\(\int x^m (d-c^2 d x^2) (a+b \arccos (c x)) \, dx\) [147]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [F]
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 23, antiderivative size = 129 \[ \int x^m \left (d-c^2 d x^2\right ) (a+b \arccos (c x)) \, dx=-\frac {b c d x^{2+m} \sqrt {1-c^2 x^2}}{(3+m)^2}+\frac {d x^{1+m} (a+b \arccos (c x))}{1+m}-\frac {c^2 d x^{3+m} (a+b \arccos (c x))}{3+m}-\frac {b c d (7+3 m) x^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},c^2 x^2\right )}{(1+m) (2+m) (3+m)^2} \] Output:

-b*c*d*x^(2+m)*(-c^2*x^2+1)^(1/2)/(3+m)^2+d*x^(1+m)*(a+b*arccos(c*x))/(1+m 
)-c^2*d*x^(3+m)*(a+b*arccos(c*x))/(3+m)-b*c*d*(7+3*m)*x^(2+m)*hypergeom([1 
/2, 1+1/2*m],[2+1/2*m],c^2*x^2)/(1+m)/(2+m)/(3+m)^2
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.98 \[ \int x^m \left (d-c^2 d x^2\right ) (a+b \arccos (c x)) \, dx=d x^{1+m} \left (\frac {b c x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1+\frac {m}{2},2+\frac {m}{2},c^2 x^2\right )}{2+3 m+m^2}-\frac {\frac {\left (-3+c^2 x^2+m \left (-1+c^2 x^2\right )\right ) (a+b \arccos (c x))}{1+m}+\frac {b c^3 x^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},2+\frac {m}{2},3+\frac {m}{2},c^2 x^2\right )}{4+m}}{3+m}\right ) \] Input:

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

Output:

d*x^(1 + m)*((b*c*x*Hypergeometric2F1[1/2, 1 + m/2, 2 + m/2, c^2*x^2])/(2 
+ 3*m + m^2) - (((-3 + c^2*x^2 + m*(-1 + c^2*x^2))*(a + b*ArcCos[c*x]))/(1 
 + m) + (b*c^3*x^3*Hypergeometric2F1[1/2, 2 + m/2, 3 + m/2, c^2*x^2])/(4 + 
 m))/(3 + m))
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {5193, 27, 363, 278}

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.

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

\(\Big \downarrow \) 5193

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 363

\(\displaystyle b c d \left (\frac {(3 m+7) \int \frac {x^{m+1}}{\sqrt {1-c^2 x^2}}dx}{(m+1) (m+3)^2}+\frac {\sqrt {1-c^2 x^2} x^{m+2}}{(m+3)^2}\right )-\frac {c^2 d x^{m+3} (a+b \arccos (c x))}{m+3}+\frac {d x^{m+1} (a+b \arccos (c x))}{m+1}\)

\(\Big \downarrow \) 278

\(\displaystyle -\frac {c^2 d x^{m+3} (a+b \arccos (c x))}{m+3}+\frac {d x^{m+1} (a+b \arccos (c x))}{m+1}+b c d \left (\frac {(3 m+7) x^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},c^2 x^2\right )}{(m+1) (m+2) (m+3)^2}+\frac {\sqrt {1-c^2 x^2} x^{m+2}}{(m+3)^2}\right )\)

Input:

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

Output:

(d*x^(1 + m)*(a + b*ArcCos[c*x]))/(1 + m) - (c^2*d*x^(3 + m)*(a + b*ArcCos 
[c*x]))/(3 + m) + b*c*d*((x^(2 + m)*Sqrt[1 - c^2*x^2])/(3 + m)^2 + ((7 + 3 
*m)*x^(2 + m)*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, c^2*x^2])/((1 + 
 m)*(2 + m)*(3 + m)^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 278
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( 
c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( 
-b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && (ILtQ[p, 0 
] || GtQ[a, 0])
 

rule 363
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), 
 x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3))   Int[(e*x)^ 
m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d 
, 0] && NeQ[m + 2*p + 3, 0]
 

rule 5193
Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_) 
^2)^(p_.), x_Symbol] :> With[{u = IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Simp[ 
(a + b*ArcCos[c*x])   u, x] + Simp[b*c   Int[SimplifyIntegrand[u/Sqrt[1 - c 
^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0 
] && IGtQ[p, 0]
 
Maple [F]

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

Input:

int(x^m*(-c^2*d*x^2+d)*(a+b*arccos(c*x)),x)
 

Output:

int(x^m*(-c^2*d*x^2+d)*(a+b*arccos(c*x)),x)
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

-d*(Integral(-a*x**m, x) + Integral(-b*x**m*acos(c*x), x) + Integral(a*c** 
2*x**2*x**m, x) + Integral(b*c**2*x**2*x**m*acos(c*x), x))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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