Integrand size = 8, antiderivative size = 97 \[ \int \frac {x}{\arcsin (a x)^4} \, dx=-\frac {x \sqrt {1-a^2 x^2}}{3 a \arcsin (a x)^3}-\frac {1}{6 a^2 \arcsin (a x)^2}+\frac {x^2}{3 \arcsin (a x)^2}+\frac {2 x \sqrt {1-a^2 x^2}}{3 a \arcsin (a x)}-\frac {2 \operatorname {CosIntegral}(2 \arcsin (a x))}{3 a^2} \]
-1/6/a^2/arcsin(a*x)^2+1/3*x^2/arcsin(a*x)^2-2/3*Ci(2*arcsin(a*x))/a^2-1/3 *x*(-a^2*x^2+1)^(1/2)/a/arcsin(a*x)^3+2/3*x*(-a^2*x^2+1)^(1/2)/a/arcsin(a* x)
Time = 0.15 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.89 \[ \int \frac {x}{\arcsin (a x)^4} \, dx=\frac {-2 a x \sqrt {1-a^2 x^2}+\left (-1+2 a^2 x^2\right ) \arcsin (a x)+4 a x \sqrt {1-a^2 x^2} \arcsin (a x)^2-4 \arcsin (a x)^3 \operatorname {CosIntegral}(2 \arcsin (a x))}{6 a^2 \arcsin (a x)^3} \]
(-2*a*x*Sqrt[1 - a^2*x^2] + (-1 + 2*a^2*x^2)*ArcSin[a*x] + 4*a*x*Sqrt[1 - a^2*x^2]*ArcSin[a*x]^2 - 4*ArcSin[a*x]^3*CosIntegral[2*ArcSin[a*x]])/(6*a^ 2*ArcSin[a*x]^3)
Time = 0.62 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {5144, 5152, 5222, 5142, 3042, 3783}
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.
\(\displaystyle \int \frac {x}{\arcsin (a x)^4} \, dx\) |
\(\Big \downarrow \) 5144 |
\(\displaystyle \frac {\int \frac {1}{\sqrt {1-a^2 x^2} \arcsin (a x)^3}dx}{3 a}-\frac {2}{3} a \int \frac {x^2}{\sqrt {1-a^2 x^2} \arcsin (a x)^3}dx-\frac {x \sqrt {1-a^2 x^2}}{3 a \arcsin (a x)^3}\) |
\(\Big \downarrow \) 5152 |
\(\displaystyle -\frac {2}{3} a \int \frac {x^2}{\sqrt {1-a^2 x^2} \arcsin (a x)^3}dx-\frac {x \sqrt {1-a^2 x^2}}{3 a \arcsin (a x)^3}-\frac {1}{6 a^2 \arcsin (a x)^2}\) |
\(\Big \downarrow \) 5222 |
\(\displaystyle -\frac {2}{3} a \left (\frac {\int \frac {x}{\arcsin (a x)^2}dx}{a}-\frac {x^2}{2 a \arcsin (a x)^2}\right )-\frac {x \sqrt {1-a^2 x^2}}{3 a \arcsin (a x)^3}-\frac {1}{6 a^2 \arcsin (a x)^2}\) |
\(\Big \downarrow \) 5142 |
\(\displaystyle -\frac {2}{3} a \left (\frac {\frac {\int \frac {\cos (2 \arcsin (a x))}{\arcsin (a x)}d\arcsin (a x)}{a^2}-\frac {x \sqrt {1-a^2 x^2}}{a \arcsin (a x)}}{a}-\frac {x^2}{2 a \arcsin (a x)^2}\right )-\frac {x \sqrt {1-a^2 x^2}}{3 a \arcsin (a x)^3}-\frac {1}{6 a^2 \arcsin (a x)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2}{3} a \left (\frac {\frac {\int \frac {\sin \left (2 \arcsin (a x)+\frac {\pi }{2}\right )}{\arcsin (a x)}d\arcsin (a x)}{a^2}-\frac {x \sqrt {1-a^2 x^2}}{a \arcsin (a x)}}{a}-\frac {x^2}{2 a \arcsin (a x)^2}\right )-\frac {x \sqrt {1-a^2 x^2}}{3 a \arcsin (a x)^3}-\frac {1}{6 a^2 \arcsin (a x)^2}\) |
\(\Big \downarrow \) 3783 |
\(\displaystyle -\frac {2}{3} a \left (\frac {\frac {\operatorname {CosIntegral}(2 \arcsin (a x))}{a^2}-\frac {x \sqrt {1-a^2 x^2}}{a \arcsin (a x)}}{a}-\frac {x^2}{2 a \arcsin (a x)^2}\right )-\frac {x \sqrt {1-a^2 x^2}}{3 a \arcsin (a x)^3}-\frac {1}{6 a^2 \arcsin (a x)^2}\) |
-1/3*(x*Sqrt[1 - a^2*x^2])/(a*ArcSin[a*x]^3) - 1/(6*a^2*ArcSin[a*x]^2) - ( 2*a*(-1/2*x^2/(a*ArcSin[a*x]^2) + (-((x*Sqrt[1 - a^2*x^2])/(a*ArcSin[a*x]) ) + CosIntegral[2*ArcSin[a*x]]/a^2)/a))/3
3.1.70.3.1 Defintions of rubi rules used
Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosInte gral[e - Pi/2 + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 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] - Simp [1/(b^2*c^(m + 1)*(n + 1)) Subst[Int[ExpandTrigReduce[x^(n + 1), Sin[-a/b + x/b]^(m - 1)*(m - (m + 1)*Sin[-a/b + x/b]^2), x], x], x, a + b*ArcSin[c* x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && GeQ[n, -2] && LtQ[n, -1]
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_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_S ymbol] :> Simp[(1/(b*c*(n + 1)))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSin[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e, 0] && NeQ[n, -1]
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 = 0.03 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.62
method | result | size |
derivativedivides | \(\frac {-\frac {\sin \left (2 \arcsin \left (a x \right )\right )}{6 \arcsin \left (a x \right )^{3}}-\frac {\cos \left (2 \arcsin \left (a x \right )\right )}{6 \arcsin \left (a x \right )^{2}}+\frac {\sin \left (2 \arcsin \left (a x \right )\right )}{3 \arcsin \left (a x \right )}-\frac {2 \,\operatorname {Ci}\left (2 \arcsin \left (a x \right )\right )}{3}}{a^{2}}\) | \(60\) |
default | \(\frac {-\frac {\sin \left (2 \arcsin \left (a x \right )\right )}{6 \arcsin \left (a x \right )^{3}}-\frac {\cos \left (2 \arcsin \left (a x \right )\right )}{6 \arcsin \left (a x \right )^{2}}+\frac {\sin \left (2 \arcsin \left (a x \right )\right )}{3 \arcsin \left (a x \right )}-\frac {2 \,\operatorname {Ci}\left (2 \arcsin \left (a x \right )\right )}{3}}{a^{2}}\) | \(60\) |
1/a^2*(-1/6/arcsin(a*x)^3*sin(2*arcsin(a*x))-1/6/arcsin(a*x)^2*cos(2*arcsi n(a*x))+1/3/arcsin(a*x)*sin(2*arcsin(a*x))-2/3*Ci(2*arcsin(a*x)))
\[ \int \frac {x}{\arcsin (a x)^4} \, dx=\int { \frac {x}{\arcsin \left (a x\right )^{4}} \,d x } \]
\[ \int \frac {x}{\arcsin (a x)^4} \, dx=\int \frac {x}{\operatorname {asin}^{4}{\left (a x \right )}}\, dx \]
\[ \int \frac {x}{\arcsin (a x)^4} \, dx=\int { \frac {x}{\arcsin \left (a x\right )^{4}} \,d x } \]
-1/6*(6*a^2*arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1))^3*integrate(2/3*(2* a^2*x^2 - 1)*sqrt(a*x + 1)*sqrt(-a*x + 1)/((a^3*x^2 - a)*arctan2(a*x, sqrt (a*x + 1)*sqrt(-a*x + 1))), x) - 2*(2*a*x*arctan2(a*x, sqrt(a*x + 1)*sqrt( -a*x + 1))^2 - a*x)*sqrt(a*x + 1)*sqrt(-a*x + 1) - (2*a^2*x^2 - 1)*arctan2 (a*x, sqrt(a*x + 1)*sqrt(-a*x + 1)))/(a^2*arctan2(a*x, sqrt(a*x + 1)*sqrt( -a*x + 1))^3)
Time = 0.27 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.95 \[ \int \frac {x}{\arcsin (a x)^4} \, dx=\frac {2 \, \sqrt {-a^{2} x^{2} + 1} x}{3 \, a \arcsin \left (a x\right )} - \frac {2 \, \operatorname {Ci}\left (2 \, \arcsin \left (a x\right )\right )}{3 \, a^{2}} - \frac {\sqrt {-a^{2} x^{2} + 1} x}{3 \, a \arcsin \left (a x\right )^{3}} + \frac {a^{2} x^{2} - 1}{3 \, a^{2} \arcsin \left (a x\right )^{2}} + \frac {1}{6 \, a^{2} \arcsin \left (a x\right )^{2}} \]
2/3*sqrt(-a^2*x^2 + 1)*x/(a*arcsin(a*x)) - 2/3*cos_integral(2*arcsin(a*x)) /a^2 - 1/3*sqrt(-a^2*x^2 + 1)*x/(a*arcsin(a*x)^3) + 1/3*(a^2*x^2 - 1)/(a^2 *arcsin(a*x)^2) + 1/6/(a^2*arcsin(a*x)^2)
Timed out. \[ \int \frac {x}{\arcsin (a x)^4} \, dx=\int \frac {x}{{\mathrm {asin}\left (a\,x\right )}^4} \,d x \]