\(\int \frac {(a+b \arcsin (c x))^2}{x^5} \, dx\) [20]

Optimal result

Integrand size = 14, antiderivative size = 107 \[ \int \frac {(a+b \arcsin (c x))^2}{x^5} \, dx=-\frac {b^2 c^2}{12 x^2}-\frac {b c \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{6 x^3}-\frac {b c^3 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{3 x}-\frac {(a+b \arcsin (c x))^2}{4 x^4}+\frac {1}{3} b^2 c^4 \log (x) \] Output:

-1/12*b^2*c^2/x^2-1/6*b*c*(-c^2*x^2+1)^(1/2)*(a+b*arcsin(c*x))/x^3-1/3*b*c 
^3*(-c^2*x^2+1)^(1/2)*(a+b*arcsin(c*x))/x-1/4*(a+b*arcsin(c*x))^2/x^4+1/3* 
b^2*c^4*ln(x)
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.21 \[ \int \frac {(a+b \arcsin (c x))^2}{x^5} \, dx=-\frac {3 a^2+b^2 c^2 x^2+2 a b c x \sqrt {1-c^2 x^2}+4 a b c^3 x^3 \sqrt {1-c^2 x^2}+2 b \left (3 a+b c x \sqrt {1-c^2 x^2} \left (1+2 c^2 x^2\right )\right ) \arcsin (c x)+3 b^2 \arcsin (c x)^2-4 b^2 c^4 x^4 \log (x)}{12 x^4} \] Input:

Integrate[(a + b*ArcSin[c*x])^2/x^5,x]
 

Output:

-1/12*(3*a^2 + b^2*c^2*x^2 + 2*a*b*c*x*Sqrt[1 - c^2*x^2] + 4*a*b*c^3*x^3*S 
qrt[1 - c^2*x^2] + 2*b*(3*a + b*c*x*Sqrt[1 - c^2*x^2]*(1 + 2*c^2*x^2))*Arc 
Sin[c*x] + 3*b^2*ArcSin[c*x]^2 - 4*b^2*c^4*x^4*Log[x])/x^4
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {5138, 5204, 15, 5186, 14}

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.

\(\displaystyle \int \frac {(a+b \arcsin (c x))^2}{x^5} \, dx\)

\(\Big \downarrow \) 5138

\(\displaystyle \frac {1}{2} b c \int \frac {a+b \arcsin (c x)}{x^4 \sqrt {1-c^2 x^2}}dx-\frac {(a+b \arcsin (c x))^2}{4 x^4}\)

\(\Big \downarrow \) 5204

\(\displaystyle \frac {1}{2} b c \left (\frac {2}{3} c^2 \int \frac {a+b \arcsin (c x)}{x^2 \sqrt {1-c^2 x^2}}dx+\frac {1}{3} b c \int \frac {1}{x^3}dx-\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{3 x^3}\right )-\frac {(a+b \arcsin (c x))^2}{4 x^4}\)

\(\Big \downarrow \) 15

\(\displaystyle \frac {1}{2} b c \left (\frac {2}{3} c^2 \int \frac {a+b \arcsin (c x)}{x^2 \sqrt {1-c^2 x^2}}dx-\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{3 x^3}-\frac {b c}{6 x^2}\right )-\frac {(a+b \arcsin (c x))^2}{4 x^4}\)

\(\Big \downarrow \) 5186

\(\displaystyle \frac {1}{2} b c \left (\frac {2}{3} c^2 \left (b c \int \frac {1}{x}dx-\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{x}\right )-\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{3 x^3}-\frac {b c}{6 x^2}\right )-\frac {(a+b \arcsin (c x))^2}{4 x^4}\)

\(\Big \downarrow \) 14

\(\displaystyle \frac {1}{2} b c \left (\frac {2}{3} c^2 \left (b c \log (x)-\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{x}\right )-\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{3 x^3}-\frac {b c}{6 x^2}\right )-\frac {(a+b \arcsin (c x))^2}{4 x^4}\)

