\(\int \frac {\sqrt {e-c e x} (a+b \arcsin (c x))^2}{\sqrt {d+c d x}} \, dx\) [543]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [F]
   Sympy [F]
   Maxima [F(-2)]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 32, antiderivative size = 230 \[ \int \frac {\sqrt {e-c e x} (a+b \arcsin (c x))^2}{\sqrt {d+c d x}} \, dx=-\frac {2 a b e x \sqrt {1-c^2 x^2}}{\sqrt {d+c d x} \sqrt {e-c e x}}-\frac {2 b^2 e \left (1-c^2 x^2\right )}{c \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {2 b^2 e x \sqrt {1-c^2 x^2} \arcsin (c x)}{\sqrt {d+c d x} \sqrt {e-c e x}}+\frac {e \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{c \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {e \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^3}{3 b c \sqrt {d+c d x} \sqrt {e-c e x}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {4763, 4847, 4737, 4767, 4715, 267} \[ \int \frac {\sqrt {e-c e x} (a+b \arcsin (c x))^2}{\sqrt {d+c d x}} \, dx=\frac {e \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^3}{3 b c \sqrt {c d x+d} \sqrt {e-c e x}}+\frac {e \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{c \sqrt {c d x+d} \sqrt {e-c e x}}-\frac {2 a b e x \sqrt {1-c^2 x^2}}{\sqrt {c d x+d} \sqrt {e-c e x}}-\frac {2 b^2 e x \sqrt {1-c^2 x^2} \arcsin (c x)}{\sqrt {c d x+d} \sqrt {e-c e x}}-\frac {2 b^2 e \left (1-c^2 x^2\right )}{c \sqrt {c d x+d} \sqrt {e-c e x}} \]

[In]

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

[Out]

(-2*a*b*e*x*Sqrt[1 - c^2*x^2])/(Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) - (2*b^2*e*(1 - c^2*x^2))/(c*Sqrt[d + c*d*x]*
Sqrt[e - c*e*x]) - (2*b^2*e*x*Sqrt[1 - c^2*x^2]*ArcSin[c*x])/(Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) + (e*(1 - c^2*x
^2)*(a + b*ArcSin[c*x])^2)/(c*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) + (e*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^3)/(
3*b*c*Sqrt[d + c*d*x]*Sqrt[e - c*e*x])

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 4715

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*ArcSin[c*x])^n, x] - Dist[b*c*n, Int[
x*((a + b*ArcSin[c*x])^(n - 1)/Sqrt[1 - c^2*x^2]), x], x] /; FreeQ[{a, b, c}, x] && GtQ[n, 0]

Rule 4737

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(1/(b*c*(n + 1)))*Si
mp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSin[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c
^2*d + e, 0] && NeQ[n, -1]

Rule 4763

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_))^(p_)*((f_) + (g_.)*(x_))^(q_), x_Symbol] :> D
ist[(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 4767

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x^2)^(
p + 1)*((a + b*ArcSin[c*x])^n/(2*e*(p + 1))), x] + Dist[b*(n/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p
], Int[(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*
d + e, 0] && GtQ[n, 0] && NeQ[p, -1]

Rule 4847

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))

