\(\int \frac {\sqrt {e-c e x} (a+b \arcsin (c x))^2}{(d+c d x)^{3/2}} \, dx\) [73]

Optimal result

Integrand size = 32, antiderivative size = 530 \[ \int \frac {\sqrt {e-c e x} (a+b \arcsin (c x))^2}{(d+c d x)^{3/2}} \, dx=-\frac {2 e^2 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {2 e^2 x \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{(d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {2 i e^2 \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))^2}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {e^2 \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))^3}{3 b c (d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {8 i b e^2 \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x)) \arctan \left (e^{i \arcsin (c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {4 b e^2 \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x)) \log \left (1+e^{2 i \arcsin (c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {4 i b^2 e^2 \left (1-c^2 x^2\right )^{3/2} \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {4 i b^2 e^2 \left (1-c^2 x^2\right )^{3/2} \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {2 i b^2 e^2 \left (1-c^2 x^2\right )^{3/2} \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}} \] Output:

-2*e^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*e^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*e^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*e^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*e^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*e^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*e^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*e^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*e^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)
 

Mathematica [A] (verified)

Time = 4.43 (sec) , antiderivative size = 547, normalized size of antiderivative = 1.03 \[ \int \frac {\sqrt {e-c e x} (a+b \arcsin (c x))^2}{(d+c d 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 \sqrt {d+c d x} \sqrt {e-c e x} \left (\cos \left (\frac {1}{2} \arcsin (c x)\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 )+\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 ) \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 )}+\frac {b^2 \sqrt {d+c d x} \sqrt {e-c e x} \left ((-6-6 i) \arcsin (c x)^2 \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+i \sin \left (\frac {1}{2} \arcsin (c x)\right )\right )-\arcsin (c x)^3 \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )+6 \arcsin (c x) \left (i \pi +4 \log \left (1-i e^{i \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 )+12 \pi \left (2 \log \left (1+e^{-i \arcsin (c x)}\right )+\log \left (1-i e^{i \arcsin (c x)}\right )-2 \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )\right )-\log \left (\sin \left (\frac {1}{4} (\pi +2 \arcsin (c x))\right )\right )\right ) \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )-24 i \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right ) \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )\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 )}}{3 c d^2} \] Input:

Integrate[(Sqrt[e - c*e*x]*(a + b*ArcSin[c*x])^2)/(d + c*d*x)^(3/2),x]
 

Output:

((-6*a^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x])/(1 + c*x) + 3*a^2*Sqrt[d]*Sqrt[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*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*(Cos[ArcSin[c*x]/2]*(ArcSi 
n[c*x]*(4 + ArcSin[c*x]) - 8*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]]) 
 + ((-4 + ArcSin[c*x])*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[ArcSin[c*x]/2] + S 
in[ArcSin[c*x]/2])) + (b^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*((-6 - 6*I)*Arc 
Sin[c*x]^2*(Cos[ArcSin[c*x]/2] + I*Sin[ArcSin[c*x]/2]) - ArcSin[c*x]^3*(Co 
s[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]) + 6*ArcSin[c*x]*(I*Pi + 4*Log[1 - I 
*E^(I*ArcSin[c*x])])*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]) + 12*Pi*(2* 
Log[1 + E^((-I)*ArcSin[c*x])] + Log[1 - I*E^(I*ArcSin[c*x])] - 2*Log[Cos[A 
rcSin[c*x]/2]] - Log[Sin[(Pi + 2*ArcSin[c*x])/4]])*(Cos[ArcSin[c*x]/2] + S 
in[ArcSin[c*x]/2]) - (24*I)*PolyLog[2, I*E^(I*ArcSin[c*x])]*(Cos[ArcSin[c* 
x]/2] + Sin[ArcSin[c*x]/2])))/(Sqrt[1 - c^2*x^2]*(Cos[ArcSin[c*x]/2] + Sin 
[ArcSin[c*x]/2])))/(3*c*d^2)
 

Rubi [A] (verified)

Time = 1.20 (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.

