Integrand size = 23, antiderivative size = 138 \[ \int x \left (d-c^2 d x^2\right ) (a+b \arcsin (c x))^2 \, dx=-\frac {5}{32} b^2 d x^2+\frac {1}{32} b^2 c^2 d x^4+\frac {3 b d x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{16 c}+\frac {b d x \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))}{8 c}+\frac {3 d (a+b \arcsin (c x))^2}{32 c^2}-\frac {d \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))^2}{4 c^2} \]
-5/32*b^2*d*x^2+1/32*b^2*c^2*d*x^4+1/8*b*d*x*(-c^2*x^2+1)^(3/2)*(a+b*arcsi n(c*x))/c+3/32*d*(a+b*arcsin(c*x))^2/c^2-1/4*d*(-c^2*x^2+1)^2*(a+b*arcsin( c*x))^2/c^2+3/16*b*d*x*(a+b*arcsin(c*x))*(-c^2*x^2+1)^(1/2)/c
Time = 0.17 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.20 \[ \int x \left (d-c^2 d x^2\right ) (a+b \arcsin (c x))^2 \, dx=-\frac {d \left (b^2 c^2 x^2 \left (5-c^2 x^2\right )+2 a b c x \sqrt {1-c^2 x^2} \left (-5+2 c^2 x^2\right )+a^2 \left (5-16 c^2 x^2+8 c^4 x^4\right )+2 b \left (b c x \sqrt {1-c^2 x^2} \left (-5+2 c^2 x^2\right )+a \left (5-16 c^2 x^2+8 c^4 x^4\right )\right ) \arcsin (c x)+b^2 \left (5-16 c^2 x^2+8 c^4 x^4\right ) \arcsin (c x)^2\right )}{32 c^2} \]
-1/32*(d*(b^2*c^2*x^2*(5 - c^2*x^2) + 2*a*b*c*x*Sqrt[1 - c^2*x^2]*(-5 + 2* c^2*x^2) + a^2*(5 - 16*c^2*x^2 + 8*c^4*x^4) + 2*b*(b*c*x*Sqrt[1 - c^2*x^2] *(-5 + 2*c^2*x^2) + a*(5 - 16*c^2*x^2 + 8*c^4*x^4))*ArcSin[c*x] + b^2*(5 - 16*c^2*x^2 + 8*c^4*x^4)*ArcSin[c*x]^2))/c^2
Time = 0.55 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {5182, 5158, 244, 2009, 5156, 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.
| \(\displaystyle \int x \left (d-c^2 d x^2\right ) (a+b \arcsin (c x))^2 \, dx\) |
| \(\Big \downarrow \) 5182 |
| \(\displaystyle \frac {b d \int \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))dx}{2 c}-\frac {d \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))^2}{4 c^2}\) |
| \(\Big \downarrow \) 5158 |
| \(\displaystyle \frac {b d \left (\frac {3}{4} \int \sqrt {1-c^2 x^2} (a+b \arcsin (c x))dx-\frac {1}{4} b c \int x \left (1-c^2 x^2\right )dx+\frac {1}{4} x \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))\right )}{2 c}-\frac {d \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))^2}{4 c^2}\) |
| \(\Big \downarrow \) 244 |
| \(\displaystyle \frac {b d \left (\frac {3}{4} \int \sqrt {1-c^2 x^2} (a+b \arcsin (c x))dx-\frac {1}{4} b c \int \left (x-c^2 x^3\right )dx+\frac {1}{4} x \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))\right )}{2 c}-\frac {d \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))^2}{4 c^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b d \left (\frac {3}{4} \int \sqrt {1-c^2 x^2} (a+b \arcsin (c x))dx+\frac {1}{4} x \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))-\frac {1}{4} b c \left (\frac {x^2}{2}-\frac {c^2 x^4}{4}\right )\right )}{2 c}-\frac {d \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))^2}{4 c^2}\) |
| \(\Big \downarrow \) 5156 |
| \(\displaystyle \frac {b d \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {a+b \arcsin (c x)}{\sqrt {1-c^2 x^2}}dx-\frac {1}{2} b c \int xdx+\frac {1}{2} x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))\right )+\frac {1}{4} x \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))-\frac {1}{4} b c \left (\frac {x^2}{2}-\frac {c^2 x^4}{4}\right )\right )}{2 c}-\frac {d \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))^2}{4 c^2}\) |
