\(\int \frac {(d-c^2 d x^2)^{3/2} (a+b \arcsin (c x))}{x^6} \, dx\) [73]

Optimal result

Integrand size = 27, antiderivative size = 154 \[ \int \frac {\left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{x^6} \, dx=-\frac {b c d \sqrt {d-c^2 d x^2}}{20 x^4 \sqrt {1-c^2 x^2}}+\frac {b c^3 d \sqrt {d-c^2 d x^2}}{5 x^2 \sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x))}{5 d x^5}+\frac {b c^5 d \sqrt {d-c^2 d x^2} \log (x)}{5 \sqrt {1-c^2 x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.01 \[ \int \frac {\left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{x^6} \, dx=\frac {d \sqrt {d-c^2 d x^2} \left (-3 b c x+12 b c^3 x^3-25 b c^5 x^5-12 a \sqrt {1-c^2 x^2}+24 a c^2 x^2 \sqrt {1-c^2 x^2}-12 a c^4 x^4 \sqrt {1-c^2 x^2}-12 b \left (1-c^2 x^2\right )^{5/2} \arcsin (c x)+12 b c^5 x^5 \log (x)\right )}{60 x^5 \sqrt {1-c^2 x^2}} \] Input:

Integrate[((d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x]))/x^6,x]
 

Output:

(d*Sqrt[d - c^2*d*x^2]*(-3*b*c*x + 12*b*c^3*x^3 - 25*b*c^5*x^5 - 12*a*Sqrt 
[1 - c^2*x^2] + 24*a*c^2*x^2*Sqrt[1 - c^2*x^2] - 12*a*c^4*x^4*Sqrt[1 - c^2 
*x^2] - 12*b*(1 - c^2*x^2)^(5/2)*ArcSin[c*x] + 12*b*c^5*x^5*Log[x]))/(60*x 
^5*Sqrt[1 - c^2*x^2])
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.61, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {5186, 243, 49, 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.

\(\displaystyle \int \frac {\left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{x^6} \, dx\)

\(\Big \downarrow \) 5186

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

\(\Big \downarrow \) 243

\(\displaystyle \frac {b c d \sqrt {d-c^2 d x^2} \int \frac {\left (1-c^2 x^2\right )^2}{x^6}dx^2}{10 \sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x))}{5 d x^5}\)

\(\Big \downarrow \) 49

\(\displaystyle \frac {b c d \sqrt {d-c^2 d x^2} \int \left (\frac {c^4}{x^2}-\frac {2 c^2}{x^4}+\frac {1}{x^6}\right )dx^2}{10 \sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x))}{5 d x^5}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b c d \sqrt {d-c^2 d x^2} \left (c^4 \log \left (x^2\right )+\frac {2 c^2}{x^2}-\frac {1}{2 x^4}\right )}{10 \sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x))}{5 d x^5}\)

Input:

Int[((d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x]))/x^6,x]
 

Output:

-1/5*((d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x]))/(d*x^5) + (b*c*d*Sqrt[d - 
 c^2*d*x^2]*(-1/2*1/x^4 + (2*c^2)/x^2 + c^4*Log[x^2]))/(10*Sqrt[1 - c^2*x^ 
2])
 

Defintions of rubi rules used

rule 49

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

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 2009

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

rule 5186

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_. 
)*(x_)^2)^(p_), x_Symbol] :> Simp[(f*x)^(m + 1)*(d + e*x^2)^(p + 1)*((a + b 
*ArcSin[c*x])^n/(d*f*(m + 1))), x] - Simp[b*c*(n/(f*(m + 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*A 
rcSin[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[c^ 
2*d + e, 0] && GtQ[n, 0] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.58 (sec) , antiderivative size = 2350, normalized size of antiderivative = 15.26

method	result	size
default	\(\text {Expression too large to display}\)	\(2350\)
parts	\(\text {Expression too large to display}\)	\(2350\)

Input:

int((-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))/x^6,x,method=_RETURNVERBOSE)
 

Output:

