3.11 \(\int \frac {\sin ^{-1}(a x)}{x^6} \, dx\)

Optimal. Leaf size=80 \[ -\frac {a \sqrt {1-a^2 x^2}}{20 x^4}-\frac {3}{40} a^5 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )-\frac {3 a^3 \sqrt {1-a^2 x^2}}{40 x^2}-\frac {\sin ^{-1}(a x)}{5 x^5} \]

[Out]

-1/5*arcsin(a*x)/x^5-3/40*a^5*arctanh((-a^2*x^2+1)^(1/2))-1/20*a*(-a^2*x^2+1)^(1/2)/x^4-3/40*a^3*(-a^2*x^2+1)^
(1/2)/x^2

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Rubi [A]  time = 0.05, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \[ -\frac {3 a^3 \sqrt {1-a^2 x^2}}{40 x^2}-\frac {a \sqrt {1-a^2 x^2}}{20 x^4}-\frac {3}{40} a^5 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )-\frac {\sin ^{-1}(a x)}{5 x^5} \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[a*x]/x^6,x]

[Out]

-(a*Sqrt[1 - a^2*x^2])/(20*x^4) - (3*a^3*Sqrt[1 - a^2*x^2])/(40*x^2) - ArcSin[a*x]/(5*x^5) - (3*a^5*ArcTanh[Sq
rt[1 - a^2*x^2]])/40

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^6} \, dx &=-\frac {\sin ^{-1}(a x)}{5 x^5}+\frac {1}{5} a \int \frac {1}{x^5 \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {\sin ^{-1}(a x)}{5 x^5}+\frac {1}{10} a \operatorname {Subst}\left (\int \frac {1}{x^3 \sqrt {1-a^2 x}} \, dx,x,x^2\right )\\ &=-\frac {a \sqrt {1-a^2 x^2}}{20 x^4}-\frac {\sin ^{-1}(a x)}{5 x^5}+\frac {1}{40} \left (3 a^3\right ) \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}}{20 x^4}-\frac {3 a^3 \sqrt {1-a^2 x^2}}{40 x^2}-\frac {\sin ^{-1}(a x)}{5 x^5}+\frac {1}{80} \left (3 a^5\right ) \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}}{20 x^4}-\frac {3 a^3 \sqrt {1-a^2 x^2}}{40 x^2}-\frac {\sin ^{-1}(a x)}{5 x^5}-\frac {1}{40} \left (3 a^3\right ) \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}}{20 x^4}-\frac {3 a^3 \sqrt {1-a^2 x^2}}{40 x^2}-\frac {\sin ^{-1}(a x)}{5 x^5}-\frac {3}{40} a^5 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 51, normalized size = 0.64 \[ -\frac {1}{5} a^5 \sqrt {1-a^2 x^2} \, _2F_1\left (\frac {1}{2},3;\frac {3}{2};1-a^2 x^2\right )-\frac {\sin ^{-1}(a x)}{5 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[a*x]/x^6,x]

[Out]

-1/5*ArcSin[a*x]/x^5 - (a^5*Sqrt[1 - a^2*x^2]*Hypergeometric2F1[1/2, 3, 3/2, 1 - a^2*x^2])/5

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fricas [A]  time = 0.58, size = 85, normalized size = 1.06 \[ -\frac {3 \, a^{5} x^{5} \log \left (\sqrt {-a^{2} x^{2} + 1} + 1\right ) - 3 \, a^{5} x^{5} \log \left (\sqrt {-a^{2} x^{2} + 1} - 1\right ) + 2 \, {\left (3 \, a^{3} x^{3} + 2 \, a x\right )} \sqrt {-a^{2} x^{2} + 1} + 16 \, \arcsin \left (a x\right )}{80 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(a*x)/x^6,x, algorithm="fricas")

[Out]

-1/80*(3*a^5*x^5*log(sqrt(-a^2*x^2 + 1) + 1) - 3*a^5*x^5*log(sqrt(-a^2*x^2 + 1) - 1) + 2*(3*a^3*x^3 + 2*a*x)*s
qrt(-a^2*x^2 + 1) + 16*arcsin(a*x))/x^5

