\(\int \frac {\sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{x^6} \, dx\) [60]

Optimal result

Integrand size = 27, antiderivative size = 187 \[ \int \frac {\sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{x^6} \, dx=-\frac {b c \sqrt {d-c^2 d x^2}}{20 x^4 \sqrt {1-c^2 x^2}}+\frac {b c^3 \sqrt {d-c^2 d x^2}}{30 x^2 \sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{5 d x^5}-\frac {2 c^2 \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{15 d x^3}-\frac {2 b c^5 \sqrt {d-c^2 d x^2} \log (x)}{15 \sqrt {1-c^2 x^2}} \] Output:

-1/20*b*c*(-c^2*d*x^2+d)^(1/2)/x^4/(-c^2*x^2+1)^(1/2)+1/30*b*c^3*(-c^2*d*x 
^2+d)^(1/2)/x^2/(-c^2*x^2+1)^(1/2)-1/5*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c* 
x))/d/x^5-2/15*c^2*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))/d/x^3-2/15*b*c^5 
*(-c^2*d*x^2+d)^(1/2)*ln(x)/(-c^2*x^2+1)^(1/2)
 

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.92 \[ \int \frac {\sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{x^6} \, dx=\frac {\sqrt {d-c^2 d x^2} \left (-9 b c x+6 b c^3 x^3+50 b c^5 x^5-36 a \sqrt {1-c^2 x^2}+12 a c^2 x^2 \sqrt {1-c^2 x^2}+24 a c^4 x^4 \sqrt {1-c^2 x^2}+12 b \sqrt {1-c^2 x^2} \left (-3+c^2 x^2+2 c^4 x^4\right ) \arcsin (c x)-24 b c^5 x^5 \log (x)\right )}{180 x^5 \sqrt {1-c^2 x^2}} \] Input:

Integrate[(Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/x^6,x]
 

Output:

(Sqrt[d - c^2*d*x^2]*(-9*b*c*x + 6*b*c^3*x^3 + 50*b*c^5*x^5 - 36*a*Sqrt[1 
- c^2*x^2] + 12*a*c^2*x^2*Sqrt[1 - c^2*x^2] + 24*a*c^4*x^4*Sqrt[1 - c^2*x^ 
2] + 12*b*Sqrt[1 - c^2*x^2]*(-3 + c^2*x^2 + 2*c^4*x^4)*ArcSin[c*x] - 24*b* 
c^5*x^5*Log[x]))/(180*x^5*Sqrt[1 - c^2*x^2])
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.70, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {5194, 27, 1433, 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.

\(\displaystyle \int \frac {\sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{x^6} \, dx\)

\(\Big \downarrow \) 5194

\(\displaystyle -\frac {b c \sqrt {d-c^2 d x^2} \int -\frac {-2 c^4 x^4-c^2 x^2+3}{15 x^5}dx}{\sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{5 d x^5}-\frac {2 c^2 \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{15 d x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b c \sqrt {d-c^2 d x^2} \int \frac {-2 c^4 x^4-c^2 x^2+3}{x^5}dx}{15 \sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{5 d x^5}-\frac {2 c^2 \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{15 d x^3}\)

\(\Big \downarrow \) 1433

\(\displaystyle \frac {b c \sqrt {d-c^2 d x^2} \int \left (-\frac {2 c^4}{x}-\frac {c^2}{x^3}+\frac {3}{x^5}\right )dx}{15 \sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{5 d x^5}-\frac {2 c^2 \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{15 d x^3}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{5 d x^5}-\frac {2 c^2 \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{15 d x^3}+\frac {b c \sqrt {d-c^2 d x^2} \left (-2 c^4 \log (x)+\frac {c^2}{2 x^2}-\frac {3}{4 x^4}\right )}{15 \sqrt {1-c^2 x^2}}\)

Input:

Int[(Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/x^6,x]
 

Output:

-1/5*((d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x]))/(d*x^5) - (2*c^2*(d - c^2 
*d*x^2)^(3/2)*(a + b*ArcSin[c*x]))/(15*d*x^3) + (b*c*Sqrt[d - c^2*d*x^2]*( 
-3/(4*x^4) + c^2/(2*x^2) - 2*c^4*Log[x]))/(15*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 1433

Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] 
 :> Int[ExpandIntegrand[(d*x)^m*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, 
b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] ||  !IntegerQ[(m + 1)/2])
 

rule 2009

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

rule 5194

Int[((a_.) + ArcSin[(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*ArcSin 
[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 = 0.53 (sec) , antiderivative size = 1903, normalized size of antiderivative = 10.18

method	result	size
default	\(\text {Expression too large to display}\)	\(1903\)
parts	\(\text {Expression too large to display}\)	\(1903\)

Input:

int((-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))/x^6,x,method=_RETURNVERBOSE)
 