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {1-c^2 x^2} \int \frac {(e-c e x) (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}} \, dx}{\sqrt {d+c d x} \sqrt {e-c e x}} \\ & = \frac {\sqrt {1-c^2 x^2} \int \left (\frac {e (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}-\frac {c e x (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}\right ) \, dx}{\sqrt {d+c d x} \sqrt {e-c e x}} \\ & = \frac {\left (e \sqrt {1-c^2 x^2}\right ) \int \frac {(a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}} \, dx}{\sqrt {d+c d x} \sqrt {e-c e x}}-\frac {\left (c e \sqrt {1-c^2 x^2}\right ) \int \frac {x (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}} \, dx}{\sqrt {d+c d x} \sqrt {e-c e x}} \\ & = \frac {e \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{c \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {e \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^3}{3 b c \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {\left (2 b e \sqrt {1-c^2 x^2}\right ) \int (a+b \arcsin (c x)) \, dx}{\sqrt {d+c d x} \sqrt {e-c e x}} \\ & = -\frac {2 a b e x \sqrt {1-c^2 x^2}}{\sqrt {d+c d x} \sqrt {e-c e x}}+\frac {e \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{c \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {e \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^3}{3 b c \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {\left (2 b^2 e \sqrt {1-c^2 x^2}\right ) \int \arcsin (c x) \, dx}{\sqrt {d+c d x} \sqrt {e-c e x}} \\ & = -\frac {2 a b e x \sqrt {1-c^2 x^2}}{\sqrt {d+c d x} \sqrt {e-c e x}}-\frac {2 b^2 e x \sqrt {1-c^2 x^2} \arcsin (c x)}{\sqrt {d+c d x} \sqrt {e-c e x}}+\frac {e \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{c \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {e \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^3}{3 b c \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {\left (2 b^2 c e \sqrt {1-c^2 x^2}\right ) \int \frac {x}{\sqrt {1-c^2 x^2}} \, dx}{\sqrt {d+c d x} \sqrt {e-c e x}} \\ & = -\frac {2 a b e x \sqrt {1-c^2 x^2}}{\sqrt {d+c d x} \sqrt {e-c e x}}-\frac {2 b^2 e \left (1-c^2 x^2\right )}{c \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {2 b^2 e x \sqrt {1-c^2 x^2} \arcsin (c x)}{\sqrt {d+c d x} \sqrt {e-c e x}}+\frac {e \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{c \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {e \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^3}{3 b c \sqrt {d+c d x} \sqrt {e-c e x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.97 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.29 \[ \int \frac {\sqrt {e-c e x} (a+b \arcsin (c x))^2}{\sqrt {d+c d x}} \, dx=\frac {3 \sqrt {d+c d x} \sqrt {e-c e x} \left (-2 a b c x+a^2 \sqrt {1-c^2 x^2}-2 b^2 \sqrt {1-c^2 x^2}\right )-6 b \sqrt {d+c d x} \sqrt {e-c e x} \left (b c x-a \sqrt {1-c^2 x^2}\right ) \arcsin (c x)+3 b \sqrt {d+c d x} \sqrt {e-c e x} \left (a+b \sqrt {1-c^2 x^2}\right ) \arcsin (c x)^2+b^2 \sqrt {d+c d x} \sqrt {e-c e x} \arcsin (c x)^3-3 a^2 \sqrt {d} \sqrt {e} \sqrt {1-c^2 x^2} \arctan \left (\frac {c x \sqrt {d+c d x} \sqrt {e-c e x}}{\sqrt {d} \sqrt {e} \left (-1+c^2 x^2\right )}\right )}{3 c d \sqrt {1-c^2 x^2}} \]

[In]

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

[Out]

(3*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*(-2*a*b*c*x + a^2*Sqrt[1 - c^2*x^2] - 2*b^2*Sqrt[1 - c^2*x^2]) - 6*b*Sqrt[d
 + c*d*x]*Sqrt[e - c*e*x]*(b*c*x - a*Sqrt[1 - c^2*x^2])*ArcSin[c*x] + 3*b*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*(a +
 b*Sqrt[1 - c^2*x^2])*ArcSin[c*x]^2 + b^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*ArcSin[c*x]^3 - 3*a^2*Sqrt[d]*Sqrt[e
]*Sqrt[1 - c^2*x^2]*ArcTan[(c*x*Sqrt[d + c*d*x]*Sqrt[e - c*e*x])/(Sqrt[d]*Sqrt[e]*(-1 + c^2*x^2))])/(3*c*d*Sqr
t[1 - c^2*x^2])

Maple [F]

\[\int \frac {\left (a +b \arcsin \left (c x \right )\right )^{2} \sqrt {-c e x +e}}{\sqrt {c d x +d}}d x\]

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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