\(\int x (d+e x^2) (a+b \arccos (c x)) \, dx\) [18]

Optimal result

Integrand size = 17, antiderivative size = 120 \[ \int x \left (d+e x^2\right ) (a+b \arccos (c x)) \, dx=-\frac {b \left (8 c^2 d+3 e\right ) x \sqrt {1-c^2 x^2}}{32 c^3}-\frac {b e x^3 \sqrt {1-c^2 x^2}}{16 c}+\frac {\left (d+e x^2\right )^2 (a+b \arccos (c x))}{4 e}+\frac {b \left (8 c^4 d^2+8 c^2 d e+3 e^2\right ) \arcsin (c x)}{32 c^4 e} \] Output:

-1/32*b*(8*c^2*d+3*e)*x*(-c^2*x^2+1)^(1/2)/c^3-1/16*b*e*x^3*(-c^2*x^2+1)^( 
1/2)/c+1/4*(e*x^2+d)^2*(a+b*arccos(c*x))/e+1/32*b*(8*c^4*d^2+8*c^2*d*e+3*e 
^2)*arcsin(c*x)/c^4/e
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.09 \[ \int x \left (d+e x^2\right ) (a+b \arccos (c x)) \, dx=\frac {1}{2} a d x^2+\frac {1}{4} a e x^4-\frac {b d x \sqrt {1-c^2 x^2}}{4 c}+b e \sqrt {1-c^2 x^2} \left (-\frac {3 x}{32 c^3}-\frac {x^3}{16 c}\right )+\frac {1}{2} b d x^2 \arccos (c x)+\frac {1}{4} b e x^4 \arccos (c x)+\frac {b d \arcsin (c x)}{4 c^2}+\frac {3 b e \arcsin (c x)}{32 c^4} \] Input:

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

Output:

(a*d*x^2)/2 + (a*e*x^4)/4 - (b*d*x*Sqrt[1 - c^2*x^2])/(4*c) + b*e*Sqrt[1 - 
 c^2*x^2]*((-3*x)/(32*c^3) - x^3/(16*c)) + (b*d*x^2*ArcCos[c*x])/2 + (b*e* 
x^4*ArcCos[c*x])/4 + (b*d*ArcSin[c*x])/(4*c^2) + (3*b*e*ArcSin[c*x])/(32*c 
^4)
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.13, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {5229, 318, 25, 299, 223}

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.

\(\displaystyle \int x \left (d+e x^2\right ) (a+b \arccos (c x)) \, dx\)

\(\Big \downarrow \) 5229

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

\(\Big \downarrow \) 318

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 299

\(\displaystyle \frac {b c \left (\frac {\frac {\left (8 c^4 d^2+8 c^2 d e+3 e^2\right ) \int \frac {1}{\sqrt {1-c^2 x^2}}dx}{2 c^2}-\frac {3 e x \sqrt {1-c^2 x^2} \left (2 c^2 d+e\right )}{2 c^2}}{4 c^2}-\frac {e x \sqrt {1-c^2 x^2} \left (d+e x^2\right )}{4 c^2}\right )}{4 e}+\frac {\left (d+e x^2\right )^2 (a+b \arccos (c x))}{4 e}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {\left (d+e x^2\right )^2 (a+b \arccos (c x))}{4 e}+\frac {b c \left (\frac {\frac {\arcsin (c x) \left (8 c^4 d^2+8 c^2 d e+3 e^2\right )}{2 c^3}-\frac {3 e x \sqrt {1-c^2 x^2} \left (2 c^2 d+e\right )}{2 c^2}}{4 c^2}-\frac {e x \sqrt {1-c^2 x^2} \left (d+e x^2\right )}{4 c^2}\right )}{4 e}\)

Input:

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

Output:

