\(\int \frac {(a+b \arcsin (c x))^2}{x^4} \, dx\) [19]

Optimal result

Integrand size = 14, antiderivative size = 139 \[ \int \frac {(a+b \arcsin (c x))^2}{x^4} \, dx=-\frac {b^2 c^2}{3 x}-\frac {b c \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{3 x^2}-\frac {(a+b \arcsin (c x))^2}{3 x^3}-\frac {2}{3} b c^3 (a+b \arcsin (c x)) \text {arctanh}\left (e^{i \arcsin (c x)}\right )+\frac {1}{3} i b^2 c^3 \operatorname {PolyLog}\left (2,-e^{i \arcsin (c x)}\right )-\frac {1}{3} i b^2 c^3 \operatorname {PolyLog}\left (2,e^{i \arcsin (c x)}\right ) \] Output:

-1/3*b^2*c^2/x-1/3*b*c*(-c^2*x^2+1)^(1/2)*(a+b*arcsin(c*x))/x^2-1/3*(a+b*a 
rcsin(c*x))^2/x^3-2/3*b*c^3*(a+b*arcsin(c*x))*arctanh(I*c*x+(-c^2*x^2+1)^( 
1/2))+1/3*I*b^2*c^3*polylog(2,-I*c*x-(-c^2*x^2+1)^(1/2))-1/3*I*b^2*c^3*pol 
ylog(2,I*c*x+(-c^2*x^2+1)^(1/2))
 

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.55 \[ \int \frac {(a+b \arcsin (c x))^2}{x^4} \, dx=-\frac {a^2+b^2 c^2 x^2+a b c x \sqrt {1-c^2 x^2}+2 a b \arcsin (c x)+b^2 c x \sqrt {1-c^2 x^2} \arcsin (c x)+b^2 \arcsin (c x)^2+a b c^3 x^3 \text {arctanh}\left (\sqrt {1-c^2 x^2}\right )-b^2 c^3 x^3 \arcsin (c x) \log \left (1-e^{i \arcsin (c x)}\right )+b^2 c^3 x^3 \arcsin (c x) \log \left (1+e^{i \arcsin (c x)}\right )-i b^2 c^3 x^3 \operatorname {PolyLog}\left (2,-e^{i \arcsin (c x)}\right )+i b^2 c^3 x^3 \operatorname {PolyLog}\left (2,e^{i \arcsin (c x)}\right )}{3 x^3} \] Input:

Integrate[(a + b*ArcSin[c*x])^2/x^4,x]
 

Output:

-1/3*(a^2 + b^2*c^2*x^2 + a*b*c*x*Sqrt[1 - c^2*x^2] + 2*a*b*ArcSin[c*x] + 
b^2*c*x*Sqrt[1 - c^2*x^2]*ArcSin[c*x] + b^2*ArcSin[c*x]^2 + a*b*c^3*x^3*Ar 
cTanh[Sqrt[1 - c^2*x^2]] - b^2*c^3*x^3*ArcSin[c*x]*Log[1 - E^(I*ArcSin[c*x 
])] + b^2*c^3*x^3*ArcSin[c*x]*Log[1 + E^(I*ArcSin[c*x])] - I*b^2*c^3*x^3*P 
olyLog[2, -E^(I*ArcSin[c*x])] + I*b^2*c^3*x^3*PolyLog[2, E^(I*ArcSin[c*x]) 
])/x^3
 

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.92, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {5138, 5204, 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.

\(\displaystyle \int \frac {(a+b \arcsin (c x))^2}{x^4} \, dx\)

\(\Big \downarrow \) 5138

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

\(\Big \downarrow \) 5204

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

\(\Big \downarrow \) 15

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

\(\Big \downarrow \) 5218

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4671

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

\(\Big \downarrow \) 2715

\(\displaystyle -\frac {(a+b \arcsin (c x))^2}{3 x^3}+\frac {2}{3} b c \left (\frac {1}{2} c^2 \left (i b \int e^{-i \arcsin (c x)} \log \left (1-e^{i \arcsin (c x)}\right )de^{i \arcsin (c x)}-i b \int e^{-i \arcsin (c x)} \log \left (1+e^{i \arcsin (c x)}\right )de^{i \arcsin (c x)}-2 \text {arctanh}\left (e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))\right )-\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{2 x^2}-\frac {b c}{2 x}\right )\)

\(\Big \downarrow \) 2838

\(\displaystyle -\frac {(a+b \arcsin (c x))^2}{3 x^3}+\frac {2}{3} b c \left (\frac {1}{2} c^2 \left (-2 \text {arctanh}\left (e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))+i b \operatorname {PolyLog}\left (2,-e^{i \arcsin (c x)}\right )-i b \operatorname {PolyLog}\left (2,e^{i \arcsin (c x)}\right )\right )-\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{2 x^2}-\frac {b c}{2 x}\right )\)

Input:

Int[(a + b*ArcSin[c*x])^2/x^4,x]
 

Output:

-1/3*(a + b*ArcSin[c*x])^2/x^3 + (2*b*c*(-1/2*(b*c)/x - (Sqrt[1 - c^2*x^2] 
*(a + b*ArcSin[c*x]))/(2*x^2) + (c^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*A 
rcSin[c*x])]))/2))/3
 

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 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 5204