\(\displaystyle \int \frac {\sqrt {e-c e x} (a+b \arcsin (c x))^2}{(c d x+d)^{3/2}} \, dx\)

\(\Big \downarrow \) 5178

\(\displaystyle \frac {\left (1-c^2 x^2\right )^{3/2} \int \frac {e^2 (1-c x)^2 (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{3/2}}dx}{(c d x+d)^{3/2} (e-c e x)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {e^2 \left (1-c^2 x^2\right )^{3/2} \int \frac {(1-c x)^2 (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{3/2}}dx}{(c d x+d)^{3/2} (e-c e x)^{3/2}}\)

\(\Big \downarrow \) 5274

\(\displaystyle \frac {e^2 \left (1-c^2 x^2\right )^{3/2} \int \left (\frac {2 (1-c x) (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{3/2}}-\frac {(a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}\right )dx}{(c d x+d)^{3/2} (e-c e x)^{3/2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {e^2 \left (1-c^2 x^2\right )^{3/2} \left (-\frac {8 i b \arctan \left (e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{c}+\frac {2 x (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}-\frac {2 (a+b \arcsin (c x))^2}{c \sqrt {1-c^2 x^2}}-\frac {(a+b \arcsin (c x))^3}{3 b c}-\frac {2 i (a+b \arcsin (c x))^2}{c}+\frac {4 b \log \left (1+e^{2 i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{c}+\frac {4 i b^2 \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )}{c}-\frac {4 i b^2 \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{c}-\frac {2 i b^2 \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (c x)}\right )}{c}\right )}{(c d x+d)^{3/2} (e-c e x)^{3/2}}\)

Input:

Int[(Sqrt[e - c*e*x]*(a + b*ArcSin[c*x])^2)/(d + c*d*x)^(3/2),x]
 

Output:

(e^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))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 5178

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]
 

rule 5274

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]
 

Maple [A] (verified)

Time = 4.38 (sec) , antiderivative size = 318, normalized size of antiderivative = 0.60

method	result	size
default	\(\frac {\sqrt {-c^{2} x^{2}+1}\, \sqrt {d \left (c x +1\right )}\, \sqrt {-e \left (c x -1\right )}\, \left (a +b \arcsin \left (c x \right )\right )^{3}}{3 \left (c x +1\right ) \left (c x -1\right ) d^{2} c b}-\frac {2 \left (\arcsin \left (c x \right )^{2} b^{2}+2 \arcsin \left (c x \right ) a b +a^{2}\right ) \left (i \sqrt {-c^{2} x^{2}+1}+c x -1\right ) \sqrt {-e \left (c x -1\right )}\, \sqrt {d \left (c x +1\right )}}{d^{2} c \left (c^{2} x^{2}-1\right )}+\frac {4 i \sqrt {-c^{2} x^{2}+1}\, \sqrt {d \left (c x +1\right )}\, \sqrt {-e \left (c x -1\right )}\, b \left (2 i \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right ) b +\arcsin \left (c x \right )^{2} b +2 i a \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right )-2 i a \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )+2 \operatorname {polylog}\left (2, i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right ) b \right )}{\left (c x +1\right ) \left (c x -1\right ) d^{2} c}\)	\(318\)

Input:

int((-c*e*x+e)^(1/2)*(a+b*arcsin(c*x))^2/(c*d*x+d)^(3/2),x,method=_RETURNV 
ERBOSE)
 

Output:

1/3*(-c^2*x^2+1)^(1/2)*(d*(c*x+1))^(1/2)*(-e*(c*x-1))^(1/2)/(c*x+1)/(c*x-1 
)/d^2/c*(a+b*arcsin(c*x))^3/b-2*(arcsin(c*x)^2*b^2+2*arcsin(c*x)*a*b+a^2)* 
(I*(-c^2*x^2+1)^(1/2)+c*x-1)*(-e*(c*x-1))^(1/2)*(d*(c*x+1))^(1/2)/d^2/c/(c 
^2*x^2-1)+4*I*(-c^2*x^2+1)^(1/2)*(d*(c*x+1))^(1/2)*(-e*(c*x-1))^(1/2)/(c*x 
+1)/(c*x-1)/d^2/c*b*(2*I*arcsin(c*x)*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))*b+ 
arcsin(c*x)^2*b+2*I*a*ln(I*c*x+(-c^2*x^2+1)^(1/2)+I)-2*I*a*ln(I*c*x+(-c^2* 
x^2+1)^(1/2))+2*polylog(2,I*(I*c*x+(-c^2*x^2+1)^(1/2)))*b)
 