((d + e*x^2)^2*(a + b*ArcCos[c*x]))/(4*e) + (b*c*(-1/4*(e*x*Sqrt[1 - c^2*x 
^2]*(d + e*x^2))/c^2 + ((-3*e*(2*c^2*d + e)*x*Sqrt[1 - c^2*x^2])/(2*c^2) + 
 ((8*c^4*d^2 + 8*c^2*d*e + 3*e^2)*ArcSin[c*x])/(2*c^3))/(4*c^2)))/(4*e)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 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 318

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S 
imp[1/(b*(2*(p + q) + 1))   Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b 
*c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 
 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G 
tQ[q, 1] && NeQ[2*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntBinomialQ[a, b, c, 
d, 2, p, q, x]
 

rule 5229

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

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.34

method	result	size
parts	\(\frac {a \left (e \,x^{2}+d \right )^{2}}{4 e}+\frac {b \left (\frac {c^{2} e \arccos \left (c x \right ) x^{4}}{4}+\frac {\arccos \left (c x \right ) c^{2} x^{2} d}{2}+\frac {c^{2} \arccos \left (c x \right ) d^{2}}{4 e}+\frac {\arcsin \left (c x \right ) c^{4} d^{2}+e^{2} \left (-\frac {c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}{4}-\frac {3 c x \sqrt {-c^{2} x^{2}+1}}{8}+\frac {3 \arcsin \left (c x \right )}{8}\right )+2 d \,c^{2} e \left (-\frac {c x \sqrt {-c^{2} x^{2}+1}}{2}+\frac {\arcsin \left (c x \right )}{2}\right )}{4 c^{2} e}\right )}{c^{2}}\)	\(161\)
derivativedivides	\(\frac {\frac {a \left (c^{2} e \,x^{2}+c^{2} d \right )^{2}}{4 c^{2} e}+\frac {b \left (\frac {\arccos \left (c x \right ) c^{4} d^{2}}{4 e}+\frac {\arccos \left (c x \right ) c^{4} d \,x^{2}}{2}+\frac {\arccos \left (c x \right ) c^{4} e \,x^{4}}{4}+\frac {\arcsin \left (c x \right ) c^{4} d^{2}+e^{2} \left (-\frac {c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}{4}-\frac {3 c x \sqrt {-c^{2} x^{2}+1}}{8}+\frac {3 \arcsin \left (c x \right )}{8}\right )+2 d \,c^{2} e \left (-\frac {c x \sqrt {-c^{2} x^{2}+1}}{2}+\frac {\arcsin \left (c x \right )}{2}\right )}{4 e}\right )}{c^{2}}}{c^{2}}\)	\(172\)
default	\(\frac {\frac {a \left (c^{2} e \,x^{2}+c^{2} d \right )^{2}}{4 c^{2} e}+\frac {b \left (\frac {\arccos \left (c x \right ) c^{4} d^{2}}{4 e}+\frac {\arccos \left (c x \right ) c^{4} d \,x^{2}}{2}+\frac {\arccos \left (c x \right ) c^{4} e \,x^{4}}{4}+\frac {\arcsin \left (c x \right ) c^{4} d^{2}+e^{2} \left (-\frac {c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}{4}-\frac {3 c x \sqrt {-c^{2} x^{2}+1}}{8}+\frac {3 \arcsin \left (c x \right )}{8}\right )+2 d \,c^{2} e \left (-\frac {c x \sqrt {-c^{2} x^{2}+1}}{2}+\frac {\arcsin \left (c x \right )}{2}\right )}{4 e}\right )}{c^{2}}}{c^{2}}\)	\(172\)
orering	\(\frac {\left (14 x^{6} c^{4} e^{2}+50 x^{4} c^{4} d e +24 x^{2} d^{2} c^{4}+3 x^{4} e^{2} c^{2}-31 c^{2} d e \,x^{2}-16 d^{2} c^{2}-12 e^{2} x^{2}-6 d e \right ) \left (a +b \arccos \left (c x \right )\right )}{32 c^{4} \left (e \,x^{2}+d \right )}-\frac {\left (2 c^{2} e \,x^{2}+8 c^{2} d +3 e \right ) \left (c x -1\right ) \left (c x +1\right ) \left (\left (e \,x^{2}+d \right ) \left (a +b \arccos \left (c x \right )\right )+2 e \,x^{2} \left (a +b \arccos \left (c x \right )\right )-\frac {x \left (e \,x^{2}+d \right ) b c}{\sqrt {-c^{2} x^{2}+1}}\right )}{32 c^{4} \left (e \,x^{2}+d \right )}\)	\(196\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.86 \[ \int x \left (d+e x^2\right ) (a+b \arccos (c x)) \, dx=\frac {8 \, a c^{4} e x^{4} + 16 \, a c^{4} d x^{2} + {\left (8 \, b c^{4} e x^{4} + 16 \, b c^{4} d x^{2} - 8 \, b c^{2} d - 3 \, b e\right )} \arccos \left (c x\right ) - {\left (2 \, b c^{3} e x^{3} + {\left (8 \, b c^{3} d + 3 \, b c e\right )} x\right )} \sqrt {-c^{2} x^{2} + 1}}{32 \, c^{4}} \] Input:

