Integrand size = 10, antiderivative size = 83 \[ \int \frac {x^3}{\arcsin (a x)^3} \, dx=-\frac {x^3 \sqrt {1-a^2 x^2}}{2 a \arcsin (a x)^2}-\frac {3 x^2}{2 a^2 \arcsin (a x)}+\frac {2 x^4}{\arcsin (a x)}-\frac {\text {Si}(2 \arcsin (a x))}{2 a^4}+\frac {\text {Si}(4 \arcsin (a x))}{a^4} \]
-3/2*x^2/a^2/arcsin(a*x)+2*x^4/arcsin(a*x)-1/2*Si(2*arcsin(a*x))/a^4+Si(4* arcsin(a*x))/a^4-1/2*x^3*(-a^2*x^2+1)^(1/2)/a/arcsin(a*x)^2
Time = 0.18 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.88 \[ \int \frac {x^3}{\arcsin (a x)^3} \, dx=\frac {\frac {a^2 x^2 \left (-a x \sqrt {1-a^2 x^2}+\left (-3+4 a^2 x^2\right ) \arcsin (a x)\right )}{\arcsin (a x)^2}-\text {Si}(2 \arcsin (a x))+2 \text {Si}(4 \arcsin (a x))}{2 a^4} \]
((a^2*x^2*(-(a*x*Sqrt[1 - a^2*x^2]) + (-3 + 4*a^2*x^2)*ArcSin[a*x]))/ArcSi n[a*x]^2 - SinIntegral[2*ArcSin[a*x]] + 2*SinIntegral[4*ArcSin[a*x]])/(2*a ^4)
Time = 0.72 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.33, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {5144, 5222, 5146, 4906, 27, 2009, 3042, 3780}
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.
| \(\displaystyle \int \frac {x^3}{\arcsin (a x)^3} \, dx\) |
| \(\Big \downarrow \) 5144 |
| \(\displaystyle \frac {3 \int \frac {x^2}{\sqrt {1-a^2 x^2} \arcsin (a x)^2}dx}{2 a}-2 a \int \frac {x^4}{\sqrt {1-a^2 x^2} \arcsin (a x)^2}dx-\frac {x^3 \sqrt {1-a^2 x^2}}{2 a \arcsin (a x)^2}\) |
| \(\Big \downarrow \) 5222 |
| \(\displaystyle \frac {3 \left (\frac {2 \int \frac {x}{\arcsin (a x)}dx}{a}-\frac {x^2}{a \arcsin (a x)}\right )}{2 a}-2 a \left (\frac {4 \int \frac {x^3}{\arcsin (a x)}dx}{a}-\frac {x^4}{a \arcsin (a x)}\right )-\frac {x^3 \sqrt {1-a^2 x^2}}{2 a \arcsin (a x)^2}\) |
| \(\Big \downarrow \) 5146 |
| \(\displaystyle \frac {3 \left (\frac {2 \int \frac {a x \sqrt {1-a^2 x^2}}{\arcsin (a x)}d\arcsin (a x)}{a^3}-\frac {x^2}{a \arcsin (a x)}\right )}{2 a}-2 a \left (\frac {4 \int \frac {a^3 x^3 \sqrt {1-a^2 x^2}}{\arcsin (a x)}d\arcsin (a x)}{a^5}-\frac {x^4}{a \arcsin (a x)}\right )-\frac {x^3 \sqrt {1-a^2 x^2}}{2 a \arcsin (a x)^2}\) |
| \(\Big \downarrow \) 4906 |
| \(\displaystyle -2 a \left (\frac {4 \int \left (\frac {\sin (2 \arcsin (a x))}{4 \arcsin (a x)}-\frac {\sin (4 \arcsin (a x))}{8 \arcsin (a x)}\right )d\arcsin (a x)}{a^5}-\frac {x^4}{a \arcsin (a x)}\right )+\frac {3 \left (\frac {2 \int \frac {\sin (2 \arcsin (a x))}{2 \arcsin (a x)}d\arcsin (a x)}{a^3}-\frac {x^2}{a \arcsin (a x)}\right )}{2 a}-\frac {x^3 \sqrt {1-a^2 x^2}}{2 a \arcsin (a x)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -2 a \left (\frac {4 \int \left (\frac {\sin (2 \arcsin (a x))}{4 \arcsin (a x)}-\frac {\sin (4 \arcsin (a x))}{8 \arcsin (a x)}\right )d\arcsin (a x)}{a^5}-\frac {x^4}{a \arcsin (a x)}\right )+\frac {3 \left (\frac {\int \frac {\sin (2 \arcsin (a x))}{\arcsin (a x)}d\arcsin (a x)}{a^3}-\frac {x^2}{a \arcsin (a x)}\right )}{2 a}-\frac {x^3 \sqrt {1-a^2 x^2}}{2 a \arcsin (a x)^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {3 \left (\frac {\int \frac {\sin (2 \arcsin (a x))}{\arcsin (a x)}d\arcsin (a x)}{a^3}-\frac {x^2}{a \arcsin (a x)}\right )}{2 a}-2 a \left (\frac {4 \left (\frac {1}{4} \text {Si}(2 \arcsin (a x))-\frac {1}{8} \text {Si}(4 \arcsin (a x))\right )}{a^5}-\frac {x^4}{a \arcsin (a x)}\right )-\frac {x^3 \sqrt {1-a^2 x^2}}{2 a \arcsin (a x)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 \left (\frac {\int \frac {\sin (2 \arcsin (a x))}{\arcsin (a x)}d\arcsin (a x)}{a^3}-\frac {x^2}{a \arcsin (a x)}\right )}{2 a}-2 a \left (\frac {4 \left (\frac {1}{4} \text {Si}(2 \arcsin (a x))-\frac {1}{8} \text {Si}(4 \arcsin (a x))\right )}{a^5}-\frac {x^4}{a \arcsin (a x)}\right )-\frac {x^3 \sqrt {1-a^2 x^2}}{2 a \arcsin (a x)^2}\) |
| \(\Big \downarrow \) 3780 |
| \(\displaystyle -2 a \left (\frac {4 \left (\frac {1}{4} \text {Si}(2 \arcsin (a x))-\frac {1}{8} \text {Si}(4 \arcsin (a x))\right )}{a^5}-\frac {x^4}{a \arcsin (a x)}\right )+\frac {3 \left (\frac {\text {Si}(2 \arcsin (a x))}{a^3}-\frac {x^2}{a \arcsin (a x)}\right )}{2 a}-\frac {x^3 \sqrt {1-a^2 x^2}}{2 a \arcsin (a x)^2}\) |
-1/2*(x^3*Sqrt[1 - a^2*x^2])/(a*ArcSin[a*x]^2) + (3*(-(x^2/(a*ArcSin[a*x]) ) + SinIntegral[2*ArcSin[a*x]]/a^3))/(2*a) - 2*a*(-(x^4/(a*ArcSin[a*x])) + (4*(SinIntegral[2*ArcSin[a*x]]/4 - SinIntegral[4*ArcSin[a*x]]/8))/a^5)
3.1.61.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinInte gral[e + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*e - c*f, 0]
Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b _.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin[a + b*x ]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IG tQ[p, 0]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[x ^m*Sqrt[1 - c^2*x^2]*((a + b*ArcSin[c*x])^(n + 1)/(b*c*(n + 1))), x] + (Sim p[c*((m + 1)/(b*(n + 1))) Int[x^(m + 1)*((a + b*ArcSin[c*x])^(n + 1)/Sqrt [1 - c^2*x^2]), x], x] - Simp[m/(b*c*(n + 1)) Int[x^(m - 1)*((a + b*ArcSi n[c*x])^(n + 1)/Sqrt[1 - c^2*x^2]), x], x]) /; FreeQ[{a, b, c}, x] && IGtQ[ m, 0] && LtQ[n, -2]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[1 /(b*c^(m + 1)) Subst[Int[x^n*Sin[-a/b + x/b]^m*Cos[-a/b + x/b], x], x, a + b*ArcSin[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]
Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[((f*x)^m/(b*c*(n + 1)))*Simp[Sqrt[1 - c^ 2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSin[c*x])^(n + 1), x] - Simp[f*(m/(b*c*(n + 1)))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]] Int[(f*x)^(m - 1)*(a + b* ArcSin[c*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2* d + e, 0] && LtQ[n, -1]
Time = 0.04 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.99
| method | result | size |
