3.4.25 \(\int \frac {(1-a^2-2 a b x-b^2 x^2)^{3/2}}{\arcsin (a+b x)^3} \, dx\) [325]

3.4.25.1 Optimal result

Integrand size = 33, antiderivative size = 90 \[ \int \frac {\left (1-a^2-2 a b x-b^2 x^2\right )^{3/2}}{\arcsin (a+b x)^3} \, dx=-\frac {\left (1-(a+b x)^2\right )^2}{2 b \arcsin (a+b x)^2}+\frac {2 (a+b x) \left (1-(a+b x)^2\right )^{3/2}}{b \arcsin (a+b x)}-\frac {\operatorname {CosIntegral}(2 \arcsin (a+b x))}{b}-\frac {\operatorname {CosIntegral}(4 \arcsin (a+b x))}{b} \]

output

-1/2*(1-(b*x+a)^2)^2/b/arcsin(b*x+a)^2+2*(b*x+a)*(1-(b*x+a)^2)^(3/2)/b/arc 
sin(b*x+a)-Ci(2*arcsin(b*x+a))/b-Ci(4*arcsin(b*x+a))/b
 

3.4.25.2 Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.22 \[ \int \frac {\left (1-a^2-2 a b x-b^2 x^2\right )^{3/2}}{\arcsin (a+b x)^3} \, dx=-\frac {\frac {\left (-1+a^2+2 a b x+b^2 x^2\right ) \left (-1+a^2+2 a b x+b^2 x^2+4 (a+b x) \sqrt {1-a^2-2 a b x-b^2 x^2} \arcsin (a+b x)\right )}{\arcsin (a+b x)^2}+2 \operatorname {CosIntegral}(2 \arcsin (a+b x))+2 \operatorname {CosIntegral}(4 \arcsin (a+b x))}{2 b} \]

input

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

output

-1/2*(((-1 + a^2 + 2*a*b*x + b^2*x^2)*(-1 + a^2 + 2*a*b*x + b^2*x^2 + 4*(a 
 + b*x)*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2]*ArcSin[a + b*x]))/ArcSin[a + b*x 
]^2 + 2*CosIntegral[2*ArcSin[a + b*x]] + 2*CosIntegral[4*ArcSin[a + b*x]]) 
/b
 

3.4.25.3 Rubi [A] (verified)

Time = 0.93 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.27, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {5306, 5166, 5214, 5168, 3042, 3793, 2009, 5224, 4906, 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.

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

\(\Big \downarrow \) 5306

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

\(\Big \downarrow \) 5166

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

\(\Big \downarrow \) 5214

\(\displaystyle \frac {-2 \left (\int \frac {\sqrt {1-(a+b x)^2}}{\arcsin (a+b x)}d(a+b x)-4 \int \frac {(a+b x)^2 \sqrt {1-(a+b x)^2}}{\arcsin (a+b x)}d(a+b x)-\frac {(a+b x) \left (1-(a+b x)^2\right )^{3/2}}{\arcsin (a+b x)}\right )-\frac {\left (1-(a+b x)^2\right )^2}{2 \arcsin (a+b x)^2}}{b}\)

\(\Big \downarrow \) 5168

\(\displaystyle \frac {-2 \left (-4 \int \frac {(a+b x)^2 \sqrt {1-(a+b x)^2}}{\arcsin (a+b x)}d(a+b x)+\int \frac {1-(a+b x)^2}{\arcsin (a+b x)}d\arcsin (a+b x)-\frac {(a+b x) \left (1-(a+b x)^2\right )^{3/2}}{\arcsin (a+b x)}\right )-\frac {\left (1-(a+b x)^2\right )^2}{2 \arcsin (a+b x)^2}}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-2 \left (-4 \int \frac {(a+b x)^2 \sqrt {1-(a+b x)^2}}{\arcsin (a+b x)}d(a+b x)+\int \frac {\sin \left (\arcsin (a+b x)+\frac {\pi }{2}\right )^2}{\arcsin (a+b x)}d\arcsin (a+b x)-\frac {(a+b x) \left (1-(a+b x)^2\right )^{3/2}}{\arcsin (a+b x)}\right )-\frac {\left (1-(a+b x)^2\right )^2}{2 \arcsin (a+b x)^2}}{b}\)