Output:

a*(-1/5/d/x^5*(-c^2*d*x^2+d)^(3/2)-2/15*c^2/d/x^3*(-c^2*d*x^2+d)^(3/2))+9/ 
5*b*(-d*(c^2*x^2-1))^(1/2)/(15*c^6*x^6-5*c^4*x^4-15*c^2*x^2+9)/x^5/(c^2*x^ 
2-1)*arcsin(c*x)+2/15*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/(c^2*x^2 
-1)*ln((I*c*x+(-c^2*x^2+1)^(1/2))^2-1)*c^5+1/4*b*(-d*(c^2*x^2-1))^(1/2)/(1 
5*c^6*x^6-5*c^4*x^4-15*c^2*x^2+9)/(c^2*x^2-1)*c^5*(-c^2*x^2+1)^(1/2)-2*I*b 
*(-d*(c^2*x^2-1))^(1/2)/(15*c^6*x^6-5*c^4*x^4-15*c^2*x^2+9)*x^2/(c^2*x^2-1 
)*(-c^2*x^2+1)^(1/2)*arcsin(c*x)*c^7+2*I*b*(-d*(c^2*x^2-1))^(1/2)/(15*c^6* 
x^6-5*c^4*x^4-15*c^2*x^2+9)*x^6/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*arcsin(c*x) 
*c^11-2/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(15*c^6*x^6-5*c^4*x^4-15*c^2*x^2+9)*x 
^4/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*arcsin(c*x)*c^9-3/10*I*b*(-d*(c^2*x^2-1) 
)^(1/2)/(15*c^6*x^6-5*c^4*x^4-15*c^2*x^2+9)*x^3/(c^2*x^2-1)*(-c^2*x^2+1)*c 
^8-2/15*I*b*(-d*(c^2*x^2-1))^(1/2)/(15*c^6*x^6-5*c^4*x^4-15*c^2*x^2+9)*x^5 
/(c^2*x^2-1)*(-c^2*x^2+1)*c^10+6/5*I*b*(-d*(c^2*x^2-1))^(1/2)/(15*c^6*x^6- 
5*c^4*x^4-15*c^2*x^2+9)/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*arcsin(c*x)*c^5+3/1 
0*I*b*(-d*(c^2*x^2-1))^(1/2)/(15*c^6*x^6-5*c^4*x^4-15*c^2*x^2+9)*x/(c^2*x^ 
2-1)*(-c^2*x^2+1)*c^6+2/15*I*b*(-d*(c^2*x^2-1))^(1/2)/(15*c^6*x^6-5*c^4*x^ 
4-15*c^2*x^2+9)*x^7/(c^2*x^2-1)*(-c^2*x^2+1)*c^12-5/3*b*(-d*(c^2*x^2-1))^( 
1/2)/(15*c^6*x^6-5*c^4*x^4-15*c^2*x^2+9)*x^5/(c^2*x^2-1)*arcsin(c*x)*c^10- 
1/2*b*(-d*(c^2*x^2-1))^(1/2)/(15*c^6*x^6-5*c^4*x^4-15*c^2*x^2+9)*x^4/(c^2* 
x^2-1)*(-c^2*x^2+1)^(1/2)*c^9-17/3*b*(-d*(c^2*x^2-1))^(1/2)/(15*c^6*x^6...
 

Fricas [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 502, normalized size of antiderivative = 2.68 \[ \int \frac {\sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{x^6} \, dx=\left [\frac {4 \, {\left (b c^{7} x^{7} - b c^{5} x^{5}\right )} \sqrt {d} \log \left (\frac {c^{2} d x^{6} + c^{2} d x^{2} - d x^{4} + \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} {\left (x^{4} - 1\right )} \sqrt {d} - d}{c^{2} x^{4} - x^{2}}\right ) - {\left (2 \, b c^{3} x^{3} - {\left (2 \, b c^{3} - 3 \, b c\right )} x^{5} - 3 \, b c x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} + 4 \, {\left (2 \, a c^{6} x^{6} - a c^{4} x^{4} - 4 \, a c^{2} x^{2} + {\left (2 \, b c^{6} x^{6} - b c^{4} x^{4} - 4 \, b c^{2} x^{2} + 3 \, b\right )} \arcsin \left (c x\right ) + 3 \, a\right )} \sqrt {-c^{2} d x^{2} + d}}{60 \, {\left (c^{2} x^{7} - x^{5}\right )}}, -\frac {8 \, {\left (b c^{7} x^{7} - b c^{5} x^{5}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} {\left (x^{2} - 1\right )} \sqrt {-d}}{c^{2} d x^{4} + {\left (c^{2} - 1\right )} d x^{2} - d}\right ) + {\left (2 \, b c^{3} x^{3} - {\left (2 \, b c^{3} - 3 \, b c\right )} x^{5} - 3 \, b c x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} - 4 \, {\left (2 \, a c^{6} x^{6} - a c^{4} x^{4} - 4 \, a c^{2} x^{2} + {\left (2 \, b c^{6} x^{6} - b c^{4} x^{4} - 4 \, b c^{2} x^{2} + 3 \, b\right )} \arcsin \left (c x\right ) + 3 \, a\right )} \sqrt {-c^{2} d x^{2} + d}}{60 \, {\left (c^{2} x^{7} - x^{5}\right )}}\right ] \] Input:

integrate((-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))/x^6,x, algorithm="fricas" 
)
 

