Integrand size = 32, antiderivative size = 709 \[ \int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{5/2} (e-c e x)^{3/2}} \, dx=-\frac {b^2 e \left (1-c^2 x^2\right )^2}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {b^2 e x \left (1-c^2 x^2\right )^2}{3 (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {b e \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {b e x \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))}{3 (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {e \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {e x \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{3 (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {2 e x \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))^2}{3 (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {2 i e \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))^2}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {2 i b e \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x)) \arctan \left (e^{i \arcsin (c x)}\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {4 b e \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x)) \log \left (1+e^{2 i \arcsin (c x)}\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {i b^2 e \left (1-c^2 x^2\right )^{5/2} \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {i b^2 e \left (1-c^2 x^2\right )^{5/2} \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {2 i b^2 e \left (1-c^2 x^2\right )^{5/2} \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (c x)}\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}} \]
-1/3*b^2*e*(-c^2*x^2+1)^2/c/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)+1/3*b^2*e*x*( -c^2*x^2+1)^2/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)-1/3*b*e*(-c^2*x^2+1)^(3/2)* (a+b*arcsin(c*x))/c/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)+1/3*b*e*x*(-c^2*x^2+1 )^(3/2)*(a+b*arcsin(c*x))/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)-1/3*e*(-c^2*x^2 +1)*(a+b*arcsin(c*x))^2/c/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)+1/3*e*x*(-c^2*x ^2+1)*(a+b*arcsin(c*x))^2/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)+2/3*e*x*(-c^2*x ^2+1)^2*(a+b*arcsin(c*x))^2/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)-2/3*I*e*(-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)-2/3*I *b*e*(-c^2*x^2+1)^(5/2)*(a+b*arcsin(c*x))*arctan(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*e*(-c^2*x^2+1)^(5/2)*(a+b*arcsin (c*x))*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)/c/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/ 2)+1/3*I*b^2*e*(-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)-1/3*I*b^2*e*(-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)-2/3*I* b^2*e*(-c^2*x^2+1)^(5/2)*polylog(2,-(I*c*x+(-c^2*x^2+1)^(1/2))^2)/c/(c*d*x +d)^(5/2)/(-c*e*x+e)^(5/2)