\(\Big \downarrow \) 3793

\(\displaystyle \frac {-2 \left (-4 \int \frac {(a+b x)^2 \sqrt {1-(a+b x)^2}}{\arcsin (a+b x)}d(a+b x)+\int \left (\frac {\cos (2 \arcsin (a+b x))}{2 \arcsin (a+b x)}+\frac {1}{2 \arcsin (a+b x)}\right )d\arcsin (a+b x)-\frac {(a+b x) \left (1-(a+b x)^2\right )^{3/2}}{\arcsin (a+b x)}\right )-\frac {\left (1-(a+b x)^2\right )^2}{2 \arcsin (a+b x)^2}}{b}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-2 \left (-4 \int \frac {(a+b x)^2 \sqrt {1-(a+b x)^2}}{\arcsin (a+b x)}d(a+b x)+\frac {1}{2} \operatorname {CosIntegral}(2 \arcsin (a+b x))-\frac {(a+b x) \left (1-(a+b x)^2\right )^{3/2}}{\arcsin (a+b x)}+\frac {1}{2} \log (\arcsin (a+b x))\right )-\frac {\left (1-(a+b x)^2\right )^2}{2 \arcsin (a+b x)^2}}{b}\)

\(\Big \downarrow \) 5224

\(\displaystyle \frac {-2 \left (-4 \int \frac {(a+b x)^2 \left (1-(a+b x)^2\right )}{\arcsin (a+b x)}d\arcsin (a+b x)+\frac {1}{2} \operatorname {CosIntegral}(2 \arcsin (a+b x))-\frac {(a+b x) \left (1-(a+b x)^2\right )^{3/2}}{\arcsin (a+b x)}+\frac {1}{2} \log (\arcsin (a+b x))\right )-\frac {\left (1-(a+b x)^2\right )^2}{2 \arcsin (a+b x)^2}}{b}\)

\(\Big \downarrow \) 4906

\(\displaystyle \frac {-2 \left (-4 \int \left (\frac {1}{8 \arcsin (a+b x)}-\frac {\cos (4 \arcsin (a+b x))}{8 \arcsin (a+b x)}\right )d\arcsin (a+b x)+\frac {1}{2} \operatorname {CosIntegral}(2 \arcsin (a+b x))-\frac {(a+b x) \left (1-(a+b x)^2\right )^{3/2}}{\arcsin (a+b x)}+\frac {1}{2} \log (\arcsin (a+b x))\right )-\frac {\left (1-(a+b x)^2\right )^2}{2 \arcsin (a+b x)^2}}{b}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-2 \left (\frac {1}{2} \operatorname {CosIntegral}(2 \arcsin (a+b x))-4 \left (\frac {1}{8} \log (\arcsin (a+b x))-\frac {1}{8} \operatorname {CosIntegral}(4 \arcsin (a+b x))\right )-\frac {(a+b x) \left (1-(a+b x)^2\right )^{3/2}}{\arcsin (a+b x)}+\frac {1}{2} \log (\arcsin (a+b x))\right )-\frac {\left (1-(a+b x)^2\right )^2}{2 \arcsin (a+b x)^2}}{b}\)

input

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

output

(-1/2*(1 - (a + b*x)^2)^2/ArcSin[a + b*x]^2 - 2*(-(((a + b*x)*(1 - (a + b* 
x)^2)^(3/2))/ArcSin[a + b*x]) + CosIntegral[2*ArcSin[a + b*x]]/2 - 4*(-1/8 
*CosIntegral[4*ArcSin[a + b*x]] + Log[ArcSin[a + b*x]]/8) + Log[ArcSin[a + 
 b*x]]/2))/b
 

3.4.25.3.1 Defintions of rubi rules used

rule 2009

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3793

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In 
t[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 4906

