\(\int \frac {a+b \arcsin (c x)}{x \sqrt {d-c^2 d x^2}} \, dx\) [113]

Optimal result

Integrand size = 27, antiderivative size = 145 \[ \int \frac {a+b \arcsin (c x)}{x \sqrt {d-c^2 d x^2}} \, dx=-\frac {2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x)) \text {arctanh}\left (e^{i \arcsin (c x)}\right )}{\sqrt {d-c^2 d x^2}}+\frac {i b \sqrt {1-c^2 x^2} \operatorname {PolyLog}\left (2,-e^{i \arcsin (c x)}\right )}{\sqrt {d-c^2 d x^2}}-\frac {i b \sqrt {1-c^2 x^2} \operatorname {PolyLog}\left (2,e^{i \arcsin (c x)}\right )}{\sqrt {d-c^2 d x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.01 \[ \int \frac {a+b \arcsin (c x)}{x \sqrt {d-c^2 d x^2}} \, dx=\frac {a \log (x)}{\sqrt {d}}-\frac {a \log \left (d+\sqrt {d} \sqrt {-d \left (-1+c^2 x^2\right )}\right )}{\sqrt {d}}+\frac {b \sqrt {1-c^2 x^2} \left (\arcsin (c x) \left (\log \left (1-e^{i \arcsin (c x)}\right )-\log \left (1+e^{i \arcsin (c x)}\right )\right )+i \operatorname {PolyLog}\left (2,-e^{i \arcsin (c x)}\right )-i \operatorname {PolyLog}\left (2,e^{i \arcsin (c x)}\right )\right )}{\sqrt {d \left (1-c^2 x^2\right )}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.61, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {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.

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

\(\Big \downarrow \) 5218

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4671

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

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

Output:

(Sqrt[1 - c^2*x^2]*(-2*(a + b*ArcSin[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])]))/Sqr 
t[d - c^2*d*x^2]
 

Defintions of rubi rules used

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 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 = 0.42 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.24

method	result	size
default	\(-\frac {a \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {-c^{2} d \,x^{2}+d}}{x}\right )}{\sqrt {d}}-\frac {i b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )-i \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )+\operatorname {polylog}\left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )-\operatorname {polylog}\left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d \left (c^{2} x^{2}-1\right )}\)	\(180\)
parts	\(-\frac {a \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {-c^{2} d \,x^{2}+d}}{x}\right )}{\sqrt {d}}-\frac {i b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )-i \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )+\operatorname {polylog}\left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )-\operatorname {polylog}\left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d \left (c^{2} x^{2}-1\right )}\)	\(180\)

Input:

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

Output:

-a/d^(1/2)*ln((2*d+2*d^(1/2)*(-c^2*d*x^2+d)^(1/2))/x)-I*b*(-c^2*x^2+1)^(1/ 
2)*(-d*(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)))/d/(c^2*x^2-1)
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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