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giac [A]  time = 0.14, size = 101, normalized size = 1.26 \[ -\frac {3 \, a^{6} \log \left (\sqrt {-a^{2} x^{2} + 1} + 1\right ) - 3 \, a^{6} \log \left (-\sqrt {-a^{2} x^{2} + 1} + 1\right ) - \frac {2 \, {\left (3 \, {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} a^{6} - 5 \, \sqrt {-a^{2} x^{2} + 1} a^{6}\right )}}{a^{4} x^{4}}}{80 \, a} - \frac {\arcsin \left (a x\right )}{5 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(a*x)/x^6,x, algorithm="giac")

[Out]

-1/80*(3*a^6*log(sqrt(-a^2*x^2 + 1) + 1) - 3*a^6*log(-sqrt(-a^2*x^2 + 1) + 1) - 2*(3*(-a^2*x^2 + 1)^(3/2)*a^6
- 5*sqrt(-a^2*x^2 + 1)*a^6)/(a^4*x^4))/a - 1/5*arcsin(a*x)/x^5

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maple [A]  time = 0.00, size = 73, normalized size = 0.91 \[ a^{5} \left (-\frac {\arcsin \left (a x \right )}{5 a^{5} x^{5}}-\frac {\sqrt {-a^{2} x^{2}+1}}{20 a^{4} x^{4}}-\frac {3 \sqrt {-a^{2} x^{2}+1}}{40 a^{2} x^{2}}-\frac {3 \arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )}{40}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsin(a*x)/x^6,x)

[Out]

a^5*(-1/5*arcsin(a*x)/a^5/x^5-1/20/a^4/x^4*(-a^2*x^2+1)^(1/2)-3/40/a^2/x^2*(-a^2*x^2+1)^(1/2)-3/40*arctanh(1/(
-a^2*x^2+1)^(1/2)))

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maxima [A]  time = 0.42, size = 82, normalized size = 1.02 \[ -\frac {1}{40} \, {\left (3 \, a^{4} \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + \frac {3 \, \sqrt {-a^{2} x^{2} + 1} a^{2}}{x^{2}} + \frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{x^{4}}\right )} a - \frac {\arcsin \left (a x\right )}{5 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(a*x)/x^6,x, algorithm="maxima")

[Out]

-1/40*(3*a^4*log(2*sqrt(-a^2*x^2 + 1)/abs(x) + 2/abs(x)) + 3*sqrt(-a^2*x^2 + 1)*a^2/x^2 + 2*sqrt(-a^2*x^2 + 1)
/x^4)*a - 1/5*arcsin(a*x)/x^5

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {asin}\left (a\,x\right )}{x^6} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin(a*x)/x^6,x)

[Out]

int(asin(a*x)/x^6, x)

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sympy [A]  time = 4.23, size = 182, normalized size = 2.28 \[ \frac {a \left (\begin {cases} - \frac {3 a^{4} \operatorname {acosh}{\left (\frac {1}{a x} \right )}}{8} + \frac {3 a^{3}}{8 x \sqrt {-1 + \frac {1}{a^{2} x^{2}}}} - \frac {a}{8 x^{3} \sqrt {-1 + \frac {1}{a^{2} x^{2}}}} - \frac {1}{4 a x^{5} \sqrt {-1 + \frac {1}{a^{2} x^{2}}}} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\\frac {3 i a^{4} \operatorname {asin}{\left (\frac {1}{a x} \right )}}{8} - \frac {3 i a^{3}}{8 x \sqrt {1 - \frac {1}{a^{2} x^{2}}}} + \frac {i a}{8 x^{3} \sqrt {1 - \frac {1}{a^{2} x^{2}}}} + \frac {i}{4 a x^{5} \sqrt {1 - \frac {1}{a^{2} x^{2}}}} & \text {otherwise} \end {cases}\right )}{5} - \frac {\operatorname {asin}{\left (a x \right )}}{5 x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asin(a*x)/x**6,x)

[Out]

a*Piecewise((-3*a**4*acosh(1/(a*x))/8 + 3*a**3/(8*x*sqrt(-1 + 1/(a**2*x**2))) - a/(8*x**3*sqrt(-1 + 1/(a**2*x*
*2))) - 1/(4*a*x**5*sqrt(-1 + 1/(a**2*x**2))), 1/Abs(a**2*x**2) > 1), (3*I*a**4*asin(1/(a*x))/8 - 3*I*a**3/(8*
x*sqrt(1 - 1/(a**2*x**2))) + I*a/(8*x**3*sqrt(1 - 1/(a**2*x**2))) + I/(4*a*x**5*sqrt(1 - 1/(a**2*x**2))), True
))/5 - asin(a*x)/(5*x**5)

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