3.9 \(\int \frac {\sin ^{-1}(a x)}{x^4} \, dx\)

Optimal. Leaf size=56 \[ -\frac {a \sqrt {1-a^2 x^2}}{6 x^2}-\frac {1}{6} a^3 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )-\frac {\sin ^{-1}(a x)}{3 x^3} \]

[Out]

-1/3*arcsin(a*x)/x^3-1/6*a^3*arctanh((-a^2*x^2+1)^(1/2))-1/6*a*(-a^2*x^2+1)^(1/2)/x^2

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Rubi [A]  time = 0.03, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {4627, 266, 51, 63, 208} \[ -\frac {a \sqrt {1-a^2 x^2}}{6 x^2}-\frac {1}{6} a^3 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )-\frac {\sin ^{-1}(a x)}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[a*x]/x^4,x]

[Out]

-(a*Sqrt[1 - a^2*x^2])/(6*x^2) - ArcSin[a*x]/(3*x^3) - (a^3*ArcTanh[Sqrt[1 - a^2*x^2]])/6

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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]

Rubi steps

\begin {align*} \int \frac {\sin ^{-1}(a x)}{x^4} \, dx &=-\frac {\sin ^{-1}(a x)}{3 x^3}+\frac {1}{3} a \int \frac {1}{x^3 \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {\sin ^{-1}(a x)}{3 x^3}+\frac {1}{6} a \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {1-a^2 x}} \, dx,x,x^2\right )\\ &=-\frac {a \sqrt {1-a^2 x^2}}{6 x^2}-\frac {\sin ^{-1}(a x)}{3 x^3}+\frac {1}{12} a^3 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-a^2 x}} \, dx,x,x^2\right )\\ &=-\frac {a \sqrt {1-a^2 x^2}}{6 x^2}-\frac {\sin ^{-1}(a x)}{3 x^3}-\frac {1}{6} a \operatorname {Subst}\left (\int \frac {1}{\frac {1}{a^2}-\frac {x^2}{a^2}} \, dx,x,\sqrt {1-a^2 x^2}\right )\\ &=-\frac {a \sqrt {1-a^2 x^2}}{6 x^2}-\frac {\sin ^{-1}(a x)}{3 x^3}-\frac {1}{6} a^3 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 53, normalized size = 0.95 \[ -\frac {a x \sqrt {1-a^2 x^2}+a^3 x^3 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )+2 \sin ^{-1}(a x)}{6 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[a*x]/x^4,x]

[Out]

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

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fricas [A]  time = 0.72, size = 73, normalized size = 1.30 \[ -\frac {a^{3} x^{3} \log \left (\sqrt {-a^{2} x^{2} + 1} + 1\right ) - a^{3} x^{3} \log \left (\sqrt {-a^{2} x^{2} + 1} - 1\right ) + 2 \, \sqrt {-a^{2} x^{2} + 1} a x + 4 \, \arcsin \left (a x\right )}{12 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(a*x)/x^4,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 77, normalized size = 1.38 \[ -\frac {a^{4} \log \left (\sqrt {-a^{2} x^{2} + 1} + 1\right ) - a^{4} \log \left (-\sqrt {-a^{2} x^{2} + 1} + 1\right ) + \frac {2 \, \sqrt {-a^{2} x^{2} + 1} a^{2}}{x^{2}}}{12 \, a} - \frac {\arcsin \left (a x\right )}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(a*x)/x^4,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.00, size = 53, normalized size = 0.95 \[ a^{3} \left (-\frac {\arcsin \left (a x \right )}{3 a^{3} x^{3}}-\frac {\sqrt {-a^{2} x^{2}+1}}{6 a^{2} x^{2}}-\frac {\arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )}{6}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsin(a*x)/x^4,x)

[Out]

a^3*(-1/3*arcsin(a*x)/a^3/x^3-1/6/a^2/x^2*(-a^2*x^2+1)^(1/2)-1/6*arctanh(1/(-a^2*x^2+1)^(1/2)))

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maxima [A]  time = 0.42, size = 60, normalized size = 1.07 \[ -\frac {1}{6} \, {\left (a^{2} \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + \frac {\sqrt {-a^{2} x^{2} + 1}}{x^{2}}\right )} a - \frac {\arcsin \left (a x\right )}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(a*x)/x^4,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {asin}\left (a\,x\right )}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin(a*x)/x^4,x)

[Out]

int(asin(a*x)/x^4, x)

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sympy [A]  time = 2.33, size = 109, normalized size = 1.95 \[ \frac {a \left (\begin {cases} - \frac {a^{2} \operatorname {acosh}{\left (\frac {1}{a x} \right )}}{2} - \frac {a \sqrt {-1 + \frac {1}{a^{2} x^{2}}}}{2 x} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\\frac {i a^{2} \operatorname {asin}{\left (\frac {1}{a x} \right )}}{2} - \frac {i a}{2 x \sqrt {1 - \frac {1}{a^{2} x^{2}}}} + \frac {i}{2 a x^{3} \sqrt {1 - \frac {1}{a^{2} x^{2}}}} & \text {otherwise} \end {cases}\right )}{3} - \frac {\operatorname {asin}{\left (a x \right )}}{3 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asin(a*x)/x**4,x)

[Out]

a*Piecewise((-a**2*acosh(1/(a*x))/2 - a*sqrt(-1 + 1/(a**2*x**2))/(2*x), 1/Abs(a**2*x**2) > 1), (I*a**2*asin(1/
(a*x))/2 - I*a/(2*x*sqrt(1 - 1/(a**2*x**2))) + I/(2*a*x**3*sqrt(1 - 1/(a**2*x**2))), True))/3 - asin(a*x)/(3*x
**3)

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