Input:

Int[(a + b*ArcSin[c*x])^2/x^5,x]
 

Output:

-1/4*(a + b*ArcSin[c*x])^2/x^4 + (b*c*(-1/6*(b*c)/x^2 - (Sqrt[1 - c^2*x^2] 
*(a + b*ArcSin[c*x]))/(3*x^3) + (2*c^2*(-((Sqrt[1 - c^2*x^2]*(a + b*ArcSin 
[c*x]))/x) + b*c*Log[x]))/3))/2
 

Defintions of rubi rules used

rule 14

Int[(a_.)/(x_), x_Symbol] :> Simp[a*Log[x], x] /; FreeQ[a, x]
 

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

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 5186

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[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*A 
rcSin[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[c^ 
2*d + e, 0] && GtQ[n, 0] && EqQ[m + 2*p + 3, 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]
 

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.44

method	result	size
parts	\(-\frac {a^{2}}{4 x^{4}}+b^{2} c^{4} \left (-\frac {\arcsin \left (c x \right )^{2}}{4 c^{4} x^{4}}-\frac {\arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}}{6 c^{3} x^{3}}-\frac {1}{12 c^{2} x^{2}}-\frac {\arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}}{3 c x}+\frac {\ln \left (c x \right )}{3}\right )+2 a b \,c^{4} \left (-\frac {\arcsin \left (c x \right )}{4 c^{4} x^{4}}-\frac {\sqrt {-c^{2} x^{2}+1}}{12 c^{3} x^{3}}-\frac {\sqrt {-c^{2} x^{2}+1}}{6 c x}\right )\)	\(154\)
derivativedivides	\(c^{4} \left (-\frac {a^{2}}{4 c^{4} x^{4}}+b^{2} \left (-\frac {\arcsin \left (c x \right )^{2}}{4 c^{4} x^{4}}-\frac {\arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}}{6 c^{3} x^{3}}-\frac {1}{12 c^{2} x^{2}}-\frac {\arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}}{3 c x}+\frac {\ln \left (c x \right )}{3}\right )+2 a b \left (-\frac {\arcsin \left (c x \right )}{4 c^{4} x^{4}}-\frac {\sqrt {-c^{2} x^{2}+1}}{12 c^{3} x^{3}}-\frac {\sqrt {-c^{2} x^{2}+1}}{6 c x}\right )\right )\)	\(155\)
default	\(c^{4} \left (-\frac {a^{2}}{4 c^{4} x^{4}}+b^{2} \left (-\frac {\arcsin \left (c x \right )^{2}}{4 c^{4} x^{4}}-\frac {\arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}}{6 c^{3} x^{3}}-\frac {1}{12 c^{2} x^{2}}-\frac {\arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}}{3 c x}+\frac {\ln \left (c x \right )}{3}\right )+2 a b \left (-\frac {\arcsin \left (c x \right )}{4 c^{4} x^{4}}-\frac {\sqrt {-c^{2} x^{2}+1}}{12 c^{3} x^{3}}-\frac {\sqrt {-c^{2} x^{2}+1}}{6 c x}\right )\right )\)	\(155\)

Input:

int((a+b*arcsin(c*x))^2/x^5,x,method=_RETURNVERBOSE)
 

Output:

-1/4*a^2/x^4+b^2*c^4*(-1/4/c^4/x^4*arcsin(c*x)^2-1/6*arcsin(c*x)*(-c^2*x^2 
+1)^(1/2)/c^3/x^3-1/12/c^2/x^2-1/3*arcsin(c*x)/c/x*(-c^2*x^2+1)^(1/2)+1/3* 
ln(c*x))+2*a*b*c^4*(-1/4/c^4/x^4*arcsin(c*x)-1/12/c^3/x^3*(-c^2*x^2+1)^(1/ 
2)-1/6/c/x*(-c^2*x^2+1)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b \arcsin (c x))^2}{x^5} \, dx=\frac {4 \, b^{2} c^{4} x^{4} \log \left (x\right ) - b^{2} c^{2} x^{2} - 3 \, b^{2} \arcsin \left (c x\right )^{2} - 6 \, a b \arcsin \left (c x\right ) - 3 \, a^{2} - 2 \, {\left (2 \, a b c^{3} x^{3} + a b c x + {\left (2 \, b^{2} c^{3} x^{3} + b^{2} c x\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} x^{2} + 1}}{12 \, x^{4}} \] Input:

