\(\int \frac {a+b \arccos (c x)}{x^2 (d-c^2 d x^2)^{5/2}} \, dx\) [139]

Optimal result

Integrand size = 27, antiderivative size = 224 \[ \int \frac {a+b \arccos (c x)}{x^2 \left (d-c^2 d x^2\right )^{5/2}} \, dx=-\frac {b c \sqrt {d-c^2 d x^2}}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {a+b \arccos (c x)}{d x \left (d-c^2 d x^2\right )^{3/2}}+\frac {4 c^2 x (a+b \arccos (c x))}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {8 c^2 x (a+b \arccos (c x))}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {b c \sqrt {d-c^2 d x^2} \log (x)}{d^3 \sqrt {1-c^2 x^2}}+\frac {5 b c \sqrt {d-c^2 d x^2} \log \left (1-c^2 x^2\right )}{6 d^3 \sqrt {1-c^2 x^2}} \] Output:

-1/6*b*c*(-c^2*d*x^2+d)^(1/2)/d^3/(-c^2*x^2+1)^(3/2)-(a+b*arccos(c*x))/d/x 
/(-c^2*d*x^2+d)^(3/2)+4/3*c^2*x*(a+b*arccos(c*x))/d/(-c^2*d*x^2+d)^(3/2)+8 
/3*c^2*x*(a+b*arccos(c*x))/d^2/(-c^2*d*x^2+d)^(1/2)+b*c*(-c^2*d*x^2+d)^(1/ 
2)*ln(x)/d^3/(-c^2*x^2+1)^(1/2)+5/6*b*c*(-c^2*d*x^2+d)^(1/2)*ln(-c^2*x^2+1 
)/d^3/(-c^2*x^2+1)^(1/2)
 

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.06 \[ \int \frac {a+b \arccos (c x)}{x^2 \left (d-c^2 d x^2\right )^{5/2}} \, dx=\frac {\sqrt {d-c^2 d x^2} \left (b c x-b c^3 x^3-6 a \sqrt {1-c^2 x^2}+24 a c^2 x^2 \sqrt {1-c^2 x^2}-16 a c^4 x^4 \sqrt {1-c^2 x^2}-2 b \sqrt {1-c^2 x^2} \left (3-12 c^2 x^2+8 c^4 x^4\right ) \arccos (c x)+3 b c x \left (-1+c^2 x^2\right )^2 \log \left (1-\frac {1}{c^2 x^2}\right )-8 b c x \log \left (1-c^2 x^2\right )+16 b c^3 x^3 \log \left (1-c^2 x^2\right )-8 b c^5 x^5 \log \left (1-c^2 x^2\right )\right )}{6 d^3 x \left (1-c^2 x^2\right )^{5/2}} \] Input:

Integrate[(a + b*ArcCos[c*x])/(x^2*(d - c^2*d*x^2)^(5/2)),x]
 

Output:

(Sqrt[d - c^2*d*x^2]*(b*c*x - b*c^3*x^3 - 6*a*Sqrt[1 - c^2*x^2] + 24*a*c^2 
*x^2*Sqrt[1 - c^2*x^2] - 16*a*c^4*x^4*Sqrt[1 - c^2*x^2] - 2*b*Sqrt[1 - c^2 
*x^2]*(3 - 12*c^2*x^2 + 8*c^4*x^4)*ArcCos[c*x] + 3*b*c*x*(-1 + c^2*x^2)^2* 
Log[1 - 1/(c^2*x^2)] - 8*b*c*x*Log[1 - c^2*x^2] + 16*b*c^3*x^3*Log[1 - c^2 
*x^2] - 8*b*c^5*x^5*Log[1 - c^2*x^2]))/(6*d^3*x*(1 - c^2*x^2)^(5/2))
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.75, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {5195, 27, 1578, 1195, 2009}

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.

