Integrand size = 32, antiderivative size = 398 \[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))^2}{\sqrt {e-c e x}} \, dx=\frac {4 b^2 d^2 \left (1-c^2 x^2\right )}{c \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {b^2 d^2 x \left (1-c^2 x^2\right )}{4 \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {b^2 d^2 \sqrt {1-c^2 x^2} \arcsin (c x)}{4 c \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {4 b d^2 x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{\sqrt {d+c d x} \sqrt {e-c e x}}+\frac {b c d^2 x^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{2 \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {2 d^2 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{c \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {d^2 x \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{2 \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {d^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^3}{2 b c \sqrt {d+c d x} \sqrt {e-c e x}} \]
4*b^2*d^2*(-c^2*x^2+1)/c/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)+1/4*b^2*d^2*x*(- c^2*x^2+1)/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-2*d^2*(-c^2*x^2+1)*(a+b*arcsin (c*x))^2/c/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-1/2*d^2*x*(-c^2*x^2+1)*(a+b*ar csin(c*x))^2/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-1/4*b^2*d^2*arcsin(c*x)*(-c^ 2*x^2+1)^(1/2)/c/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)+4*b*d^2*x*(a+b*arcsin(c* x))*(-c^2*x^2+1)^(1/2)/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)+1/2*b*c*d^2*x^2*(a +b*arcsin(c*x))*(-c^2*x^2+1)^(1/2)/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)+1/2*d^ 2*(a+b*arcsin(c*x))^3*(-c^2*x^2+1)^(1/2)/b/c/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1 /2)
Time = 5.76 (sec) , antiderivative size = 344, normalized size of antiderivative = 0.86 \[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))^2}{\sqrt {e-c e x}} \, dx=\frac {b d \sqrt {d+c d x} \sqrt {e-c e x} \left (-4 a (4+c x) \sqrt {1-c^2 x^2}+b \left (-1+16 c x+2 c^2 x^2\right )\right ) \arcsin (c x)-2 b d \sqrt {d+c d x} \sqrt {e-c e x} \left (-3 a+b (4+c x) \sqrt {1-c^2 x^2}\right ) \arcsin (c x)^2+2 b^2 d \sqrt {d+c d x} \sqrt {e-c e x} \arcsin (c x)^3-6 a^2 d^{3/2} \sqrt {e} \sqrt {1-c^2 x^2} \arctan \left (\frac {c x \sqrt {d+c d x} \sqrt {e-c e x}}{\sqrt {d} \sqrt {e} \left (-1+c^2 x^2\right )}\right )+d \sqrt {d+c d x} \sqrt {e-c e x} \left (16 a b c x-2 a^2 (4+c x) \sqrt {1-c^2 x^2}+b^2 (16+c x) \sqrt {1-c^2 x^2}-a b \cos (2 \arcsin (c x))\right )}{4 c e \sqrt {1-c^2 x^2}} \]
(b*d*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*(-4*a*(4 + c*x)*Sqrt[1 - c^2*x^2] + b *(-1 + 16*c*x + 2*c^2*x^2))*ArcSin[c*x] - 2*b*d*Sqrt[d + c*d*x]*Sqrt[e - c *e*x]*(-3*a + b*(4 + c*x)*Sqrt[1 - c^2*x^2])*ArcSin[c*x]^2 + 2*b^2*d*Sqrt[ d + c*d*x]*Sqrt[e - c*e*x]*ArcSin[c*x]^3 - 6*a^2*d^(3/2)*Sqrt[e]*Sqrt[1 - c^2*x^2]*ArcTan[(c*x*Sqrt[d + c*d*x]*Sqrt[e - c*e*x])/(Sqrt[d]*Sqrt[e]*(-1 + c^2*x^2))] + d*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*(16*a*b*c*x - 2*a^2*(4 + c*x)*Sqrt[1 - c^2*x^2] + b^2*(16 + c*x)*Sqrt[1 - c^2*x^2] - a*b*Cos[2*Arc Sin[c*x]]))/(4*c*e*Sqrt[1 - c^2*x^2])
Time = 0.73 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.55, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {5178, 27, 5272, 3042, 3798, 2009}
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.
