Integrand size = 27, antiderivative size = 124 \[ \int \frac {x^2 (a+b \arcsin (c x))}{\sqrt {d-c^2 d x^2}} \, dx=\frac {b x^2 \sqrt {1-c^2 x^2}}{4 c \sqrt {d-c^2 d x^2}}-\frac {x \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{2 c^2 d}+\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{4 b c^3 \sqrt {d-c^2 d x^2}} \]
1/4*b*x^2*(-c^2*x^2+1)^(1/2)/c/(-c^2*d*x^2+d)^(1/2)+1/4*(a+b*arcsin(c*x))^ 2*(-c^2*x^2+1)^(1/2)/b/c^3/(-c^2*d*x^2+d)^(1/2)-1/2*x*(a+b*arcsin(c*x))*(- c^2*d*x^2+d)^(1/2)/c^2/d
Time = 0.96 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.08 \[ \int \frac {x^2 (a+b \arcsin (c x))}{\sqrt {d-c^2 d x^2}} \, dx=-\frac {\frac {4 a c x \sqrt {d-c^2 d x^2}}{d}+\frac {4 a \arctan \left (\frac {c x \sqrt {d-c^2 d x^2}}{\sqrt {d} \left (-1+c^2 x^2\right )}\right )}{\sqrt {d}}+\frac {b \sqrt {1-c^2 x^2} \left (-2 \arcsin (c x)^2+\cos (2 \arcsin (c x))+2 \arcsin (c x) \sin (2 \arcsin (c x))\right )}{\sqrt {d-c^2 d x^2}}}{8 c^3} \]
-1/8*((4*a*c*x*Sqrt[d - c^2*d*x^2])/d + (4*a*ArcTan[(c*x*Sqrt[d - c^2*d*x^ 2])/(Sqrt[d]*(-1 + c^2*x^2))])/Sqrt[d] + (b*Sqrt[1 - c^2*x^2]*(-2*ArcSin[c *x]^2 + Cos[2*ArcSin[c*x]] + 2*ArcSin[c*x]*Sin[2*ArcSin[c*x]]))/Sqrt[d - c ^2*d*x^2])/c^3
Time = 0.36 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {5210, 15, 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.
| \(\displaystyle \int \frac {x^2 (a+b \arcsin (c x))}{\sqrt {d-c^2 d x^2}} \, dx\) |
| \(\Big \downarrow \) 5210 |
| \(\displaystyle \frac {\int \frac {a+b \arcsin (c x)}{\sqrt {d-c^2 d x^2}}dx}{2 c^2}+\frac {b \sqrt {1-c^2 x^2} \int xdx}{2 c \sqrt {d-c^2 d x^2}}-\frac {x \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{2 c^2 d}\) |
| \(\Big \downarrow \) 15 |
| \(\displaystyle \frac {\int \frac {a+b \arcsin (c x)}{\sqrt {d-c^2 d x^2}}dx}{2 c^2}-\frac {x \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{2 c^2 d}+\frac {b x^2 \sqrt {1-c^2 x^2}}{4 c \sqrt {d-c^2 d x^2}}\) |
| \(\Big \downarrow \) 5152 |
| \(\displaystyle -\frac {x \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{2 c^2 d}+\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{4 b c^3 \sqrt {d-c^2 d x^2}}+\frac {b x^2 \sqrt {1-c^2 x^2}}{4 c \sqrt {d-c^2 d x^2}}\) |
(b*x^2*Sqrt[1 - c^2*x^2])/(4*c*Sqrt[d - c^2*d*x^2]) - (x*Sqrt[d - c^2*d*x^ 2]*(a + b*ArcSin[c*x]))/(2*c^2*d) + (Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]) ^2)/(4*b*c^3*Sqrt[d - c^2*d*x^2])
3.2.12.3.1 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_.) + 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_.)*((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]
Leaf count of result is larger than twice the leaf count of optimal. \(267\) vs. \(2(108)=216\).