integrate((a+b*arcsin(c*x))^2/x^5,x, algorithm="fricas")
 

Output:

1/12*(4*b^2*c^4*x^4*log(x) - b^2*c^2*x^2 - 3*b^2*arcsin(c*x)^2 - 6*a*b*arc 
sin(c*x) - 3*a^2 - 2*(2*a*b*c^3*x^3 + a*b*c*x + (2*b^2*c^3*x^3 + b^2*c*x)* 
arcsin(c*x))*sqrt(-c^2*x^2 + 1))/x^4
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {(a+b \arcsin (c x))^2}{x^5} \, dx=\int \frac {\left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}{x^{5}}\, dx \] Input:

integrate((a+b*asin(c*x))**2/x**5,x)
 

Output:

Integral((a + b*asin(c*x))**2/x**5, x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.34 \[ \int \frac {(a+b \arcsin (c x))^2}{x^5} \, dx=-\frac {1}{6} \, {\left ({\left (\frac {2 \, \sqrt {-c^{2} x^{2} + 1} c^{2}}{x} + \frac {\sqrt {-c^{2} x^{2} + 1}}{x^{3}}\right )} c + \frac {3 \, \arcsin \left (c x\right )}{x^{4}}\right )} a b + \frac {1}{12} \, {\left ({\left (4 \, c^{2} \log \left (x\right ) - \frac {1}{x^{2}}\right )} c^{2} - 2 \, {\left (\frac {2 \, \sqrt {-c^{2} x^{2} + 1} c^{2}}{x} + \frac {\sqrt {-c^{2} x^{2} + 1}}{x^{3}}\right )} c \arcsin \left (c x\right )\right )} b^{2} - \frac {b^{2} \arcsin \left (c x\right )^{2}}{4 \, x^{4}} - \frac {a^{2}}{4 \, x^{4}} \] Input:

integrate((a+b*arcsin(c*x))^2/x^5,x, algorithm="maxima")
 

Output:

-1/6*((2*sqrt(-c^2*x^2 + 1)*c^2/x + sqrt(-c^2*x^2 + 1)/x^3)*c + 3*arcsin(c 
*x)/x^4)*a*b + 1/12*((4*c^2*log(x) - 1/x^2)*c^2 - 2*(2*sqrt(-c^2*x^2 + 1)* 
c^2/x + sqrt(-c^2*x^2 + 1)/x^3)*c*arcsin(c*x))*b^2 - 1/4*b^2*arcsin(c*x)^2 
/x^4 - 1/4*a^2/x^4
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 735 vs. \(2 (93) = 186\).

Time = 0.24 (sec) , antiderivative size = 735, normalized size of antiderivative = 6.87 \[ \int \frac {(a+b \arcsin (c x))^2}{x^5} \, dx =\text {Too large to display} \] Input:

integrate((a+b*arcsin(c*x))^2/x^5,x, algorithm="giac")
 

Output:

-1/64*b^2*c^8*x^4*arcsin(c*x)^2/(sqrt(-c^2*x^2 + 1) + 1)^4 - 1/32*a*b*c^8* 
x^4*arcsin(c*x)/(sqrt(-c^2*x^2 + 1) + 1)^4 - 1/64*a^2*c^8*x^4/(sqrt(-c^2*x 
^2 + 1) + 1)^4 + 1/48*b^2*c^7*x^3*arcsin(c*x)/(sqrt(-c^2*x^2 + 1) + 1)^3 - 
 1/16*b^2*c^6*x^2*arcsin(c*x)^2/(sqrt(-c^2*x^2 + 1) + 1)^2 + 1/48*a*b*c^7* 
x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 - 1/8*a*b*c^6*x^2*arcsin(c*x)/(sqrt(-c^2*x^ 
2 + 1) + 1)^2 - 1/16*a^2*c^6*x^2/(sqrt(-c^2*x^2 + 1) + 1)^2 - 1/48*b^2*c^6 
*x^2/(sqrt(-c^2*x^2 + 1) + 1)^2 + 3/16*b^2*c^5*x*arcsin(c*x)/(sqrt(-c^2*x^ 
2 + 1) + 1) - 3/32*b^2*c^4*arcsin(c*x)^2 + 3/16*a*b*c^5*x/(sqrt(-c^2*x^2 + 
 1) + 1) - 3/16*a*b*c^4*arcsin(c*x) + 2/3*b^2*c^4*log(2) + 1/3*b^2*c^4*log 
(abs(c)*abs(x)) - 1/3*b^2*c^4*log(2*sqrt(-c^2*x^2 + 1) + 2) + 1/3*b^2*c^4* 
log(sqrt(-c^2*x^2 + 1) + 1) - 3/32*a^2*c^4 - 1/24*b^2*c^4 - 3/16*b^2*c^3*( 
sqrt(-c^2*x^2 + 1) + 1)*arcsin(c*x)/x - 1/16*b^2*c^2*(sqrt(-c^2*x^2 + 1) + 
 1)^2*arcsin(c*x)^2/x^2 - 3/16*a*b*c^3*(sqrt(-c^2*x^2 + 1) + 1)/x - 1/8*a* 
b*c^2*(sqrt(-c^2*x^2 + 1) + 1)^2*arcsin(c*x)/x^2 - 1/16*a^2*c^2*(sqrt(-c^2 
*x^2 + 1) + 1)^2/x^2 - 1/48*b^2*c^2*(sqrt(-c^2*x^2 + 1) + 1)^2/x^2 - 1/48* 
b^2*c*(sqrt(-c^2*x^2 + 1) + 1)^3*arcsin(c*x)/x^3 - 1/64*b^2*(sqrt(-c^2*x^2 
 + 1) + 1)^4*arcsin(c*x)^2/x^4 - 1/48*a*b*c*(sqrt(-c^2*x^2 + 1) + 1)^3/x^3 
 - 1/32*a*b*(sqrt(-c^2*x^2 + 1) + 1)^4*arcsin(c*x)/x^4 - 1/64*a^2*(sqrt(-c 
^2*x^2 + 1) + 1)^4/x^4
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \arcsin (c x))^2}{x^5} \, dx=\int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2}{x^5} \,d x \] Input:

int((a + b*asin(c*x))^2/x^5,x)
 

Output:

int((a + b*asin(c*x))^2/x^5, x)
 

Reduce [F]

\[ \int \frac {(a+b \arcsin (c x))^2}{x^5} \, dx=\frac {-6 \mathit {asin} \left (c x \right ) a b -4 \sqrt {-c^{2} x^{2}+1}\, a b \,c^{3} x^{3}-2 \sqrt {-c^{2} x^{2}+1}\, a b c x +12 \left (\int \frac {\mathit {asin} \left (c x \right )^{2}}{x^{5}}d x \right ) b^{2} x^{4}-3 a^{2}}{12 x^{4}} \] Input:

int((a+b*asin(c*x))^2/x^5,x)
 

Output:

( - 6*asin(c*x)*a*b - 4*sqrt( - c**2*x**2 + 1)*a*b*c**3*x**3 - 2*sqrt( - c 
**2*x**2 + 1)*a*b*c*x + 12*int(asin(c*x)**2/x**5,x)*b**2*x**4 - 3*a**2)/(1 
2*x**4)