Fricas [F]

\[ \int \frac {\sqrt {e-c e x} (a+b \arcsin (c x))^2}{(d+c d x)^{3/2}} \, dx=\int { \frac {\sqrt {-c e x + e} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (c d x + d\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((-c*e*x+e)^(1/2)*(a+b*arcsin(c*x))^2/(c*d*x+d)^(3/2),x, algorith 
m="fricas")
 

Output:

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*d^2*x^2 + 2*c*d^2*x + d^2), x)
 

Sympy [F]

\[ \int \frac {\sqrt {e-c e x} (a+b \arcsin (c x))^2}{(d+c d x)^{3/2}} \, dx=\int \frac {\sqrt {- e \left (c x - 1\right )} \left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}{\left (d \left (c x + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((-c*e*x+e)**(1/2)*(a+b*asin(c*x))**2/(c*d*x+d)**(3/2),x)
 

Output:

Integral(sqrt(-e*(c*x - 1))*(a + b*asin(c*x))**2/(d*(c*x + 1))**(3/2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {e-c e x} (a+b \arcsin (c x))^2}{(d+c d x)^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((-c*e*x+e)^(1/2)*(a+b*arcsin(c*x))^2/(c*d*x+d)^(3/2),x, algorith 
m="maxima")
 

Output:

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
 

Giac [F]

\[ \int \frac {\sqrt {e-c e x} (a+b \arcsin (c x))^2}{(d+c d x)^{3/2}} \, dx=\int { \frac {\sqrt {-c e x + e} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (c d x + d\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((-c*e*x+e)^(1/2)*(a+b*arcsin(c*x))^2/(c*d*x+d)^(3/2),x, algorith 
m="giac")
 

Output:

integrate(sqrt(-c*e*x + e)*(b*arcsin(c*x) + a)^2/(c*d*x + d)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {e-c e x} (a+b \arcsin (c x))^2}{(d+c d x)^{3/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,\sqrt {e-c\,e\,x}}{{\left (d+c\,d\,x\right )}^{3/2}} \,d x \] Input:

int(((a + b*asin(c*x))^2*(e - c*e*x)^(1/2))/(d + c*d*x)^(3/2),x)
 

Output:

int(((a + b*asin(c*x))^2*(e - c*e*x)^(1/2))/(d + c*d*x)^(3/2), x)
 

Reduce [F]

\[ \int \frac {\sqrt {e-c e x} (a+b \arcsin (c x))^2}{(d+c d x)^{3/2}} \, dx=\frac {\sqrt {e}\, \left (2 \sqrt {c x +1}\, \mathit {asin} \left (\frac {\sqrt {-c x +1}}{\sqrt {2}}\right ) a^{2}-2 \sqrt {-c x +1}\, a^{2}+2 \sqrt {c x +1}\, \left (\int \frac {\sqrt {-c x +1}\, \mathit {asin} \left (c x \right )}{\sqrt {c x +1}\, c x +\sqrt {c x +1}}d x \right ) a b c +\sqrt {c x +1}\, \left (\int \frac {\sqrt {-c x +1}\, \mathit {asin} \left (c x \right )^{2}}{\sqrt {c x +1}\, c x +\sqrt {c x +1}}d x \right ) b^{2} c \right )}{\sqrt {d}\, \sqrt {c x +1}\, c d} \] Input:

int((-c*e*x+e)^(1/2)*(a+b*asin(c*x))^2/(c*d*x+d)^(3/2),x)
 

Output:

(sqrt(e)*(2*sqrt(c*x + 1)*asin(sqrt( - c*x + 1)/sqrt(2))*a**2 - 2*sqrt( - 
c*x + 1)*a**2 + 2*sqrt(c*x + 1)*int((sqrt( - c*x + 1)*asin(c*x))/(sqrt(c*x 
 + 1)*c*x + sqrt(c*x + 1)),x)*a*b*c + sqrt(c*x + 1)*int((sqrt( - c*x + 1)* 
asin(c*x)**2)/(sqrt(c*x + 1)*c*x + sqrt(c*x + 1)),x)*b**2*c))/(sqrt(d)*sqr 
t(c*x + 1)*c*d)