\(\int x^m (d-c^2 d x^2)^{3/2} (a+b \arccos (c x)) \, dx\) [152]

Optimal result
Mathematica [F]
Rubi [A] (verified)
Maple [F]
Fricas [F]
Sympy [F(-1)]
Maxima [F]
Giac [F(-2)]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 27, antiderivative size = 399 \[ \int x^m \left (d-c^2 d x^2\right )^{3/2} (a+b \arccos (c x)) \, dx=-\frac {3 b c d x^{2+m} \sqrt {d-c^2 d x^2}}{(2+m)^2 (4+m) \sqrt {1-c^2 x^2}}-\frac {b c d x^{2+m} \sqrt {d-c^2 d x^2}}{\left (8+6 m+m^2\right ) \sqrt {1-c^2 x^2}}+\frac {b c^3 d x^{4+m} \sqrt {d-c^2 d x^2}}{(4+m)^2 \sqrt {1-c^2 x^2}}+\frac {3 d x^{1+m} \sqrt {d-c^2 d x^2} (a+b \arccos (c x))}{8+6 m+m^2}+\frac {x^{1+m} \left (d-c^2 d x^2\right )^{3/2} (a+b \arccos (c x))}{4+m}+\frac {3 d x^{1+m} \sqrt {d-c^2 d x^2} (a+b \arccos (c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},c^2 x^2\right )}{\left (8+14 m+7 m^2+m^3\right ) \sqrt {1-c^2 x^2}}-\frac {3 b c d x^{2+m} \sqrt {d-c^2 d x^2} \, _3F_2\left (1,1+\frac {m}{2},1+\frac {m}{2};\frac {3}{2}+\frac {m}{2},2+\frac {m}{2};c^2 x^2\right )}{(1+m) (2+m)^2 (4+m) \sqrt {1-c^2 x^2}} \] Output:

-3*b*c*d*x^(2+m)*(-c^2*d*x^2+d)^(1/2)/(2+m)^2/(4+m)/(-c^2*x^2+1)^(1/2)-b*c 
*d*x^(2+m)*(-c^2*d*x^2+d)^(1/2)/(m^2+6*m+8)/(-c^2*x^2+1)^(1/2)+b*c^3*d*x^( 
4+m)*(-c^2*d*x^2+d)^(1/2)/(4+m)^2/(-c^2*x^2+1)^(1/2)+3*d*x^(1+m)*(-c^2*d*x 
^2+d)^(1/2)*(a+b*arccos(c*x))/(m^2+6*m+8)+x^(1+m)*(-c^2*d*x^2+d)^(3/2)*(a+ 
b*arccos(c*x))/(4+m)+3*d*x^(1+m)*(-c^2*d*x^2+d)^(1/2)*(a+b*arccos(c*x))*hy 
pergeom([1/2, 1/2+1/2*m],[3/2+1/2*m],c^2*x^2)/(m^3+7*m^2+14*m+8)/(-c^2*x^2 
+1)^(1/2)-3*b*c*d*x^(2+m)*(-c^2*d*x^2+d)^(1/2)*hypergeom([1, 1+1/2*m, 1+1/ 
2*m],[2+1/2*m, 3/2+1/2*m],c^2*x^2)/(1+m)/(2+m)^2/(4+m)/(-c^2*x^2+1)^(1/2)
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 324, normalized size of antiderivative = 0.81, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {5203, 244, 2009, 5199, 15, 5221}

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.

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

\(\Big \downarrow \) 5203

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

\(\Big \downarrow \) 244

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 5199

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

\(\Big \downarrow \) 15

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

\(\Big \downarrow \) 5221

\(\displaystyle \frac {3 d \left (\frac {\sqrt {d-c^2 d x^2} \left (\frac {b c x^{m+2} \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;c^2 x^2\right )}{m^2+3 m+2}+\frac {x^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},c^2 x^2\right ) (a+b \arccos (c x))}{m+1}\right )}{(m+2) \sqrt {1-c^2 x^2}}+\frac {x^{m+1} \sqrt {d-c^2 d x^2} (a+b \arccos (c x))}{m+2}+\frac {b c x^{m+2} \sqrt {d-c^2 d x^2}}{(m+2)^2 \sqrt {1-c^2 x^2}}\right )}{m+4}+\frac {x^{m+1} \left (d-c^2 d x^2\right )^{3/2} (a+b \arccos (c x))}{m+4}+\frac {b c d \sqrt {d-c^2 d x^2} \left (\frac {x^{m+2}}{m+2}-\frac {c^2 x^{m+4}}{m+4}\right )}{(m+4) \sqrt {1-c^2 x^2}}\)

