Integrand size = 32, antiderivative size = 530 \[ \int \frac {\sqrt {d+c d x} (a+b \arcsin (c x))^2}{(e-c e x)^{3/2}} \, dx=\frac {2 d^2 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {2 d^2 x \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{(d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {2 i d^2 \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))^2}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {d^2 \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))^3}{3 b c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {8 i b d^2 \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x)) \arctan \left (e^{i \arcsin (c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {4 b d^2 \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x)) \log \left (1+e^{2 i \arcsin (c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {4 i b^2 d^2 \left (1-c^2 x^2\right )^{3/2} \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {4 i b^2 d^2 \left (1-c^2 x^2\right )^{3/2} \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {2 i b^2 d^2 \left (1-c^2 x^2\right )^{3/2} \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}} \]
2*d^2*(-c^2*x^2+1)*(a+b*arcsin(c*x))^2/c/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)+ 2*d^2*x*(-c^2*x^2+1)*(a+b*arcsin(c*x))^2/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)- 2*I*d^2*(-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)-1/3*d^2*(-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)+8*I*b*d^2*(-c^2*x^2+1)^(3/2)*(a+b*arcsin(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)+4*b*d^2*(-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)-4*I*b^2*d^2*(-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*I*b^2*d^2*(-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)-2*I*b^2*d^2*(-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)
Time = 6.73 (sec) , antiderivative size = 513, normalized size of antiderivative = 0.97 \[ \int \frac {\sqrt {d+c d x} (a+b \arcsin (c x))^2}{(e-c e x)^{3/2}} \, dx=-\frac {\frac {6 a^2 \sqrt {d+c d x} \sqrt {e-c e x}}{-1+c x}-3 a^2 \sqrt {d} \sqrt {e} \arctan \left (\frac {c x \sqrt {d+c d x} \sqrt {e-c e x}}{\sqrt {d} \sqrt {e} \left (-1+c^2 x^2\right )}\right )+\frac {3 a b (1+c x) \sqrt {d+c d x} \sqrt {e-c e x} \left (\cos \left (\frac {1}{2} \arcsin (c x)\right ) \left ((-4+\arcsin (c x)) \arcsin (c x)-8 \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )\right )-\left (\arcsin (c x) (4+\arcsin (c x))-8 \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )\right ) \sin \left (\frac {1}{2} \arcsin (c x)\right )\right )}{\sqrt {1-c^2 x^2} \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right ) \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )^2}+\frac {b^2 (1+c x) \sqrt {d+c d x} \sqrt {e-c e x} \left (-18 i \pi \arcsin (c x)-(6-6 i) \arcsin (c x)^2+\arcsin (c x)^3-24 \pi \log \left (1+e^{-i \arcsin (c x)}\right )+12 (\pi -2 \arcsin (c x)) \log \left (1+i e^{i \arcsin (c x)}\right )+24 \pi \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )\right )-12 \pi \log \left (-\cos \left (\frac {1}{4} (\pi +2 \arcsin (c x))\right )\right )+24 i \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )-\frac {12 \arcsin (c x)^2 \sin \left (\frac {1}{2} \arcsin (c x)\right )}{\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )}\right )}{\sqrt {1-c^2 x^2} \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )^2}}{3 c e^2} \]
-1/3*((6*a^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x])/(-1 + c*x) - 3*a^2*Sqrt[d]*S qrt[e]*ArcTan[(c*x*Sqrt[d + c*d*x]*Sqrt[e - c*e*x])/(Sqrt[d]*Sqrt[e]*(-1 + c^2*x^2))] + (3*a*b*(1 + c*x)*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*(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]))/(Sqrt[1 - c^2*x^2]*(Cos[Arc Sin[c*x]/2] - Sin[ArcSin[c*x]/2])*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2] )^2) + (b^2*(1 + c*x)*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*((-18*I)*Pi*ArcSin[c *x] - (6 - 6*I)*ArcSin[c*x]^2 + ArcSin[c*x]^3 - 24*Pi*Log[1 + E^((-I)*ArcS in[c*x])] + 12*(Pi - 2*ArcSin[c*x])*Log[1 + I*E^(I*ArcSin[c*x])] + 24*Pi*L og[Cos[ArcSin[c*x]/2]] - 12*Pi*Log[-Cos[(Pi + 2*ArcSin[c*x])/4]] + (24*I)* PolyLog[2, (-I)*E^(I*ArcSin[c*x])] - (12*ArcSin[c*x]^2*Sin[ArcSin[c*x]/2]) /(Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2])))/(Sqrt[1 - c^2*x^2]*(Cos[ArcSi n[c*x]/2] + Sin[ArcSin[c*x]/2])^2))/(c*e^2)
Time = 1.03 (sec) , antiderivative size = 263, normalized size of antiderivative = 0.50, 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.