Time = 10.31 (sec) , antiderivative size = 739, normalized size of antiderivative = 1.04 \[ \int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{5/2} (e-c e x)^{3/2}} \, dx=\frac {\sqrt {-e (-1+c x)} \sqrt {d (1+c x)} \left (-\frac {a^2}{4 d^3 e^2 (-1+c x)}-\frac {a^2}{6 d^3 e^2 (1+c x)^2}-\frac {5 a^2}{12 d^3 e^2 (1+c x)}\right )}{c}+\frac {a b \sqrt {d+c d x} \sqrt {e-c e x} \left (2 \arcsin (c x) (-2 c x+\cos (2 \arcsin (c x)))-\sqrt {1-c^2 x^2} \left (-1+3 \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )+5 \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )+c x \left (3 \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )+5 \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )\right )\right )\right )}{3 c d^2 e \sqrt {(-d-c d x) (e-c e x)} \sqrt {-d e \left (1-c^2 x^2\right )} \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )^2}+\frac {b^2 \sqrt {d+c d x} \sqrt {e-c e x} \sqrt {1-c^2 x^2} \left (-7 i \pi \arcsin (c x)+(1+4 i) \arcsin (c x)^2-16 \pi \log \left (1+e^{-i \arcsin (c x)}\right )-5 (\pi +2 \arcsin (c x)) \log \left (1-i e^{i \arcsin (c x)}\right )+3 (\pi -2 \arcsin (c x)) \log \left (1+i e^{i \arcsin (c x)}\right )+16 \pi \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )\right )-3 \pi \log \left (-\cos \left (\frac {1}{4} (\pi +2 \arcsin (c x))\right )\right )+5 \pi \log \left (\sin \left (\frac {1}{4} (\pi +2 \arcsin (c x))\right )\right )+6 i \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )+10 i \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )-\frac {3 \arcsin (c x)^2 \sin \left (\frac {1}{2} \arcsin (c x)\right )}{\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )}-\frac {2 \arcsin (c x)^2 \sin \left (\frac {1}{2} \arcsin (c x)\right )}{\left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )^3}+\frac {\arcsin (c x) (2+\arcsin (c x))}{\left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )^2}-\frac {\left (4+5 \arcsin (c x)^2\right ) \sin \left (\frac {1}{2} \arcsin (c x)\right )}{\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )}\right )}{6 c d^2 e \sqrt {(-d-c d x) (e-c e x)} \sqrt {-d e \left (1-c^2 x^2\right )}} \]
(Sqrt[-(e*(-1 + c*x))]*Sqrt[d*(1 + c*x)]*(-1/4*a^2/(d^3*e^2*(-1 + c*x)) - a^2/(6*d^3*e^2*(1 + c*x)^2) - (5*a^2)/(12*d^3*e^2*(1 + c*x))))/c + (a*b*Sq rt[d + c*d*x]*Sqrt[e - c*e*x]*(2*ArcSin[c*x]*(-2*c*x + Cos[2*ArcSin[c*x]]) - Sqrt[1 - c^2*x^2]*(-1 + 3*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]] + 5*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]] + c*x*(3*Log[Cos[ArcSin[c *x]/2] - Sin[ArcSin[c*x]/2]] + 5*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/ 2]]))))/(3*c*d^2*e*Sqrt[(-d - c*d*x)*(e - c*e*x)]*Sqrt[-(d*e*(1 - c^2*x^2) )]*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2])^2) + (b^2*Sqrt[d + c*d*x]*Sqr t[e - c*e*x]*Sqrt[1 - c^2*x^2]*((-7*I)*Pi*ArcSin[c*x] + (1 + 4*I)*ArcSin[c *x]^2 - 16*Pi*Log[1 + E^((-I)*ArcSin[c*x])] - 5*(Pi + 2*ArcSin[c*x])*Log[1 - I*E^(I*ArcSin[c*x])] + 3*(Pi - 2*ArcSin[c*x])*Log[1 + I*E^(I*ArcSin[c*x ])] + 16*Pi*Log[Cos[ArcSin[c*x]/2]] - 3*Pi*Log[-Cos[(Pi + 2*ArcSin[c*x])/4 ]] + 5*Pi*Log[Sin[(Pi + 2*ArcSin[c*x])/4]] + (6*I)*PolyLog[2, (-I)*E^(I*Ar cSin[c*x])] + (10*I)*PolyLog[2, I*E^(I*ArcSin[c*x])] - (3*ArcSin[c*x]^2*Si n[ArcSin[c*x]/2])/(Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]) - (2*ArcSin[c* x]^2*Sin[ArcSin[c*x]/2])/(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2])^3 + (Ar cSin[c*x]*(2 + ArcSin[c*x]))/(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2])^2 - ((4 + 5*ArcSin[c*x]^2)*Sin[ArcSin[c*x]/2])/(Cos[ArcSin[c*x]/2] + Sin[ArcS in[c*x]/2])))/(6*c*d^2*e*Sqrt[(-d - c*d*x)*(e - c*e*x)]*Sqrt[-(d*e*(1 - c^ 2*x^2))])
Time = 1.05 (sec) , antiderivative size = 386, normalized size of antiderivative = 0.54, 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.
