Integrand size = 33, antiderivative size = 115 \[ \int \frac {\sqrt {1-a^2-2 a b x-b^2 x^2}}{\arcsin (a+b x)^4} \, dx=-\frac {1-(a+b x)^2}{3 b \arcsin (a+b x)^3}+\frac {(a+b x) \sqrt {1-(a+b x)^2}}{3 b \arcsin (a+b x)^2}+\frac {1}{3 b \arcsin (a+b x)}-\frac {2 (a+b x)^2}{3 b \arcsin (a+b x)}+\frac {2 \text {Si}(2 \arcsin (a+b x))}{3 b} \]
1/3*(-1+(b*x+a)^2)/b/arcsin(b*x+a)^3+1/3/b/arcsin(b*x+a)-2/3*(b*x+a)^2/b/a rcsin(b*x+a)+2/3*Si(2*arcsin(b*x+a))/b+1/3*(b*x+a)*(1-(b*x+a)^2)^(1/2)/b/a rcsin(b*x+a)^2
Time = 0.07 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.02 \[ \int \frac {\sqrt {1-a^2-2 a b x-b^2 x^2}}{\arcsin (a+b x)^4} \, dx=\frac {-1+a^2+2 a b x+b^2 x^2+(a+b x) \sqrt {1-a^2-2 a b x-b^2 x^2} \arcsin (a+b x)-\left (-1+2 a^2+4 a b x+2 b^2 x^2\right ) \arcsin (a+b x)^2+2 \arcsin (a+b x)^3 \text {Si}(2 \arcsin (a+b x))}{3 b \arcsin (a+b x)^3} \]
(-1 + a^2 + 2*a*b*x + b^2*x^2 + (a + b*x)*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2 ]*ArcSin[a + b*x] - (-1 + 2*a^2 + 4*a*b*x + 2*b^2*x^2)*ArcSin[a + b*x]^2 + 2*ArcSin[a + b*x]^3*SinIntegral[2*ArcSin[a + b*x]])/(3*b*ArcSin[a + b*x]^ 3)
Time = 0.74 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.90, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {5306, 5166, 5144, 5152, 5222, 5146, 4906, 27, 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.
| \(\displaystyle \int \frac {\sqrt {-a^2-2 a b x-b^2 x^2+1}}{\arcsin (a+b x)^4} \, dx\) |
| \(\Big \downarrow \) 5306 |
| \(\displaystyle \frac {\int \frac {\sqrt {1-(a+b x)^2}}{\arcsin (a+b x)^4}d(a+b x)}{b}\) |
| \(\Big \downarrow \) 5166 |
| \(\displaystyle \frac {-\frac {2}{3} \int \frac {a+b x}{\arcsin (a+b x)^3}d(a+b x)-\frac {1-(a+b x)^2}{3 \arcsin (a+b x)^3}}{b}\) |
| \(\Big \downarrow \) 5144 |
| \(\displaystyle \frac {-\frac {2}{3} \left (\frac {1}{2} \int \frac {1}{\sqrt {1-(a+b x)^2} \arcsin (a+b x)^2}d(a+b x)-\int \frac {(a+b x)^2}{\sqrt {1-(a+b x)^2} \arcsin (a+b x)^2}d(a+b x)-\frac {\sqrt {1-(a+b x)^2} (a+b x)}{2 \arcsin (a+b x)^2}\right )-\frac {1-(a+b x)^2}{3 \arcsin (a+b x)^3}}{b}\) |
| \(\Big \downarrow \) 5152 |
| \(\displaystyle \frac {-\frac {2}{3} \left (-\int \frac {(a+b x)^2}{\sqrt {1-(a+b x)^2} \arcsin (a+b x)^2}d(a+b x)-\frac {\sqrt {1-(a+b x)^2} (a+b x)}{2 \arcsin (a+b x)^2}-\frac {1}{2 \arcsin (a+b x)}\right )-\frac {1-(a+b x)^2}{3 \arcsin (a+b x)^3}}{b}\) |
| \(\Big \downarrow \) 5222 |
| \(\displaystyle \frac {-\frac {2}{3} \left (-2 \int \frac {a+b x}{\arcsin (a+b x)}d(a+b x)+\frac {(a+b x)^2}{\arcsin (a+b x)}-\frac {\sqrt {1-(a+b x)^2} (a+b x)}{2 \arcsin (a+b x)^2}-\frac {1}{2 \arcsin (a+b x)}\right )-\frac {1-(a+b x)^2}{3 \arcsin (a+b x)^3}}{b}\) |
| \(\Big \downarrow \) 5146 |
| \(\displaystyle \frac {-\frac {2}{3} \left (-2 \int \frac {(a+b x) \sqrt {1-(a+b x)^2}}{\arcsin (a+b x)}d\arcsin (a+b x)+\frac {(a+b x)^2}{\arcsin (a+b x)}-\frac {\sqrt {1-(a+b x)^2} (a+b x)}{2 \arcsin (a+b x)^2}-\frac {1}{2 \arcsin (a+b x)}\right )-\frac {1-(a+b x)^2}{3 \arcsin (a+b x)^3}}{b}\) |
| \(\Big \downarrow \) 4906 |
| \(\displaystyle \frac {-\frac {2}{3} \left (-2 \int \frac {\sin (2 \arcsin (a+b x))}{2 \arcsin (a+b x)}d\arcsin (a+b x)+\frac {(a+b x)^2}{\arcsin (a+b x)}-\frac {\sqrt {1-(a+b x)^2} (a+b x)}{2 \arcsin (a+b x)^2}-\frac {1}{2 \arcsin (a+b x)}\right )-\frac {1-(a+b x)^2}{3 \arcsin (a+b x)^3}}{b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {2}{3} \left (-\int \frac {\sin (2 \arcsin (a+b x))}{\arcsin (a+b x)}d\arcsin (a+b x)+\frac {(a+b x)^2}{\arcsin (a+b x)}-\frac {\sqrt {1-(a+b x)^2} (a+b x)}{2 \arcsin (a+b x)^2}-\frac {1}{2 \arcsin (a+b x)}\right )-\frac {1-(a+b x)^2}{3 \arcsin (a+b x)^3}}{b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {2}{3} \left (-\int \frac {\sin (2 \arcsin (a+b x))}{\arcsin (a+b x)}d\arcsin (a+b x)+\frac {(a+b x)^2}{\arcsin (a+b x)}-\frac {\sqrt {1-(a+b x)^2} (a+b x)}{2 \arcsin (a+b x)^2}-\frac {1}{2 \arcsin (a+b x)}\right )-\frac {1-(a+b x)^2}{3 \arcsin (a+b x)^3}}{b}\) |
