Integrand size = 25, antiderivative size = 211 \[ \int x^2 \left (d-c^2 d x^2\right ) (a+b \arcsin (c x))^2 \, dx=-\frac {52 b^2 d x}{225 c^2}-\frac {26}{675} b^2 d x^3+\frac {2}{125} b^2 c^2 d x^5+\frac {8 b d \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{45 c^3}+\frac {4 b d x^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{45 c}+\frac {2 b d \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))}{15 c^3}-\frac {2 b d \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))}{25 c^3}+\frac {2}{15} d x^3 (a+b \arcsin (c x))^2+\frac {1}{5} d x^3 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2 \]
-52/225*b^2*d*x/c^2-26/675*b^2*d*x^3+2/125*b^2*c^2*d*x^5+2/15*b*d*(-c^2*x^ 2+1)^(3/2)*(a+b*arcsin(c*x))/c^3-2/25*b*d*(-c^2*x^2+1)^(5/2)*(a+b*arcsin(c *x))/c^3+2/15*d*x^3*(a+b*arcsin(c*x))^2+1/5*d*x^3*(-c^2*x^2+1)*(a+b*arcsin (c*x))^2+8/45*b*d*(a+b*arcsin(c*x))*(-c^2*x^2+1)^(1/2)/c^3+4/45*b*d*x^2*(a +b*arcsin(c*x))*(-c^2*x^2+1)^(1/2)/c
Time = 0.15 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.85 \[ \int x^2 \left (d-c^2 d x^2\right ) (a+b \arcsin (c x))^2 \, dx=-\frac {d \left (225 a^2 c^3 x^3 \left (-5+3 c^2 x^2\right )+30 a b \sqrt {1-c^2 x^2} \left (-26-13 c^2 x^2+9 c^4 x^4\right )+b^2 \left (780 c x+130 c^3 x^3-54 c^5 x^5\right )+30 b \left (15 a c^3 x^3 \left (-5+3 c^2 x^2\right )+b \sqrt {1-c^2 x^2} \left (-26-13 c^2 x^2+9 c^4 x^4\right )\right ) \arcsin (c x)+225 b^2 c^3 x^3 \left (-5+3 c^2 x^2\right ) \arcsin (c x)^2\right )}{3375 c^3} \]
-1/3375*(d*(225*a^2*c^3*x^3*(-5 + 3*c^2*x^2) + 30*a*b*Sqrt[1 - c^2*x^2]*(- 26 - 13*c^2*x^2 + 9*c^4*x^4) + b^2*(780*c*x + 130*c^3*x^3 - 54*c^5*x^5) + 30*b*(15*a*c^3*x^3*(-5 + 3*c^2*x^2) + b*Sqrt[1 - c^2*x^2]*(-26 - 13*c^2*x^ 2 + 9*c^4*x^4))*ArcSin[c*x] + 225*b^2*c^3*x^3*(-5 + 3*c^2*x^2)*ArcSin[c*x] ^2))/c^3
Time = 0.98 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.14, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {5202, 5138, 5194, 27, 2009, 5210, 15, 5182, 24}
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.