\(\displaystyle \int \frac {a+b \arccos (c x)}{x^2 \left (d-c^2 d x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 5195

\(\displaystyle \frac {b c \sqrt {d-c^2 d x^2} \int -\frac {8 c^4 x^4-12 c^2 x^2+3}{3 d^3 x \left (1-c^2 x^2\right )^2}dx}{\sqrt {1-c^2 x^2}}+\frac {8 c^2 x (a+b \arccos (c x))}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {4 c^2 x (a+b \arccos (c x))}{3 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {a+b \arccos (c x)}{d x \left (d-c^2 d x^2\right )^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {b c \sqrt {d-c^2 d x^2} \int \frac {8 c^4 x^4-12 c^2 x^2+3}{x \left (1-c^2 x^2\right )^2}dx}{3 d^3 \sqrt {1-c^2 x^2}}+\frac {8 c^2 x (a+b \arccos (c x))}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {4 c^2 x (a+b \arccos (c x))}{3 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {a+b \arccos (c x)}{d x \left (d-c^2 d x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1578

\(\displaystyle -\frac {b c \sqrt {d-c^2 d x^2} \int \frac {8 c^4 x^4-12 c^2 x^2+3}{x^2 \left (1-c^2 x^2\right )^2}dx^2}{6 d^3 \sqrt {1-c^2 x^2}}+\frac {8 c^2 x (a+b \arccos (c x))}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {4 c^2 x (a+b \arccos (c x))}{3 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {a+b \arccos (c x)}{d x \left (d-c^2 d x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1195

\(\displaystyle -\frac {b c \sqrt {d-c^2 d x^2} \int \left (\frac {5 c^2}{c^2 x^2-1}-\frac {c^2}{\left (c^2 x^2-1\right )^2}+\frac {3}{x^2}\right )dx^2}{6 d^3 \sqrt {1-c^2 x^2}}+\frac {8 c^2 x (a+b \arccos (c x))}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {4 c^2 x (a+b \arccos (c x))}{3 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {a+b \arccos (c x)}{d x \left (d-c^2 d x^2\right )^{3/2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {8 c^2 x (a+b \arccos (c x))}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {4 c^2 x (a+b \arccos (c x))}{3 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {a+b \arccos (c x)}{d x \left (d-c^2 d x^2\right )^{3/2}}-\frac {b c \sqrt {d-c^2 d x^2} \left (\frac {1}{c^2 x^2-1}+5 \log \left (1-c^2 x^2\right )+3 \log \left (x^2\right )\right )}{6 d^3 \sqrt {1-c^2 x^2}}\)

Input:

Int[(a + b*ArcCos[c*x])/(x^2*(d - c^2*d*x^2)^(5/2)),x]
 

Output:

-((a + b*ArcCos[c*x])/(d*x*(d - c^2*d*x^2)^(3/2))) + (4*c^2*x*(a + b*ArcCo 
s[c*x]))/(3*d*(d - c^2*d*x^2)^(3/2)) + (8*c^2*x*(a + b*ArcCos[c*x]))/(3*d^ 
2*Sqrt[d - c^2*d*x^2]) - (b*c*Sqrt[d - c^2*d*x^2]*((-1 + c^2*x^2)^(-1) + 3 
*Log[x^2] + 5*Log[1 - c^2*x^2]))/(6*d^3*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 1195

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x 
_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + 
 g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x 
] && IGtQ[p, 0]
 

rule 1578

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_ 
)^4)^(p_.), x_Symbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a 
+ b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && Int 
egerQ[(m - 1)/2]
 

rule 2009

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

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 = 0.88 (sec) , antiderivative size = 1347, normalized size of antiderivative = 6.01

method	result	size
default	\(\text {Expression too large to display}\)	\(1347\)
parts	\(\text {Expression too large to display}\)	\(1347\)