| \(\Big \downarrow \) 3780 |
| \(\displaystyle \frac {-\frac {2}{3} \left (-\text {Si}(2 \arcsin (a+b x))+\frac {(a+b x)^2}{\arcsin (a+b x)}-\frac {\sqrt {1-(a+b x)^2} (a+b x)}{2 \arcsin (a+b x)^2}-\frac {1}{2 \arcsin (a+b x)}\right )-\frac {1-(a+b x)^2}{3 \arcsin (a+b x)^3}}{b}\) |
(-1/3*(1 - (a + b*x)^2)/ArcSin[a + b*x]^3 - (2*(-1/2*((a + b*x)*Sqrt[1 - ( a + b*x)^2])/ArcSin[a + b*x]^2 - 1/(2*ArcSin[a + b*x]) + (a + b*x)^2/ArcSi n[a + b*x] - SinIntegral[2*ArcSin[a + b*x]]))/3)/b
3.4.19.3.1 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_.)/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_)*((d_) + (e_.)*(x_)^2)^(p_.), x_ Symbol] :> Simp[Sqrt[1 - c^2*x^2]*(d + e*x^2)^p*((a + b*ArcSin[c*x])^(n + 1 )/(b*c*(n + 1))), x] + Simp[c*((2*p + 1)/(b*(n + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p] Int[x*(1 - c^2*x^2)^(p - 1/2)*(a + b*ArcSin[c*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && LtQ[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]
Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((A_.) + (B_.)*(x_) + ( C_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/d Subst[Int[(-C/d^2 + (C/d^2)*x^2 )^p*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, A, B, C, n, p}, x] && EqQ[B*(1 - c^2) + 2*A*c*d, 0] && EqQ[2*c*C - B*d, 0]
Time = 1.71 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.70
| method | result | size |
| default | \(\frac {4 \,\operatorname {Si}\left (2 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )^{3}+2 \cos \left (2 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )^{2}+\sin \left (2 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )-\cos \left (2 \arcsin \left (b x +a \right )\right )-1}{6 b \arcsin \left (b x +a \right )^{3}}\) | \(81\) |
1/6/b*(4*Si(2*arcsin(b*x+a))*arcsin(b*x+a)^3+2*cos(2*arcsin(b*x+a))*arcsin (b*x+a)^2+sin(2*arcsin(b*x+a))*arcsin(b*x+a)-cos(2*arcsin(b*x+a))-1)/arcsi n(b*x+a)^3
\[ \int \frac {\sqrt {1-a^2-2 a b x-b^2 x^2}}{\arcsin (a+b x)^4} \, dx=\int { \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{\arcsin \left (b x + a\right )^{4}} \,d x } \]
\[ \int \frac {\sqrt {1-a^2-2 a b x-b^2 x^2}}{\arcsin (a+b x)^4} \, dx=\int \frac {\sqrt {- \left (a + b x - 1\right ) \left (a + b x + 1\right )}}{\operatorname {asin}^{4}{\left (a + b x \right )}}\, dx \]
Timed out. \[ \int \frac {\sqrt {1-a^2-2 a b x-b^2 x^2}}{\arcsin (a+b x)^4} \, dx=\text {Timed out} \]
Time = 0.37 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.11 \[ \int \frac {\sqrt {1-a^2-2 a b x-b^2 x^2}}{\arcsin (a+b x)^4} \, dx=\frac {2 \, \operatorname {Si}\left (2 \, \arcsin \left (b x + a\right )\right )}{3 \, b} - \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )}}{3 \, b \arcsin \left (b x + a\right )} + \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{3 \, b \arcsin \left (b x + a\right )^{2}} - \frac {1}{3 \, b \arcsin \left (b x + a\right )} + \frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{3 \, b \arcsin \left (b x + a\right )^{3}} \]
2/3*sin_integral(2*arcsin(b*x + a))/b - 2/3*(b^2*x^2 + 2*a*b*x + a^2 - 1)/ (b*arcsin(b*x + a)) + 1/3*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(b*x + a)/(b* arcsin(b*x + a)^2) - 1/3/(b*arcsin(b*x + a)) + 1/3*(b^2*x^2 + 2*a*b*x + a^ 2 - 1)/(b*arcsin(b*x + a)^3)
Timed out. \[ \int \frac {\sqrt {1-a^2-2 a b x-b^2 x^2}}{\arcsin (a+b x)^4} \, dx=\int \frac {\sqrt {-a^2-2\,a\,b\,x-b^2\,x^2+1}}{{\mathrm {asin}\left (a+b\,x\right )}^4} \,d x \]