\(\int \frac {\sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{x^3} \, dx\) [66]

Optimal result

Integrand size = 27, antiderivative size = 225 \[ \int \frac {\sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{x^3} \, dx=-\frac {b c \sqrt {d-c^2 d x^2}}{2 x \sqrt {1-c^2 x^2}}-\frac {\sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{2 x^2}+\frac {c^2 \sqrt {d-c^2 d x^2} (a+b \arcsin (c x)) \text {arctanh}\left (e^{i \arcsin (c x)}\right )}{\sqrt {1-c^2 x^2}}-\frac {i b c^2 \sqrt {d-c^2 d x^2} \operatorname {PolyLog}\left (2,-e^{i \arcsin (c x)}\right )}{2 \sqrt {1-c^2 x^2}}+\frac {i b c^2 \sqrt {d-c^2 d x^2} \operatorname {PolyLog}\left (2,e^{i \arcsin (c x)}\right )}{2 \sqrt {1-c^2 x^2}} \] Output:

-1/2*b*c*(-c^2*d*x^2+d)^(1/2)/x/(-c^2*x^2+1)^(1/2)-1/2*(-c^2*d*x^2+d)^(1/2 
)*(a+b*arcsin(c*x))/x^2+c^2*(-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))*arctanh 
(I*c*x+(-c^2*x^2+1)^(1/2))/(-c^2*x^2+1)^(1/2)-1/2*I*b*c^2*(-c^2*d*x^2+d)^( 
1/2)*polylog(2,-I*c*x-(-c^2*x^2+1)^(1/2))/(-c^2*x^2+1)^(1/2)+1/2*I*b*c^2*( 
-c^2*d*x^2+d)^(1/2)*polylog(2,I*c*x+(-c^2*x^2+1)^(1/2))/(-c^2*x^2+1)^(1/2)
 

Mathematica [A] (verified)

Time = 1.50 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.06 \[ \int \frac {\sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{x^3} \, dx=\frac {1}{8} \left (-\frac {4 a \sqrt {d-c^2 d x^2}}{x^2}-4 a c^2 \sqrt {d} \log (x)+4 a c^2 \sqrt {d} \log \left (d+\sqrt {d} \sqrt {d-c^2 d x^2}\right )+\frac {b c^2 d \sqrt {1-c^2 x^2} \left (-2 \cot \left (\frac {1}{2} \arcsin (c x)\right )-\arcsin (c x) \csc ^2\left (\frac {1}{2} \arcsin (c x)\right )-4 \arcsin (c x) \log \left (1-e^{i \arcsin (c x)}\right )+4 \arcsin (c x) \log \left (1+e^{i \arcsin (c x)}\right )-4 i \operatorname {PolyLog}\left (2,-e^{i \arcsin (c x)}\right )+4 i \operatorname {PolyLog}\left (2,e^{i \arcsin (c x)}\right )+\arcsin (c x) \sec ^2\left (\frac {1}{2} \arcsin (c x)\right )-2 \tan \left (\frac {1}{2} \arcsin (c x)\right )\right )}{\sqrt {d-c^2 d x^2}}\right ) \] Input:

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

Output:

((-4*a*Sqrt[d - c^2*d*x^2])/x^2 - 4*a*c^2*Sqrt[d]*Log[x] + 4*a*c^2*Sqrt[d] 
*Log[d + Sqrt[d]*Sqrt[d - c^2*d*x^2]] + (b*c^2*d*Sqrt[1 - c^2*x^2]*(-2*Cot 
[ArcSin[c*x]/2] - ArcSin[c*x]*Csc[ArcSin[c*x]/2]^2 - 4*ArcSin[c*x]*Log[1 - 
 E^(I*ArcSin[c*x])] + 4*ArcSin[c*x]*Log[1 + E^(I*ArcSin[c*x])] - (4*I)*Pol 
yLog[2, -E^(I*ArcSin[c*x])] + (4*I)*PolyLog[2, E^(I*ArcSin[c*x])] + ArcSin 
[c*x]*Sec[ArcSin[c*x]/2]^2 - 2*Tan[ArcSin[c*x]/2]))/Sqrt[d - c^2*d*x^2])/8
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.72, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {5196, 15, 5218, 3042, 4671, 2715, 2838}

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.

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

\(\Big \downarrow \) 5196

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

\(\Big \downarrow \) 15

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

\(\Big \downarrow \) 5218

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

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {c^2 \sqrt {d-c^2 d x^2} \int (a+b \arcsin (c x)) \csc (\arcsin (c x))d\arcsin (c x)}{2 \sqrt {1-c^2 x^2}}-\frac {\sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{2 x^2}-\frac {b c \sqrt {d-c^2 d x^2}}{2 x \sqrt {1-c^2 x^2}}\)

\(\Big \downarrow \) 4671

\(\displaystyle -\frac {c^2 \sqrt {d-c^2 d x^2} \left (-b \int \log \left (1-e^{i \arcsin (c x)}\right )d\arcsin (c x)+b \int \log \left (1+e^{i \arcsin (c x)}\right )d\arcsin (c x)-2 \text {arctanh}\left (e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))\right )}{2 \sqrt {1-c^2 x^2}}-\frac {\sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{2 x^2}-\frac {b c \sqrt {d-c^2 d x^2}}{2 x \sqrt {1-c^2 x^2}}\)

