Integrand size = 32, antiderivative size = 918 \[ \int \frac {(d+c d x)^{5/2} (a+b \arcsin (c x))^2}{(e-c e x)^{3/2}} \, dx=-\frac {8 a b d^4 x \left (1-c^2 x^2\right )^{3/2}}{(d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {8 b^2 d^4 \left (1-c^2 x^2\right )^2}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {b^2 d^4 x \left (1-c^2 x^2\right )^2}{4 (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {b^2 d^4 \left (1-c^2 x^2\right )^{3/2} \arcsin (c x)}{4 c (d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {8 b^2 d^4 x \left (1-c^2 x^2\right )^{3/2} \arcsin (c x)}{(d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {b c d^4 x^2 \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))}{2 (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {8 d^4 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {8 d^4 x \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{(d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {8 i d^4 \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))^2}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {4 d^4 \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))^2}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {d^4 x \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))^2}{2 (d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {5 d^4 \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))^3}{2 b c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {32 i b d^4 \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x)) \arctan \left (e^{i \arcsin (c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {16 b d^4 \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x)) \log \left (1+e^{2 i \arcsin (c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {16 i b^2 d^4 \left (1-c^2 x^2\right )^{3/2} \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {16 i b^2 d^4 \left (1-c^2 x^2\right )^{3/2} \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {8 i b^2 d^4 \left (1-c^2 x^2\right )^{3/2} \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}} \]
-8*a*b*d^4*x*(-c^2*x^2+1)^(3/2)/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)-8*b^2*d^4 *(-c^2*x^2+1)^2/c/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)-1/4*b^2*d^4*x*(-c^2*x^2 +1)^2/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)+1/4*b^2*d^4*(-c^2*x^2+1)^(3/2)*arcs in(c*x)/c/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)-8*b^2*d^4*x*(-c^2*x^2+1)^(3/2)* arcsin(c*x)/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)-1/2*b*c*d^4*x^2*(-c^2*x^2+1)^ (3/2)*(a+b*arcsin(c*x))/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)+8*d^4*(-c^2*x^2+1 )*(a+b*arcsin(c*x))^2/c/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)+8*d^4*x*(-c^2*x^2 +1)*(a+b*arcsin(c*x))^2/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)+16*I*b^2*d^4*(-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)+4*d^4*(-c^2*x^2+1)^2*(a+b*arcsin(c*x))^2/c/(c*d*x+d)^(3/2) /(-c*e*x+e)^(3/2)+1/2*d^4*x*(-c^2*x^2+1)^2*(a+b*arcsin(c*x))^2/(c*d*x+d)^( 3/2)/(-c*e*x+e)^(3/2)-5/2*d^4*(-c^2*x^2+1)^(3/2)*(a+b*arcsin(c*x))^3/b/c/( c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)-16*I*b^2*d^4*(-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)+16*b*d^4 *(-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)+32*I*b*d^4*(-c^2*x^2+1)^(3/2)*(a+b*arcsi n(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 )-8*I*d^4*(-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)-8*I*b^2*d^4*(-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)
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(2041\) vs. \(2(918)=1836\).
Time = 20.33 (sec) , antiderivative size = 2041, normalized size of antiderivative = 2.22 \[ \int \frac {(d+c d x)^{5/2} (a+b \arcsin (c x))^2}{(e-c e x)^{3/2}} \, dx=\text {Result too large to show} \]
(Sqrt[-(e*(-1 + c*x))]*Sqrt[d*(1 + c*x)]*((4*a^2*d^2)/e^2 + (a^2*c*d^2*x)/ (2*e^2) - (8*a^2*d^2)/(e^2*(-1 + c*x))))/c + (15*a^2*d^(5/2)*ArcTan[(c*x*S qrt[-(e*(-1 + c*x))]*Sqrt[d*(1 + c*x)])/(Sqrt[d]*Sqrt[e]*(-1 + c*x)*(1 + c *x))])/(2*c*e^(3/2)) - (a*b*d^2*(1 + c*x)*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]* Sqrt[-(d*e*(1 - c^2*x^2))]*(Cos[ArcSin[c*x]/2]*((-4 + ArcSin[c*x])*ArcSin[ c*x] - 8*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]]) - (ArcSin[c*x]*(4 + ArcSin[c*x]) - 8*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]])*Sin[ArcSin [c*x]/2]))/(c*e^2*Sqrt[(-d - c*d*x)*(e - c*e*x)]*Sqrt[1 - c^2*x^2]*(Cos[Ar cSin[c*x]/2] - Sin[ArcSin[c*x]/2])*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2 ])^2) + (4*a*b*d^2*(1 + c*x)*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*Sqrt[-(d*e*(1 - c^2*x^2))]*(Cos[ArcSin[c*x]/2]*(-(c*x) + 2*ArcSin[c*x] + Sqrt[1 - c^2*x ^2]*ArcSin[c*x] - ArcSin[c*x]^2 + 4*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c* x]/2]]) + (c*x + 2*ArcSin[c*x] - Sqrt[1 - c^2*x^2]*ArcSin[c*x] + ArcSin[c* x]^2 - 4*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]])*Sin[ArcSin[c*x]/2]) )/(c*e^2*Sqrt[(-d - c*d*x)*(e - c*e*x)]*Sqrt[1 - c^2*x^2]*(Cos[ArcSin[c*x] /2] - Sin[ArcSin[c*x]/2])*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2])^2) - ( b^2*d^2*(1 + c*x)*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*Sqrt[-(d*e*(1 - c^2*x^2) )]*((-18*I)*Pi*ArcSin[c*x] - (6 - 6*I)*ArcSin[c*x]^2 + ArcSin[c*x]^3 - 24* Pi*Log[1 + E^((-I)*ArcSin[c*x])] + 12*(Pi - 2*ArcSin[c*x])*Log[1 + I*E^(I* ArcSin[c*x])] + 24*Pi*Log[Cos[ArcSin[c*x]/2]] - 12*Pi*Log[-Cos[(Pi + 2*...