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 + 1)*((a + b 
*ArcSin[c*x])^n/(d*f*(m + 1))), x] + (Simp[c^2*((m + 2*p + 3)/(f^2*(m + 1)) 
)   Int[(f*x)^(m + 2)*(d + e*x^2)^p*(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, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && ILtQ[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 = 0.22 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.55

method	result	size
parts	\(-\frac {a^{2}}{3 x^{3}}+b^{2} c^{3} \left (-\frac {\arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, c x +\arcsin \left (c x \right )^{2}+c^{2} x^{2}}{3 c^{3} x^{3}}+\frac {\arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )}{3}-\frac {i \operatorname {polylog}\left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )}{3}-\frac {\arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{3}+\frac {i \operatorname {polylog}\left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )}{3}\right )+2 a b \,c^{3} \left (-\frac {\arcsin \left (c x \right )}{3 c^{3} x^{3}}-\frac {\sqrt {-c^{2} x^{2}+1}}{6 c^{2} x^{2}}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )}{6}\right )\)	\(216\)
derivativedivides	\(c^{3} \left (-\frac {a^{2}}{3 c^{3} x^{3}}+b^{2} \left (-\frac {\arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, c x +\arcsin \left (c x \right )^{2}+c^{2} x^{2}}{3 c^{3} x^{3}}+\frac {\arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )}{3}-\frac {i \operatorname {polylog}\left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )}{3}-\frac {\arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{3}+\frac {i \operatorname {polylog}\left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )}{3}\right )+2 a b \left (-\frac {\arcsin \left (c x \right )}{3 c^{3} x^{3}}-\frac {\sqrt {-c^{2} x^{2}+1}}{6 c^{2} x^{2}}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )}{6}\right )\right )\)	\(217\)
default	\(c^{3} \left (-\frac {a^{2}}{3 c^{3} x^{3}}+b^{2} \left (-\frac {\arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, c x +\arcsin \left (c x \right )^{2}+c^{2} x^{2}}{3 c^{3} x^{3}}+\frac {\arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )}{3}-\frac {i \operatorname {polylog}\left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )}{3}-\frac {\arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{3}+\frac {i \operatorname {polylog}\left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )}{3}\right )+2 a b \left (-\frac {\arcsin \left (c x \right )}{3 c^{3} x^{3}}-\frac {\sqrt {-c^{2} x^{2}+1}}{6 c^{2} x^{2}}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )}{6}\right )\right )\)	\(217\)

Input:

int((a+b*arcsin(c*x))^2/x^4,x,method=_RETURNVERBOSE)
 

Output:

-1/3*a^2/x^3+b^2*c^3*(-1/3*(arcsin(c*x)*(-c^2*x^2+1)^(1/2)*c*x+arcsin(c*x) 
^2+c^2*x^2)/c^3/x^3+1/3*arcsin(c*x)*ln(1-I*c*x-(-c^2*x^2+1)^(1/2))-1/3*I*p 
olylog(2,I*c*x+(-c^2*x^2+1)^(1/2))-1/3*arcsin(c*x)*ln(1+I*c*x+(-c^2*x^2+1) 
^(1/2))+1/3*I*polylog(2,-I*c*x-(-c^2*x^2+1)^(1/2)))+2*a*b*c^3*(-1/3/c^3/x^ 
3*arcsin(c*x)-1/6/c^2/x^2*(-c^2*x^2+1)^(1/2)-1/6*arctanh(1/(-c^2*x^2+1)^(1 
/2)))
 

Fricas [F]

\[ \int \frac {(a+b \arcsin (c x))^2}{x^4} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{x^{4}} \,d x } \] Input:

integrate((a+b*arcsin(c*x))^2/x^4,x, algorithm="fricas")
 

Output:

integral((b^2*arcsin(c*x)^2 + 2*a*b*arcsin(c*x) + a^2)/x^4, x)
 

Sympy [F]

\[ \int \frac {(a+b \arcsin (c x))^2}{x^4} \, dx=\int \frac {\left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}{x^{4}}\, dx \] Input:

integrate((a+b*asin(c*x))**2/x**4,x)
 

Output:

Integral((a + b*asin(c*x))**2/x**4, x)
 

Maxima [F]

\[ \int \frac {(a+b \arcsin (c x))^2}{x^4} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{x^{4}} \,d x } \] Input:

integrate((a+b*arcsin(c*x))^2/x^4,x, algorithm="maxima")
 

Output:

-1/3*((c^2*log(2*sqrt(-c^2*x^2 + 1)/abs(x) + 2/abs(x)) + sqrt(-c^2*x^2 + 1 
)/x^2)*c + 2*arcsin(c*x)/x^3)*a*b - 1/3*(6*c*x^3*integrate(1/3*sqrt(c*x + 
1)*sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/(c^2*x^5 - x^ 
3), x) + arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2)*b^2/x^3 - 1/3*a^2/x 
^3
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(a+b \arcsin (c x))^2}{x^4} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((a+b*arcsin(c*x))^2/x^4,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. \[ \int \frac {(a+b \arcsin (c x))^2}{x^4} \, dx=\int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2}{x^4} \,d x \] Input:

int((a + b*asin(c*x))^2/x^4,x)
                                                                                    
                                                                                    
 

Output:

int((a + b*asin(c*x))^2/x^4, x)
 

Reduce [F]

\[ \int \frac {(a+b \arcsin (c x))^2}{x^4} \, dx=\frac {-2 \mathit {asin} \left (c x \right ) a b -\sqrt {-c^{2} x^{2}+1}\, a b c x +3 \left (\int \frac {\mathit {asin} \left (c x \right )^{2}}{x^{4}}d x \right ) b^{2} x^{3}+\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (c x \right )}{2}\right )\right ) a b \,c^{3} x^{3}-a^{2}}{3 x^{3}} \] Input:

int((a+b*asin(c*x))^2/x^4,x)
 

Output:

( - 2*asin(c*x)*a*b - sqrt( - c**2*x**2 + 1)*a*b*c*x + 3*int(asin(c*x)**2/ 
x**4,x)*b**2*x**3 + log(tan(asin(c*x)/2))*a*b*c**3*x**3 - a**2)/(3*x**3)