\(\Big \downarrow \) 2715

\(\displaystyle -\frac {c^2 \sqrt {d-c^2 d x^2} \left (i b \int e^{-i \arcsin (c x)} \log \left (1-e^{i \arcsin (c x)}\right )de^{i \arcsin (c x)}-i b \int e^{-i \arcsin (c x)} \log \left (1+e^{i \arcsin (c x)}\right )de^{i \arcsin (c x)}-2 \text {arctanh}\left (e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))\right )}{2 \sqrt {1-c^2 x^2}}-\frac {\sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{2 x^2}-\frac {b c \sqrt {d-c^2 d x^2}}{2 x \sqrt {1-c^2 x^2}}\)

\(\Big \downarrow \) 2838

\(\displaystyle -\frac {c^2 \sqrt {d-c^2 d x^2} \left (-2 \text {arctanh}\left (e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))+i b \operatorname {PolyLog}\left (2,-e^{i \arcsin (c x)}\right )-i b \operatorname {PolyLog}\left (2,e^{i \arcsin (c x)}\right )\right )}{2 \sqrt {1-c^2 x^2}}-\frac {\sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{2 x^2}-\frac {b c \sqrt {d-c^2 d x^2}}{2 x \sqrt {1-c^2 x^2}}\)

Input:

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

Output:

-1/2*(b*c*Sqrt[d - c^2*d*x^2])/(x*Sqrt[1 - c^2*x^2]) - (Sqrt[d - c^2*d*x^2 
]*(a + b*ArcSin[c*x]))/(2*x^2) - (c^2*Sqrt[d - c^2*d*x^2]*(-2*(a + b*ArcSi 
n[c*x])*ArcTanh[E^(I*ArcSin[c*x])] + I*b*PolyLog[2, -E^(I*ArcSin[c*x])] - 
I*b*PolyLog[2, E^(I*ArcSin[c*x])]))/(2*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 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
 

rule 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4671

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[- 
2*(c + d*x)^m*(ArcTanh[E^(I*(e + f*x))]/f), x] + (-Simp[d*(m/f)   Int[(c + 
d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Simp[d*(m/f)   Int[(c + d*x 
)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IG 
tQ[m, 0]
 

rule 5196

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

rule 5218

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

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.28

method	result	size
default	\(a \left (-\frac {\left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{2 d \,x^{2}}-\frac {c^{2} \left (\sqrt {-c^{2} d \,x^{2}+d}-\sqrt {d}\, \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {-c^{2} d \,x^{2}+d}}{x}\right )\right )}{2}\right )+b \left (-\frac {\left (c^{2} x^{2} \arcsin \left (c x \right )-c x \sqrt {-c^{2} x^{2}+1}-\arcsin \left (c x \right )\right ) \sqrt {-d \left (c^{2} x^{2}-1\right )}}{2 \left (c^{2} x^{2}-1\right ) x^{2}}+\frac {i \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \left (i \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )-i \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )+\operatorname {polylog}\left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )-\operatorname {polylog}\left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )\right ) c^{2}}{2 c^{2} x^{2}-2}\right )\)	\(289\)
parts	\(a \left (-\frac {\left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{2 d \,x^{2}}-\frac {c^{2} \left (\sqrt {-c^{2} d \,x^{2}+d}-\sqrt {d}\, \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {-c^{2} d \,x^{2}+d}}{x}\right )\right )}{2}\right )+b \left (-\frac {\left (c^{2} x^{2} \arcsin \left (c x \right )-c x \sqrt {-c^{2} x^{2}+1}-\arcsin \left (c x \right )\right ) \sqrt {-d \left (c^{2} x^{2}-1\right )}}{2 \left (c^{2} x^{2}-1\right ) x^{2}}+\frac {i \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \left (i \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )-i \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )+\operatorname {polylog}\left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )-\operatorname {polylog}\left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )\right ) c^{2}}{2 c^{2} x^{2}-2}\right )\)	\(289\)

Input:

int((-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))/x^3,x,method=_RETURNVERBOSE)
 

Output:

a*(-1/2/d/x^2*(-c^2*d*x^2+d)^(3/2)-1/2*c^2*((-c^2*d*x^2+d)^(1/2)-d^(1/2)*l 
n((2*d+2*d^(1/2)*(-c^2*d*x^2+d)^(1/2))/x)))+b*(-1/2*(c^2*x^2*arcsin(c*x)-c 
*x*(-c^2*x^2+1)^(1/2)-arcsin(c*x))*(-d*(c^2*x^2-1))^(1/2)/(c^2*x^2-1)/x^2+ 
I*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)*(I*arcsin(c*x)*ln(1+I*c*x+(-c^ 
2*x^2+1)^(1/2))-I*arcsin(c*x)*ln(1-I*c*x-(-c^2*x^2+1)^(1/2))+polylog(2,-I* 
c*x-(-c^2*x^2+1)^(1/2))-polylog(2,I*c*x+(-c^2*x^2+1)^(1/2)))*c^2/(2*c^2*x^ 
2-2))
 

Fricas [F]

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

integrate((-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))/x^3,x, algorithm="fricas" 
)
 

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate((-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))/x^3,x, algorithm="maxima" 
)
                                                                                    
                                                                                    
 

Output:

b*sqrt(d)*integrate(sqrt(c*x + 1)*sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1 
)*sqrt(-c*x + 1))/x^3, x) + 1/2*(c^2*sqrt(d)*log(2*sqrt(-c^2*d*x^2 + d)*sq 
rt(d)/abs(x) + 2*d/abs(x)) - sqrt(-c^2*d*x^2 + d)*c^2 - (-c^2*d*x^2 + d)^( 
3/2)/(d*x^2))*a
 

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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