Integrand size = 8, antiderivative size = 80 \[ \int \frac {\arcsin (a x)}{x^6} \, dx=-\frac {a \sqrt {1-a^2 x^2}}{20 x^4}-\frac {3 a^3 \sqrt {1-a^2 x^2}}{40 x^2}-\frac {\arcsin (a x)}{5 x^5}-\frac {3}{40} a^5 \text {arctanh}\left (\sqrt {1-a^2 x^2}\right ) \]
-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
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.02 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.64 \[ \int \frac {\arcsin (a x)}{x^6} \, dx=-\frac {\arcsin (a x)}{5 x^5}-\frac {1}{5} a^5 \sqrt {1-a^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3,\frac {3}{2},1-a^2 x^2\right ) \]
-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
Time = 0.21 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {5138, 243, 52, 52, 73, 221}
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.
\(\displaystyle \int \frac {\arcsin (a x)}{x^6} \, dx\) |
\(\Big \downarrow \) 5138 |
\(\displaystyle \frac {1}{5} a \int \frac {1}{x^5 \sqrt {1-a^2 x^2}}dx-\frac {\arcsin (a x)}{5 x^5}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{10} a \int \frac {1}{x^6 \sqrt {1-a^2 x^2}}dx^2-\frac {\arcsin (a x)}{5 x^5}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{10} a \left (\frac {3}{4} a^2 \int \frac {1}{x^4 \sqrt {1-a^2 x^2}}dx^2-\frac {\sqrt {1-a^2 x^2}}{2 x^4}\right )-\frac {\arcsin (a x)}{5 x^5}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{10} a \left (\frac {3}{4} a^2 \left (\frac {1}{2} a^2 \int \frac {1}{x^2 \sqrt {1-a^2 x^2}}dx^2-\frac {\sqrt {1-a^2 x^2}}{x^2}\right )-\frac {\sqrt {1-a^2 x^2}}{2 x^4}\right )-\frac {\arcsin (a x)}{5 x^5}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{10} a \left (\frac {3}{4} a^2 \left (-\int \frac {1}{\frac {1}{a^2}-\frac {x^4}{a^2}}d\sqrt {1-a^2 x^2}-\frac {\sqrt {1-a^2 x^2}}{x^2}\right )-\frac {\sqrt {1-a^2 x^2}}{2 x^4}\right )-\frac {\arcsin (a x)}{5 x^5}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{10} a \left (\frac {3}{4} a^2 \left (a^2 \left (-\text {arctanh}\left (\sqrt {1-a^2 x^2}\right )\right )-\frac {\sqrt {1-a^2 x^2}}{x^2}\right )-\frac {\sqrt {1-a^2 x^2}}{2 x^4}\right )-\frac {\arcsin (a x)}{5 x^5}\) |
-1/5*ArcSin[a*x]/x^5 + (a*(-1/2*Sqrt[1 - a^2*x^2]/x^4 + (3*a^2*(-(Sqrt[1 - a^2*x^2]/x^2) - a^2*ArcTanh[Sqrt[1 - a^2*x^2]]))/4))/10
3.1.11.3.1 Defintions of rubi rules used
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] - Simp[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] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
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]
Time = 0.01 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.91
method | result | size |
derivativedivides | \(a^{5} \left (-\frac {\arcsin \left (a x \right )}{5 a^{5} x^{5}}-\frac {\sqrt {-a^{2} x^{2}+1}}{20 a^{4} x^{4}}-\frac {3 \sqrt {-a^{2} x^{2}+1}}{40 a^{2} x^{2}}-\frac {3 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )}{40}\right )\) | \(73\) |
default | \(a^{5} \left (-\frac {\arcsin \left (a x \right )}{5 a^{5} x^{5}}-\frac {\sqrt {-a^{2} x^{2}+1}}{20 a^{4} x^{4}}-\frac {3 \sqrt {-a^{2} x^{2}+1}}{40 a^{2} x^{2}}-\frac {3 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )}{40}\right )\) | \(73\) |
parts | \(-\frac {\arcsin \left (a x \right )}{5 x^{5}}+\frac {a \left (-\frac {\sqrt {-a^{2} x^{2}+1}}{4 x^{4}}+\frac {3 a^{2} \left (-\frac {\sqrt {-a^{2} x^{2}+1}}{2 x^{2}}-\frac {a^{2} \operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )}{2}\right )}{4}\right )}{5}\) | \(73\) |
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)))
Time = 0.26 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.06 \[ \int \frac {\arcsin (a x)}{x^6} \, dx=-\frac {3 \, a^{5} x^{5} \log \left (\sqrt {-a^{2} x^{2} + 1} + 1\right ) - 3 \, a^{5} x^{5} \log \left (\sqrt {-a^{2} x^{2} + 1} - 1\right ) + 2 \, {\left (3 \, a^{3} x^{3} + 2 \, a x\right )} \sqrt {-a^{2} x^{2} + 1} + 16 \, \arcsin \left (a x\right )}{80 \, x^{5}} \]
-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)*sqrt(-a^2*x^2 + 1) + 16*arcsin(a*x))/x^ 5
Time = 2.79 (sec) , antiderivative size = 182, normalized size of antiderivative = 2.28 \[ \int \frac {\arcsin (a x)}{x^6} \, dx=\frac {a \left (\begin {cases} - \frac {3 a^{4} \operatorname {acosh}{\left (\frac {1}{a x} \right )}}{8} + \frac {3 a^{3}}{8 x \sqrt {-1 + \frac {1}{a^{2} x^{2}}}} - \frac {a}{8 x^{3} \sqrt {-1 + \frac {1}{a^{2} x^{2}}}} - \frac {1}{4 a x^{5} \sqrt {-1 + \frac {1}{a^{2} x^{2}}}} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\\frac {3 i a^{4} \operatorname {asin}{\left (\frac {1}{a x} \right )}}{8} - \frac {3 i a^{3}}{8 x \sqrt {1 - \frac {1}{a^{2} x^{2}}}} + \frac {i a}{8 x^{3} \sqrt {1 - \frac {1}{a^{2} x^{2}}}} + \frac {i}{4 a x^{5} \sqrt {1 - \frac {1}{a^{2} x^{2}}}} & \text {otherwise} \end {cases}\right )}{5} - \frac {\operatorname {asin}{\left (a x \right )}}{5 x^{5}} \]
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)
Time = 0.29 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.02 \[ \int \frac {\arcsin (a x)}{x^6} \, dx=-\frac {1}{40} \, {\left (3 \, a^{4} \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + \frac {3 \, \sqrt {-a^{2} x^{2} + 1} a^{2}}{x^{2}} + \frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{x^{4}}\right )} a - \frac {\arcsin \left (a x\right )}{5 \, x^{5}} \]
-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
Time = 0.28 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.26 \[ \int \frac {\arcsin (a x)}{x^6} \, dx=-\frac {3 \, a^{6} \log \left (\sqrt {-a^{2} x^{2} + 1} + 1\right ) - 3 \, a^{6} \log \left (-\sqrt {-a^{2} x^{2} + 1} + 1\right ) - \frac {2 \, {\left (3 \, {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} a^{6} - 5 \, \sqrt {-a^{2} x^{2} + 1} a^{6}\right )}}{a^{4} x^{4}}}{80 \, a} - \frac {\arcsin \left (a x\right )}{5 \, x^{5}} \]
-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
Timed out. \[ \int \frac {\arcsin (a x)}{x^6} \, dx=\int \frac {\mathrm {asin}\left (a\,x\right )}{x^6} \,d x \]