Integrand size = 33, antiderivative size = 111 \[ \int \sqrt {1-a^2-2 a b x-b^2 x^2} \arcsin (a+b x)^2 \, dx=-\frac {(a+b x) \sqrt {1-(a+b x)^2}}{4 b}+\frac {\arcsin (a+b x)}{4 b}-\frac {(a+b x)^2 \arcsin (a+b x)}{2 b}+\frac {(a+b x) \sqrt {1-(a+b x)^2} \arcsin (a+b x)^2}{2 b}+\frac {\arcsin (a+b x)^3}{6 b} \]
1/4*arcsin(b*x+a)/b-1/2*(b*x+a)^2*arcsin(b*x+a)/b+1/6*arcsin(b*x+a)^3/b-1/ 4*(b*x+a)*(1-(b*x+a)^2)^(1/2)/b+1/2*(b*x+a)*arcsin(b*x+a)^2*(1-(b*x+a)^2)^ (1/2)/b
Time = 0.07 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.05 \[ \int \sqrt {1-a^2-2 a b x-b^2 x^2} \arcsin (a+b x)^2 \, dx=\frac {-3 (a+b x) \sqrt {1-a^2-2 a b x-b^2 x^2}-3 \left (-1+2 a^2+4 a b x+2 b^2 x^2\right ) \arcsin (a+b x)+6 (a+b x) \sqrt {1-a^2-2 a b x-b^2 x^2} \arcsin (a+b x)^2+2 \arcsin (a+b x)^3}{12 b} \]
(-3*(a + b*x)*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2] - 3*(-1 + 2*a^2 + 4*a*b*x + 2*b^2*x^2)*ArcSin[a + b*x] + 6*(a + b*x)*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^ 2]*ArcSin[a + b*x]^2 + 2*ArcSin[a + b*x]^3)/(12*b)
Time = 0.42 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {5306, 5156, 5138, 262, 223, 5152}
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.
\(\displaystyle \int \sqrt {-a^2-2 a b x-b^2 x^2+1} \arcsin (a+b x)^2 \, dx\) |
\(\Big \downarrow \) 5306 |
\(\displaystyle \frac {\int \sqrt {1-(a+b x)^2} \arcsin (a+b x)^2d(a+b x)}{b}\) |
\(\Big \downarrow \) 5156 |
\(\displaystyle \frac {-\int (a+b x) \arcsin (a+b x)d(a+b x)+\frac {1}{2} \int \frac {\arcsin (a+b x)^2}{\sqrt {1-(a+b x)^2}}d(a+b x)+\frac {1}{2} (a+b x) \sqrt {1-(a+b x)^2} \arcsin (a+b x)^2}{b}\) |
\(\Big \downarrow \) 5138 |
\(\displaystyle \frac {\frac {1}{2} \int \frac {\arcsin (a+b x)^2}{\sqrt {1-(a+b x)^2}}d(a+b x)+\frac {1}{2} \int \frac {(a+b x)^2}{\sqrt {1-(a+b x)^2}}d(a+b x)-\frac {1}{2} (a+b x)^2 \arcsin (a+b x)+\frac {1}{2} \sqrt {1-(a+b x)^2} (a+b x) \arcsin (a+b x)^2}{b}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {\frac {1}{2} \int \frac {\arcsin (a+b x)^2}{\sqrt {1-(a+b x)^2}}d(a+b x)+\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{\sqrt {1-(a+b x)^2}}d(a+b x)-\frac {1}{2} (a+b x) \sqrt {1-(a+b x)^2}\right )-\frac {1}{2} (a+b x)^2 \arcsin (a+b x)+\frac {1}{2} \sqrt {1-(a+b x)^2} (a+b x) \arcsin (a+b x)^2}{b}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {\frac {1}{2} \int \frac {\arcsin (a+b x)^2}{\sqrt {1-(a+b x)^2}}d(a+b x)-\frac {1}{2} (a+b x)^2 \arcsin (a+b x)+\frac {1}{2} \sqrt {1-(a+b x)^2} (a+b x) \arcsin (a+b x)^2+\frac {1}{2} \left (\frac {1}{2} \arcsin (a+b x)-\frac {1}{2} (a+b x) \sqrt {1-(a+b x)^2}\right )}{b}\) |
\(\Big \downarrow \) 5152 |
\(\displaystyle \frac {\frac {1}{6} \arcsin (a+b x)^3+\frac {1}{2} (a+b x) \sqrt {1-(a+b x)^2} \arcsin (a+b x)^2-\frac {1}{2} (a+b x)^2 \arcsin (a+b x)+\frac {1}{2} \left (\frac {1}{2} \arcsin (a+b x)-\frac {1}{2} (a+b x) \sqrt {1-(a+b x)^2}\right )}{b}\) |
((-1/2*((a + b*x)*Sqrt[1 - (a + b*x)^2]) + ArcSin[a + b*x]/2)/2 - ((a + b* x)^2*ArcSin[a + b*x])/2 + ((a + b*x)*Sqrt[1 - (a + b*x)^2]*ArcSin[a + b*x] ^2)/2 + ArcSin[a + b*x]^3/6)/b
3.4.14.3.1 Defintions of rubi rules used
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcSin[c*x])^n/(d*(m + 1))), x] - Simp[b*c*(n /(d*(m + 1))) Int[(d*x)^(m + 1)*((a + b*ArcSin[c*x])^(n - 1)/Sqrt[1 - c^2 *x^2]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]
