3.21 \(\int \frac {\sin ^{-1}(a x)^2}{x^5} \, dx\)

Optimal. Leaf size=87 \[ \frac {1}{3} a^4 \log (x)-\frac {a^2}{12 x^2}-\frac {a \sqrt {1-a^2 x^2} \sin ^{-1}(a x)}{6 x^3}-\frac {a^3 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)}{3 x}-\frac {\sin ^{-1}(a x)^2}{4 x^4} \]

[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

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Rubi [A]  time = 0.14, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4627, 4701, 4681, 29, 30} \[ -\frac {a^2}{12 x^2}-\frac {a^3 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)}{3 x}-\frac {a \sqrt {1-a^2 x^2} \sin ^{-1}(a x)}{6 x^3}+\frac {1}{3} a^4 \log (x)-\frac {\sin ^{-1}(a x)^2}{4 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-a^2/(12*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 4627

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 4681

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*d^IntPart[p]*(d + e*x
^2)^FracPart[p])/(f*(m + 1)*(1 - c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSi
n[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 4701

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*d^IntPart[p]*(d + e*x^2)^F
racPart[p])/(f*(m + 1)*(1 - c^2*x^2)^FracPart[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] && LtQ[m, -1] && Inte
gerQ[m]

Rubi steps

\begin {align*} \int \frac {\sin ^{-1}(a x)^2}{x^5} \, dx &=-\frac {\sin ^{-1}(a x)^2}{4 x^4}+\frac {1}{2} a \int \frac {\sin ^{-1}(a x)}{x^4 \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {a \sqrt {1-a^2 x^2} \sin ^{-1}(a x)}{6 x^3}-\frac {\sin ^{-1}(a x)^2}{4 x^4}+\frac {1}{6} a^2 \int \frac {1}{x^3} \, dx+\frac {1}{3} a^3 \int \frac {\sin ^{-1}(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} \sin ^{-1}(a x)}{6 x^3}-\frac {a^3 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)}{3 x}-\frac {\sin ^{-1}(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} \sin ^{-1}(a x)}{6 x^3}-\frac {a^3 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)}{3 x}-\frac {\sin ^{-1}(a x)^2}{4 x^4}+\frac {1}{3} a^4 \log (x)\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 69, normalized size = 0.79 \[ \frac {1}{3} a^4 \log (x)-\frac {a^2}{12 x^2}-\frac {a \sqrt {1-a^2 x^2} \left (2 a^2 x^2+1\right ) \sin ^{-1}(a x)}{6 x^3}-\frac {\sin ^{-1}(a x)^2}{4 x^4} \]

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.76, size = 62, normalized size = 0.71 \[ \frac {4 \, a^{4} x^{4} \log \relax (x) - a^{2} x^{2} - 2 \, {\left (2 \, a^{3} x^{3} + a x\right )} \sqrt {-a^{2} x^{2} + 1} \arcsin \left (a x\right ) - 3 \, \arcsin \left (a x\right )^{2}}{12 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [B]  time = 0.78, size = 185, normalized size = 2.13 \[ \frac {1}{48} \, {\left ({\left (\frac {{\left (a^{4} + \frac {9 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2}}{x^{2}}\right )} a^{6} x^{3}}{{\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3} {\left | a \right |}} - \frac {\frac {9 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a^{4}}{x} + \frac {{\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3}}{x^{3}}}{a^{2} {\left | a \right |}}\right )} \arcsin \left (a x\right ) + \frac {4 \, {\left (2 \, a^{4} \log \left (a^{2} x^{2}\right ) - \frac {2 \, {\left (a^{2} x^{2} - 1\right )} a^{4} + 3 \, a^{4}}{a^{2} x^{2}}\right )}}{a}\right )} a - \frac {\arcsin \left (a x\right )^{2}}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.05, size = 76, normalized size = 0.87 \[ -\frac {\arcsin \left (a x \right )^{2}}{4 x^{4}}-\frac {a \arcsin \left (a x \right ) \sqrt {-a^{2} x^{2}+1}}{6 x^{3}}-\frac {a^{2}}{12 x^{2}}-\frac {a^{3} \arcsin \left (a x \right ) \sqrt {-a^{2} x^{2}+1}}{3 x}+\frac {a^{4} \ln \left (a x \right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.49, size = 74, normalized size = 0.85 \[ \frac {1}{12} \, {\left (4 \, a^{2} \log \relax (x) - \frac {1}{x^{2}}\right )} a^{2} - \frac {1}{6} \, {\left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1} a^{2}}{x} + \frac {\sqrt {-a^{2} x^{2} + 1}}{x^{3}}\right )} a \arcsin \left (a x\right ) - \frac {\arcsin \left (a x\right )^{2}}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {asin}\left (a\,x\right )}^2}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(asin(a*x)^2/x^5, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asin}^{2}{\left (a x \right )}}{x^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asin(a*x)**2/x**5, x)

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