Integrand size = 35, antiderivative size = 250 \[ \int \frac {x^2 (a+b \arcsin (c x))^2}{\sqrt {d+c d x} \sqrt {e-c e x}} \, dx=\frac {b^2 x \left (1-c^2 x^2\right )}{4 c^2 \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {b^2 \sqrt {1-c^2 x^2} \arcsin (c x)}{4 c^3 \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {b x^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{2 c \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {x \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{2 c^2 \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^3}{6 b c^3 \sqrt {d+c d x} \sqrt {e-c e x}} \]
1/4*b^2*x*(-c^2*x^2+1)/c^2/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-1/2*x*(-c^2*x^ 2+1)*(a+b*arcsin(c*x))^2/c^2/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-1/4*b^2*arcs in(c*x)*(-c^2*x^2+1)^(1/2)/c^3/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)+1/2*b*x^2* (a+b*arcsin(c*x))*(-c^2*x^2+1)^(1/2)/c/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)+1/ 6*(a+b*arcsin(c*x))^3*(-c^2*x^2+1)^(1/2)/b/c^3/(c*d*x+d)^(1/2)/(-c*e*x+e)^ (1/2)
Time = 1.47 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.30 \[ \int \frac {x^2 (a+b \arcsin (c x))^2}{\sqrt {d+c d x} \sqrt {e-c e x}} \, dx=\frac {12 b \sqrt {d} \sqrt {e} \left (a \sqrt {1-c^2 x^2}+b c x \left (-1+c^2 x^2\right )\right ) \arcsin (c x)^2+4 b^2 \sqrt {d} \sqrt {e} \sqrt {1-c^2 x^2} \arcsin (c x)^3-12 a^2 \sqrt {d+c d x} \sqrt {e-c e x} \arctan \left (\frac {c x \sqrt {d+c d x} \sqrt {e-c e x}}{\sqrt {d} \sqrt {e} \left (-1+c^2 x^2\right )}\right )-3 \sqrt {d} \sqrt {e} \left (a b \sqrt {1-c^2 x^2}+2 b^2 c x \left (-1+c^2 x^2\right )+a^2 \left (4 c x-4 c^3 x^3\right )+a b \cos (3 \arcsin (c x))\right )-3 b \sqrt {d} \sqrt {e} \arcsin (c x) \left (2 a c x+b \sqrt {1-c^2 x^2}+b \cos (3 \arcsin (c x))+2 a \sin (3 \arcsin (c x))\right )}{24 c^3 \sqrt {d} \sqrt {e} \sqrt {d+c d x} \sqrt {e-c e x}} \]
(12*b*Sqrt[d]*Sqrt[e]*(a*Sqrt[1 - c^2*x^2] + b*c*x*(-1 + c^2*x^2))*ArcSin[ c*x]^2 + 4*b^2*Sqrt[d]*Sqrt[e]*Sqrt[1 - c^2*x^2]*ArcSin[c*x]^3 - 12*a^2*Sq rt[d + c*d*x]*Sqrt[e - c*e*x]*ArcTan[(c*x*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) /(Sqrt[d]*Sqrt[e]*(-1 + c^2*x^2))] - 3*Sqrt[d]*Sqrt[e]*(a*b*Sqrt[1 - c^2*x ^2] + 2*b^2*c*x*(-1 + c^2*x^2) + a^2*(4*c*x - 4*c^3*x^3) + a*b*Cos[3*ArcSi n[c*x]]) - 3*b*Sqrt[d]*Sqrt[e]*ArcSin[c*x]*(2*a*c*x + b*Sqrt[1 - c^2*x^2] + b*Cos[3*ArcSin[c*x]] + 2*a*Sin[3*ArcSin[c*x]]))/(24*c^3*Sqrt[d]*Sqrt[e]* Sqrt[d + c*d*x]*Sqrt[e - c*e*x])
Time = 1.00 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.60, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {5238, 5210, 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.