Time = 1.31 (sec) , antiderivative size = 411, normalized size of antiderivative = 0.45, 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.
| \(\displaystyle \int \frac {(c d x+d)^{5/2} (a+b \arcsin (c x))^2}{(e-c e x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 5178 |
| \(\displaystyle \frac {\left (1-c^2 x^2\right )^{3/2} \int \frac {d^4 (c x+1)^4 (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{3/2}}dx}{(c d x+d)^{3/2} (e-c e x)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d^4 \left (1-c^2 x^2\right )^{3/2} \int \frac {(c x+1)^4 (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{3/2}}dx}{(c d x+d)^{3/2} (e-c e x)^{3/2}}\) |
| \(\Big \downarrow \) 5274 |
| \(\displaystyle \frac {d^4 \left (1-c^2 x^2\right )^{3/2} \int \left (-\frac {c^2 x^2 (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}-\frac {4 c x (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}-\frac {7 (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}+\frac {8 (c x+1) (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{3/2}}\right )dx}{(c d x+d)^{3/2} (e-c e x)^{3/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^4 \left (1-c^2 x^2\right )^{3/2} \left (\frac {32 i b \arctan \left (e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{c}+\frac {1}{2} x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2+\frac {4 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{c}+\frac {8 x (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}+\frac {8 (a+b \arcsin (c x))^2}{c \sqrt {1-c^2 x^2}}-\frac {1}{2} b c x^2 (a+b \arcsin (c x))-\frac {5 (a+b \arcsin (c x))^3}{2 b c}-\frac {8 i (a+b \arcsin (c x))^2}{c}+\frac {16 b \log \left (1+e^{2 i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{c}-8 a b x-\frac {16 i b^2 \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )}{c}+\frac {16 i b^2 \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{c}-\frac {8 i b^2 \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (c x)}\right )}{c}-8 b^2 x \arcsin (c x)+\frac {b^2 \arcsin (c x)}{4 c}-\frac {1}{4} b^2 x \sqrt {1-c^2 x^2}-\frac {8 b^2 \sqrt {1-c^2 x^2}}{c}\right )}{(c d x+d)^{3/2} (e-c e x)^{3/2}}\) |
(d^4*(1 - c^2*x^2)^(3/2)*(-8*a*b*x - (8*b^2*Sqrt[1 - c^2*x^2])/c - (b^2*x* Sqrt[1 - c^2*x^2])/4 + (b^2*ArcSin[c*x])/(4*c) - 8*b^2*x*ArcSin[c*x] - (b* c*x^2*(a + b*ArcSin[c*x]))/2 - ((8*I)*(a + b*ArcSin[c*x])^2)/c + (8*(a + b *ArcSin[c*x])^2)/(c*Sqrt[1 - c^2*x^2]) + (8*x*(a + b*ArcSin[c*x])^2)/Sqrt[ 1 - c^2*x^2] + (4*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2)/c + (x*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2)/2 - (5*(a + b*ArcSin[c*x])^3)/(2*b*c) + ( (32*I)*b*(a + b*ArcSin[c*x])*ArcTan[E^(I*ArcSin[c*x])])/c + (16*b*(a + b*A rcSin[c*x])*Log[1 + E^((2*I)*ArcSin[c*x])])/c - ((16*I)*b^2*PolyLog[2, (-I )*E^(I*ArcSin[c*x])])/c + ((16*I)*b^2*PolyLog[2, I*E^(I*ArcSin[c*x])])/c - ((8*I)*b^2*PolyLog[2, -E^((2*I)*ArcSin[c*x])])/c))/((d + c*d*x)^(3/2)*(e - c*e*x)^(3/2))
3.6.64.3.1 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]
\[\int \frac {\left (c d x +d \right )^{\frac {5}{2}} \left (a +b \arcsin \left (c x \right )\right )^{2}}{\left (-c e x +e \right )^{\frac {3}{2}}}d x\]
\[ \int \frac {(d+c d x)^{5/2} (a+b \arcsin (c x))^2}{(e-c e x)^{3/2}} \, dx=\int { \frac {{\left (c d x + d\right )}^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (-c e x + e\right )}^{\frac {3}{2}}} \,d x } \]
integral((a^2*c^2*d^2*x^2 + 2*a^2*c*d^2*x + a^2*d^2 + (b^2*c^2*d^2*x^2 + 2 *b^2*c*d^2*x + b^2*d^2)*arcsin(c*x)^2 + 2*(a*b*c^2*d^2*x^2 + 2*a*b*c*d^2*x + a*b*d^2)*arcsin(c*x))*sqrt(c*d*x + d)*sqrt(-c*e*x + e)/(c^2*e^2*x^2 - 2 *c*e^2*x + e^2), x)
Timed out. \[ \int \frac {(d+c d x)^{5/2} (a+b \arcsin (c x))^2}{(e-c e x)^{3/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(d+c d x)^{5/2} (a+b \arcsin (c x))^2}{(e-c e x)^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {(d+c d x)^{5/2} (a+b \arcsin (c x))^2}{(e-c e x)^{3/2}} \, dx=\int { \frac {{\left (c d x + d\right )}^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (-c e x + e\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(d+c d x)^{5/2} (a+b \arcsin (c x))^2}{(e-c e x)^{3/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,{\left (d+c\,d\,x\right )}^{5/2}}{{\left (e-c\,e\,x\right )}^{3/2}} \,d x \]