\(\displaystyle \int \frac {(c d x+d)^{3/2} (a+b \arcsin (c x))^2}{\sqrt {e-c e x}} \, dx\) |
\(\Big \downarrow \) 5178 |
\(\displaystyle \frac {\sqrt {1-c^2 x^2} \int \frac {d^2 (c x+1)^2 (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}dx}{\sqrt {c d x+d} \sqrt {e-c e x}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {d^2 \sqrt {1-c^2 x^2} \int \frac {(c x+1)^2 (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}dx}{\sqrt {c d x+d} \sqrt {e-c e x}}\) |
\(\Big \downarrow \) 5272 |
\(\displaystyle \frac {d^2 \sqrt {1-c^2 x^2} \int \left (x c^2+c\right )^2 (a+b \arcsin (c x))^2d\arcsin (c x)}{c^3 \sqrt {c d x+d} \sqrt {e-c e x}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {d^2 \sqrt {1-c^2 x^2} \int (a+b \arcsin (c x))^2 (\sin (\arcsin (c x)) c+c)^2d\arcsin (c x)}{c^3 \sqrt {c d x+d} \sqrt {e-c e x}}\) |
\(\Big \downarrow \) 3798 |
\(\displaystyle \frac {d^2 \sqrt {1-c^2 x^2} \int \left (x^2 (a+b \arcsin (c x))^2 c^4+2 x (a+b \arcsin (c x))^2 c^3+(a+b \arcsin (c x))^2 c^2\right )d\arcsin (c x)}{c^3 \sqrt {c d x+d} \sqrt {e-c e x}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {d^2 \sqrt {1-c^2 x^2} \left (\frac {1}{2} b c^4 x^2 (a+b \arcsin (c x))+4 b c^3 x (a+b \arcsin (c x))-2 c^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2+\frac {c^2 (a+b \arcsin (c x))^3}{2 b}-\frac {1}{2} c^3 x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2-\frac {1}{4} b^2 c^2 \arcsin (c x)+4 b^2 c^2 \sqrt {1-c^2 x^2}+\frac {1}{4} b^2 c^3 x \sqrt {1-c^2 x^2}\right )}{c^3 \sqrt {c d x+d} \sqrt {e-c e x}}\) |
(d^2*Sqrt[1 - c^2*x^2]*(4*b^2*c^2*Sqrt[1 - c^2*x^2] + (b^2*c^3*x*Sqrt[1 - c^2*x^2])/4 - (b^2*c^2*ArcSin[c*x])/4 + 4*b*c^3*x*(a + b*ArcSin[c*x]) + (b *c^4*x^2*(a + b*ArcSin[c*x]))/2 - 2*c^2*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c* x])^2 - (c^3*x*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2)/2 + (c^2*(a + b*Ar cSin[c*x])^3)/(2*b)))/(c^3*Sqrt[d + c*d*x]*Sqrt[e - c*e*x])
3.6.59.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.) , x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (a + b*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[n, 0] && (EqQ[n, 1] || IGtQ[ m, 0] || NeQ[a^2 - b^2, 0])
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_))^(p_)*((f_) + (g_.)*(x_))^(q_), x_Symbol] :> Simp[(d + e*x)^q*((f + g*x)^q/(1 - c^2*x^ 2)^q) Int[(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, n}, x] && EqQ[e*f + d*g, 0] && EqQ[c^2*d^2 - e^2, 0] && HalfIntegerQ[p, q] && GeQ[p - q, 0]
Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.))/Sq rt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[1/(c^(m + 1)*Sqrt[d]) Subst[In t[(a + b*x)^n*(c*f + g*Sin[x])^m, x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c , d, e, f, g, n}, x] && EqQ[c^2*d + e, 0] && IntegerQ[m] && GtQ[d, 0] && (G tQ[m, 0] || IGtQ[n, 0])
\[\int \frac {\left (c d x +d \right )^{\frac {3}{2}} \left (a +b \arcsin \left (c x \right )\right )^{2}}{\sqrt {-c e x +e}}d x\]
\[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))^2}{\sqrt {e-c e x}} \, dx=\int { \frac {{\left (c d x + d\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{\sqrt {-c e x + e}} \,d x } \]
integral(-(a^2*c*d*x + a^2*d + (b^2*c*d*x + b^2*d)*arcsin(c*x)^2 + 2*(a*b* c*d*x + a*b*d)*arcsin(c*x))*sqrt(c*d*x + d)*sqrt(-c*e*x + e)/(c*e*x - e), x)
\[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))^2}{\sqrt {e-c e x}} \, dx=\int \frac {\left (d \left (c x + 1\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}{\sqrt {- e \left (c x - 1\right )}}\, dx \]
Exception generated. \[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))^2}{\sqrt {e-c e x}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))^2}{\sqrt {e-c e x}} \, dx=\int { \frac {{\left (c d x + d\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{\sqrt {-c e x + e}} \,d x } \]
Timed out. \[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))^2}{\sqrt {e-c e x}} \, dx=\int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,{\left (d+c\,d\,x\right )}^{3/2}}{\sqrt {e-c\,e\,x}} \,d x \]