3.2.52 \(\int \frac {(a+b \arcsin (c x))^2}{x^2} \, dx\) [152]

3.2.52.1 Optimal result
3.2.52.2 Mathematica [A] (verified)
3.2.52.3 Rubi [A] (verified)
3.2.52.4 Maple [A] (verified)
3.2.52.5 Fricas [F]
3.2.52.6 Sympy [F]
3.2.52.7 Maxima [F]
3.2.52.8 Giac [F]
3.2.52.9 Mupad [F(-1)]

3.2.52.1 Optimal result

Integrand size = 14, antiderivative size = 81 \[ \int \frac {(a+b \arcsin (c x))^2}{x^2} \, dx=-\frac {(a+b \arcsin (c x))^2}{x}-4 b c (a+b \arcsin (c x)) \text {arctanh}\left (e^{i \arcsin (c x)}\right )+2 i b^2 c \operatorname {PolyLog}\left (2,-e^{i \arcsin (c x)}\right )-2 i b^2 c \operatorname {PolyLog}\left (2,e^{i \arcsin (c x)}\right ) \]

output
-(a+b*arcsin(c*x))^2/x-4*b*c*(a+b*arcsin(c*x))*arctanh(I*c*x+(-c^2*x^2+1)^ 
(1/2))+2*I*b^2*c*polylog(2,-I*c*x-(-c^2*x^2+1)^(1/2))-2*I*b^2*c*polylog(2, 
I*c*x+(-c^2*x^2+1)^(1/2))
 
3.2.52.2 Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.56 \[ \int \frac {(a+b \arcsin (c x))^2}{x^2} \, dx=-\frac {a^2+2 a b \left (\arcsin (c x)+c x \text {arctanh}\left (\sqrt {1-c^2 x^2}\right )\right )-i b^2 \left (i \arcsin (c x) \left (\arcsin (c x)+2 c x \left (-\log \left (1-e^{i \arcsin (c x)}\right )+\log \left (1+e^{i \arcsin (c x)}\right )\right )\right )+2 c x \operatorname {PolyLog}\left (2,-e^{i \arcsin (c x)}\right )-2 c x \operatorname {PolyLog}\left (2,e^{i \arcsin (c x)}\right )\right )}{x} \]

input
Integrate[(a + b*ArcSin[c*x])^2/x^2,x]
 
output
-((a^2 + 2*a*b*(ArcSin[c*x] + c*x*ArcTanh[Sqrt[1 - c^2*x^2]]) - I*b^2*(I*A 
rcSin[c*x]*(ArcSin[c*x] + 2*c*x*(-Log[1 - E^(I*ArcSin[c*x])] + Log[1 + E^( 
I*ArcSin[c*x])])) + 2*c*x*PolyLog[2, -E^(I*ArcSin[c*x])] - 2*c*x*PolyLog[2 
, E^(I*ArcSin[c*x])]))/x)
 
3.2.52.3 Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5138, 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.

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

\(\Big \downarrow \) 5138

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

\(\Big \downarrow \) 5218

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4671

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

input
Int[(a + b*ArcSin[c*x])^2/x^2,x]
 
output
-((a + b*ArcSin[c*x])^2/x) + 2*b*c*(-2*(a + b*ArcSin[c*x])*ArcTanh[E^(I*Ar 
cSin[c*x])] + I*b*PolyLog[2, -E^(I*ArcSin[c*x])] - I*b*PolyLog[2, E^(I*Arc 
Sin[c*x])])
 

3.2.52.3.1 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 5138
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] 
:> Simp[(d*x)^(m + 1)*((a + b*ArcSin[c*x])^n/(d*(m + 1))), x] - Simp[b*c*(n 
/(d*(m + 1)))   Int[(d*x)^(m + 1)*((a + b*ArcSin[c*x])^(n - 1)/Sqrt[1 - c^2 
*x^2]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[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]
 
3.2.52.4 Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 165, normalized size of antiderivative = 2.04

method result size
parts \(-\frac {a^{2}}{x}+b^{2} c \left (-\frac {\arcsin \left (c x \right )^{2}}{c x}+2 \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )-2 \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )+2 i \operatorname {dilog}\left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )-2 i \operatorname {dilog}\left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )\right )+2 a b c \left (-\frac {\arcsin \left (c x \right )}{c x}-\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )\right )\) \(165\)
derivativedivides \(c \left (-\frac {a^{2}}{c x}+b^{2} \left (-\frac {\arcsin \left (c x \right )^{2}}{c x}+2 \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )-2 \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )+2 i \operatorname {dilog}\left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )-2 i \operatorname {dilog}\left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )\right )+2 a b \left (-\frac {\arcsin \left (c x \right )}{c x}-\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )\right )\right )\) \(168\)
default \(c \left (-\frac {a^{2}}{c x}+b^{2} \left (-\frac {\arcsin \left (c x \right )^{2}}{c x}+2 \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )-2 \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )+2 i \operatorname {dilog}\left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )-2 i \operatorname {dilog}\left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )\right )+2 a b \left (-\frac {\arcsin \left (c x \right )}{c x}-\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )\right )\right )\) \(168\)

input
int((a+b*arcsin(c*x))^2/x^2,x,method=_RETURNVERBOSE)
 
output
-a^2/x+b^2*c*(-1/c/x*arcsin(c*x)^2+2*arcsin(c*x)*ln(1-I*c*x-(-c^2*x^2+1)^( 
1/2))-2*arcsin(c*x)*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))+2*I*dilog(1+I*c*x+(-c^2 
*x^2+1)^(1/2))-2*I*dilog(1-I*c*x-(-c^2*x^2+1)^(1/2)))+2*a*b*c*(-1/c/x*arcs 
in(c*x)-arctanh(1/(-c^2*x^2+1)^(1/2)))
 
3.2.52.5 Fricas [F]

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

input
integrate((a+b*arcsin(c*x))^2/x^2,x, algorithm="fricas")
 
output
integral((b^2*arcsin(c*x)^2 + 2*a*b*arcsin(c*x) + a^2)/x^2, x)
 
3.2.52.6 Sympy [F]

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

input
integrate((a+b*asin(c*x))**2/x**2,x)
 
output
Integral((a + b*asin(c*x))**2/x**2, x)
 
3.2.52.7 Maxima [F]

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

input
integrate((a+b*arcsin(c*x))^2/x^2,x, algorithm="maxima")
 
output
-2*(c*log(2*sqrt(-c^2*x^2 + 1)/abs(x) + 2/abs(x)) + arcsin(c*x)/x)*a*b - ( 
2*c*x*integrate(sqrt(c*x + 1)*sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1)*sq 
rt(-c*x + 1))/(c^2*x^3 - x), x) + arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1 
))^2)*b^2/x - a^2/x
 
3.2.52.8 Giac [F]

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

input
integrate((a+b*arcsin(c*x))^2/x^2,x, algorithm="giac")
 
output
integrate((b*arcsin(c*x) + a)^2/x^2, x)
 
3.2.52.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \arcsin (c x))^2}{x^2} \, dx=\int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2}{x^2} \,d x \]

input
int((a + b*asin(c*x))^2/x^2,x)
 
output
int((a + b*asin(c*x))^2/x^2, x)