Integrand size = 32, antiderivative size = 729 \[ \int \frac {(e-c e x)^{5/2} (a+b \arcsin (c x))^2}{(d+c d x)^{5/2}} \, dx=-\frac {2 a b e^5 x \left (1-c^2 x^2\right )^{5/2}}{(d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {2 b^2 e^5 \left (1-c^2 x^2\right )^3}{c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {2 b^2 e^5 x \left (1-c^2 x^2\right )^{5/2} \arcsin (c x)}{(d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {28 i e^5 \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))^2}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {e^5 \left (1-c^2 x^2\right )^3 (a+b \arcsin (c x))^2}{c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {5 e^5 \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))^3}{3 b c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {16 b^2 e^5 \left (1-c^2 x^2\right )^{5/2} \cot \left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {28 e^5 \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))^2 \cot \left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {8 b e^5 \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x)) \csc ^2\left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {4 e^5 \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))^2 \cot \left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right ) \csc ^2\left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {112 b e^5 \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x)) \log \left (1-i e^{i \arcsin (c x)}\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {112 i b^2 e^5 \left (1-c^2 x^2\right )^{5/2} \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}} \]
-2*a*b*e^5*x*(-c^2*x^2+1)^(5/2)/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)-2*b^2*e^5 *(-c^2*x^2+1)^3/c/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)-2*b^2*e^5*x*(-c^2*x^2+1 )^(5/2)*arcsin(c*x)/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)+28/3*I*e^5*(-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)+e^5*(-c^2* x^2+1)^3*(a+b*arcsin(c*x))^2/c/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)+5/3*e^5*(- 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)- 16/3*b^2*e^5*(-c^2*x^2+1)^(5/2)*cot(1/4*Pi+1/2*arcsin(c*x))/c/(c*d*x+d)^(5 /2)/(-c*e*x+e)^(5/2)+28/3*e^5*(-c^2*x^2+1)^(5/2)*(a+b*arcsin(c*x))^2*cot(1 /4*Pi+1/2*arcsin(c*x))/c/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)-8/3*b*e^5*(-c^2* x^2+1)^(5/2)*(a+b*arcsin(c*x))*csc(1/4*Pi+1/2*arcsin(c*x))^2/c/(c*d*x+d)^( 5/2)/(-c*e*x+e)^(5/2)-4/3*e^5*(-c^2*x^2+1)^(5/2)*(a+b*arcsin(c*x))^2*cot(1 /4*Pi+1/2*arcsin(c*x))*csc(1/4*Pi+1/2*arcsin(c*x))^2/c/(c*d*x+d)^(5/2)/(-c *e*x+e)^(5/2)-112/3*b*e^5*(-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)+112/3*I*b^2*e^5 *(-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)
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(2338\) vs. \(2(729)=1458\).
Time = 24.67 (sec) , antiderivative size = 2338, normalized size of antiderivative = 3.21 \[ \int \frac {(e-c e x)^{5/2} (a+b \arcsin (c x))^2}{(d+c d x)^{5/2}} \, dx=\text {Result too large to show} \]
(Sqrt[-(e*(-1 + c*x))]*Sqrt[d*(1 + c*x)]*((a^2*e^2)/d^3 - (8*a^2*e^2)/(3*d ^3*(1 + c*x)^2) + (28*a^2*e^2)/(3*d^3*(1 + c*x))))/c - (5*a^2*e^(5/2)*ArcT an[(c*x*Sqrt[-(e*(-1 + c*x))]*Sqrt[d*(1 + c*x)])/(Sqrt[d]*Sqrt[e]*(-1 + c* x)*(1 + c*x))])/(c*d^(5/2)) - (a*b*e^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*Sqr t[-(d*e*(1 - c^2*x^2))]*(Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2])*(Cos[Arc Sin[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]*((14 - 3*ArcSin[c*x])* 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 + Sqrt[1 - c^2*x^2]*(ArcSin[c*x]*(14 + 3*Ar cSin[c*x]) - 28*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]]) - 56*Log[Cos [ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]])*Sin[ArcSin[c*x]/2]))/(3*c*d^3*(-1 + c*x)*Sqrt[(-d - c*d*x)*(e - c*e*x)]*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x] /2])^4) - (a*b*e^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*Sqrt[-(d*e*(1 - c^2*x^2 ))]*(Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2])*(Cos[(3*ArcSin[c*x])/2]*(Arc Sin[c*x] + 2*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]]) - Cos[ArcSin[c* x]/2]*(4 + 3*ArcSin[c*x] + 6*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[ArcSi n[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*d^3*(-1 + c*x)*Sqrt[(- d - c*d*x)*(e - c*e*x)]*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2])^4) - ...
