\(\int \frac {\sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{x} \, dx\) [65]

Optimal result

Integrand size = 27, antiderivative size = 203 \[ \int \frac {\sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{x} \, dx=-\frac {b c x \sqrt {d-c^2 d x^2}}{\sqrt {1-c^2 x^2}}+\sqrt {d-c^2 d x^2} (a+b \arcsin (c x))-\frac {2 \sqrt {d-c^2 d x^2} (a+b \arcsin (c x)) \text {arctanh}\left (e^{i \arcsin (c x)}\right )}{\sqrt {1-c^2 x^2}}+\frac {i b \sqrt {d-c^2 d x^2} \operatorname {PolyLog}\left (2,-e^{i \arcsin (c x)}\right )}{\sqrt {1-c^2 x^2}}-\frac {i b \sqrt {d-c^2 d x^2} \operatorname {PolyLog}\left (2,e^{i \arcsin (c x)}\right )}{\sqrt {1-c^2 x^2}} \] Output:

-b*c*x*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)+(-c^2*d*x^2+d)^(1/2)*(a+b*a 
rcsin(c*x))-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)+I*b*(-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)-I*b*(-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 = 0.37 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.92 \[ \int \frac {\sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{x} \, dx=a \sqrt {d-c^2 d x^2}+a \sqrt {d} \log (x)-a \sqrt {d} \log \left (d+\sqrt {d} \sqrt {d-c^2 d x^2}\right )+\frac {b \sqrt {d-c^2 d x^2} \left (-c x+\sqrt {1-c^2 x^2} \arcsin (c x)+\arcsin (c x) \log \left (1-e^{i \arcsin (c x)}\right )-\arcsin (c x) \log \left (1+e^{i \arcsin (c x)}\right )+i \operatorname {PolyLog}\left (2,-e^{i \arcsin (c x)}\right )-i \operatorname {PolyLog}\left (2,e^{i \arcsin (c x)}\right )\right )}{\sqrt {1-c^2 x^2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 147, 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 = {5198, 24, 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.

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

\(\Big \downarrow \) 5198

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

\(\Big \downarrow \) 24

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

\(\Big \downarrow \) 5218

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4671

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

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 5198

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 + 2))), x] + (Simp[(1/(m + 2))*Simp[Sqrt[d + e*x^2]/Sqrt[1 
 - c^2*x^2]]   Int[(f*x)^m*((a + b*ArcSin[c*x])^n/Sqrt[1 - c^2*x^2]), x], x 
] - Simp[b*c*(n/(f*(m + 2)))*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]) /; FreeQ[{a, b, c, d, e, 
 f, m}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && (IGtQ[m, -2] || EqQ[n, 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 = 0.43 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.55

method	result	size
default	\(-\sqrt {d}\, \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {-c^{2} d \,x^{2}+d}}{x}\right ) a +\sqrt {-c^{2} d \,x^{2}+d}\, a +b \left (\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (c^{2} x^{2}-i \sqrt {-c^{2} x^{2}+1}\, c x -1\right ) \left (\arcsin \left (c x \right )+i\right )}{2 c^{2} x^{2}-2}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i \sqrt {-c^{2} x^{2}+1}\, c x +c^{2} x^{2}-1\right ) \left (\arcsin \left (c x \right )-i\right )}{2 c^{2} x^{2}-2}-\frac {i \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \left (i \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )-i \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )+\operatorname {polylog}\left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )-\operatorname {polylog}\left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c^{2} x^{2}-1}\right )\)	\(314\)
parts	\(-\sqrt {d}\, \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {-c^{2} d \,x^{2}+d}}{x}\right ) a +\sqrt {-c^{2} d \,x^{2}+d}\, a +b \left (\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (c^{2} x^{2}-i \sqrt {-c^{2} x^{2}+1}\, c x -1\right ) \left (\arcsin \left (c x \right )+i\right )}{2 c^{2} x^{2}-2}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i \sqrt {-c^{2} x^{2}+1}\, c x +c^{2} x^{2}-1\right ) \left (\arcsin \left (c x \right )-i\right )}{2 c^{2} x^{2}-2}-\frac {i \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \left (i \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )-i \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )+\operatorname {polylog}\left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )-\operatorname {polylog}\left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c^{2} x^{2}-1}\right )\)	\(314\)

Input:

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

Output:

-d^(1/2)*ln((2*d+2*d^(1/2)*(-c^2*d*x^2+d)^(1/2))/x)*a+(-c^2*d*x^2+d)^(1/2) 
*a+b*(1/2*(-d*(c^2*x^2-1))^(1/2)*(c^2*x^2-I*c*x*(-c^2*x^2+1)^(1/2)-1)*(arc 
sin(c*x)+I)/(c^2*x^2-1)+1/2*(-d*(c^2*x^2-1))^(1/2)*(I*(-c^2*x^2+1)^(1/2)*c 
*x+c^2*x^2-1)*(arcsin(c*x)-I)/(c^2*x^2-1)-I*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x 
^2+1)^(1/2)/(c^2*x^2-1)*(I*arcsin(c*x)*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))-I*ar 
csin(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))))
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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