Integrand size = 32, antiderivative size = 454 \[ \int \frac {(a+b \arcsin (c x))^2}{\sqrt {d+c d x} (e-c e x)^{3/2}} \, dx=\frac {d \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {d x \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{(d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {i d \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))^2}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {4 i b d \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x)) \arctan \left (e^{i \arcsin (c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {2 b d \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x)) \log \left (1+e^{2 i \arcsin (c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {2 i b^2 d \left (1-c^2 x^2\right )^{3/2} \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {2 i b^2 d \left (1-c^2 x^2\right )^{3/2} \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {i b^2 d \left (1-c^2 x^2\right )^{3/2} \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}} \]
d*(-c^2*x^2+1)*(a+b*arcsin(c*x))^2/c/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)+d*x* (-c^2*x^2+1)*(a+b*arcsin(c*x))^2/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)-I*d*(-c^ 2*x^2+1)^(3/2)*(a+b*arcsin(c*x))^2/c/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)+4*I* b*d*(-c^2*x^2+1)^(3/2)*(a+b*arcsin(c*x))*arctan(I*c*x+(-c^2*x^2+1)^(1/2))/ c/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)+2*b*d*(-c^2*x^2+1)^(3/2)*(a+b*arcsin(c* x))*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)/c/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)- 2*I*b^2*d*(-c^2*x^2+1)^(3/2)*polylog(2,-I*(I*c*x+(-c^2*x^2+1)^(1/2)))/c/(c *d*x+d)^(3/2)/(-c*e*x+e)^(3/2)+2*I*b^2*d*(-c^2*x^2+1)^(3/2)*polylog(2,I*(I *c*x+(-c^2*x^2+1)^(1/2)))/c/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)-I*b^2*d*(-c^2 *x^2+1)^(3/2)*polylog(2,-(I*c*x+(-c^2*x^2+1)^(1/2))^2)/c/(c*d*x+d)^(3/2)/( -c*e*x+e)^(3/2)
Time = 3.25 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.49 \[ \int \frac {(a+b \arcsin (c x))^2}{\sqrt {d+c d x} (e-c e x)^{3/2}} \, dx=-\frac {\sqrt {d+c d x} \sqrt {e-c e x} \left (a \left (a+a c x+4 b \sqrt {1-c^2 x^2} \log \left (\cos \left (\frac {1}{4} (\pi +2 \arcsin (c x))\right )\right )\right )-4 i b^2 \sqrt {1-c^2 x^2} \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )+b^2 \sqrt {1-c^2 x^2} \arcsin (c x)^2 \left (-i+\tan \left (\frac {1}{4} (\pi +2 \arcsin (c x))\right )\right )+2 b \sqrt {1-c^2 x^2} \arcsin (c x) \left (2 b \log \left (1+i e^{i \arcsin (c x)}\right )+a \tan \left (\frac {1}{4} (\pi +2 \arcsin (c x))\right )\right )\right )}{c d e^2 (-1+c x) (1+c x)} \]
-((Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*(a*(a + a*c*x + 4*b*Sqrt[1 - c^2*x^2]*L og[Cos[(Pi + 2*ArcSin[c*x])/4]]) - (4*I)*b^2*Sqrt[1 - c^2*x^2]*PolyLog[2, (-I)*E^(I*ArcSin[c*x])] + b^2*Sqrt[1 - c^2*x^2]*ArcSin[c*x]^2*(-I + Tan[(P i + 2*ArcSin[c*x])/4]) + 2*b*Sqrt[1 - c^2*x^2]*ArcSin[c*x]*(2*b*Log[1 + I* E^(I*ArcSin[c*x])] + a*Tan[(Pi + 2*ArcSin[c*x])/4])))/(c*d*e^2*(-1 + c*x)* (1 + c*x)))
Time = 0.83 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.53, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5178, 27, 5262, 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.