| \(\displaystyle \int x^2 \left (d-c^2 d x^2\right ) (a+b \arcsin (c x))^2 \, dx\) |
| \(\Big \downarrow \) 5202 |
| \(\displaystyle -\frac {2}{5} b c d \int x^3 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))dx+\frac {2}{5} d \int x^2 (a+b \arcsin (c x))^2dx+\frac {1}{5} d x^3 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2\) |
| \(\Big \downarrow \) 5138 |
| \(\displaystyle \frac {2}{5} d \left (\frac {1}{3} x^3 (a+b \arcsin (c x))^2-\frac {2}{3} b c \int \frac {x^3 (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}dx\right )-\frac {2}{5} b c d \int x^3 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))dx+\frac {1}{5} d x^3 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2\) |
| \(\Big \downarrow \) 5194 |
| \(\displaystyle \frac {2}{5} d \left (\frac {1}{3} x^3 (a+b \arcsin (c x))^2-\frac {2}{3} b c \int \frac {x^3 (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}dx\right )-\frac {2}{5} b c d \left (-b c \int -\frac {-3 c^4 x^4+c^2 x^2+2}{15 c^4}dx+\frac {\left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))}{5 c^4}-\frac {\left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))}{3 c^4}\right )+\frac {1}{5} d x^3 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2}{5} d \left (\frac {1}{3} x^3 (a+b \arcsin (c x))^2-\frac {2}{3} b c \int \frac {x^3 (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}dx\right )-\frac {2}{5} b c d \left (\frac {b \int \left (-3 c^4 x^4+c^2 x^2+2\right )dx}{15 c^3}+\frac {\left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))}{5 c^4}-\frac {\left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))}{3 c^4}\right )+\frac {1}{5} d x^3 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2}{5} d \left (\frac {1}{3} x^3 (a+b \arcsin (c x))^2-\frac {2}{3} b c \int \frac {x^3 (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}dx\right )+\frac {1}{5} d x^3 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2-\frac {2}{5} b c d \left (\frac {\left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))}{5 c^4}-\frac {\left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))}{3 c^4}+\frac {b \left (-\frac {3}{5} c^4 x^5+\frac {c^2 x^3}{3}+2 x\right )}{15 c^3}\right )\) |
| \(\Big \downarrow \) 5210 |
| \(\displaystyle \frac {2}{5} d \left (\frac {1}{3} x^3 (a+b \arcsin (c x))^2-\frac {2}{3} b c \left (\frac {2 \int \frac {x (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}dx}{3 c^2}+\frac {b \int x^2dx}{3 c}-\frac {x^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{3 c^2}\right )\right )+\frac {1}{5} d x^3 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2-\frac {2}{5} b c d \left (\frac {\left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))}{5 c^4}-\frac {\left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))}{3 c^4}+\frac {b \left (-\frac {3}{5} c^4 x^5+\frac {c^2 x^3}{3}+2 x\right )}{15 c^3}\right )\) |
| \(\Big \downarrow \) 15 |