-1/5*a/d/x^5*(-c^2*d*x^2+d)^(5/2)+1/5*b*(-d*(c^2*x^2-1))^(1/2)*d/(5*c^8*x^ 
8-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)/x^5/(c^2*x^2-1)*arcsin(c*x)+3/2*b*(-d 
*(c^2*x^2-1))^(1/2)*d/(5*c^8*x^8-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)/(c^2*x 
^2-1)*c^5*(-c^2*x^2+1)^(1/2)-1/5*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)*d*c^5+1/5*I*b*(-d*(c^2*x 
^2-1))^(1/2)*d/(5*c^8*x^8-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*x^9/(c^2*x^2- 
1)*c^14-13/20*I*b*(-d*(c^2*x^2-1))^(1/2)*d/(5*c^8*x^8-10*c^6*x^6+10*c^4*x^ 
4-5*c^2*x^2+1)*x^7/(c^2*x^2-1)*c^12+3/4*I*b*(-d*(c^2*x^2-1))^(1/2)*d/(5*c^ 
8*x^8-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*x^5/(c^2*x^2-1)*c^10+9/4*b*(-d*(c 
^2*x^2-1))^(1/2)*d/(5*c^8*x^8-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*x^4/(c^2* 
x^2-1)*(-c^2*x^2+1)^(1/2)*c^9+14*b*(-d*(c^2*x^2-1))^(1/2)*d/(5*c^8*x^8-10* 
c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*x^3/(c^2*x^2-1)*arcsin(c*x)*c^8-5/2*b*(-d* 
(c^2*x^2-1))^(1/2)*d/(5*c^8*x^8-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*x^2/(c^ 
2*x^2-1)*c^7*(-c^2*x^2+1)^(1/2)-56/5*b*(-d*(c^2*x^2-1))^(1/2)*d/(5*c^8*x^8 
-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*x/(c^2*x^2-1)*arcsin(c*x)*c^6+28/5*b*( 
-d*(c^2*x^2-1))^(1/2)*d/(5*c^8*x^8-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)/x/(c 
^2*x^2-1)*arcsin(c*x)*c^4-9/20*b*(-d*(c^2*x^2-1))^(1/2)*d/(5*c^8*x^8-10*c^ 
6*x^6+10*c^4*x^4-5*c^2*x^2+1)/x^2/(c^2*x^2-1)*c^3*(-c^2*x^2+1)^(1/2)+2*I*b 
*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)*arcsin(c*x)*d*c^5/(5*c^2*x^2-5) 
-8/5*b*(-d*(c^2*x^2-1))^(1/2)*d/(5*c^8*x^8-10*c^6*x^6+10*c^4*x^4-5*c^2*...
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 526, normalized size of antiderivative = 3.42 \[ \int \frac {\left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{x^6} \, dx=\left [\frac {2 \, {\left (b c^{7} d x^{7} - b c^{5} d x^{5}\right )} \sqrt {d} \log \left (\frac {c^{2} d x^{6} + c^{2} d x^{2} - d x^{4} - \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} {\left (x^{4} - 1\right )} \sqrt {d} - d}{c^{2} x^{4} - x^{2}}\right ) - {\left (4 \, b c^{3} d x^{3} - {\left (4 \, b c^{3} - b c\right )} d x^{5} - b c d x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} - 4 \, {\left (a c^{6} d x^{6} - 3 \, a c^{4} d x^{4} + 3 \, a c^{2} d x^{2} - a d + {\left (b c^{6} d x^{6} - 3 \, b c^{4} d x^{4} + 3 \, b c^{2} d x^{2} - b d\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} d x^{2} + d}}{20 \, {\left (c^{2} x^{7} - x^{5}\right )}}, \frac {4 \, {\left (b c^{7} d x^{7} - b c^{5} d x^{5}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} {\left (x^{2} - 1\right )} \sqrt {-d}}{c^{2} d x^{4} + {\left (c^{2} - 1\right )} d x^{2} - d}\right ) - {\left (4 \, b c^{3} d x^{3} - {\left (4 \, b c^{3} - b c\right )} d x^{5} - b c d x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} - 4 \, {\left (a c^{6} d x^{6} - 3 \, a c^{4} d x^{4} + 3 \, a c^{2} d x^{2} - a d + {\left (b c^{6} d x^{6} - 3 \, b c^{4} d x^{4} + 3 \, b c^{2} d x^{2} - b d\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} d x^{2} + d}}{20 \, {\left (c^{2} x^{7} - x^{5}\right )}}\right ] \] Input:

integrate((-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))/x^6,x, algorithm="fricas" 
)
 

Output:

