Integrand size = 32, antiderivative size = 544 \[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))^2}{(e-c e x)^{5/2}} \, dx=-\frac {8 i d^4 \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))^2}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {d^4 \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))^3}{3 b c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {32 b d^4 \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x)) \log \left (1-i e^{-i \arcsin (c x)}\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {32 i b^2 d^4 \left (1-c^2 x^2\right )^{5/2} \operatorname {PolyLog}\left (2,i e^{-i \arcsin (c x)}\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {4 b d^4 \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x)) \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {8 b^2 d^4 \left (1-c^2 x^2\right )^{5/2} \tan \left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {8 d^4 \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))^2 \tan \left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {2 d^4 \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))^2 \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}} \]
-8/3*I*d^4*(-c^2*x^2+1)^(5/2)*(a+b*arcsin(c*x))^2/c/(c*d*x+d)^(5/2)/(-c*e* x+e)^(5/2)+1/3*d^4*(-c^2*x^2+1)^(5/2)*(a+b*arcsin(c*x))^3/b/c/(c*d*x+d)^(5 /2)/(-c*e*x+e)^(5/2)-32/3*b*d^4*(-c^2*x^2+1)^(5/2)*(a+b*arcsin(c*x))*ln(1- I/(I*c*x+(-c^2*x^2+1)^(1/2)))/c/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)-32/3*I*b^ 2*d^4*(-c^2*x^2+1)^(5/2)*polylog(2,I/(I*c*x+(-c^2*x^2+1)^(1/2)))/c/(c*d*x+ d)^(5/2)/(-c*e*x+e)^(5/2)-4/3*b*d^4*(-c^2*x^2+1)^(5/2)*(a+b*arcsin(c*x))*s ec(1/4*Pi+1/2*arcsin(c*x))^2/c/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)+8/3*b^2*d^ 4*(-c^2*x^2+1)^(5/2)*tan(1/4*Pi+1/2*arcsin(c*x))/c/(c*d*x+d)^(5/2)/(-c*e*x +e)^(5/2)-8/3*d^4*(-c^2*x^2+1)^(5/2)*(a+b*arcsin(c*x))^2*tan(1/4*Pi+1/2*ar csin(c*x))/c/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)+2/3*d^4*(-c^2*x^2+1)^(5/2)*( a+b*arcsin(c*x))^2*sec(1/4*Pi+1/2*arcsin(c*x))^2*tan(1/4*Pi+1/2*arcsin(c*x ))/c/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(1419\) vs. \(2(544)=1088\).
Time = 15.57 (sec) , antiderivative size = 1419, normalized size of antiderivative = 2.61 \[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))^2}{(e-c e x)^{5/2}} \, dx =\text {Too large to display} \]
(Sqrt[-(e*(-1 + c*x))]*Sqrt[d*(1 + c*x)]*((4*a^2*d)/(3*e^3*(-1 + c*x)^2) + (8*a^2*d)/(3*e^3*(-1 + c*x))))/c - (a^2*d^(3/2)*ArcTan[(c*x*Sqrt[-(e*(-1 + c*x))]*Sqrt[d*(1 + c*x)])/(Sqrt[d]*Sqrt[e]*(-1 + c*x)*(1 + c*x))])/(c*e^ (5/2)) + (a*b*d*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*Sqrt[-(d*e*(1 - c^2*x^2))] *(Cos[ArcSin[c*x]/2]*(-4 + 3*ArcSin[c*x] - 6*Log[Cos[ArcSin[c*x]/2] - Sin[ ArcSin[c*x]/2]]) - Cos[(3*ArcSin[c*x])/2]*(ArcSin[c*x] - 2*Log[Cos[ArcSin[ c*x]/2] - Sin[ArcSin[c*x]/2]]) + 2*(2 + 2*ArcSin[c*x] + Sqrt[1 - c^2*x^2]* ArcSin[c*x] + 4*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]] + 2*Sqrt[1 - c^2*x^2]*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]])*Sin[ArcSin[c*x]/2]) )/(3*c*e^3*Sqrt[(-d - c*d*x)*(e - c*e*x)]*(Cos[ArcSin[c*x]/2] - Sin[ArcSin [c*x]/2])^4*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2])) + (a*b*d*Sqrt[d + c *d*x]*Sqrt[e - c*e*x]*Sqrt[-(d*e*(1 - c^2*x^2))]*(Cos[ArcSin[c*x]/2]*(-8 - 6*ArcSin[c*x] + 9*ArcSin[c*x]^2 - 84*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[ c*x]/2]]) + Cos[(3*ArcSin[c*x])/2]*(-(ArcSin[c*x]*(14 + 3*ArcSin[c*x])) + 28*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]]) + 2*(4 + 4*ArcSin[c*x] - 6*ArcSin[c*x]^2 + 56*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]] + Sqrt[1 - c^2*x^2]*((14 - 3*ArcSin[c*x])*ArcSin[c*x] + 28*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]]))*Sin[ArcSin[c*x]/2]))/(6*c*e^3*Sqrt[(-d - c*d*x)*(e - c*e*x)]*(Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2])^4*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2])) + (b^2*d*(1 + c*x)*Sqrt[d + c*d*x]*Sqrt[e - c*e...
