Integrand size = 33, antiderivative size = 57 \[ \int \frac {\left (1-a^2-2 a b x-b^2 x^2\right )^{3/2}}{\arcsin (a+b x)^2} \, dx=-\frac {\left (1-(a+b x)^2\right )^2}{b \arcsin (a+b x)}-\frac {\text {Si}(2 \arcsin (a+b x))}{b}-\frac {\text {Si}(4 \arcsin (a+b x))}{2 b} \]
Time = 0.23 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.23 \[ \int \frac {\left (1-a^2-2 a b x-b^2 x^2\right )^{3/2}}{\arcsin (a+b x)^2} \, dx=-\frac {2 \left (-1+a^2+2 a b x+b^2 x^2\right )^2+2 \arcsin (a+b x) \text {Si}(2 \arcsin (a+b x))+\arcsin (a+b x) \text {Si}(4 \arcsin (a+b x))}{2 b \arcsin (a+b x)} \]
-1/2*(2*(-1 + a^2 + 2*a*b*x + b^2*x^2)^2 + 2*ArcSin[a + b*x]*SinIntegral[2 *ArcSin[a + b*x]] + ArcSin[a + b*x]*SinIntegral[4*ArcSin[a + b*x]])/(b*Arc Sin[a + b*x])
Time = 0.43 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {5306, 5166, 5224, 4906, 2009}
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.
\(\displaystyle \int \frac {\left (-a^2-2 a b x-b^2 x^2+1\right )^{3/2}}{\arcsin (a+b x)^2} \, dx\) |
\(\Big \downarrow \) 5306 |
\(\displaystyle \frac {\int \frac {\left (1-(a+b x)^2\right )^{3/2}}{\arcsin (a+b x)^2}d(a+b x)}{b}\) |
\(\Big \downarrow \) 5166 |
\(\displaystyle \frac {-4 \int \frac {(a+b x) \left (1-(a+b x)^2\right )}{\arcsin (a+b x)}d(a+b x)-\frac {\left (1-(a+b x)^2\right )^2}{\arcsin (a+b x)}}{b}\) |
\(\Big \downarrow \) 5224 |
\(\displaystyle \frac {-4 \int \frac {(a+b x) \left (1-(a+b x)^2\right )^{3/2}}{\arcsin (a+b x)}d\arcsin (a+b x)-\frac {\left (1-(a+b x)^2\right )^2}{\arcsin (a+b x)}}{b}\) |
\(\Big \downarrow \) 4906 |
\(\displaystyle \frac {-4 \int \left (\frac {\sin (2 \arcsin (a+b x))}{4 \arcsin (a+b x)}+\frac {\sin (4 \arcsin (a+b x))}{8 \arcsin (a+b x)}\right )d\arcsin (a+b x)-\frac {\left (1-(a+b x)^2\right )^2}{\arcsin (a+b x)}}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-4 \left (\frac {1}{4} \text {Si}(2 \arcsin (a+b x))+\frac {1}{8} \text {Si}(4 \arcsin (a+b x))\right )-\frac {\left (1-(a+b x)^2\right )^2}{\arcsin (a+b x)}}{b}\) |
(-((1 - (a + b*x)^2)^2/ArcSin[a + b*x]) - 4*(SinIntegral[2*ArcSin[a + b*x] ]/4 + SinIntegral[4*ArcSin[a + b*x]]/8))/b
3.4.24.3.1 Defintions of rubi rules used
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_)*((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_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^ 2)^(p_.), x_Symbol] :> Simp[(1/(b*c^(m + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x ^2)^p] Subst[Int[x^n*Sin[-a/b + x/b]^m*Cos[-a/b + x/b]^(2*p + 1), x], x, a + b*ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e, 0] && IGtQ[2*p + 2, 0] && IGtQ[m, 0]
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 = 1.78 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.23
method | result | size |
default | \(-\frac {8 \,\operatorname {Si}\left (2 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )+4 \,\operatorname {Si}\left (4 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )+4 \cos \left (2 \arcsin \left (b x +a \right )\right )+\cos \left (4 \arcsin \left (b x +a \right )\right )+3}{8 b \arcsin \left (b x +a \right )}\) | \(70\) |
-1/8/b*(8*Si(2*arcsin(b*x+a))*arcsin(b*x+a)+4*Si(4*arcsin(b*x+a))*arcsin(b *x+a)+4*cos(2*arcsin(b*x+a))+cos(4*arcsin(b*x+a))+3)/arcsin(b*x+a)
\[ \int \frac {\left (1-a^2-2 a b x-b^2 x^2\right )^{3/2}}{\arcsin (a+b x)^2} \, dx=\int { \frac {{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}}}{\arcsin \left (b x + a\right )^{2}} \,d x } \]
\[ \int \frac {\left (1-a^2-2 a b x-b^2 x^2\right )^{3/2}}{\arcsin (a+b x)^2} \, dx=\int \frac {\left (- \left (a + b x - 1\right ) \left (a + b x + 1\right )\right )^{\frac {3}{2}}}{\operatorname {asin}^{2}{\left (a + b x \right )}}\, dx \]
\[ \int \frac {\left (1-a^2-2 a b x-b^2 x^2\right )^{3/2}}{\arcsin (a+b x)^2} \, dx=\int { \frac {{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}}}{\arcsin \left (b x + a\right )^{2}} \,d x } \]
-(b^4*x^4 + 4*a*b^3*x^3 + 2*(3*a^2 - 1)*b^2*x^2 + a^4 + 4*(a^3 - a)*b*x - b*arctan2(b*x + a, sqrt(b*x + a + 1)*sqrt(-b*x - a + 1))*integrate(4*(b^3* x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2 - 1)*b*x - a)/arctan2(b*x + a, sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)), x) - 2*a^2 + 1)/(b*arctan2(b*x + a, sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)))
Time = 0.35 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.07 \[ \int \frac {\left (1-a^2-2 a b x-b^2 x^2\right )^{3/2}}{\arcsin (a+b x)^2} \, dx=-\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )}^{2}}{b \arcsin \left (b x + a\right )} - \frac {\operatorname {Si}\left (4 \, \arcsin \left (b x + a\right )\right )}{2 \, b} - \frac {\operatorname {Si}\left (2 \, \arcsin \left (b x + a\right )\right )}{b} \]
-(b^2*x^2 + 2*a*b*x + a^2 - 1)^2/(b*arcsin(b*x + a)) - 1/2*sin_integral(4* arcsin(b*x + a))/b - sin_integral(2*arcsin(b*x + a))/b
Timed out. \[ \int \frac {\left (1-a^2-2 a b x-b^2 x^2\right )^{3/2}}{\arcsin (a+b x)^2} \, dx=\int \frac {{\left (-a^2-2\,a\,b\,x-b^2\,x^2+1\right )}^{3/2}}{{\mathrm {asin}\left (a+b\,x\right )}^2} \,d x \]