| \(\Big \downarrow \) 15 |
| \(\displaystyle \frac {b d \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {a+b \arcsin (c x)}{\sqrt {1-c^2 x^2}}dx+\frac {1}{2} x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))-\frac {1}{4} b c x^2\right )+\frac {1}{4} x \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))-\frac {1}{4} b c \left (\frac {x^2}{2}-\frac {c^2 x^4}{4}\right )\right )}{2 c}-\frac {d \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))^2}{4 c^2}\) |
| \(\Big \downarrow \) 5152 |
| \(\displaystyle \frac {b d \left (\frac {1}{4} x \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))+\frac {3}{4} \left (\frac {1}{2} x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))+\frac {(a+b \arcsin (c x))^2}{4 b c}-\frac {1}{4} b c x^2\right )-\frac {1}{4} b c \left (\frac {x^2}{2}-\frac {c^2 x^4}{4}\right )\right )}{2 c}-\frac {d \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))^2}{4 c^2}\) |
-1/4*(d*(1 - c^2*x^2)^2*(a + b*ArcSin[c*x])^2)/c^2 + (b*d*(-1/4*(b*c*(x^2/ 2 - (c^2*x^4)/4)) + (x*(1 - c^2*x^2)^(3/2)*(a + b*ArcSin[c*x]))/4 + (3*(-1 /4*(b*c*x^2) + (x*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/2 + (a + b*ArcSin [c*x])^2/(4*b*c)))/4))/(2*c)
3.2.59.3.1 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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
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_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x _Symbol] :> Simp[x*(d + e*x^2)^p*((a + b*ArcSin[c*x])^n/(2*p + 1)), x] + (S imp[2*d*(p/(2*p + 1)) Int[(d + e*x^2)^(p - 1)*(a + b*ArcSin[c*x])^n, x], x] - Simp[b*c*(n/(2*p + 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}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[p, 0]
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]
Time = 0.22 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.39
| method | result | size |
| derivativedivides | \(\frac {-\frac {d \,a^{2} \left (c^{2} x^{2}-1\right )^{2}}{4}-d \,b^{2} \left (\frac {\arcsin \left (c x \right )^{2} \left (c^{2} x^{2}-1\right )^{2}}{4}-\frac {\arcsin \left (c x \right ) \left (-2 c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}+5 c x \sqrt {-c^{2} x^{2}+1}+3 \arcsin \left (c x \right )\right )}{16}+\frac {3 \arcsin \left (c x \right )^{2}}{32}-\frac {\left (2 c^{2} x^{2}-5\right )^{2}}{128}\right )-2 d a b \left (\frac {c^{4} x^{4} \arcsin \left (c x \right )}{4}-\frac {c^{2} x^{2} \arcsin \left (c x \right )}{2}+\frac {5 \arcsin \left (c x \right )}{32}+\frac {c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}{16}-\frac {5 c x \sqrt {-c^{2} x^{2}+1}}{32}\right )}{c^{2}}\) | \(192\) |
| default | \(\frac {-\frac {d \,a^{2} \left (c^{2} x^{2}-1\right )^{2}}{4}-d \,b^{2} \left (\frac {\arcsin \left (c x \right )^{2} \left (c^{2} x^{2}-1\right )^{2}}{4}-\frac {\arcsin \left (c x \right ) \left (-2 c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}+5 c x \sqrt {-c^{2} x^{2}+1}+3 \arcsin \left (c x \right )\right )}{16}+\frac {3 \arcsin \left (c x \right )^{2}}{32}-\frac {\left (2 c^{2} x^{2}-5\right )^{2}}{128}\right )-2 d a b \left (\frac {c^{4} x^{4} \arcsin \left (c x \right )}{4}-\frac {c^{2} x^{2} \arcsin \left (c x \right )}{2}+\frac {5 \arcsin \left (c x \right )}{32}+\frac {c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}{16}-\frac {5 c x \sqrt {-c^{2} x^{2}+1}}{32}\right )}{c^{2}}\) | \(192\) |