| \(\displaystyle \int \frac {(a+b \arcsin (c x))^2}{(c d x+d)^{5/2} (e-c e x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 5178 |
| \(\displaystyle \frac {\left (1-c^2 x^2\right )^{5/2} \int \frac {e (1-c x) (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {e \left (1-c^2 x^2\right )^{5/2} \int \frac {(1-c x) (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\) |
| \(\Big \downarrow \) 5262 |
| \(\displaystyle \frac {e \left (1-c^2 x^2\right )^{5/2} \int \left (\frac {(a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{5/2}}-\frac {c x (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{5/2}}\right )dx}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {e \left (1-c^2 x^2\right )^{5/2} \left (-\frac {2 i b \arctan \left (e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{3 c}+\frac {b x (a+b \arcsin (c x))}{3 \left (1-c^2 x^2\right )}-\frac {b (a+b \arcsin (c x))}{3 c \left (1-c^2 x^2\right )}+\frac {2 x (a+b \arcsin (c x))^2}{3 \sqrt {1-c^2 x^2}}+\frac {x (a+b \arcsin (c x))^2}{3 \left (1-c^2 x^2\right )^{3/2}}-\frac {(a+b \arcsin (c x))^2}{3 c \left (1-c^2 x^2\right )^{3/2}}-\frac {2 i (a+b \arcsin (c x))^2}{3 c}+\frac {4 b \log \left (1+e^{2 i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{3 c}+\frac {i b^2 \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )}{3 c}-\frac {i b^2 \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{3 c}-\frac {2 i b^2 \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (c x)}\right )}{3 c}+\frac {b^2 x}{3 \sqrt {1-c^2 x^2}}-\frac {b^2}{3 c \sqrt {1-c^2 x^2}}\right )}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\) |
(e*(1 - c^2*x^2)^(5/2)*(-1/3*b^2/(c*Sqrt[1 - c^2*x^2]) + (b^2*x)/(3*Sqrt[1 - c^2*x^2]) - (b*(a + b*ArcSin[c*x]))/(3*c*(1 - c^2*x^2)) + (b*x*(a + b*A rcSin[c*x]))/(3*(1 - c^2*x^2)) - (((2*I)/3)*(a + b*ArcSin[c*x])^2)/c - (a + b*ArcSin[c*x])^2/(3*c*(1 - c^2*x^2)^(3/2)) + (x*(a + b*ArcSin[c*x])^2)/( 3*(1 - c^2*x^2)^(3/2)) + (2*x*(a + b*ArcSin[c*x])^2)/(3*Sqrt[1 - c^2*x^2]) - (((2*I)/3)*b*(a + b*ArcSin[c*x])*ArcTan[E^(I*ArcSin[c*x])])/c + (4*b*(a + b*ArcSin[c*x])*Log[1 + E^((2*I)*ArcSin[c*x])])/(3*c) + ((I/3)*b^2*PolyL og[2, (-I)*E^(I*ArcSin[c*x])])/c - ((I/3)*b^2*PolyLog[2, I*E^(I*ArcSin[c*x ])])/c - (((2*I)/3)*b^2*PolyLog[2, -E^((2*I)*ArcSin[c*x])])/c))/((d + c*d* x)^(5/2)*(e - c*e*x)^(5/2))
3.6.69.3.1 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))
\[\int \frac {\left (a +b \arcsin \left (c x \right )\right )^{2}}{\left (c d x +d \right )^{\frac {5}{2}} \left (-c e x +e \right )^{\frac {3}{2}}}d x\]
\[ \int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{5/2} (e-c e x)^{3/2}} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (c d x + d\right )}^{\frac {5}{2}} {\left (-c e x + e\right )}^{\frac {3}{2}}} \,d x } \]
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^5*d^3*e^2*x^5 + c^4*d^3*e^2*x^4 - 2*c^3*d^3*e^2*x^3 - 2*c ^2*d^3*e^2*x^2 + c*d^3*e^2*x + d^3*e^2), x)
Timed out. \[ \int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{5/2} (e-c e x)^{3/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{5/2} (e-c e x)^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{5/2} (e-c e x)^{3/2}} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (c d x + d\right )}^{\frac {5}{2}} {\left (-c e x + e\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{5/2} (e-c e x)^{3/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2}{{\left (d+c\,d\,x\right )}^{5/2}\,{\left (e-c\,e\,x\right )}^{3/2}} \,d x \]