| \(\displaystyle \int \frac {(a+b \arcsin (c x))^2}{\sqrt {c d x+d} (e-c e x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 5178 |
| \(\displaystyle \frac {\left (1-c^2 x^2\right )^{3/2} \int \frac {d (c x+1) (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{3/2}}dx}{(c d x+d)^{3/2} (e-c e x)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d \left (1-c^2 x^2\right )^{3/2} \int \frac {(c x+1) (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{3/2}}dx}{(c d x+d)^{3/2} (e-c e x)^{3/2}}\) |
| \(\Big \downarrow \) 5262 |
| \(\displaystyle \frac {d \left (1-c^2 x^2\right )^{3/2} \int \left (\frac {c x (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{3/2}}+\frac {(a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{3/2}}\right )dx}{(c d x+d)^{3/2} (e-c e x)^{3/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d \left (1-c^2 x^2\right )^{3/2} \left (\frac {4 i b \arctan \left (e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{c}+\frac {x (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}+\frac {(a+b \arcsin (c x))^2}{c \sqrt {1-c^2 x^2}}-\frac {i (a+b \arcsin (c x))^2}{c}+\frac {2 b \log \left (1+e^{2 i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{c}-\frac {2 i b^2 \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )}{c}+\frac {2 i b^2 \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{c}-\frac {i b^2 \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (c x)}\right )}{c}\right )}{(c d x+d)^{3/2} (e-c e x)^{3/2}}\) |
(d*(1 - c^2*x^2)^(3/2)*(((-I)*(a + b*ArcSin[c*x])^2)/c + (a + b*ArcSin[c*x ])^2/(c*Sqrt[1 - c^2*x^2]) + (x*(a + b*ArcSin[c*x])^2)/Sqrt[1 - c^2*x^2] + ((4*I)*b*(a + b*ArcSin[c*x])*ArcTan[E^(I*ArcSin[c*x])])/c + (2*b*(a + b*A rcSin[c*x])*Log[1 + E^((2*I)*ArcSin[c*x])])/c - ((2*I)*b^2*PolyLog[2, (-I) *E^(I*ArcSin[c*x])])/c + ((2*I)*b^2*PolyLog[2, I*E^(I*ArcSin[c*x])])/c - ( I*b^2*PolyLog[2, -E^((2*I)*ArcSin[c*x])])/c))/((d + c*d*x)^(3/2)*(e - c*e* x)^(3/2))
3.6.67.3.1 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[((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_.)*((d_ ) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] & & EqQ[c^2*d + e, 0] && IGtQ[m, 0] && IntegerQ[p + 1/2] && GtQ[d, 0] && IGtQ [n, 0] && (m == 1 || p > 0 || (n == 1 && p > -1) || (m == 2 && p < -2))
\[\int \frac {\left (a +b \arcsin \left (c x \right )\right )^{2}}{\sqrt {c d x +d}\, \left (-c e x +e \right )^{\frac {3}{2}}}d x\]
\[ \int \frac {(a+b \arcsin (c x))^2}{\sqrt {d+c d x} (e-c e x)^{3/2}} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{\sqrt {c d x + d} {\left (-c e x + e\right )}^{\frac {3}{2}}} \,d x } \]
integral((b^2*arcsin(c*x)^2 + 2*a*b*arcsin(c*x) + a^2)*sqrt(c*d*x + d)*sqr t(-c*e*x + e)/(c^3*d*e^2*x^3 - c^2*d*e^2*x^2 - c*d*e^2*x + d*e^2), x)
\[ \int \frac {(a+b \arcsin (c x))^2}{\sqrt {d+c d x} (e-c e x)^{3/2}} \, dx=\int \frac {\left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}{\sqrt {d \left (c x + 1\right )} \left (- e \left (c x - 1\right )\right )^{\frac {3}{2}}}\, dx \]
Exception generated. \[ \int \frac {(a+b \arcsin (c x))^2}{\sqrt {d+c d x} (e-c e x)^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {(a+b \arcsin (c x))^2}{\sqrt {d+c d x} (e-c e x)^{3/2}} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{\sqrt {c d x + d} {\left (-c e x + e\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+b \arcsin (c x))^2}{\sqrt {d+c d x} (e-c e x)^{3/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2}{\sqrt {d+c\,d\,x}\,{\left (e-c\,e\,x\right )}^{3/2}} \,d x \]