[1/20*(2*(b*c^7*d*x^7 - b*c^5*d*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)) - (4*b*c^3*d*x^3 - (4*b*c^3 - b*c)*d*x^5 - b*c*d*x)*sqrt(-c^ 
2*d*x^2 + d)*sqrt(-c^2*x^2 + 1) - 4*(a*c^6*d*x^6 - 3*a*c^4*d*x^4 + 3*a*c^2 
*d*x^2 - a*d + (b*c^6*d*x^6 - 3*b*c^4*d*x^4 + 3*b*c^2*d*x^2 - b*d)*arcsin( 
c*x))*sqrt(-c^2*d*x^2 + d))/(c^2*x^7 - x^5), 1/20*(4*(b*c^7*d*x^7 - b*c^5* 
d*x^5)*sqrt(-d)*arctan(sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1)*(x^2 - 1)*s 
qrt(-d)/(c^2*d*x^4 + (c^2 - 1)*d*x^2 - d)) - (4*b*c^3*d*x^3 - (4*b*c^3 - b 
*c)*d*x^5 - b*c*d*x)*sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1) - 4*(a*c^6*d* 
x^6 - 3*a*c^4*d*x^4 + 3*a*c^2*d*x^2 - a*d + (b*c^6*d*x^6 - 3*b*c^4*d*x^4 + 
 3*b*c^2*d*x^2 - b*d)*arcsin(c*x))*sqrt(-c^2*d*x^2 + d))/(c^2*x^7 - x^5)]
 

Sympy [F]

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

integrate((-c**2*d*x**2+d)**(3/2)*(a+b*asin(c*x))/x**6,x)
 

Output:

Integral((-d*(c*x - 1)*(c*x + 1))**(3/2)*(a + b*asin(c*x))/x**6, x)
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.12 \[ \int \frac {\left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{x^6} \, dx=-\frac {{\left (2 \, \left (-1\right )^{-2 \, c^{2} d x^{2} + 2 \, d} c^{4} d^{\frac {5}{2}} \log \left (-2 \, c^{2} d + \frac {2 \, d}{x^{2}}\right ) + 2 \, c^{4} d^{\frac {5}{2}} \log \left (x^{2} - \frac {1}{c^{2}}\right ) - \frac {3 \, \sqrt {c^{4} d x^{4} - 2 \, c^{2} d x^{2} + d} c^{2} d^{2}}{x^{2}} + \frac {\sqrt {c^{4} d x^{4} - 2 \, c^{2} d x^{2} + d} d^{2}}{x^{4}}\right )} b c}{20 \, d} - \frac {{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} b \arcsin \left (c x\right )}{5 \, d x^{5}} - \frac {{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} a}{5 \, d x^{5}} \] Input:

integrate((-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))/x^6,x, algorithm="maxima" 
)
 

Output:

-1/20*(2*(-1)^(-2*c^2*d*x^2 + 2*d)*c^4*d^(5/2)*log(-2*c^2*d + 2*d/x^2) + 2 
*c^4*d^(5/2)*log(x^2 - 1/c^2) - 3*sqrt(c^4*d*x^4 - 2*c^2*d*x^2 + d)*c^2*d^ 
2/x^2 + sqrt(c^4*d*x^4 - 2*c^2*d*x^2 + d)*d^2/x^4)*b*c/d - 1/5*(-c^2*d*x^2 
 + d)^(5/2)*b*arcsin(c*x)/(d*x^5) - 1/5*(-c^2*d*x^2 + d)^(5/2)*a/(d*x^5)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{x^6} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((-c^2*d*x^2+d)^(3/2)*(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. \[ \int \frac {\left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{x^6} \, dx=\int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d-c^2\,d\,x^2\right )}^{3/2}}{x^6} \,d x \] Input:

int(((a + b*asin(c*x))*(d - c^2*d*x^2)^(3/2))/x^6,x)
 

Output:

int(((a + b*asin(c*x))*(d - c^2*d*x^2)^(3/2))/x^6, x)
 

Reduce [F]

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

int((-c^2*d*x^2+d)^(3/2)*(a+b*asin(c*x))/x^6,x)
                                                                                    
                                                                                    
 

Output:

(sqrt(d)*d*( - sqrt( - c**2*x**2 + 1)*a*c**4*x**4 + 2*sqrt( - c**2*x**2 + 
1)*a*c**2*x**2 - sqrt( - c**2*x**2 + 1)*a + 5*int((sqrt( - c**2*x**2 + 1)* 
asin(c*x))/x**6,x)*b*x**5 - 5*int((sqrt( - c**2*x**2 + 1)*asin(c*x))/x**4, 
x)*b*c**2*x**5))/(5*x**5)