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

Output:

1/32*(8*a*c^4*e*x^4 + 16*a*c^4*d*x^2 + (8*b*c^4*e*x^4 + 16*b*c^4*d*x^2 - 8 
*b*c^2*d - 3*b*e)*arccos(c*x) - (2*b*c^3*e*x^3 + (8*b*c^3*d + 3*b*c*e)*x)* 
sqrt(-c^2*x^2 + 1))/c^4
                                                                                    
                                                                                    
 

Sympy [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.32 \[ \int x \left (d+e x^2\right ) (a+b \arccos (c x)) \, dx=\begin {cases} \frac {a d x^{2}}{2} + \frac {a e x^{4}}{4} + \frac {b d x^{2} \operatorname {acos}{\left (c x \right )}}{2} + \frac {b e x^{4} \operatorname {acos}{\left (c x \right )}}{4} - \frac {b d x \sqrt {- c^{2} x^{2} + 1}}{4 c} - \frac {b e x^{3} \sqrt {- c^{2} x^{2} + 1}}{16 c} - \frac {b d \operatorname {acos}{\left (c x \right )}}{4 c^{2}} - \frac {3 b e x \sqrt {- c^{2} x^{2} + 1}}{32 c^{3}} - \frac {3 b e \operatorname {acos}{\left (c x \right )}}{32 c^{4}} & \text {for}\: c \neq 0 \\\left (a + \frac {\pi b}{2}\right ) \left (\frac {d x^{2}}{2} + \frac {e x^{4}}{4}\right ) & \text {otherwise} \end {cases} \] Input:

integrate(x*(e*x**2+d)*(a+b*acos(c*x)),x)
 

Output:

Piecewise((a*d*x**2/2 + a*e*x**4/4 + b*d*x**2*acos(c*x)/2 + b*e*x**4*acos( 
c*x)/4 - b*d*x*sqrt(-c**2*x**2 + 1)/(4*c) - b*e*x**3*sqrt(-c**2*x**2 + 1)/ 
(16*c) - b*d*acos(c*x)/(4*c**2) - 3*b*e*x*sqrt(-c**2*x**2 + 1)/(32*c**3) - 
 3*b*e*acos(c*x)/(32*c**4), Ne(c, 0)), ((a + pi*b/2)*(d*x**2/2 + e*x**4/4) 
, True))
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.03 \[ \int x \left (d+e x^2\right ) (a+b \arccos (c x)) \, dx=\frac {1}{4} \, a e x^{4} + \frac {1}{2} \, a d x^{2} + \frac {1}{4} \, {\left (2 \, x^{2} \arccos \left (c x\right ) - c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x}{c^{2}} - \frac {\arcsin \left (c x\right )}{c^{3}}\right )}\right )} b d + \frac {1}{32} \, {\left (8 \, x^{4} \arccos \left (c x\right ) - {\left (\frac {2 \, \sqrt {-c^{2} x^{2} + 1} x^{3}}{c^{2}} + \frac {3 \, \sqrt {-c^{2} x^{2} + 1} x}{c^{4}} - \frac {3 \, \arcsin \left (c x\right )}{c^{5}}\right )} c\right )} b e \] Input:

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

Output:

1/4*a*e*x^4 + 1/2*a*d*x^2 + 1/4*(2*x^2*arccos(c*x) - c*(sqrt(-c^2*x^2 + 1) 
*x/c^2 - arcsin(c*x)/c^3))*b*d + 1/32*(8*x^4*arccos(c*x) - (2*sqrt(-c^2*x^ 
2 + 1)*x^3/c^2 + 3*sqrt(-c^2*x^2 + 1)*x/c^4 - 3*arcsin(c*x)/c^5)*c)*b*e
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.01 \[ \int x \left (d+e x^2\right ) (a+b \arccos (c x)) \, dx=\frac {1}{4} \, b e x^{4} \arccos \left (c x\right ) + \frac {1}{4} \, a e x^{4} + \frac {1}{2} \, b d x^{2} \arccos \left (c x\right ) - \frac {\sqrt {-c^{2} x^{2} + 1} b e x^{3}}{16 \, c} + \frac {1}{2} \, a d x^{2} - \frac {\sqrt {-c^{2} x^{2} + 1} b d x}{4 \, c} - \frac {b d \arccos \left (c x\right )}{4 \, c^{2}} - \frac {3 \, \sqrt {-c^{2} x^{2} + 1} b e x}{32 \, c^{3}} - \frac {3 \, b e \arccos \left (c x\right )}{32 \, c^{4}} \] Input:

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

Output:

1/4*b*e*x^4*arccos(c*x) + 1/4*a*e*x^4 + 1/2*b*d*x^2*arccos(c*x) - 1/16*sqr 
t(-c^2*x^2 + 1)*b*e*x^3/c + 1/2*a*d*x^2 - 1/4*sqrt(-c^2*x^2 + 1)*b*d*x/c - 
 1/4*b*d*arccos(c*x)/c^2 - 3/32*sqrt(-c^2*x^2 + 1)*b*e*x/c^3 - 3/32*b*e*ar 
ccos(c*x)/c^4
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.08 \[ \int x \left (d+e x^2\right ) (a+b \arccos (c x)) \, dx=\frac {16 \mathit {acos} \left (c x \right ) b \,c^{4} d \,x^{2}+8 \mathit {acos} \left (c x \right ) b \,c^{4} e \,x^{4}+8 \mathit {asin} \left (c x \right ) b \,c^{2} d +3 \mathit {asin} \left (c x \right ) b e -8 \sqrt {-c^{2} x^{2}+1}\, b \,c^{3} d x -2 \sqrt {-c^{2} x^{2}+1}\, b \,c^{3} e \,x^{3}-3 \sqrt {-c^{2} x^{2}+1}\, b c e x +16 a \,c^{4} d \,x^{2}+8 a \,c^{4} e \,x^{4}}{32 c^{4}} \] Input:

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

Output:

(16*acos(c*x)*b*c**4*d*x**2 + 8*acos(c*x)*b*c**4*e*x**4 + 8*asin(c*x)*b*c* 
*2*d + 3*asin(c*x)*b*e - 8*sqrt( - c**2*x**2 + 1)*b*c**3*d*x - 2*sqrt( - c 
**2*x**2 + 1)*b*c**3*e*x**3 - 3*sqrt( - c**2*x**2 + 1)*b*c*e*x + 16*a*c**4 
*d*x**2 + 8*a*c**4*e*x**4)/(32*c**4)