Input:

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

Output:

(b*c*d*Sqrt[d - c^2*d*x^2]*(x^(2 + m)/(2 + m) - (c^2*x^(4 + m))/(4 + m)))/ 
((4 + m)*Sqrt[1 - c^2*x^2]) + (x^(1 + m)*(d - c^2*d*x^2)^(3/2)*(a + b*ArcC 
os[c*x]))/(4 + m) + (3*d*((b*c*x^(2 + m)*Sqrt[d - c^2*d*x^2])/((2 + m)^2*S 
qrt[1 - c^2*x^2]) + (x^(1 + m)*Sqrt[d - c^2*d*x^2]*(a + b*ArcCos[c*x]))/(2 
 + m) + (Sqrt[d - c^2*d*x^2]*((x^(1 + m)*(a + b*ArcCos[c*x])*Hypergeometri 
c2F1[1/2, (1 + m)/2, (3 + m)/2, c^2*x^2])/(1 + m) + (b*c*x^(2 + m)*Hyperge 
ometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, c^2*x^2])/(2 + 3*m 
 + m^2)))/((2 + m)*Sqrt[1 - c^2*x^2])))/(4 + m)
 

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 244
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand 
Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p 
, 0]
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 5199
Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*Sqrt[(d_) + 
(e_.)*(x_)^2], x_Symbol] :> Simp[(f*x)^(m + 1)*Sqrt[d + e*x^2]*((a + b*ArcC 
os[c*x])^n/(f*(m + 2))), x] + (Simp[(1/(m + 2))*Simp[Sqrt[d + e*x^2]/Sqrt[1 
 - c^2*x^2]]   Int[(f*x)^m*((a + b*ArcCos[c*x])^n/Sqrt[1 - c^2*x^2]), x], x 
] + Simp[b*c*(n/(f*(m + 2)))*Simp[Sqrt[d + e*x^2]/Sqrt[1 - c^2*x^2]]   Int[ 
(f*x)^(m + 1)*(a + b*ArcCos[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, 
 f, m}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && (IGtQ[m, -2] || EqQ[n, 1])
 

rule 5203
Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_. 
)*(x_)^2)^(p_.), x_Symbol] :> Simp[(f*x)^(m + 1)*(d + e*x^2)^p*((a + b*ArcC 
os[c*x])^n/(f*(m + 2*p + 1))), x] + (Simp[2*d*(p/(m + 2*p + 1))   Int[(f*x) 
^m*(d + e*x^2)^(p - 1)*(a + b*ArcCos[c*x])^n, x], x] + Simp[b*c*(n/(f*(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, m}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[p, 0] &&  !LtQ[m, -1]
 

rule 5221
Int[(((a_.) + ArcCos[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_. 
)*(x_)^2], x_Symbol] :> Simp[((f*x)^(m + 1)/(f*(m + 1)))*(a + b*ArcCos[c*x] 
)*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]]*Hypergeometric2F1[1/2, (1 + m)/2, 
 (3 + m)/2, c^2*x^2], x] + Simp[b*c*((f*x)^(m + 2)/(f^2*(m + 1)*(m + 2)))*S 
imp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]]*HypergeometricPFQ[{1, 1 + m/2, 1 + m 
/2}, {3/2 + m/2, 2 + m/2}, c^2*x^2], x] /; FreeQ[{a, b, c, d, e, f, m}, x] 
&& EqQ[c^2*d + e, 0] &&  !IntegerQ[m]
 
Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

integrate(x^m*(-c^2*d*x^2+d)^(3/2)*(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))*sqrt(-c^2* 
d*x^2 + d)*x^m, x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

sqrt(d)*d*( - int(x**m*sqrt( - c**2*x**2 + 1)*acos(c*x)*x**2,x)*b*c**2 + i 
nt(x**m*sqrt( - c**2*x**2 + 1)*acos(c*x),x)*b - int(x**m*sqrt( - c**2*x**2 
 + 1)*x**2,x)*a*c**2 + int(x**m*sqrt( - c**2*x**2 + 1),x)*a)