| \(\displaystyle \frac {2}{5} d \left (\frac {1}{3} x^3 (a+b \arcsin (c x))^2-\frac {2}{3} b c \left (\frac {2 \int \frac {x (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}dx}{3 c^2}-\frac {x^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{3 c^2}+\frac {b x^3}{9 c}\right )\right )+\frac {1}{5} d x^3 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2-\frac {2}{5} b c d \left (\frac {\left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))}{5 c^4}-\frac {\left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))}{3 c^4}+\frac {b \left (-\frac {3}{5} c^4 x^5+\frac {c^2 x^3}{3}+2 x\right )}{15 c^3}\right )\) |
| \(\Big \downarrow \) 5182 |
| \(\displaystyle \frac {2}{5} d \left (\frac {1}{3} x^3 (a+b \arcsin (c x))^2-\frac {2}{3} b c \left (\frac {2 \left (\frac {b \int 1dx}{c}-\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{c^2}\right )}{3 c^2}-\frac {x^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{3 c^2}+\frac {b x^3}{9 c}\right )\right )+\frac {1}{5} d x^3 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2-\frac {2}{5} b c d \left (\frac {\left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))}{5 c^4}-\frac {\left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))}{3 c^4}+\frac {b \left (-\frac {3}{5} c^4 x^5+\frac {c^2 x^3}{3}+2 x\right )}{15 c^3}\right )\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{5} d x^3 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2+\frac {2}{5} d \left (\frac {1}{3} x^3 (a+b \arcsin (c x))^2-\frac {2}{3} b c \left (-\frac {x^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{3 c^2}+\frac {2 \left (\frac {b x}{c}-\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{c^2}\right )}{3 c^2}+\frac {b x^3}{9 c}\right )\right )-\frac {2}{5} b c d \left (\frac {\left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))}{5 c^4}-\frac {\left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))}{3 c^4}+\frac {b \left (-\frac {3}{5} c^4 x^5+\frac {c^2 x^3}{3}+2 x\right )}{15 c^3}\right )\) |
(d*x^3*(1 - c^2*x^2)*(a + b*ArcSin[c*x])^2)/5 - (2*b*c*d*((b*(2*x + (c^2*x ^3)/3 - (3*c^4*x^5)/5))/(15*c^3) - ((1 - c^2*x^2)^(3/2)*(a + b*ArcSin[c*x] ))/(3*c^4) + ((1 - c^2*x^2)^(5/2)*(a + b*ArcSin[c*x]))/(5*c^4)))/5 + (2*d* ((x^3*(a + b*ArcSin[c*x])^2)/3 - (2*b*c*((b*x^3)/(9*c) - (x^2*Sqrt[1 - c^2 *x^2]*(a + b*ArcSin[c*x]))/(3*c^2) + (2*((b*x)/c - (Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/c^2))/(3*c^2)))/3))/5
3.2.58.3.1 Defintions of rubi rules used
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_ .), x_Symbol] :> Simp[(d + e*x^2)^(p + 1)*((a + b*ArcSin[c*x])^n/(2*e*(p + 1))), x] + Simp[b*(n/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p] I nt[(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] && GtQ[n, 0] && NeQ[p, -1]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(p_) , x_Symbol] :> With[{u = IntHide[x^m*(d + e*x^2)^p, x]}, Simp[(a + b*ArcSin [c*x]) u, x] - Simp[b*c*Simp[Sqrt[d + e*x^2]/Sqrt[1 - c^2*x^2]] Int[Sim plifyIntegrand[u/Sqrt[d + e*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IntegerQ[p - 1/2] && NeQ[p, -2^(-1)] && (IGtQ[(m + 1)/2, 0] || ILtQ[(m + 2*p + 3)/2, 0])