| parts | \(-\frac {d \,a^{2} \left (c^{2} x^{2}-1\right )^{2}}{4 c^{2}}-\frac {d \,b^{2} \left (\frac {\arcsin \left (c x \right )^{2} \left (c^{2} x^{2}-1\right )^{2}}{4}-\frac {\arcsin \left (c x \right ) \left (-2 c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}+5 c x \sqrt {-c^{2} x^{2}+1}+3 \arcsin \left (c x \right )\right )}{16}+\frac {3 \arcsin \left (c x \right )^{2}}{32}-\frac {\left (2 c^{2} x^{2}-5\right )^{2}}{128}\right )}{c^{2}}-\frac {2 d a b \left (\frac {c^{4} x^{4} \arcsin \left (c x \right )}{4}-\frac {c^{2} x^{2} \arcsin \left (c x \right )}{2}+\frac {5 \arcsin \left (c x \right )}{32}+\frac {c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}{16}-\frac {5 c x \sqrt {-c^{2} x^{2}+1}}{32}\right )}{c^{2}}\) | \(197\) |
1/c^2*(-1/4*d*a^2*(c^2*x^2-1)^2-d*b^2*(1/4*arcsin(c*x)^2*(c^2*x^2-1)^2-1/1 6*arcsin(c*x)*(-2*c^3*x^3*(-c^2*x^2+1)^(1/2)+5*c*x*(-c^2*x^2+1)^(1/2)+3*ar csin(c*x))+3/32*arcsin(c*x)^2-1/128*(2*c^2*x^2-5)^2)-2*d*a*b*(1/4*c^4*x^4* arcsin(c*x)-1/2*c^2*x^2*arcsin(c*x)+5/32*arcsin(c*x)+1/16*c^3*x^3*(-c^2*x^ 2+1)^(1/2)-5/32*c*x*(-c^2*x^2+1)^(1/2)))
Time = 0.27 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.28 \[ \int x \left (d-c^2 d x^2\right ) (a+b \arcsin (c x))^2 \, dx=-\frac {{\left (8 \, a^{2} - b^{2}\right )} c^{4} d x^{4} - {\left (16 \, a^{2} - 5 \, b^{2}\right )} c^{2} d x^{2} + {\left (8 \, b^{2} c^{4} d x^{4} - 16 \, b^{2} c^{2} d x^{2} + 5 \, b^{2} d\right )} \arcsin \left (c x\right )^{2} + 2 \, {\left (8 \, a b c^{4} d x^{4} - 16 \, a b c^{2} d x^{2} + 5 \, a b d\right )} \arcsin \left (c x\right ) + 2 \, {\left (2 \, a b c^{3} d x^{3} - 5 \, a b c d x + {\left (2 \, b^{2} c^{3} d x^{3} - 5 \, b^{2} c d x\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} x^{2} + 1}}{32 \, c^{2}} \]
-1/32*((8*a^2 - b^2)*c^4*d*x^4 - (16*a^2 - 5*b^2)*c^2*d*x^2 + (8*b^2*c^4*d *x^4 - 16*b^2*c^2*d*x^2 + 5*b^2*d)*arcsin(c*x)^2 + 2*(8*a*b*c^4*d*x^4 - 16 *a*b*c^2*d*x^2 + 5*a*b*d)*arcsin(c*x) + 2*(2*a*b*c^3*d*x^3 - 5*a*b*c*d*x + (2*b^2*c^3*d*x^3 - 5*b^2*c*d*x)*arcsin(c*x))*sqrt(-c^2*x^2 + 1))/c^2
Leaf count of result is larger than twice the leaf count of optimal. 269 vs. \(2 (129) = 258\).
Time = 0.40 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.95 \[ \int x \left (d-c^2 d x^2\right ) (a+b \arcsin (c x))^2 \, dx=\begin {cases} - \frac {a^{2} c^{2} d x^{4}}{4} + \frac {a^{2} d x^{2}}{2} - \frac {a b c^{2} d x^{4} \operatorname {asin}{\left (c x \right )}}{2} - \frac {a b c d x^{3} \sqrt {- c^{2} x^{2} + 1}}{8} + a b d x^{2} \operatorname {asin}{\left (c x \right )} + \frac {5 a b d x \sqrt {- c^{2} x^{2} + 1}}{16 c} - \frac {5 a b d \operatorname {asin}{\left (c x \right )}}{16 c^{2}} - \frac {b^{2} c^{2} d x^{4} \operatorname {asin}^{2}{\left (c x \right )}}{4} + \frac {b^{2} c^{2} d x^{4}}{32} - \frac {b^{2} c d x^{3} \sqrt {- c^{2} x^{2} + 1} \operatorname {asin}{\left (c x \right )}}{8} + \frac {b^{2} d x^{2} \operatorname {asin}^{2}{\left (c x \right )}}{2} - \frac {5 b^{2} d x^{2}}{32} + \frac {5 b^{2} d x \sqrt {- c^{2} x^{2} + 1} \operatorname {asin}{\left (c x \right )}}{16 c} - \frac {5 b^{2} d \operatorname {asin}^{2}{\left (c x \right )}}{32 c^{2}} & \text {for}\: c \neq 0 \\\frac {a^{2} d x^{2}}{2} & \text {otherwise} \end {cases} \]