Time = 0.12 (sec) , antiderivative size = 268, normalized size of antiderivative = 2.16
| method | result | size |
| default | \(-\frac {a x \sqrt {-c^{2} d \,x^{2}+d}}{2 c^{2} d}+\frac {a \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{2 c^{2} \sqrt {c^{2} d}}+b \left (-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right )^{2}}{4 c^{3} d \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-c^{2} x^{2}+1}}{16 c^{3} \sqrt {-d \left (c^{2} x^{2}-1\right )}}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) x}{8 c^{2} d \left (c^{2} x^{2}-1\right )}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \cos \left (3 \arcsin \left (c x \right )\right )}{16 c^{3} d \left (c^{2} x^{2}-1\right )}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) \sin \left (3 \arcsin \left (c x \right )\right )}{8 c^{3} d \left (c^{2} x^{2}-1\right )}\right )\) | \(268\) |
| parts | \(-\frac {a x \sqrt {-c^{2} d \,x^{2}+d}}{2 c^{2} d}+\frac {a \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{2 c^{2} \sqrt {c^{2} d}}+b \left (-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right )^{2}}{4 c^{3} d \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-c^{2} x^{2}+1}}{16 c^{3} \sqrt {-d \left (c^{2} x^{2}-1\right )}}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) x}{8 c^{2} d \left (c^{2} x^{2}-1\right )}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \cos \left (3 \arcsin \left (c x \right )\right )}{16 c^{3} d \left (c^{2} x^{2}-1\right )}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) \sin \left (3 \arcsin \left (c x \right )\right )}{8 c^{3} d \left (c^{2} x^{2}-1\right )}\right )\) | \(268\) |
-1/2*a*x/c^2/d*(-c^2*d*x^2+d)^(1/2)+1/2*a/c^2/(c^2*d)^(1/2)*arctan((c^2*d) ^(1/2)*x/(-c^2*d*x^2+d)^(1/2))+b*(-1/4*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1) ^(1/2)/c^3/d/(c^2*x^2-1)*arcsin(c*x)^2-1/16/c^3/(-d*(c^2*x^2-1))^(1/2)*(-c ^2*x^2+1)^(1/2)+1/8*(-d*(c^2*x^2-1))^(1/2)/c^2/d/(c^2*x^2-1)*arcsin(c*x)*x +1/16*(-d*(c^2*x^2-1))^(1/2)/c^3/d/(c^2*x^2-1)*cos(3*arcsin(c*x))+1/8*(-d* (c^2*x^2-1))^(1/2)/c^3/d/(c^2*x^2-1)*arcsin(c*x)*sin(3*arcsin(c*x)))
\[ \int \frac {x^2 (a+b \arcsin (c x))}{\sqrt {d-c^2 d x^2}} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )} x^{2}}{\sqrt {-c^{2} d x^{2} + d}} \,d x } \]
\[ \int \frac {x^2 (a+b \arcsin (c x))}{\sqrt {d-c^2 d x^2}} \, dx=\int \frac {x^{2} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{\sqrt {- d \left (c x - 1\right ) \left (c x + 1\right )}}\, dx \]
\[ \int \frac {x^2 (a+b \arcsin (c x))}{\sqrt {d-c^2 d x^2}} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )} x^{2}}{\sqrt {-c^{2} d x^{2} + d}} \,d x } \]
-1/2*a*(sqrt(-c^2*d*x^2 + d)*x/(c^2*d) - arcsin(c*x)/(c^3*sqrt(d))) + b*in tegrate(x^2*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/(sqrt(c*x + 1)*sqrt (-c*x + 1)), x)/sqrt(d)
\[ \int \frac {x^2 (a+b \arcsin (c x))}{\sqrt {d-c^2 d x^2}} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )} x^{2}}{\sqrt {-c^{2} d x^{2} + d}} \,d x } \]
Timed out. \[ \int \frac {x^2 (a+b \arcsin (c x))}{\sqrt {d-c^2 d x^2}} \, dx=\int \frac {x^2\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{\sqrt {d-c^2\,d\,x^2}} \,d x \]