| derivativedivides | \(\frac {-\frac {\sin \left (2 \arcsin \left (a x \right )\right )}{8 \arcsin \left (a x \right )^{2}}-\frac {\cos \left (2 \arcsin \left (a x \right )\right )}{4 \arcsin \left (a x \right )}-\frac {\operatorname {Si}\left (2 \arcsin \left (a x \right )\right )}{2}+\frac {\sin \left (4 \arcsin \left (a x \right )\right )}{16 \arcsin \left (a x \right )^{2}}+\frac {\cos \left (4 \arcsin \left (a x \right )\right )}{4 \arcsin \left (a x \right )}+\operatorname {Si}\left (4 \arcsin \left (a x \right )\right )}{a^{4}}\) | \(82\) |
| default | \(\frac {-\frac {\sin \left (2 \arcsin \left (a x \right )\right )}{8 \arcsin \left (a x \right )^{2}}-\frac {\cos \left (2 \arcsin \left (a x \right )\right )}{4 \arcsin \left (a x \right )}-\frac {\operatorname {Si}\left (2 \arcsin \left (a x \right )\right )}{2}+\frac {\sin \left (4 \arcsin \left (a x \right )\right )}{16 \arcsin \left (a x \right )^{2}}+\frac {\cos \left (4 \arcsin \left (a x \right )\right )}{4 \arcsin \left (a x \right )}+\operatorname {Si}\left (4 \arcsin \left (a x \right )\right )}{a^{4}}\) | \(82\) |
1/a^4*(-1/8/arcsin(a*x)^2*sin(2*arcsin(a*x))-1/4/arcsin(a*x)*cos(2*arcsin( a*x))-1/2*Si(2*arcsin(a*x))+1/16/arcsin(a*x)^2*sin(4*arcsin(a*x))+1/4/arcs in(a*x)*cos(4*arcsin(a*x))+Si(4*arcsin(a*x)))
\[ \int \frac {x^3}{\arcsin (a x)^3} \, dx=\int { \frac {x^{3}}{\arcsin \left (a x\right )^{3}} \,d x } \]
\[ \int \frac {x^3}{\arcsin (a x)^3} \, dx=\int \frac {x^{3}}{\operatorname {asin}^{3}{\left (a x \right )}}\, dx \]
\[ \int \frac {x^3}{\arcsin (a x)^3} \, dx=\int { \frac {x^{3}}{\arcsin \left (a x\right )^{3}} \,d x } \]
-1/2*(sqrt(a*x + 1)*sqrt(-a*x + 1)*a*x^3 + 2*arctan2(a*x, sqrt(a*x + 1)*sq rt(-a*x + 1))^2*integrate((8*a^2*x^3 - 3*x)/arctan2(a*x, sqrt(a*x + 1)*sqr t(-a*x + 1)), x) - (4*a^2*x^4 - 3*x^2)*arctan2(a*x, sqrt(a*x + 1)*sqrt(-a* x + 1)))/(a^2*arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1))^2)
Time = 0.29 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.51 \[ \int \frac {x^3}{\arcsin (a x)^3} \, dx=\frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x}{2 \, a^{3} \arcsin \left (a x\right )^{2}} + \frac {2 \, {\left (a^{2} x^{2} - 1\right )}^{2}}{a^{4} \arcsin \left (a x\right )} + \frac {\operatorname {Si}\left (4 \, \arcsin \left (a x\right )\right )}{a^{4}} - \frac {\operatorname {Si}\left (2 \, \arcsin \left (a x\right )\right )}{2 \, a^{4}} - \frac {\sqrt {-a^{2} x^{2} + 1} x}{2 \, a^{3} \arcsin \left (a x\right )^{2}} + \frac {5 \, {\left (a^{2} x^{2} - 1\right )}}{2 \, a^{4} \arcsin \left (a x\right )} + \frac {1}{2 \, a^{4} \arcsin \left (a x\right )} \]
1/2*(-a^2*x^2 + 1)^(3/2)*x/(a^3*arcsin(a*x)^2) + 2*(a^2*x^2 - 1)^2/(a^4*ar csin(a*x)) + sin_integral(4*arcsin(a*x))/a^4 - 1/2*sin_integral(2*arcsin(a *x))/a^4 - 1/2*sqrt(-a^2*x^2 + 1)*x/(a^3*arcsin(a*x)^2) + 5/2*(a^2*x^2 - 1 )/(a^4*arcsin(a*x)) + 1/2/(a^4*arcsin(a*x))
Timed out. \[ \int \frac {x^3}{\arcsin (a x)^3} \, dx=\int \frac {x^3}{{\mathrm {asin}\left (a\,x\right )}^3} \,d x \]