\(\int \frac {\arcsin (a x)^2}{x^5} \, dx\) [21]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 10, antiderivative size = 87 \[ \int \frac {\arcsin (a x)^2}{x^5} \, dx=-\frac {a^2}{12 x^2}-\frac {a \sqrt {1-a^2 x^2} \arcsin (a x)}{6 x^3}-\frac {a^3 \sqrt {1-a^2 x^2} \arcsin (a x)}{3 x}-\frac {\arcsin (a x)^2}{4 x^4}+\frac {1}{3} a^4 \log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4723, 4789, 4771, 29, 30} \[ \int \frac {\arcsin (a x)^2}{x^5} \, dx=\frac {1}{3} a^4 \log (x)-\frac {a \sqrt {1-a^2 x^2} \arcsin (a x)}{6 x^3}-\frac {a^2}{12 x^2}-\frac {a^3 \sqrt {1-a^2 x^2} \arcsin (a x)}{3 x}-\frac {\arcsin (a x)^2}{4 x^4} \]

[In]

Int[ArcSin[a*x]^2/x^5,x]

[Out]

-1/12*a^2/x^2 - (a*Sqrt[1 - a^2*x^2]*ArcSin[a*x])/(6*x^3) - (a^3*Sqrt[1 - a^2*x^2]*ArcSin[a*x])/(3*x) - ArcSin
[a*x]^2/(4*x^4) + (a^4*Log[x])/3

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 30

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

Rule 4723

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcSi
n[c*x])^n/(d*(m + 1))), x] - Dist[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 4771

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] - Dist[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, m, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]

Rule 4789

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] + (Dist[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] - Dist[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]) /; Free
Q[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && ILtQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {\arcsin (a x)^2}{4 x^4}+\frac {1}{2} a \int \frac {\arcsin (a x)}{x^4 \sqrt {1-a^2 x^2}} \, dx \\ & = -\frac {a \sqrt {1-a^2 x^2} \arcsin (a x)}{6 x^3}-\frac {\arcsin (a x)^2}{4 x^4}+\frac {1}{6} a^2 \int \frac {1}{x^3} \, dx+\frac {1}{3} a^3 \int \frac {\arcsin (a x)}{x^2 \sqrt {1-a^2 x^2}} \, dx \\ & = -\frac {a^2}{12 x^2}-\frac {a \sqrt {1-a^2 x^2} \arcsin (a x)}{6 x^3}-\frac {a^3 \sqrt {1-a^2 x^2} \arcsin (a x)}{3 x}-\frac {\arcsin (a x)^2}{4 x^4}+\frac {1}{3} a^4 \int \frac {1}{x} \, dx \\ & = -\frac {a^2}{12 x^2}-\frac {a \sqrt {1-a^2 x^2} \arcsin (a x)}{6 x^3}-\frac {a^3 \sqrt {1-a^2 x^2} \arcsin (a x)}{3 x}-\frac {\arcsin (a x)^2}{4 x^4}+\frac {1}{3} a^4 \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.79 \[ \int \frac {\arcsin (a x)^2}{x^5} \, dx=-\frac {a^2}{12 x^2}-\frac {a \sqrt {1-a^2 x^2} \left (1+2 a^2 x^2\right ) \arcsin (a x)}{6 x^3}-\frac {\arcsin (a x)^2}{4 x^4}+\frac {1}{3} a^4 \log (x) \]

[In]

Integrate[ArcSin[a*x]^2/x^5,x]

[Out]

-1/12*a^2/x^2 - (a*Sqrt[1 - a^2*x^2]*(1 + 2*a^2*x^2)*ArcSin[a*x])/(6*x^3) - ArcSin[a*x]^2/(4*x^4) + (a^4*Log[x
])/3

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.94

method result size
derivativedivides \(a^{4} \left (-\frac {\arcsin \left (a x \right )^{2}}{4 a^{4} x^{4}}-\frac {\arcsin \left (a x \right ) \sqrt {-a^{2} x^{2}+1}}{6 a^{3} x^{3}}-\frac {1}{12 a^{2} x^{2}}-\frac {\arcsin \left (a x \right ) \sqrt {-a^{2} x^{2}+1}}{3 a x}+\frac {\ln \left (a x \right )}{3}\right )\) \(82\)
default \(a^{4} \left (-\frac {\arcsin \left (a x \right )^{2}}{4 a^{4} x^{4}}-\frac {\arcsin \left (a x \right ) \sqrt {-a^{2} x^{2}+1}}{6 a^{3} x^{3}}-\frac {1}{12 a^{2} x^{2}}-\frac {\arcsin \left (a x \right ) \sqrt {-a^{2} x^{2}+1}}{3 a x}+\frac {\ln \left (a x \right )}{3}\right )\) \(82\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.71 \[ \int \frac {\arcsin (a x)^2}{x^5} \, dx=\frac {4 \, a^{4} x^{4} \log \left (x\right ) - a^{2} x^{2} - 2 \, {\left (2 \, a^{3} x^{3} + a x\right )} \sqrt {-a^{2} x^{2} + 1} \arcsin \left (a x\right ) - 3 \, \arcsin \left (a x\right )^{2}}{12 \, x^{4}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

integrate(asin(a*x)**2/x**5,x)

[Out]

Integral(asin(a*x)**2/x**5, x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.85 \[ \int \frac {\arcsin (a x)^2}{x^5} \, dx=\frac {1}{12} \, {\left (4 \, a^{2} \log \left (x\right ) - \frac {1}{x^{2}}\right )} a^{2} - \frac {1}{6} \, {\left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1} a^{2}}{x} + \frac {\sqrt {-a^{2} x^{2} + 1}}{x^{3}}\right )} a \arcsin \left (a x\right ) - \frac {\arcsin \left (a x\right )^{2}}{4 \, x^{4}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 185 vs. \(2 (73) = 146\).

Time = 0.33 (sec) , antiderivative size = 185, normalized size of antiderivative = 2.13 \[ \int \frac {\arcsin (a x)^2}{x^5} \, dx=\frac {1}{48} \, {\left ({\left (\frac {{\left (a^{4} + \frac {9 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2}}{x^{2}}\right )} a^{6} x^{3}}{{\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3} {\left | a \right |}} - \frac {\frac {9 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a^{4}}{x} + \frac {{\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3}}{x^{3}}}{a^{2} {\left | a \right |}}\right )} \arcsin \left (a x\right ) + \frac {4 \, {\left (2 \, a^{4} \log \left (a^{2} x^{2}\right ) - \frac {2 \, {\left (a^{2} x^{2} - 1\right )} a^{4} + 3 \, a^{4}}{a^{2} x^{2}}\right )}}{a}\right )} a - \frac {\arcsin \left (a x\right )^{2}}{4 \, x^{4}} \]

[In]

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

[Out]

1/48*(((a^4 + 9*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2/x^2)*a^6*x^3/((sqrt(-a^2*x^2 + 1)*abs(a) + a)^3*abs(a)) - (9
*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a^4/x + (sqrt(-a^2*x^2 + 1)*abs(a) + a)^3/x^3)/(a^2*abs(a)))*arcsin(a*x) + 4*
(2*a^4*log(a^2*x^2) - (2*(a^2*x^2 - 1)*a^4 + 3*a^4)/(a^2*x^2))/a)*a - 1/4*arcsin(a*x)^2/x^4

Mupad [F(-1)]

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

[In]

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

[Out]

int(asin(a*x)^2/x^5, x)