Integrand size = 33, antiderivative size = 71 \[ \int \frac {\sqrt {1-a^2-2 a b x-b^2 x^2}}{\arcsin (a+b x)^3} \, dx=\frac {-1+(a+b x)^2}{2 b \arcsin (a+b x)^2}+\frac {(a+b x) \sqrt {1-(a+b x)^2}}{b \arcsin (a+b x)}-\frac {\operatorname {CosIntegral}(2 \arcsin (a+b x))}{b} \]
1/2*(-1+(b*x+a)^2)/b/arcsin(b*x+a)^2-Ci(2*arcsin(b*x+a))/b+(b*x+a)*(1-(b*x +a)^2)^(1/2)/b/arcsin(b*x+a)
Time = 0.21 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.24 \[ \int \frac {\sqrt {1-a^2-2 a b x-b^2 x^2}}{\arcsin (a+b x)^3} \, dx=\frac {-1+a^2+2 a b x+b^2 x^2+2 (a+b x) \sqrt {1-a^2-2 a b x-b^2 x^2} \arcsin (a+b x)-2 \arcsin (a+b x)^2 \operatorname {CosIntegral}(2 \arcsin (a+b x))}{2 b \arcsin (a+b x)^2} \]
(-1 + a^2 + 2*a*b*x + b^2*x^2 + 2*(a + b*x)*Sqrt[1 - a^2 - 2*a*b*x - b^2*x ^2]*ArcSin[a + b*x] - 2*ArcSin[a + b*x]^2*CosIntegral[2*ArcSin[a + b*x]])/ (2*b*ArcSin[a + b*x]^2)
Time = 0.39 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {5306, 5166, 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.
\(\displaystyle \int \frac {\sqrt {-a^2-2 a b x-b^2 x^2+1}}{\arcsin (a+b x)^3} \, dx\) |
\(\Big \downarrow \) 5306 |
\(\displaystyle \frac {\int \frac {\sqrt {1-(a+b x)^2}}{\arcsin (a+b x)^3}d(a+b x)}{b}\) |
\(\Big \downarrow \) 5166 |
\(\displaystyle \frac {-\int \frac {a+b x}{\arcsin (a+b x)^2}d(a+b x)-\frac {1-(a+b x)^2}{2 \arcsin (a+b x)^2}}{b}\) |
\(\Big \downarrow \) 5142 |
\(\displaystyle \frac {-\int \frac {\cos (2 \arcsin (a+b x))}{\arcsin (a+b x)}d\arcsin (a+b x)+\frac {\sqrt {1-(a+b x)^2} (a+b x)}{\arcsin (a+b x)}-\frac {1-(a+b x)^2}{2 \arcsin (a+b x)^2}}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {-\int \frac {\sin \left (2 \arcsin (a+b x)+\frac {\pi }{2}\right )}{\arcsin (a+b x)}d\arcsin (a+b x)+\frac {\sqrt {1-(a+b x)^2} (a+b x)}{\arcsin (a+b x)}-\frac {1-(a+b x)^2}{2 \arcsin (a+b x)^2}}{b}\) |
\(\Big \downarrow \) 3783 |
\(\displaystyle \frac {-\operatorname {CosIntegral}(2 \arcsin (a+b x))+\frac {\sqrt {1-(a+b x)^2} (a+b x)}{\arcsin (a+b x)}-\frac {1-(a+b x)^2}{2 \arcsin (a+b x)^2}}{b}\) |
(-1/2*(1 - (a + b*x)^2)/ArcSin[a + b*x]^2 + ((a + b*x)*Sqrt[1 - (a + b*x)^ 2])/ArcSin[a + b*x] - CosIntegral[2*ArcSin[a + b*x]])/b
3.4.18.3.1 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_)*((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_) + (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 = 1.74 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.86
method | result | size |
default | \(-\frac {4 \,\operatorname {Ci}\left (2 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )^{2}-2 \sin \left (2 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )+\cos \left (2 \arcsin \left (b x +a \right )\right )+1}{4 b \arcsin \left (b x +a \right )^{2}}\) | \(61\) |
-1/4/b*(4*Ci(2*arcsin(b*x+a))*arcsin(b*x+a)^2-2*sin(2*arcsin(b*x+a))*arcsi n(b*x+a)+cos(2*arcsin(b*x+a))+1)/arcsin(b*x+a)^2
\[ \int \frac {\sqrt {1-a^2-2 a b x-b^2 x^2}}{\arcsin (a+b x)^3} \, dx=\int { \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{\arcsin \left (b x + a\right )^{3}} \,d x } \]
\[ \int \frac {\sqrt {1-a^2-2 a b x-b^2 x^2}}{\arcsin (a+b x)^3} \, dx=\int \frac {\sqrt {- \left (a + b x - 1\right ) \left (a + b x + 1\right )}}{\operatorname {asin}^{3}{\left (a + b x \right )}}\, dx \]
\[ \int \frac {\sqrt {1-a^2-2 a b x-b^2 x^2}}{\arcsin (a+b x)^3} \, dx=\int { \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{\arcsin \left (b x + a\right )^{3}} \,d x } \]
1/2*(b^2*x^2 - 2*b*arctan2(b*x + a, sqrt(b*x + a + 1)*sqrt(-b*x - a + 1))^ 2*integrate((2*b^2*x^2 + 4*a*b*x + 2*a^2 - 1)*sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)/((b^2*x^2 + 2*a*b*x + a^2 - 1)*arctan2(b*x + a, sqrt(b*x + a + 1) *sqrt(-b*x - a + 1))), x) + 2*a*b*x + 2*sqrt(b*x + a + 1)*(b*x + a)*sqrt(- b*x - a + 1)*arctan2(b*x + a, sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)) + a^2 - 1)/(b*arctan2(b*x + a, sqrt(b*x + a + 1)*sqrt(-b*x - a + 1))^2)
Time = 0.36 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.18 \[ \int \frac {\sqrt {1-a^2-2 a b x-b^2 x^2}}{\arcsin (a+b x)^3} \, dx=-\frac {\operatorname {Ci}\left (2 \, \arcsin \left (b x + a\right )\right )}{b} + \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{b \arcsin \left (b x + a\right )} + \frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{2 \, b \arcsin \left (b x + a\right )^{2}} \]
-cos_integral(2*arcsin(b*x + a))/b + sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(b *x + a)/(b*arcsin(b*x + a)) + 1/2*(b^2*x^2 + 2*a*b*x + a^2 - 1)/(b*arcsin( b*x + a)^2)
Timed out. \[ \int \frac {\sqrt {1-a^2-2 a b x-b^2 x^2}}{\arcsin (a+b x)^3} \, dx=\int \frac {\sqrt {-a^2-2\,a\,b\,x-b^2\,x^2+1}}{{\mathrm {asin}\left (a+b\,x\right )}^3} \,d x \]