Integrand size = 32, antiderivative size = 730 \[ \int \frac {(d+c d x)^{5/2} (a+b \arcsin (c x))^2}{(e-c e x)^{5/2}} \, dx=\frac {2 a b d^5 x \left (1-c^2 x^2\right )^{5/2}}{(d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {2 b^2 d^5 \left (1-c^2 x^2\right )^3}{c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {2 b^2 d^5 x \left (1-c^2 x^2\right )^{5/2} \arcsin (c x)}{(d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {28 i d^5 \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))^2}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {d^5 \left (1-c^2 x^2\right )^3 (a+b \arcsin (c x))^2}{c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {5 d^5 \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))^3}{3 b c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {112 b d^5 \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x)) \log \left (1-i e^{-i \arcsin (c x)}\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {112 i b^2 d^5 \left (1-c^2 x^2\right )^{5/2} \operatorname {PolyLog}\left (2,i e^{-i \arcsin (c x)}\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {8 b d^5 \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x)) \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {16 b^2 d^5 \left (1-c^2 x^2\right )^{5/2} \tan \left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {28 d^5 \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))^2 \tan \left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {4 d^5 \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))^2 \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}} \]
2*a*b*d^5*x*(-c^2*x^2+1)^(5/2)/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)+2*b^2*d^5* (-c^2*x^2+1)^3/c/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)+2*b^2*d^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*d^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)-d^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*d^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)-1 12/3*b*d^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*d^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)-8/3*b*d^5*(-c^2*x^2+1)^(5/2)*(a+b*arcsin(c*x))*sec(1/4*Pi+1/2*arcsin (c*x))^2/c/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)+16/3*b^2*d^5*(-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)-28/3*d^5* (-c^2*x^2+1)^(5/2)*(a+b*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)+4/3*d^5*(-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. \(2312\) vs. \(2(730)=1460\).
Time = 21.10 (sec) , antiderivative size = 2312, normalized size of antiderivative = 3.17 \[ \int \frac {(d+c d x)^{5/2} (a+b \arcsin (c x))^2}{(e-c e x)^{5/2}} \, dx=\text {Result too large to show} \]
(Sqrt[-(e*(-1 + c*x))]*Sqrt[d*(1 + c*x)]*(-((a^2*d^2)/e^3) + (8*a^2*d^2)/( 3*e^3*(-1 + c*x)^2) + (28*a^2*d^2)/(3*e^3*(-1 + c*x))))/c - (5*a^2*d^(5/2) *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^2*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*L og[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]]) + 2*(2 + 2*ArcS in[c*x] + Sqrt[1 - c^2*x^2]*ArcSin[c*x] + 4*Log[Cos[ArcSin[c*x]/2] - Sin[A rcSin[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^2*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[Co s[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])) /(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])) + (b^2*d^2*(1 + c...
Time = 1.47 (sec) , antiderivative size = 345, 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.
| \(\displaystyle \int \frac {(c d x+d)^{5/2} (a+b \arcsin (c x))^2}{(e-c e x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 5178 |
| \(\displaystyle \frac {\left (1-c^2 x^2\right )^{5/2} \int \frac {d^5 (c x+1)^5 (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d^5 \left (1-c^2 x^2\right )^{5/2} \int \frac {(c x+1)^5 (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\) |
| \(\Big \downarrow \) 5274 |
| \(\displaystyle \frac {d^5 \left (1-c^2 x^2\right )^{5/2} \int \left (\frac {c x (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}+\frac {12 (a+b \arcsin (c x))^2}{(c x-1) \sqrt {1-c^2 x^2}}+\frac {8 (a+b \arcsin (c x))^2}{(c x-1)^2 \sqrt {1-c^2 x^2}}+\frac {5 (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}\right )dx}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^5 \left (1-c^2 x^2\right )^{5/2} \left (-\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{c}+\frac {5 (a+b \arcsin (c x))^3}{3 b c}-\frac {28 i (a+b \arcsin (c x))^2}{3 c}-\frac {112 b \log \left (1-i e^{-i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{3 c}-\frac {28 \tan \left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right ) (a+b \arcsin (c x))^2}{3 c}-\frac {8 b \sec ^2\left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right ) (a+b \arcsin (c x))}{3 c}+\frac {4 \tan \left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right ) \sec ^2\left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right ) (a+b \arcsin (c x))^2}{3 c}+2 a b x-\frac {112 i b^2 \operatorname {PolyLog}\left (2,i e^{-i \arcsin (c x)}\right )}{3 c}+2 b^2 x \arcsin (c x)+\frac {16 b^2 \tan \left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right )}{3 c}+\frac {2 b^2 \sqrt {1-c^2 x^2}}{c}\right )}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\) |
(d^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) - (112*b*(a + b *ArcSin[c*x])*Log[1 - I/E^(I*ArcSin[c*x])])/(3*c) - (((112*I)/3)*b^2*PolyL og[2, I/E^(I*ArcSin[c*x])])/c - (8*b*(a + b*ArcSin[c*x])*Sec[Pi/4 + ArcSin [c*x]/2]^2)/(3*c) + (16*b^2*Tan[Pi/4 + ArcSin[c*x]/2])/(3*c) - (28*(a + b* ArcSin[c*x])^2*Tan[Pi/4 + ArcSin[c*x]/2])/(3*c) + (4*(a + b*ArcSin[c*x])^2 *Sec[Pi/4 + ArcSin[c*x]/2]^2*Tan[Pi/4 + ArcSin[c*x]/2])/(3*c)))/((d + c*d* x)^(5/2)*(e - c*e*x)^(5/2))
3.6.70.3.1 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]
\[\int \frac {\left (c d x +d \right )^{\frac {5}{2}} \left (a +b \arcsin \left (c x \right )\right )^{2}}{\left (-c e x +e \right )^{\frac {5}{2}}}d x\]
\[ \int \frac {(d+c d x)^{5/2} (a+b \arcsin (c x))^2}{(e-c e x)^{5/2}} \, dx=\int { \frac {{\left (c d x + d\right )}^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (-c e x + e\right )}^{\frac {5}{2}}} \,d x } \]
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^3*e^3*x^3 - 3*c^2*e^3*x^2 + 3*c*e^3*x - e^3), x)
Timed out. \[ \int \frac {(d+c d x)^{5/2} (a+b \arcsin (c x))^2}{(e-c e x)^{5/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(d+c d x)^{5/2} (a+b \arcsin (c x))^2}{(e-c e x)^{5/2}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {(d+c d x)^{5/2} (a+b \arcsin (c x))^2}{(e-c e x)^{5/2}} \, dx=\int { \frac {{\left (c d x + d\right )}^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (-c e x + e\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(d+c d x)^{5/2} (a+b \arcsin (c x))^2}{(e-c e x)^{5/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,{\left (d+c\,d\,x\right )}^{5/2}}{{\left (e-c\,e\,x\right )}^{5/2}} \,d x \]