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_.)*Sqrt[(d_) + (e_.)*(x_)^2], x_S ymbol] :> Simp[x*Sqrt[d + e*x^2]*((a + b*ArcSin[c*x])^n/2), x] + (Simp[(1/2 )*Simp[Sqrt[d + e*x^2]/Sqrt[1 - c^2*x^2]] Int[(a + b*ArcSin[c*x])^n/Sqrt[ 1 - c^2*x^2], x], x] - Simp[b*c*(n/2)*Simp[Sqrt[d + e*x^2]/Sqrt[1 - c^2*x^2 ]] Int[x*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x ] && EqQ[c^2*d + e, 0] && GtQ[n, 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 = 1.75 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.61
method | result | size |
default | \(\frac {6 \arcsin \left (b x +a \right )^{2} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, b x -6 \arcsin \left (b x +a \right ) b^{2} x^{2}+6 \arcsin \left (b x +a \right )^{2} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, a -12 \arcsin \left (b x +a \right ) a b x +2 \arcsin \left (b x +a \right )^{3}-6 a^{2} \arcsin \left (b x +a \right )-3 x b \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}-3 a \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}+3 \arcsin \left (b x +a \right )}{12 b}\) | \(179\) |
1/12*(6*arcsin(b*x+a)^2*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*b*x-6*arcsin(b*x+a) *b^2*x^2+6*arcsin(b*x+a)^2*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*a-12*arcsin(b*x+ a)*a*b*x+2*arcsin(b*x+a)^3-6*a^2*arcsin(b*x+a)-3*x*b*(-b^2*x^2-2*a*b*x-a^2 +1)^(1/2)-3*a*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+3*arcsin(b*x+a))/b
Time = 0.25 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.82 \[ \int \sqrt {1-a^2-2 a b x-b^2 x^2} \arcsin (a+b x)^2 \, dx=\frac {2 \, \arcsin \left (b x + a\right )^{3} - 3 \, {\left (2 \, b^{2} x^{2} + 4 \, a b x + 2 \, a^{2} - 1\right )} \arcsin \left (b x + a\right ) + 3 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (2 \, {\left (b x + a\right )} \arcsin \left (b x + a\right )^{2} - b x - a\right )}}{12 \, b} \]
1/12*(2*arcsin(b*x + a)^3 - 3*(2*b^2*x^2 + 4*a*b*x + 2*a^2 - 1)*arcsin(b*x + a) + 3*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(2*(b*x + a)*arcsin(b*x + a)^ 2 - b*x - a))/b
\[ \int \sqrt {1-a^2-2 a b x-b^2 x^2} \arcsin (a+b x)^2 \, dx=\int \sqrt {- \left (a + b x - 1\right ) \left (a + b x + 1\right )} \operatorname {asin}^{2}{\left (a + b x \right )}\, dx \]
\[ \int \sqrt {1-a^2-2 a b x-b^2 x^2} \arcsin (a+b x)^2 \, dx=\int { \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} \arcsin \left (b x + a\right )^{2} \,d x } \]
Time = 0.34 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.13 \[ \int \sqrt {1-a^2-2 a b x-b^2 x^2} \arcsin (a+b x)^2 \, dx=\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )} \arcsin \left (b x + a\right )^{2}}{2 \, b} + \frac {\arcsin \left (b x + a\right )^{3}}{6 \, b} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )} \arcsin \left (b x + a\right )}{2 \, b} - \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{4 \, b} - \frac {\arcsin \left (b x + a\right )}{4 \, b} \]
1/2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(b*x + a)*arcsin(b*x + a)^2/b + 1/6 *arcsin(b*x + a)^3/b - 1/2*(b^2*x^2 + 2*a*b*x + a^2 - 1)*arcsin(b*x + a)/b - 1/4*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(b*x + a)/b - 1/4*arcsin(b*x + a )/b
Timed out. \[ \int \sqrt {1-a^2-2 a b x-b^2 x^2} \arcsin (a+b x)^2 \, dx=\int {\mathrm {asin}\left (a+b\,x\right )}^2\,\sqrt {-a^2-2\,a\,b\,x-b^2\,x^2+1} \,d x \]