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_. )*(x_)^2)^(p_.), x_Symbol] :> Simp[(f*x)^(m + 1)*(d + e*x^2)^p*((a + b*ArcS in[c*x])^n/(f*(m + 2*p + 1))), x] + (Simp[2*d*(p/(m + 2*p + 1)) Int[(f*x) ^m*(d + e*x^2)^(p - 1)*(a + b*ArcSin[c*x])^n, x], x] - Simp[b*c*(n/(f*(m + 2*p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p] Int[(f*x)^(m + 1)*(1 - c^2 *x^2)^(p - 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[p, 0] && !LtQ[m, -1]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_. )*(x_)^2)^(p_), x_Symbol] :> Simp[f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*((a + b*ArcSin[c*x])^n/(e*(m + 2*p + 1))), x] + (Simp[f^2*((m - 1)/(c^2*(m + 2*p + 1))) Int[(f*x)^(m - 2)*(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, x], x] + S imp[b*f*(n/(c*(m + 2*p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p] Int[(f* x)^(m - 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && IGtQ[m , 1] && NeQ[m + 2*p + 1, 0]
Time = 0.32 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.32
| method | result | size |
| parts | \(-d \,a^{2} \left (\frac {1}{5} c^{2} x^{5}-\frac {1}{3} x^{3}\right )-\frac {d \,b^{2} \left (\frac {\arcsin \left (c x \right )^{2} \left (c^{2} x^{2}-3\right ) c x}{3}+\frac {4 c x}{15}-\frac {4 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}}{15}+\frac {2 \arcsin \left (c x \right ) \left (c^{2} x^{2}-1\right ) \sqrt {-c^{2} x^{2}+1}}{45}-\frac {2 \left (c^{2} x^{2}-3\right ) c x}{135}+\frac {\arcsin \left (c x \right )^{2} \left (3 c^{4} x^{4}-10 c^{2} x^{2}+15\right ) c x}{15}+\frac {2 \arcsin \left (c x \right ) \left (c^{2} x^{2}-1\right )^{2} \sqrt {-c^{2} x^{2}+1}}{25}-\frac {2 \left (3 c^{4} x^{4}-10 c^{2} x^{2}+15\right ) c x}{375}\right )}{c^{3}}-\frac {2 d a b \left (\frac {\arcsin \left (c x \right ) c^{5} x^{5}}{5}-\frac {c^{3} x^{3} \arcsin \left (c x \right )}{3}+\frac {c^{4} x^{4} \sqrt {-c^{2} x^{2}+1}}{25}-\frac {13 c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{225}-\frac {26 \sqrt {-c^{2} x^{2}+1}}{225}\right )}{c^{3}}\) | \(279\) |
| derivativedivides | \(\frac {-d \,a^{2} \left (\frac {1}{5} c^{5} x^{5}-\frac {1}{3} c^{3} x^{3}\right )-d \,b^{2} \left (\frac {\arcsin \left (c x \right )^{2} \left (c^{2} x^{2}-3\right ) c x}{3}+\frac {4 c x}{15}-\frac {4 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}}{15}+\frac {2 \arcsin \left (c x \right ) \left (c^{2} x^{2}-1\right ) \sqrt {-c^{2} x^{2}+1}}{45}-\frac {2 \left (c^{2} x^{2}-3\right ) c x}{135}+\frac {\arcsin \left (c x \right )^{2} \left (3 c^{4} x^{4}-10 c^{2} x^{2}+15\right ) c x}{15}+\frac {2 \arcsin \left (c x \right ) \left (c^{2} x^{2}-1\right )^{2} \sqrt {-c^{2} x^{2}+1}}{25}-\frac {2 \left (3 c^{4} x^{4}-10 c^{2} x^{2}+15\right ) c x}{375}\right )-2 d a b \left (\frac {\arcsin \left (c x \right ) c^{5} x^{5}}{5}-\frac {c^{3} x^{3} \arcsin \left (c x \right )}{3}+\frac {c^{4} x^{4} \sqrt {-c^{2} x^{2}+1}}{25}-\frac {13 c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{225}-\frac {26 \sqrt {-c^{2} x^{2}+1}}{225}\right )}{c^{3}}\) | \(280\) |