Time = 1.50 (sec) , antiderivative size = 344, normalized size of antiderivative = 0.47, 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.
| \(\displaystyle \int \frac {(e-c e x)^{5/2} (a+b \arcsin (c x))^2}{(c d x+d)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 5178 |
| \(\displaystyle \frac {\left (1-c^2 x^2\right )^{5/2} \int \frac {e^5 (1-c x)^5 (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {e^5 \left (1-c^2 x^2\right )^{5/2} \int \frac {(1-c x)^5 (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\) |
| \(\Big \downarrow \) 5274 |
| \(\displaystyle \frac {e^5 \left (1-c^2 x^2\right )^{5/2} \int \left (-\frac {c x (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}-\frac {12 (a+b \arcsin (c x))^2}{(c x+1) \sqrt {1-c^2 x^2}}+\frac {8 (a+b \arcsin (c x))^2}{(c x+1)^2 \sqrt {1-c^2 x^2}}+\frac {5 (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}\right )dx}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {e^5 \left (1-c^2 x^2\right )^{5/2} \left (\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{c}+\frac {5 (a+b \arcsin (c x))^3}{3 b c}+\frac {28 i (a+b \arcsin (c x))^2}{3 c}-\frac {112 b \log \left (1-i e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{3 c}+\frac {28 \cot \left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right ) (a+b \arcsin (c x))^2}{3 c}-\frac {8 b \csc ^2\left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right ) (a+b \arcsin (c x))}{3 c}-\frac {4 \cot \left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right ) \csc ^2\left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right ) (a+b \arcsin (c x))^2}{3 c}-2 a b x+\frac {112 i b^2 \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{3 c}-2 b^2 x \arcsin (c x)-\frac {16 b^2 \cot \left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right )}{3 c}-\frac {2 b^2 \sqrt {1-c^2 x^2}}{c}\right )}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\) |
(e^5*(1 - c^2*x^2)^(5/2)*(-2*a*b*x - (2*b^2*Sqrt[1 - c^2*x^2])/c - 2*b^2*x *ArcSin[c*x] + (((28*I)/3)*(a + b*ArcSin[c*x])^2)/c + (Sqrt[1 - c^2*x^2]*( a + b*ArcSin[c*x])^2)/c + (5*(a + b*ArcSin[c*x])^3)/(3*b*c) - (16*b^2*Cot[ Pi/4 + ArcSin[c*x]/2])/(3*c) + (28*(a + b*ArcSin[c*x])^2*Cot[Pi/4 + ArcSin [c*x]/2])/(3*c) - (8*b*(a + b*ArcSin[c*x])*Csc[Pi/4 + ArcSin[c*x]/2]^2)/(3 *c) - (4*(a + b*ArcSin[c*x])^2*Cot[Pi/4 + ArcSin[c*x]/2]*Csc[Pi/4 + ArcSin [c*x]/2]^2)/(3*c) - (112*b*(a + b*ArcSin[c*x])*Log[1 - I*E^(I*ArcSin[c*x]) ])/(3*c) + (((112*I)/3)*b^2*PolyLog[2, I*E^(I*ArcSin[c*x])])/c))/((d + c*d *x)^(5/2)*(e - c*e*x)^(5/2))
3.6.57.3.1 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]
\[\int \frac {\left (-c e x +e \right )^{\frac {5}{2}} \left (a +b \arcsin \left (c x \right )\right )^{2}}{\left (c d x +d \right )^{\frac {5}{2}}}d x\]
\[ \int \frac {(e-c e x)^{5/2} (a+b \arcsin (c x))^2}{(d+c d x)^{5/2}} \, dx=\int { \frac {{\left (-c e x + e\right )}^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (c d x + d\right )}^{\frac {5}{2}}} \,d x } \]
integral((a^2*c^2*e^2*x^2 - 2*a^2*c*e^2*x + a^2*e^2 + (b^2*c^2*e^2*x^2 - 2 *b^2*c*e^2*x + b^2*e^2)*arcsin(c*x)^2 + 2*(a*b*c^2*e^2*x^2 - 2*a*b*c*e^2*x + a*b*e^2)*arcsin(c*x))*sqrt(c*d*x + d)*sqrt(-c*e*x + e)/(c^3*d^3*x^3 + 3 *c^2*d^3*x^2 + 3*c*d^3*x + d^3), x)
Timed out. \[ \int \frac {(e-c e x)^{5/2} (a+b \arcsin (c x))^2}{(d+c d x)^{5/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(e-c e x)^{5/2} (a+b \arcsin (c x))^2}{(d+c d x)^{5/2}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {(e-c e x)^{5/2} (a+b \arcsin (c x))^2}{(d+c d x)^{5/2}} \, dx=\int { \frac {{\left (-c e x + e\right )}^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (c d x + d\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(e-c e x)^{5/2} (a+b \arcsin (c x))^2}{(d+c d x)^{5/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,{\left (e-c\,e\,x\right )}^{5/2}}{{\left (d+c\,d\,x\right )}^{5/2}} \,d x \]