Time = 1.26 (sec) , antiderivative size = 279, normalized size of antiderivative = 0.51, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5178, 27, 5274, 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.
| \(\displaystyle \int \frac {(c d x+d)^{3/2} (a+b \arcsin (c x))^2}{(e-c e x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 5178 |
| \(\displaystyle \frac {\left (1-c^2 x^2\right )^{5/2} \int \frac {d^4 (c x+1)^4 (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d^4 \left (1-c^2 x^2\right )^{5/2} \int \frac {(c x+1)^4 (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\) |
| \(\Big \downarrow \) 5274 |
| \(\displaystyle \frac {d^4 \left (1-c^2 x^2\right )^{5/2} \int \left (\frac {4 (a+b \arcsin (c x))^2}{(c x-1) \sqrt {1-c^2 x^2}}+\frac {4 (a+b \arcsin (c x))^2}{(c x-1)^2 \sqrt {1-c^2 x^2}}+\frac {(a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}\right )dx}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^4 \left (1-c^2 x^2\right )^{5/2} \left (\frac {(a+b \arcsin (c x))^3}{3 b c}-\frac {8 i (a+b \arcsin (c x))^2}{3 c}-\frac {32 b \log \left (1-i e^{-i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{3 c}-\frac {8 \tan \left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right ) (a+b \arcsin (c x))^2}{3 c}-\frac {4 b \sec ^2\left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right ) (a+b \arcsin (c x))}{3 c}+\frac {2 \tan \left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right ) \sec ^2\left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right ) (a+b \arcsin (c x))^2}{3 c}-\frac {32 i b^2 \operatorname {PolyLog}\left (2,i e^{-i \arcsin (c x)}\right )}{3 c}+\frac {8 b^2 \tan \left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right )}{3 c}\right )}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\) |
(d^4*(1 - c^2*x^2)^(5/2)*((((-8*I)/3)*(a + b*ArcSin[c*x])^2)/c + (a + b*Ar cSin[c*x])^3/(3*b*c) - (32*b*(a + b*ArcSin[c*x])*Log[1 - I/E^(I*ArcSin[c*x ])])/(3*c) - (((32*I)/3)*b^2*PolyLog[2, I/E^(I*ArcSin[c*x])])/c - (4*b*(a + b*ArcSin[c*x])*Sec[Pi/4 + ArcSin[c*x]/2]^2)/(3*c) + (8*b^2*Tan[Pi/4 + Ar cSin[c*x]/2])/(3*c) - (8*(a + b*ArcSin[c*x])^2*Tan[Pi/4 + ArcSin[c*x]/2])/ (3*c) + (2*(a + b*ArcSin[c*x])^2*Sec[Pi/4 + ArcSin[c*x]/2]^2*Tan[Pi/4 + Ar cSin[c*x]/2])/(3*c)))/((d + c*d*x)^(5/2)*(e - c*e*x)^(5/2))
3.6.71.3.1 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[(a + b*ArcSin[c*x] )^n/Sqrt[d + e*x^2], (f + g*x)^m*(d + e*x^2)^(p + 1/2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[c^2*d + e, 0] && IntegerQ[m] && ILtQ[p + 1/2, 0] && GtQ[d, 0] && IGtQ[n, 0]
\[\int \frac {\left (c d x +d \right )^{\frac {3}{2}} \left (a +b \arcsin \left (c x \right )\right )^{2}}{\left (-c e x +e \right )^{\frac {5}{2}}}d x\]
\[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))^2}{(e-c e x)^{5/2}} \, dx=\int { \frac {{\left (c d x + d\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (-c e x + e\right )}^{\frac {5}{2}}} \,d x } \]
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^3*e^3*x^3 - 3*c^2*e^3*x^2 + 3*c*e^3*x - e^3), x)
\[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))^2}{(e-c e x)^{5/2}} \, dx=\int \frac {\left (d \left (c x + 1\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}{\left (- e \left (c x - 1\right )\right )^{\frac {5}{2}}}\, dx \]
Exception generated. \[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))^2}{(e-c e x)^{5/2}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))^2}{(e-c e x)^{5/2}} \, dx=\int { \frac {{\left (c d x + d\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (-c e x + e\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))^2}{(e-c e x)^{5/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,{\left (d+c\,d\,x\right )}^{3/2}}{{\left (e-c\,e\,x\right )}^{5/2}} \,d x \]