| default | \(\frac {-d \,a^{2} \left (\frac {1}{5} c^{5} x^{5}-\frac {1}{3} c^{3} x^{3}\right )-d \,b^{2} \left (\frac {\arcsin \left (c x \right )^{2} \left (c^{2} x^{2}-3\right ) c x}{3}+\frac {4 c x}{15}-\frac {4 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}}{15}+\frac {2 \arcsin \left (c x \right ) \left (c^{2} x^{2}-1\right ) \sqrt {-c^{2} x^{2}+1}}{45}-\frac {2 \left (c^{2} x^{2}-3\right ) c x}{135}+\frac {\arcsin \left (c x \right )^{2} \left (3 c^{4} x^{4}-10 c^{2} x^{2}+15\right ) c x}{15}+\frac {2 \arcsin \left (c x \right ) \left (c^{2} x^{2}-1\right )^{2} \sqrt {-c^{2} x^{2}+1}}{25}-\frac {2 \left (3 c^{4} x^{4}-10 c^{2} x^{2}+15\right ) c x}{375}\right )-2 d a b \left (\frac {\arcsin \left (c x \right ) c^{5} x^{5}}{5}-\frac {c^{3} x^{3} \arcsin \left (c x \right )}{3}+\frac {c^{4} x^{4} \sqrt {-c^{2} x^{2}+1}}{25}-\frac {13 c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{225}-\frac {26 \sqrt {-c^{2} x^{2}+1}}{225}\right )}{c^{3}}\) | \(280\) |
-d*a^2*(1/5*c^2*x^5-1/3*x^3)-d*b^2/c^3*(1/3*arcsin(c*x)^2*(c^2*x^2-3)*c*x+ 4/15*c*x-4/15*arcsin(c*x)*(-c^2*x^2+1)^(1/2)+2/45*arcsin(c*x)*(c^2*x^2-1)* (-c^2*x^2+1)^(1/2)-2/135*(c^2*x^2-3)*c*x+1/15*arcsin(c*x)^2*(3*c^4*x^4-10* c^2*x^2+15)*c*x+2/25*arcsin(c*x)*(c^2*x^2-1)^2*(-c^2*x^2+1)^(1/2)-2/375*(3 *c^4*x^4-10*c^2*x^2+15)*c*x)-2*d*a*b/c^3*(1/5*arcsin(c*x)*c^5*x^5-1/3*c^3* x^3*arcsin(c*x)+1/25*c^4*x^4*(-c^2*x^2+1)^(1/2)-13/225*c^2*x^2*(-c^2*x^2+1 )^(1/2)-26/225*(-c^2*x^2+1)^(1/2))
Time = 0.26 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.92 \[ \int x^2 \left (d-c^2 d x^2\right ) (a+b \arcsin (c x))^2 \, dx=-\frac {27 \, {\left (25 \, a^{2} - 2 \, b^{2}\right )} c^{5} d x^{5} - 5 \, {\left (225 \, a^{2} - 26 \, b^{2}\right )} c^{3} d x^{3} + 780 \, b^{2} c d x + 225 \, {\left (3 \, b^{2} c^{5} d x^{5} - 5 \, b^{2} c^{3} d x^{3}\right )} \arcsin \left (c x\right )^{2} + 450 \, {\left (3 \, a b c^{5} d x^{5} - 5 \, a b c^{3} d x^{3}\right )} \arcsin \left (c x\right ) + 30 \, {\left (9 \, a b c^{4} d x^{4} - 13 \, a b c^{2} d x^{2} - 26 \, a b d + {\left (9 \, b^{2} c^{4} d x^{4} - 13 \, b^{2} c^{2} d x^{2} - 26 \, b^{2} d\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} x^{2} + 1}}{3375 \, c^{3}} \]
-1/3375*(27*(25*a^2 - 2*b^2)*c^5*d*x^5 - 5*(225*a^2 - 26*b^2)*c^3*d*x^3 + 780*b^2*c*d*x + 225*(3*b^2*c^5*d*x^5 - 5*b^2*c^3*d*x^3)*arcsin(c*x)^2 + 45 0*(3*a*b*c^5*d*x^5 - 5*a*b*c^3*d*x^3)*arcsin(c*x) + 30*(9*a*b*c^4*d*x^4 - 13*a*b*c^2*d*x^2 - 26*a*b*d + (9*b^2*c^4*d*x^4 - 13*b^2*c^2*d*x^2 - 26*b^2 *d)*arcsin(c*x))*sqrt(-c^2*x^2 + 1))/c^3