Input:

int((a+b*arccos(c*x))/x^2/(-c^2*d*x^2+d)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

a*(-1/d/x/(-c^2*d*x^2+d)^(3/2)+4*c^2*(1/3/d*x/(-c^2*d*x^2+d)^(3/2)+2/3/d^2 
*x/(-c^2*d*x^2+d)^(1/2)))+56*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^6*x^6-25*c^4*x^ 
4+26*c^2*x^2-9)/d^3*x^3*arccos(c*x)*c^4+4/3*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^ 
6*x^6-25*c^4*x^4+26*c^2*x^2-9)/d^3*x^2*(-c^2*x^2+1)^(1/2)*c^3-44*b*(-d*(c^ 
2*x^2-1))^(1/2)/(8*c^6*x^6-25*c^4*x^4+26*c^2*x^2-9)/d^3*x*arccos(c*x)*c^2+ 
b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/d^3/(c^2*x^2-1)*ln(1+(c*x+I*(- 
c^2*x^2+1)^(1/2))^2)*c+5/3*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/d^3 
/(c^2*x^2-1)*ln((c*x+I*(-c^2*x^2+1)^(1/2))^2-1)*c-80/3*I*b*(-d*(c^2*x^2-1) 
)^(1/2)/(8*c^6*x^6-25*c^4*x^4+26*c^2*x^2-9)/d^3*x^5*(-c^2*x^2+1)*c^6-4*I*b 
*(-d*(c^2*x^2-1))^(1/2)/(8*c^6*x^6-25*c^4*x^4+26*c^2*x^2-9)/d^3*x*(-c^2*x^ 
2+1)*c^2-3/2*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^6*x^6-25*c^4*x^4+26*c^2*x^2-9)/ 
d^3*(-c^2*x^2+1)^(1/2)*c+9*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^6*x^6-25*c^4*x^4+ 
26*c^2*x^2-9)/d^3/x*arccos(c*x)-16/3*I*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+ 
1)^(1/2)/d^3/(c^2*x^2-1)*arccos(c*x)*c-112/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(8 
*c^6*x^6-25*c^4*x^4+26*c^2*x^2-9)/d^3*x^7*c^8-64/3*b*(-d*(c^2*x^2-1))^(1/2 
)/(8*c^6*x^6-25*c^4*x^4+26*c^2*x^2-9)/d^3*x^5*arccos(c*x)*c^6+140/3*I*b*(- 
d*(c^2*x^2-1))^(1/2)/(8*c^6*x^6-25*c^4*x^4+26*c^2*x^2-9)/d^3*x^5*c^6+24*I* 
b*(-d*(c^2*x^2-1))^(1/2)/(8*c^6*x^6-25*c^4*x^4+26*c^2*x^2-9)/d^3*arccos(c* 
x)*(-c^2*x^2+1)^(1/2)*c-136/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^6*x^6-25*c^4 
*x^4+26*c^2*x^2-9)/d^3*x^2*arccos(c*x)*(-c^2*x^2+1)^(1/2)*c^3-24*I*b*(-...
 

Fricas [F]

\[ \int \frac {a+b \arccos (c x)}{x^2 \left (d-c^2 d x^2\right )^{5/2}} \, dx=\int { \frac {b \arccos \left (c x\right ) + a}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} x^{2}} \,d x } \] Input:

integrate((a+b*arccos(c*x))/x^2/(-c^2*d*x^2+d)^(5/2),x, algorithm="fricas" 
)
 

Output:

integral(-sqrt(-c^2*d*x^2 + d)*(b*arccos(c*x) + a)/(c^6*d^3*x^8 - 3*c^4*d^ 
3*x^6 + 3*c^2*d^3*x^4 - d^3*x^2), x)
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {a+b \arccos (c x)}{x^2 \left (d-c^2 d x^2\right )^{5/2}} \, dx=\int \frac {a + b \operatorname {acos}{\left (c x \right )}}{x^{2} \left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((a+b*acos(c*x))/x**2/(-c**2*d*x**2+d)**(5/2),x)
 

Output:

Integral((a + b*acos(c*x))/(x**2*(-d*(c*x - 1)*(c*x + 1))**(5/2)), x)
 

