \arccos (c x))}{\sqrt {d-c^2 d x^2}} \, dx=\frac {3 \mathit {asin} \left (c x \right ) a -2 \sqrt {-c^{2} x^{2}+1}\, a \,c^{3} x^{3}-3 \sqrt {-c^{2} x^{2}+1}\, a c x +8 \left (\int \frac {\mathit {acos} \left (c x \right ) x^{4}}{\sqrt {-c^{2} x^{2}+1}}d x \right ) b \,c^{5}}{8 \sqrt {d}\, c^{5}} \] Input:

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

Output:

(3*asin(c*x)*a - 2*sqrt( - c**2*x**2 + 1)*a*c**3*x**3 - 3*sqrt( - c**2*x** 
2 + 1)*a*c*x + 8*int((acos(c*x)*x**4)/sqrt( - c**2*x**2 + 1),x)*b*c**5)/(8 
*sqrt(d)*c**5)