Time = 0.49 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.48 \[ \int x^2 \left (d-c^2 d x^2\right ) (a+b \arcsin (c x))^2 \, dx=\begin {cases} - \frac {a^{2} c^{2} d x^{5}}{5} + \frac {a^{2} d x^{3}}{3} - \frac {2 a b c^{2} d x^{5} \operatorname {asin}{\left (c x \right )}}{5} - \frac {2 a b c d x^{4} \sqrt {- c^{2} x^{2} + 1}}{25} + \frac {2 a b d x^{3} \operatorname {asin}{\left (c x \right )}}{3} + \frac {26 a b d x^{2} \sqrt {- c^{2} x^{2} + 1}}{225 c} + \frac {52 a b d \sqrt {- c^{2} x^{2} + 1}}{225 c^{3}} - \frac {b^{2} c^{2} d x^{5} \operatorname {asin}^{2}{\left (c x \right )}}{5} + \frac {2 b^{2} c^{2} d x^{5}}{125} - \frac {2 b^{2} c d x^{4} \sqrt {- c^{2} x^{2} + 1} \operatorname {asin}{\left (c x \right )}}{25} + \frac {b^{2} d x^{3} \operatorname {asin}^{2}{\left (c x \right )}}{3} - \frac {26 b^{2} d x^{3}}{675} + \frac {26 b^{2} d x^{2} \sqrt {- c^{2} x^{2} + 1} \operatorname {asin}{\left (c x \right )}}{225 c} - \frac {52 b^{2} d x}{225 c^{2}} + \frac {52 b^{2} d \sqrt {- c^{2} x^{2} + 1} \operatorname {asin}{\left (c x \right )}}{225 c^{3}} & \text {for}\: c \neq 0 \\\frac {a^{2} d x^{3}}{3} & \text {otherwise} \end {cases} \]
Piecewise((-a**2*c**2*d*x**5/5 + a**2*d*x**3/3 - 2*a*b*c**2*d*x**5*asin(c* x)/5 - 2*a*b*c*d*x**4*sqrt(-c**2*x**2 + 1)/25 + 2*a*b*d*x**3*asin(c*x)/3 + 26*a*b*d*x**2*sqrt(-c**2*x**2 + 1)/(225*c) + 52*a*b*d*sqrt(-c**2*x**2 + 1 )/(225*c**3) - b**2*c**2*d*x**5*asin(c*x)**2/5 + 2*b**2*c**2*d*x**5/125 - 2*b**2*c*d*x**4*sqrt(-c**2*x**2 + 1)*asin(c*x)/25 + b**2*d*x**3*asin(c*x)* *2/3 - 26*b**2*d*x**3/675 + 26*b**2*d*x**2*sqrt(-c**2*x**2 + 1)*asin(c*x)/ (225*c) - 52*b**2*d*x/(225*c**2) + 52*b**2*d*sqrt(-c**2*x**2 + 1)*asin(c*x )/(225*c**3), Ne(c, 0)), (a**2*d*x**3/3, True))
Time = 0.28 (sec) , antiderivative size = 354, normalized size of antiderivative = 1.68 \[ \int x^2 \left (d-c^2 d x^2\right ) (a+b \arcsin (c x))^2 \, dx=-\frac {1}{5} \, b^{2} c^{2} d x^{5} \arcsin \left (c x\right )^{2} - \frac {1}{5} \, a^{2} c^{2} d x^{5} + \frac {1}{3} \, b^{2} d x^{3} \arcsin \left (c x\right )^{2} - \frac {2}{75} \, {\left (15 \, x^{5} \arcsin \left (c x\right ) + {\left (\frac {3 \, \sqrt {-c^{2} x^{2} + 1} x^{4}}{c^{2}} + \frac {4 \, \sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{4}} + \frac {8 \, \sqrt {-c^{2} x^{2} + 1}}{c^{6}}\right )} c\right )} a b c^{2} d - \frac {2}{1125} \, {\left (15 \, {\left (\frac {3 \, \sqrt {-c^{2} x^{2} + 1} x^{4}}{c^{2}} + \frac {4 \, \sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{4}} + \frac {8 \, \sqrt {-c^{2} x^{2} + 1}}{c^{6}}\right )} c \arcsin \left (c x\right ) - \frac {9 \, c^{4} x^{5} + 20 \, c^{2} x^{3} + 120 \, x}{c^{4}}\right )} b^{2} c^{2} d + \frac {1}{3} \, a^{2} d x^{3} + \frac {2}{9} \, {\left (3 \, x^{3} \arcsin \left (c x\right ) + c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{2}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} a b d + \frac {2}{27} \, {\left (3 \, c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{2}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{c^{4}}\right )} \arcsin \left (c x\right ) - \frac {c^{2} x^{3} + 6 \, x}{c^{2}}\right )} b^{2} d \]
