\(\int \frac {(d-c^2 d x^2)^{3/2} (a+b \arcsin (c x))}{x^2} \, dx\) [71]

Optimal result

Integrand size = 27, antiderivative size = 185 \[ \int \frac {\left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{x^2} \, dx=\frac {b c^3 d x^2 \sqrt {d-c^2 d x^2}}{4 \sqrt {1-c^2 x^2}}-\frac {3}{2} c^2 d x \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))-\frac {\left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{x}-\frac {3 c d \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))^2}{4 b \sqrt {1-c^2 x^2}}+\frac {b c d \sqrt {d-c^2 d x^2} \log (x)}{\sqrt {1-c^2 x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.20 \[ \int \frac {\left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{x^2} \, dx=\left (-\frac {a d}{x}-\frac {1}{2} a c^2 d x\right ) \sqrt {-d \left (-1+c^2 x^2\right )}+\frac {3}{2} a c d^{3/2} \arctan \left (\frac {c x \sqrt {-d \left (-1+c^2 x^2\right )}}{\sqrt {d} \left (-1+c^2 x^2\right )}\right )-\frac {b c d \sqrt {d \left (1-c^2 x^2\right )} \left (\frac {2 \sqrt {1-c^2 x^2} \arcsin (c x)}{c x}+\arcsin (c x)^2-2 \log (c x)\right )}{2 \sqrt {1-c^2 x^2}}-\frac {b c d \sqrt {d \left (1-c^2 x^2\right )} (\cos (2 \arcsin (c x))+2 \arcsin (c x) (\arcsin (c x)+\sin (2 \arcsin (c x))))}{8 \sqrt {1-c^2 x^2}} \] Input:

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

Output:

(-((a*d)/x) - (a*c^2*d*x)/2)*Sqrt[-(d*(-1 + c^2*x^2))] + (3*a*c*d^(3/2)*Ar 
cTan[(c*x*Sqrt[-(d*(-1 + c^2*x^2))])/(Sqrt[d]*(-1 + c^2*x^2))])/2 - (b*c*d 
*Sqrt[d*(1 - c^2*x^2)]*((2*Sqrt[1 - c^2*x^2]*ArcSin[c*x])/(c*x) + ArcSin[c 
*x]^2 - 2*Log[c*x]))/(2*Sqrt[1 - c^2*x^2]) - (b*c*d*Sqrt[d*(1 - c^2*x^2)]* 
(Cos[2*ArcSin[c*x]] + 2*ArcSin[c*x]*(ArcSin[c*x] + Sin[2*ArcSin[c*x]])))/( 
8*Sqrt[1 - c^2*x^2])
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {5200, 244, 2009, 5156, 15, 5152}

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.

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

\(\Big \downarrow \) 5200

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

\(\Big \downarrow \) 244

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 5156

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

\(\Big \downarrow \) 15

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

\(\Big \downarrow \) 5152

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

Input:

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

Output:

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

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 5152

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_S 
ymbol] :> Simp[(1/(b*c*(n + 1)))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]]*(a 
 + b*ArcSin[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d 
+ e, 0] && NeQ[n, -1]
 

rule 5156

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*Sqrt[(d_) + (e_.)*(x_)^2], x_S 
ymbol] :> Simp[x*Sqrt[d + e*x^2]*((a + b*ArcSin[c*x])^n/2), x] + (Simp[(1/2 
)*Simp[Sqrt[d + e*x^2]/Sqrt[1 - c^2*x^2]]   Int[(a + b*ArcSin[c*x])^n/Sqrt[ 
1 - c^2*x^2], x], x] - Simp[b*c*(n/2)*Simp[Sqrt[d + e*x^2]/Sqrt[1 - c^2*x^2 
]]   Int[x*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x 
] && EqQ[c^2*d + e, 0] && GtQ[n, 0]
 

rule 5200

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*((a + b*ArcS 
in[c*x])^n/(f*(m + 1))), x] + (-Simp[2*e*(p/(f^2*(m + 1)))   Int[(f*x)^(m + 
 2)*(d + e*x^2)^(p - 1)*(a + b*ArcSin[c*x])^n, x], 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*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f} 
, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.47 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.34

method	result	size
default	\(-\frac {a \left (-c^{2} d \,x^{2}+d \right )^{\frac {5}{2}}}{d x}-a \,c^{2} x \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}-\frac {3 a \,c^{2} d x \sqrt {-c^{2} d \,x^{2}+d}}{2}-\frac {3 a \,c^{2} d^{2} \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{2 \sqrt {c^{2} d}}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \left (4 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, c^{2} x^{2}-2 c^{3} x^{3}+6 \arcsin \left (c x \right )^{2} c x +8 i \arcsin \left (c x \right ) x c -8 \ln \left (\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}-1\right ) x c +8 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}+c x \right ) d}{8 \left (c^{2} x^{2}-1\right ) x}\)	\(248\)
parts	\(-\frac {a \left (-c^{2} d \,x^{2}+d \right )^{\frac {5}{2}}}{d x}-a \,c^{2} x \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}-\frac {3 a \,c^{2} d x \sqrt {-c^{2} d \,x^{2}+d}}{2}-\frac {3 a \,c^{2} d^{2} \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{2 \sqrt {c^{2} d}}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \left (4 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, c^{2} x^{2}-2 c^{3} x^{3}+6 \arcsin \left (c x \right )^{2} c x +8 i \arcsin \left (c x \right ) x c -8 \ln \left (\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}-1\right ) x c +8 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}+c x \right ) d}{8 \left (c^{2} x^{2}-1\right ) x}\)	\(248\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(-(a*c^2*d*x^2 - a*d + (b*c^2*d*x^2 - b*d)*arcsin(c*x))*sqrt(-c^2* 
d*x^2 + d)/x^2, x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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