\(\int \frac {(1-a^2-2 a b x-b^2 x^2)^{3/2}}{\arcsin (a+b x)} \, dx\) [323]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [F]
   Sympy [F]
   Maxima [F]
   Giac [A] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 33, antiderivative size = 47 \[ \int \frac {\left (1-a^2-2 a b x-b^2 x^2\right )^{3/2}}{\arcsin (a+b x)} \, dx=\frac {\operatorname {CosIntegral}(2 \arcsin (a+b x))}{2 b}+\frac {\operatorname {CosIntegral}(4 \arcsin (a+b x))}{8 b}+\frac {3 \log (\arcsin (a+b x))}{8 b} \]

[Out]

1/2*Ci(2*arcsin(b*x+a))/b+1/8*Ci(4*arcsin(b*x+a))/b+3/8*ln(arcsin(b*x+a))/b

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {4891, 4753, 3393, 3383} \[ \int \frac {\left (1-a^2-2 a b x-b^2 x^2\right )^{3/2}}{\arcsin (a+b x)} \, dx=\frac {\operatorname {CosIntegral}(2 \arcsin (a+b x))}{2 b}+\frac {\operatorname {CosIntegral}(4 \arcsin (a+b x))}{8 b}+\frac {3 \log (\arcsin (a+b x))}{8 b} \]

[In]

Int[(1 - a^2 - 2*a*b*x - b^2*x^2)^(3/2)/ArcSin[a + b*x],x]

[Out]

CosIntegral[2*ArcSin[a + b*x]]/(2*b) + CosIntegral[4*ArcSin[a + b*x]]/(8*b) + (3*Log[ArcSin[a + b*x]])/(8*b)

Rule 3383

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 4753

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

Rule 4891

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)^(p_.), x_Symbol] :> Di
st[1/d, Subst[Int[(-C/d^2 + (C/d^2)*x^2)^p*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, A, B
, C, n, p}, x] && EqQ[B*(1 - c^2) + 2*A*c*d, 0] && EqQ[2*c*C - B*d, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (1-x^2\right )^{3/2}}{\arcsin (x)} \, dx,x,a+b x\right )}{b} \\ & = \frac {\text {Subst}\left (\int \frac {\cos ^4(x)}{x} \, dx,x,\arcsin (a+b x)\right )}{b} \\ & = \frac {\text {Subst}\left (\int \left (\frac {3}{8 x}+\frac {\cos (2 x)}{2 x}+\frac {\cos (4 x)}{8 x}\right ) \, dx,x,\arcsin (a+b x)\right )}{b} \\ & = \frac {3 \log (\arcsin (a+b x))}{8 b}+\frac {\text {Subst}\left (\int \frac {\cos (4 x)}{x} \, dx,x,\arcsin (a+b x)\right )}{8 b}+\frac {\text {Subst}\left (\int \frac {\cos (2 x)}{x} \, dx,x,\arcsin (a+b x)\right )}{2 b} \\ & = \frac {\operatorname {CosIntegral}(2 \arcsin (a+b x))}{2 b}+\frac {\operatorname {CosIntegral}(4 \arcsin (a+b x))}{8 b}+\frac {3 \log (\arcsin (a+b x))}{8 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.79 \[ \int \frac {\left (1-a^2-2 a b x-b^2 x^2\right )^{3/2}}{\arcsin (a+b x)} \, dx=\frac {4 \operatorname {CosIntegral}(2 \arcsin (a+b x))+\operatorname {CosIntegral}(4 \arcsin (a+b x))+3 \log (\arcsin (a+b x))}{8 b} \]

[In]

Integrate[(1 - a^2 - 2*a*b*x - b^2*x^2)^(3/2)/ArcSin[a + b*x],x]

[Out]

(4*CosIntegral[2*ArcSin[a + b*x]] + CosIntegral[4*ArcSin[a + b*x]] + 3*Log[ArcSin[a + b*x]])/(8*b)

Maple [A] (verified)

Time = 1.79 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.77

method result size
default \(\frac {3 \ln \left (\arcsin \left (b x +a \right )\right )+4 \,\operatorname {Ci}\left (2 \arcsin \left (b x +a \right )\right )+\operatorname {Ci}\left (4 \arcsin \left (b x +a \right )\right )}{8 b}\) \(36\)

[In]

int((-b^2*x^2-2*a*b*x-a^2+1)^(3/2)/arcsin(b*x+a),x,method=_RETURNVERBOSE)

[Out]

1/8*(3*ln(arcsin(b*x+a))+4*Ci(2*arcsin(b*x+a))+Ci(4*arcsin(b*x+a)))/b

Fricas [F]

\[ \int \frac {\left (1-a^2-2 a b x-b^2 x^2\right )^{3/2}}{\arcsin (a+b x)} \, dx=\int { \frac {{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}}}{\arcsin \left (b x + a\right )} \,d x } \]

[In]

integrate((-b^2*x^2-2*a*b*x-a^2+1)^(3/2)/arcsin(b*x+a),x, algorithm="fricas")

[Out]

integral((-b^2*x^2 - 2*a*b*x - a^2 + 1)^(3/2)/arcsin(b*x + a), x)

Sympy [F]

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

[In]

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

[Out]

Integral((-(a + b*x - 1)*(a + b*x + 1))**(3/2)/asin(a + b*x), x)

Maxima [F]

\[ \int \frac {\left (1-a^2-2 a b x-b^2 x^2\right )^{3/2}}{\arcsin (a+b x)} \, dx=\int { \frac {{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}}}{\arcsin \left (b x + a\right )} \,d x } \]

[In]

integrate((-b^2*x^2-2*a*b*x-a^2+1)^(3/2)/arcsin(b*x+a),x, algorithm="maxima")

[Out]

integrate((-b^2*x^2 - 2*a*b*x - a^2 + 1)^(3/2)/arcsin(b*x + a), x)

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.87 \[ \int \frac {\left (1-a^2-2 a b x-b^2 x^2\right )^{3/2}}{\arcsin (a+b x)} \, dx=\frac {\operatorname {Ci}\left (4 \, \arcsin \left (b x + a\right )\right )}{8 \, b} + \frac {\operatorname {Ci}\left (2 \, \arcsin \left (b x + a\right )\right )}{2 \, b} + \frac {3 \, \log \left (\arcsin \left (b x + a\right )\right )}{8 \, b} \]

[In]

integrate((-b^2*x^2-2*a*b*x-a^2+1)^(3/2)/arcsin(b*x+a),x, algorithm="giac")

[Out]

1/8*cos_integral(4*arcsin(b*x + a))/b + 1/2*cos_integral(2*arcsin(b*x + a))/b + 3/8*log(arcsin(b*x + a))/b

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (1-a^2-2 a b x-b^2 x^2\right )^{3/2}}{\arcsin (a+b x)} \, dx=\int \frac {{\left (-a^2-2\,a\,b\,x-b^2\,x^2+1\right )}^{3/2}}{\mathrm {asin}\left (a+b\,x\right )} \,d x \]

[In]

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

[Out]

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