Piecewise((-a**2*c**2*d*x**4/4 + a**2*d*x**2/2 - a*b*c**2*d*x**4*asin(c*x) /2 - a*b*c*d*x**3*sqrt(-c**2*x**2 + 1)/8 + a*b*d*x**2*asin(c*x) + 5*a*b*d* x*sqrt(-c**2*x**2 + 1)/(16*c) - 5*a*b*d*asin(c*x)/(16*c**2) - b**2*c**2*d* x**4*asin(c*x)**2/4 + b**2*c**2*d*x**4/32 - b**2*c*d*x**3*sqrt(-c**2*x**2 + 1)*asin(c*x)/8 + b**2*d*x**2*asin(c*x)**2/2 - 5*b**2*d*x**2/32 + 5*b**2* d*x*sqrt(-c**2*x**2 + 1)*asin(c*x)/(16*c) - 5*b**2*d*asin(c*x)**2/(32*c**2 ), Ne(c, 0)), (a**2*d*x**2/2, True))
\[ \int x \left (d-c^2 d x^2\right ) (a+b \arcsin (c x))^2 \, dx=\int { -{\left (c^{2} d x^{2} - d\right )} {\left (b \arcsin \left (c x\right ) + a\right )}^{2} x \,d x } \]
-1/4*a^2*c^2*d*x^4 - 1/16*(8*x^4*arcsin(c*x) + (2*sqrt(-c^2*x^2 + 1)*x^3/c ^2 + 3*sqrt(-c^2*x^2 + 1)*x/c^4 - 3*arcsin(c*x)/c^5)*c)*a*b*c^2*d + 1/2*a^ 2*d*x^2 + 1/2*(2*x^2*arcsin(c*x) + c*(sqrt(-c^2*x^2 + 1)*x/c^2 - arcsin(c* x)/c^3))*a*b*d - 1/4*(b^2*c^2*d*x^4 - 2*b^2*d*x^2)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2 - integrate(1/2*(b^2*c^3*d*x^4 - 2*b^2*c*d*x^2)*sqrt (c*x + 1)*sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/(c^2*x ^2 - 1), x)
Leaf count of result is larger than twice the leaf count of optimal. 248 vs. \(2 (121) = 242\).
Time = 0.33 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.80 \[ \int x \left (d-c^2 d x^2\right ) (a+b \arcsin (c x))^2 \, dx=-\frac {1}{4} \, a^{2} c^{2} d x^{4} + \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} b^{2} d x \arcsin \left (c x\right )}{8 \, c} - \frac {{\left (c^{2} x^{2} - 1\right )}^{2} b^{2} d \arcsin \left (c x\right )^{2}}{4 \, c^{2}} + \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} a b d x}{8 \, c} + \frac {3 \, \sqrt {-c^{2} x^{2} + 1} b^{2} d x \arcsin \left (c x\right )}{16 \, c} - \frac {{\left (c^{2} x^{2} - 1\right )}^{2} a b d \arcsin \left (c x\right )}{2 \, c^{2}} + \frac {3 \, \sqrt {-c^{2} x^{2} + 1} a b d x}{16 \, c} + \frac {{\left (c^{2} x^{2} - 1\right )}^{2} b^{2} d}{32 \, c^{2}} + \frac {3 \, b^{2} d \arcsin \left (c x\right )^{2}}{32 \, c^{2}} + \frac {{\left (c^{2} x^{2} - 1\right )} a^{2} d}{2 \, c^{2}} - \frac {3 \, {\left (c^{2} x^{2} - 1\right )} b^{2} d}{32 \, c^{2}} + \frac {3 \, a b d \arcsin \left (c x\right )}{16 \, c^{2}} - \frac {15 \, b^{2} d}{256 \, c^{2}} \]
-1/4*a^2*c^2*d*x^4 + 1/8*(-c^2*x^2 + 1)^(3/2)*b^2*d*x*arcsin(c*x)/c - 1/4* (c^2*x^2 - 1)^2*b^2*d*arcsin(c*x)^2/c^2 + 1/8*(-c^2*x^2 + 1)^(3/2)*a*b*d*x /c + 3/16*sqrt(-c^2*x^2 + 1)*b^2*d*x*arcsin(c*x)/c - 1/2*(c^2*x^2 - 1)^2*a *b*d*arcsin(c*x)/c^2 + 3/16*sqrt(-c^2*x^2 + 1)*a*b*d*x/c + 1/32*(c^2*x^2 - 1)^2*b^2*d/c^2 + 3/32*b^2*d*arcsin(c*x)^2/c^2 + 1/2*(c^2*x^2 - 1)*a^2*d/c ^2 - 3/32*(c^2*x^2 - 1)*b^2*d/c^2 + 3/16*a*b*d*arcsin(c*x)/c^2 - 15/256*b^ 2*d/c^2
Timed out. \[ \int x \left (d-c^2 d x^2\right ) (a+b \arcsin (c x))^2 \, dx=\int x\,{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,\left (d-c^2\,d\,x^2\right ) \,d x \]