-1/5*b^2*c^2*d*x^5*arcsin(c*x)^2 - 1/5*a^2*c^2*d*x^5 + 1/3*b^2*d*x^3*arcsi n(c*x)^2 - 2/75*(15*x^5*arcsin(c*x) + (3*sqrt(-c^2*x^2 + 1)*x^4/c^2 + 4*sq rt(-c^2*x^2 + 1)*x^2/c^4 + 8*sqrt(-c^2*x^2 + 1)/c^6)*c)*a*b*c^2*d - 2/1125 *(15*(3*sqrt(-c^2*x^2 + 1)*x^4/c^2 + 4*sqrt(-c^2*x^2 + 1)*x^2/c^4 + 8*sqrt (-c^2*x^2 + 1)/c^6)*c*arcsin(c*x) - (9*c^4*x^5 + 20*c^2*x^3 + 120*x)/c^4)* b^2*c^2*d + 1/3*a^2*d*x^3 + 2/9*(3*x^3*arcsin(c*x) + c*(sqrt(-c^2*x^2 + 1) *x^2/c^2 + 2*sqrt(-c^2*x^2 + 1)/c^4))*a*b*d + 2/27*(3*c*(sqrt(-c^2*x^2 + 1 )*x^2/c^2 + 2*sqrt(-c^2*x^2 + 1)/c^4)*arcsin(c*x) - (c^2*x^3 + 6*x)/c^2)*b ^2*d
Time = 0.32 (sec) , antiderivative size = 356, normalized size of antiderivative = 1.69 \[ \int x^2 \left (d-c^2 d x^2\right ) (a+b \arcsin (c x))^2 \, dx=-\frac {1}{5} \, a^{2} c^{2} d x^{5} + \frac {1}{3} \, a^{2} d x^{3} - \frac {{\left (c^{2} x^{2} - 1\right )}^{2} b^{2} d x \arcsin \left (c x\right )^{2}}{5 \, c^{2}} - \frac {2 \, {\left (c^{2} x^{2} - 1\right )}^{2} a b d x \arcsin \left (c x\right )}{5 \, c^{2}} - \frac {{\left (c^{2} x^{2} - 1\right )} b^{2} d x \arcsin \left (c x\right )^{2}}{15 \, c^{2}} + \frac {2 \, {\left (c^{2} x^{2} - 1\right )}^{2} b^{2} d x}{125 \, c^{2}} - \frac {2 \, {\left (c^{2} x^{2} - 1\right )} a b d x \arcsin \left (c x\right )}{15 \, c^{2}} + \frac {2 \, b^{2} d x \arcsin \left (c x\right )^{2}}{15 \, c^{2}} - \frac {2 \, {\left (c^{2} x^{2} - 1\right )}^{2} \sqrt {-c^{2} x^{2} + 1} b^{2} d \arcsin \left (c x\right )}{25 \, c^{3}} - \frac {22 \, {\left (c^{2} x^{2} - 1\right )} b^{2} d x}{3375 \, c^{2}} + \frac {4 \, a b d x \arcsin \left (c x\right )}{15 \, c^{2}} - \frac {2 \, {\left (c^{2} x^{2} - 1\right )}^{2} \sqrt {-c^{2} x^{2} + 1} a b d}{25 \, c^{3}} + \frac {2 \, {\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} b^{2} d \arcsin \left (c x\right )}{45 \, c^{3}} - \frac {856 \, b^{2} d x}{3375 \, c^{2}} + \frac {2 \, {\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} a b d}{45 \, c^{3}} + \frac {4 \, \sqrt {-c^{2} x^{2} + 1} b^{2} d \arcsin \left (c x\right )}{15 \, c^{3}} + \frac {4 \, \sqrt {-c^{2} x^{2} + 1} a b d}{15 \, c^{3}} \]
-1/5*a^2*c^2*d*x^5 + 1/3*a^2*d*x^3 - 1/5*(c^2*x^2 - 1)^2*b^2*d*x*arcsin(c* x)^2/c^2 - 2/5*(c^2*x^2 - 1)^2*a*b*d*x*arcsin(c*x)/c^2 - 1/15*(c^2*x^2 - 1 )*b^2*d*x*arcsin(c*x)^2/c^2 + 2/125*(c^2*x^2 - 1)^2*b^2*d*x/c^2 - 2/15*(c^ 2*x^2 - 1)*a*b*d*x*arcsin(c*x)/c^2 + 2/15*b^2*d*x*arcsin(c*x)^2/c^2 - 2/25 *(c^2*x^2 - 1)^2*sqrt(-c^2*x^2 + 1)*b^2*d*arcsin(c*x)/c^3 - 22/3375*(c^2*x ^2 - 1)*b^2*d*x/c^2 + 4/15*a*b*d*x*arcsin(c*x)/c^2 - 2/25*(c^2*x^2 - 1)^2* sqrt(-c^2*x^2 + 1)*a*b*d/c^3 + 2/45*(-c^2*x^2 + 1)^(3/2)*b^2*d*arcsin(c*x) /c^3 - 856/3375*b^2*d*x/c^2 + 2/45*(-c^2*x^2 + 1)^(3/2)*a*b*d/c^3 + 4/15*s qrt(-c^2*x^2 + 1)*b^2*d*arcsin(c*x)/c^3 + 4/15*sqrt(-c^2*x^2 + 1)*a*b*d/c^ 3
Timed out. \[ \int x^2 \left (d-c^2 d x^2\right ) (a+b \arcsin (c x))^2 \, dx=\int x^2\,{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,\left (d-c^2\,d\,x^2\right ) \,d x \]