Integrand size = 32, antiderivative size = 398 \[ \int \frac {(e-c e x)^{3/2} (a+b \arcsin (c x))^2}{\sqrt {d+c d x}} \, dx=-\frac {4 b^2 e^2 \left (1-c^2 x^2\right )}{c \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {b^2 e^2 x \left (1-c^2 x^2\right )}{4 \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {b^2 e^2 \sqrt {1-c^2 x^2} \arcsin (c x)}{4 c \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {4 b e^2 x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{\sqrt {d+c d x} \sqrt {e-c e x}}+\frac {b c e^2 x^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{2 \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {2 e^2 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{c \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {e^2 x \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{2 \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {e^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^3}{2 b c \sqrt {d+c d x} \sqrt {e-c e x}} \]
-4*b^2*e^2*(-c^2*x^2+1)/c/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)+1/4*b^2*e^2*x*( -c^2*x^2+1)/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)+2*e^2*(-c^2*x^2+1)*(a+b*arcsi n(c*x))^2/c/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-1/2*e^2*x*(-c^2*x^2+1)*(a+b*a rcsin(c*x))^2/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-1/4*b^2*e^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*e^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*e^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*e ^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 = 9.07 (sec) , antiderivative size = 358, normalized size of antiderivative = 0.90 \[ \int \frac {(e-c e x)^{3/2} (a+b \arcsin (c x))^2}{\sqrt {d+c d x}} \, dx=\frac {4 b^2 e \sqrt {d+c d x} \sqrt {e-c e x} \arcsin (c x)^3-12 a^2 \sqrt {d} e^{3/2} \sqrt {1-c^2 x^2} \arctan \left (\frac {c x \sqrt {d+c d x} \sqrt {e-c e x}}{\sqrt {d} \sqrt {e} \left (-1+c^2 x^2\right )}\right )-2 b e \sqrt {d+c d x} \sqrt {e-c e x} \arcsin (c x) \left (16 b c x+4 a (-4+c x) \sqrt {1-c^2 x^2}+b \cos (2 \arcsin (c x))\right )+2 b e \sqrt {d+c d x} \sqrt {e-c e x} \arcsin (c x)^2 \left (6 a+8 b \sqrt {1-c^2 x^2}-b \sin (2 \arcsin (c x))\right )+e \sqrt {d+c d x} \sqrt {e-c e x} \left (-4 \left (8 a b c x+8 b^2 \sqrt {1-c^2 x^2}+a^2 (-4+c x) \sqrt {1-c^2 x^2}\right )-2 a b \cos (2 \arcsin (c x))+b^2 \sin (2 \arcsin (c x))\right )}{8 c d \sqrt {1-c^2 x^2}} \]
(4*b^2*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*ArcSin[c*x]^3 - 12*a^2*Sqrt[d]*e^ (3/2)*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))] - 2*b*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*ArcSi n[c*x]*(16*b*c*x + 4*a*(-4 + c*x)*Sqrt[1 - c^2*x^2] + b*Cos[2*ArcSin[c*x]] ) + 2*b*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*ArcSin[c*x]^2*(6*a + 8*b*Sqrt[1 - c^2*x^2] - b*Sin[2*ArcSin[c*x]]) + e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*(-4 *(8*a*b*c*x + 8*b^2*Sqrt[1 - c^2*x^2] + a^2*(-4 + c*x)*Sqrt[1 - c^2*x^2]) - 2*a*b*Cos[2*ArcSin[c*x]] + b^2*Sin[2*ArcSin[c*x]]))/(8*c*d*Sqrt[1 - c^2* x^2])
Time = 0.77 (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.
| \(\displaystyle \int \frac {(e-c e x)^{3/2} (a+b \arcsin (c x))^2}{\sqrt {c d x+d}} \, dx\) |
| \(\Big \downarrow \) 5178 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \int \frac {e^2 (1-c x)^2 (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}dx}{\sqrt {c d x+d} \sqrt {e-c e x}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {e^2 \sqrt {1-c^2 x^2} \int \frac {(1-c x)^2 (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}dx}{\sqrt {c d x+d} \sqrt {e-c e x}}\) |
| \(\Big \downarrow \) 5272 |
| \(\displaystyle \frac {e^2 \sqrt {1-c^2 x^2} \int \left (c-c^2 x\right )^2 (a+b \arcsin (c x))^2d\arcsin (c x)}{c^3 \sqrt {c d x+d} \sqrt {e-c e x}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {e^2 \sqrt {1-c^2 x^2} \int (a+b \arcsin (c x))^2 (c-c \sin (\arcsin (c x)))^2d\arcsin (c x)}{c^3 \sqrt {c d x+d} \sqrt {e-c e x}}\) |
| \(\Big \downarrow \) 3798 |
| \(\displaystyle \frac {e^2 \sqrt {1-c^2 x^2} \int \left (x^2 (a+b \arcsin (c x))^2 c^4-2 x (a+b \arcsin (c x))^2 c^3+(a+b \arcsin (c x))^2 c^2\right )d\arcsin (c x)}{c^3 \sqrt {c d x+d} \sqrt {e-c e x}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {e^2 \sqrt {1-c^2 x^2} \left (\frac {1}{2} b c^4 x^2 (a+b \arcsin (c x))-4 b c^3 x (a+b \arcsin (c x))+2 c^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2+\frac {c^2 (a+b \arcsin (c x))^3}{2 b}-\frac {1}{2} c^3 x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2-\frac {1}{4} b^2 c^2 \arcsin (c x)-4 b^2 c^2 \sqrt {1-c^2 x^2}+\frac {1}{4} b^2 c^3 x \sqrt {1-c^2 x^2}\right )}{c^3 \sqrt {c d x+d} \sqrt {e-c e x}}\) |
(e^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*A rcSin[c*x])^3)/(2*b)))/(c^3*Sqrt[d + c*d*x]*Sqrt[e - c*e*x])
3.6.49.3.1 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])
\[\int \frac {\left (-c e x +e \right )^{\frac {3}{2}} \left (a +b \arcsin \left (c x \right )\right )^{2}}{\sqrt {c d x +d}}d x\]
\[ \int \frac {(e-c e x)^{3/2} (a+b \arcsin (c x))^2}{\sqrt {d+c d x}} \, dx=\int { \frac {{\left (-c e x + e\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{\sqrt {c d x + d}} \,d x } \]
integral(-(a^2*c*e*x - a^2*e + (b^2*c*e*x - b^2*e)*arcsin(c*x)^2 + 2*(a*b* c*e*x - a*b*e)*arcsin(c*x))*sqrt(-c*e*x + e)/sqrt(c*d*x + d), x)
\[ \int \frac {(e-c e x)^{3/2} (a+b \arcsin (c x))^2}{\sqrt {d+c d x}} \, dx=\int \frac {\left (- e \left (c x - 1\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}{\sqrt {d \left (c x + 1\right )}}\, dx \]
Exception generated. \[ \int \frac {(e-c e x)^{3/2} (a+b \arcsin (c x))^2}{\sqrt {d+c d x}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {(e-c e x)^{3/2} (a+b \arcsin (c x))^2}{\sqrt {d+c d x}} \, dx=\int { \frac {{\left (-c e x + e\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{\sqrt {c d x + d}} \,d x } \]
Timed out. \[ \int \frac {(e-c e x)^{3/2} (a+b \arcsin (c x))^2}{\sqrt {d+c d x}} \, dx=\int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,{\left (e-c\,e\,x\right )}^{3/2}}{\sqrt {d+c\,d\,x}} \,d x \]