\(\displaystyle \int \frac {x^2 (a+b \arcsin (c x))^2}{\sqrt {c d x+d} \sqrt {e-c e x}} \, dx\) |
\(\Big \downarrow \) 5238 |
\(\displaystyle \frac {\sqrt {1-c^2 x^2} \int \frac {x^2 (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}dx}{\sqrt {c d x+d} \sqrt {e-c e x}}\) |
\(\Big \downarrow \) 5210 |
\(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (\frac {\int \frac {(a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}dx}{2 c^2}+\frac {b \int x (a+b \arcsin (c x))dx}{c}-\frac {x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{2 c^2}\right )}{\sqrt {c d x+d} \sqrt {e-c e x}}\) |
\(\Big \downarrow \) 5138 |
\(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (\frac {b \left (\frac {1}{2} x^2 (a+b \arcsin (c x))-\frac {1}{2} b c \int \frac {x^2}{\sqrt {1-c^2 x^2}}dx\right )}{c}+\frac {\int \frac {(a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}dx}{2 c^2}-\frac {x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{2 c^2}\right )}{\sqrt {c d x+d} \sqrt {e-c e x}}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (\frac {b \left (\frac {1}{2} x^2 (a+b \arcsin (c x))-\frac {1}{2} b c \left (\frac {\int \frac {1}{\sqrt {1-c^2 x^2}}dx}{2 c^2}-\frac {x \sqrt {1-c^2 x^2}}{2 c^2}\right )\right )}{c}+\frac {\int \frac {(a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}dx}{2 c^2}-\frac {x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{2 c^2}\right )}{\sqrt {c d x+d} \sqrt {e-c e x}}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (\frac {\int \frac {(a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}dx}{2 c^2}-\frac {x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{2 c^2}+\frac {b \left (\frac {1}{2} x^2 (a+b \arcsin (c x))-\frac {1}{2} b c \left (\frac {\arcsin (c x)}{2 c^3}-\frac {x \sqrt {1-c^2 x^2}}{2 c^2}\right )\right )}{c}\right )}{\sqrt {c d x+d} \sqrt {e-c e x}}\) |
\(\Big \downarrow \) 5152 |
\(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (\frac {(a+b \arcsin (c x))^3}{6 b c^3}-\frac {x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{2 c^2}+\frac {b \left (\frac {1}{2} x^2 (a+b \arcsin (c x))-\frac {1}{2} b c \left (\frac {\arcsin (c x)}{2 c^3}-\frac {x \sqrt {1-c^2 x^2}}{2 c^2}\right )\right )}{c}\right )}{\sqrt {c d x+d} \sqrt {e-c e x}}\) |
(Sqrt[1 - c^2*x^2]*(-1/2*(x*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2)/c^2 + (a + b*ArcSin[c*x])^3/(6*b*c^3) + (b*((x^2*(a + b*ArcSin[c*x]))/2 - (b*c* (-1/2*(x*Sqrt[1 - c^2*x^2])/c^2 + ArcSin[c*x]/(2*c^3)))/2))/c))/(Sqrt[d + c*d*x]*Sqrt[e - c*e*x])
3.6.86.3.1 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_.)*((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]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((h_.)*(x_))^(m_.)*((d_) + (e_ .)*(x_))^(p_)*((f_) + (g_.)*(x_))^(q_), x_Symbol] :> Simp[((-d^2)*(g/e))^In tPart[q]*(d + e*x)^FracPart[q]*((f + g*x)^FracPart[q]/(1 - c^2*x^2)^FracPar t[q]) Int[(h*x)^m*(d + e*x)^(p - q)*(1 - c^2*x^2)^q*(a + b*ArcSin[c*x])^n , x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && EqQ[e*f + d*g, 0] & & EqQ[c^2*d^2 - e^2, 0] && HalfIntegerQ[p, q] && GeQ[p - q, 0]
\[\int \frac {x^{2} \left (a +b \arcsin \left (c x \right )\right )^{2}}{\sqrt {c d x +d}\, \sqrt {-c e x +e}}d x\]
\[ \int \frac {x^2 (a+b \arcsin (c x))^2}{\sqrt {d+c d x} \sqrt {e-c e x}} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2} x^{2}}{\sqrt {c d x + d} \sqrt {-c e x + e}} \,d x } \]
integral(-(b^2*x^2*arcsin(c*x)^2 + 2*a*b*x^2*arcsin(c*x) + a^2*x^2)*sqrt(c *d*x + d)*sqrt(-c*e*x + e)/(c^2*d*e*x^2 - d*e), x)
Timed out. \[ \int \frac {x^2 (a+b \arcsin (c x))^2}{\sqrt {d+c d x} \sqrt {e-c e x}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {x^2 (a+b \arcsin (c x))^2}{\sqrt {d+c d x} \sqrt {e-c e x}} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
\[ \int \frac {x^2 (a+b \arcsin (c x))^2}{\sqrt {d+c d x} \sqrt {e-c e x}} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2} x^{2}}{\sqrt {c d x + d} \sqrt {-c e x + e}} \,d x } \]
Timed out. \[ \int \frac {x^2 (a+b \arcsin (c x))^2}{\sqrt {d+c d x} \sqrt {e-c e x}} \, dx=\int \frac {x^2\,{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2}{\sqrt {d+c\,d\,x}\,\sqrt {e-c\,e\,x}} \,d x \]