Output:

[1/60*(4*(b*c^7*x^7 - b*c^5*x^5)*sqrt(d)*log((c^2*d*x^6 + c^2*d*x^2 - d*x^ 
4 + sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1)*(x^4 - 1)*sqrt(d) - d)/(c^2*x^ 
4 - x^2)) - (2*b*c^3*x^3 - (2*b*c^3 - 3*b*c)*x^5 - 3*b*c*x)*sqrt(-c^2*d*x^ 
2 + d)*sqrt(-c^2*x^2 + 1) + 4*(2*a*c^6*x^6 - a*c^4*x^4 - 4*a*c^2*x^2 + (2* 
b*c^6*x^6 - b*c^4*x^4 - 4*b*c^2*x^2 + 3*b)*arcsin(c*x) + 3*a)*sqrt(-c^2*d* 
x^2 + d))/(c^2*x^7 - x^5), -1/60*(8*(b*c^7*x^7 - b*c^5*x^5)*sqrt(-d)*arcta 
n(sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1)*(x^2 - 1)*sqrt(-d)/(c^2*d*x^4 + 
(c^2 - 1)*d*x^2 - d)) + (2*b*c^3*x^3 - (2*b*c^3 - 3*b*c)*x^5 - 3*b*c*x)*sq 
rt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1) - 4*(2*a*c^6*x^6 - a*c^4*x^4 - 4*a*c 
^2*x^2 + (2*b*c^6*x^6 - b*c^4*x^4 - 4*b*c^2*x^2 + 3*b)*arcsin(c*x) + 3*a)* 
sqrt(-c^2*d*x^2 + d))/(c^2*x^7 - x^5)]
 

Sympy [F]

\[ \int \frac {\sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{x^6} \, dx=\int \frac {\sqrt {- d \left (c x - 1\right ) \left (c x + 1\right )} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{x^{6}}\, dx \] Input:

integrate((-c**2*d*x**2+d)**(1/2)*(a+b*asin(c*x))/x**6,x)
 

Output:

Integral(sqrt(-d*(c*x - 1)*(c*x + 1))*(a + b*asin(c*x))/x**6, x)
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.75 \[ \int \frac {\sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{x^6} \, dx=-\frac {1}{60} \, {\left (8 \, c^{4} \sqrt {d} \log \left (x\right ) - \frac {2 \, c^{2} \sqrt {d} x^{2} - 3 \, \sqrt {d}}{x^{4}}\right )} b c - \frac {1}{15} \, b {\left (\frac {2 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2}}{d x^{3}} + \frac {3 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}}}{d x^{5}}\right )} \arcsin \left (c x\right ) - \frac {1}{15} \, a {\left (\frac {2 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2}}{d x^{3}} + \frac {3 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}}}{d x^{5}}\right )} \] Input:

integrate((-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))/x^6,x, algorithm="maxima" 
)
 

Output:

-1/60*(8*c^4*sqrt(d)*log(x) - (2*c^2*sqrt(d)*x^2 - 3*sqrt(d))/x^4)*b*c - 1 
/15*b*(2*(-c^2*d*x^2 + d)^(3/2)*c^2/(d*x^3) + 3*(-c^2*d*x^2 + d)^(3/2)/(d* 
x^5))*arcsin(c*x) - 1/15*a*(2*(-c^2*d*x^2 + d)^(3/2)*c^2/(d*x^3) + 3*(-c^2 
*d*x^2 + d)^(3/2)/(d*x^5))
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{x^6} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))/x^6,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. \[ \int \frac {\sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{x^6} \, dx=\int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\sqrt {d-c^2\,d\,x^2}}{x^6} \,d x \] Input:

int(((a + b*asin(c*x))*(d - c^2*d*x^2)^(1/2))/x^6,x)
 

Output:

int(((a + b*asin(c*x))*(d - c^2*d*x^2)^(1/2))/x^6, x)
 

Reduce [F]

\[ \int \frac {\sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{x^6} \, dx=\frac {\sqrt {d}\, \left (2 \sqrt {-c^{2} x^{2}+1}\, a \,c^{4} x^{4}+\sqrt {-c^{2} x^{2}+1}\, a \,c^{2} x^{2}-3 \sqrt {-c^{2} x^{2}+1}\, a +15 \left (\int \frac {\sqrt {-c^{2} x^{2}+1}\, \mathit {asin} \left (c x \right )}{x^{6}}d x \right ) b \,x^{5}\right )}{15 x^{5}} \] Input:

int((-c^2*d*x^2+d)^(1/2)*(a+b*asin(c*x))/x^6,x)
                                                                                    
                                                                                    
 

Output:

(sqrt(d)*(2*sqrt( - c**2*x**2 + 1)*a*c**4*x**4 + sqrt( - c**2*x**2 + 1)*a* 
c**2*x**2 - 3*sqrt( - c**2*x**2 + 1)*a + 15*int((sqrt( - c**2*x**2 + 1)*as 
in(c*x))/x**6,x)*b*x**5))/(15*x**5)