| \(\displaystyle \int \frac {\sqrt {c d x+d} (a+b \arcsin (c x))^2}{(e-c e x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 5178 |
| \(\displaystyle \frac {\left (1-c^2 x^2\right )^{3/2} \int \frac {d^2 (c x+1)^2 (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{3/2}}dx}{(c d x+d)^{3/2} (e-c e x)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d^2 \left (1-c^2 x^2\right )^{3/2} \int \frac {(c x+1)^2 (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{3/2}}dx}{(c d x+d)^{3/2} (e-c e x)^{3/2}}\) |
| \(\Big \downarrow \) 5274 |
| \(\displaystyle \frac {d^2 \left (1-c^2 x^2\right )^{3/2} \int \left (\frac {2 (c x+1) (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{3/2}}-\frac {(a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}\right )dx}{(c d x+d)^{3/2} (e-c e x)^{3/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^2 \left (1-c^2 x^2\right )^{3/2} \left (\frac {8 i b \arctan \left (e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{c}+\frac {2 x (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}+\frac {2 (a+b \arcsin (c x))^2}{c \sqrt {1-c^2 x^2}}-\frac {(a+b \arcsin (c x))^3}{3 b c}-\frac {2 i (a+b \arcsin (c x))^2}{c}+\frac {4 b \log \left (1+e^{2 i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{c}-\frac {4 i b^2 \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )}{c}+\frac {4 i b^2 \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{c}-\frac {2 i b^2 \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (c x)}\right )}{c}\right )}{(c d x+d)^{3/2} (e-c e x)^{3/2}}\) |
(d^2*(1 - c^2*x^2)^(3/2)*(((-2*I)*(a + b*ArcSin[c*x])^2)/c + (2*(a + b*Arc Sin[c*x])^2)/(c*Sqrt[1 - c^2*x^2]) + (2*x*(a + b*ArcSin[c*x])^2)/Sqrt[1 - c^2*x^2] - (a + b*ArcSin[c*x])^3/(3*b*c) + ((8*I)*b*(a + b*ArcSin[c*x])*Ar cTan[E^(I*ArcSin[c*x])])/c + (4*b*(a + b*ArcSin[c*x])*Log[1 + E^((2*I)*Arc Sin[c*x])])/c - ((4*I)*b^2*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/c + ((4*I)* b^2*PolyLog[2, I*E^(I*ArcSin[c*x])])/c - ((2*I)*b^2*PolyLog[2, -E^((2*I)*A rcSin[c*x])])/c))/((d + c*d*x)^(3/2)*(e - c*e*x)^(3/2))
3.6.66.3.1 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]
\[\int \frac {\sqrt {c d x +d}\, \left (a +b \arcsin \left (c x \right )\right )^{2}}{\left (-c e x +e \right )^{\frac {3}{2}}}d x\]
\[ \int \frac {\sqrt {d+c d x} (a+b \arcsin (c x))^2}{(e-c e x)^{3/2}} \, dx=\int { \frac {\sqrt {c d x + d} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (-c e x + e\right )}^{\frac {3}{2}}} \,d x } \]
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^2*e^2*x^2 - 2*c*e^2*x + e^2), x)
\[ \int \frac {\sqrt {d+c d x} (a+b \arcsin (c x))^2}{(e-c e x)^{3/2}} \, dx=\int \frac {\sqrt {d \left (c x + 1\right )} \left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}{\left (- e \left (c x - 1\right )\right )^{\frac {3}{2}}}\, dx \]
Exception generated. \[ \int \frac {\sqrt {d+c d x} (a+b \arcsin (c x))^2}{(e-c e x)^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {\sqrt {d+c d x} (a+b \arcsin (c x))^2}{(e-c e x)^{3/2}} \, dx=\int { \frac {\sqrt {c d x + d} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (-c e x + e\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {d+c d x} (a+b \arcsin (c x))^2}{(e-c e x)^{3/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,\sqrt {d+c\,d\,x}}{{\left (e-c\,e\,x\right )}^{3/2}} \,d x \]