Maxima [F]

\[ \int \frac {a+b \arccos (c x)}{x^2 \left (d-c^2 d x^2\right )^{5/2}} \, dx=\int { \frac {b \arccos \left (c x\right ) + a}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} x^{2}} \,d x } \] Input:

integrate((a+b*arccos(c*x))/x^2/(-c^2*d*x^2+d)^(5/2),x, algorithm="maxima" 
)
 

Output:

1/3*a*(8*c^2*x/(sqrt(-c^2*d*x^2 + d)*d^2) + 4*c^2*x/((-c^2*d*x^2 + d)^(3/2 
)*d) - 3/((-c^2*d*x^2 + d)^(3/2)*d*x)) + b*integrate(arctan2(sqrt(c*x + 1) 
*sqrt(-c*x + 1), c*x)/((c^4*d^2*x^6 - 2*c^2*d^2*x^4 + d^2*x^2)*sqrt(c*x + 
1)*sqrt(-c*x + 1)), x)/sqrt(d)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {a+b \arccos (c x)}{x^2 \left (d-c^2 d x^2\right )^{5/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((a+b*arccos(c*x))/x^2/(-c^2*d*x^2+d)^(5/2),x, algorithm="giac")
 

Output:

Exception raised: RuntimeError >> an error occurred running a Giac command 
:INPUT:sage2OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const ve 
cteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \arccos (c x)}{x^2 \left (d-c^2 d x^2\right )^{5/2}} \, dx=\int \frac {a+b\,\mathrm {acos}\left (c\,x\right )}{x^2\,{\left (d-c^2\,d\,x^2\right )}^{5/2}} \,d x \] Input:

int((a + b*acos(c*x))/(x^2*(d - c^2*d*x^2)^(5/2)),x)
 

Output:

int((a + b*acos(c*x))/(x^2*(d - c^2*d*x^2)^(5/2)), x)
 

Reduce [F]

\[ \int \frac {a+b \arccos (c x)}{x^2 \left (d-c^2 d x^2\right )^{5/2}} \, dx=\frac {3 \sqrt {-c^{2} x^{2}+1}\, \left (\int \frac {\mathit {acos} \left (c x \right )}{\sqrt {-c^{2} x^{2}+1}\, c^{4} x^{6}-2 \sqrt {-c^{2} x^{2}+1}\, c^{2} x^{4}+\sqrt {-c^{2} x^{2}+1}\, x^{2}}d x \right ) b \,c^{2} x^{3}-3 \sqrt {-c^{2} x^{2}+1}\, \left (\int \frac {\mathit {acos} \left (c x \right )}{\sqrt {-c^{2} x^{2}+1}\, c^{4} x^{6}-2 \sqrt {-c^{2} x^{2}+1}\, c^{2} x^{4}+\sqrt {-c^{2} x^{2}+1}\, x^{2}}d x \right ) b x +8 a \,c^{4} x^{4}-12 a \,c^{2} x^{2}+3 a}{3 \sqrt {d}\, \sqrt {-c^{2} x^{2}+1}\, d^{2} x \left (c^{2} x^{2}-1\right )} \] Input:

int((a+b*acos(c*x))/x^2/(-c^2*d*x^2+d)^(5/2),x)
 

Output:

(3*sqrt( - c**2*x**2 + 1)*int(acos(c*x)/(sqrt( - c**2*x**2 + 1)*c**4*x**6 
- 2*sqrt( - c**2*x**2 + 1)*c**2*x**4 + sqrt( - c**2*x**2 + 1)*x**2),x)*b*c 
**2*x**3 - 3*sqrt( - c**2*x**2 + 1)*int(acos(c*x)/(sqrt( - c**2*x**2 + 1)* 
c**4*x**6 - 2*sqrt( - c**2*x**2 + 1)*c**2*x**4 + sqrt( - c**2*x**2 + 1)*x* 
*2),x)*b*x + 8*a*c**4*x**4 - 12*a*c**2*x**2 + 3*a)/(3*sqrt(d)*sqrt( - c**2 
*x**2 + 1)*d**2*x*(c**2*x**2 - 1))