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b 
_.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin[a + b*x 
]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IG 
tQ[p, 0]
 

rule 5166

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((d_) + (e_.)*(x_)^2)^(p_.), x_ 
Symbol] :> Simp[Sqrt[1 - c^2*x^2]*(d + e*x^2)^p*((a + b*ArcSin[c*x])^(n + 1 
)/(b*c*(n + 1))), x] + Simp[c*((2*p + 1)/(b*(n + 1)))*Simp[(d + e*x^2)^p/(1 
 - c^2*x^2)^p]   Int[x*(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] && LtQ[n, -1]
 

rule 5168

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x 
_Symbol] :> Simp[(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 5214

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.)*((d_) + (e_. 
)*(x_)^2)^(p_.), x_Symbol] :> Simp[(f*x)^m*Sqrt[1 - c^2*x^2]*(d + e*x^2)^p* 
((a + b*ArcSin[c*x])^(n + 1)/(b*c*(n + 1))), x] + (-Simp[f*(m/(b*c*(n + 1)) 
)*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p]   Int[(f*x)^(m - 1)*(1 - c^2*x^2)^(p 
- 1/2)*(a + b*ArcSin[c*x])^(n + 1), x], x] + Simp[c*((m + 2*p + 1)/(b*f*(n 
+ 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p]   Int[(f*x)^(m + 1)*(1 - c^2*x^2 
)^(p - 1/2)*(a + b*ArcSin[c*x])^(n + 1), x], x]) /; FreeQ[{a, b, c, d, e, f 
}, x] && EqQ[c^2*d + e, 0] && LtQ[n, -1] && IGtQ[2*p, 0] && NeQ[m + 2*p + 1 
, 0] && IGtQ[m, -3]
 

rule 5224

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^ 
2)^(p_.), x_Symbol] :> Simp[(1/(b*c^(m + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x 
^2)^p]   Subst[Int[x^n*Sin[-a/b + x/b]^m*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 + 2, 0] && IGtQ[m, 0]
 

rule 5306

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((A_.) + (B_.)*(x_) + ( 
C_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[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]
 

3.4.25.4 Maple [A] (verified)

Time = 1.88 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.20

method	result	size
default	\(-\frac {16 \,\operatorname {Ci}\left (2 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )^{2}+16 \,\operatorname {Ci}\left (4 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )^{2}-8 \sin \left (2 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )-4 \sin \left (4 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )+4 \cos \left (2 \arcsin \left (b x +a \right )\right )+\cos \left (4 \arcsin \left (b x +a \right )\right )+3}{16 b \arcsin \left (b x +a \right )^{2}}\)	\(108\)

input

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

output

-1/16/b*(16*Ci(2*arcsin(b*x+a))*arcsin(b*x+a)^2+16*Ci(4*arcsin(b*x+a))*arc 
sin(b*x+a)^2-8*sin(2*arcsin(b*x+a))*arcsin(b*x+a)-4*sin(4*arcsin(b*x+a))*a 
rcsin(b*x+a)+4*cos(2*arcsin(b*x+a))+cos(4*arcsin(b*x+a))+3)/arcsin(b*x+a)^ 
2
 

3.4.25.5 Fricas [F]

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

input

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

output

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

3.4.25.6 Sympy [F]

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

input

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

output

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

3.4.25.7 Maxima [F]

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

input

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

output

-1/2*(b^4*x^4 + 4*a*b^3*x^3 + 2*(3*a^2 - 1)*b^2*x^2 + a^4 - 2*b*arctan2(b* 
x + a, sqrt(b*x + a + 1)*sqrt(-b*x - a + 1))^2*integrate(2*(4*b^2*x^2 + 8* 
a*b*x + 4*a^2 - 1)*sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)/arctan2(b*x + a, s 
qrt(b*x + a + 1)*sqrt(-b*x - a + 1)), x) + 4*(a^3 - a)*b*x + 4*(b^3*x^3 + 
3*a*b^2*x^2 + a^3 + (3*a^2 - 1)*b*x - a)*sqrt(b*x + a + 1)*sqrt(-b*x - a + 
 1)*arctan2(b*x + a, sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)) - 2*a^2 + 1)/(b 
*arctan2(b*x + a, sqrt(b*x + a + 1)*sqrt(-b*x - a + 1))^2)
 

3.4.25.8 Giac [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.12 \[ \int \frac {\left (1-a^2-2 a b x-b^2 x^2\right )^{3/2}}{\arcsin (a+b x)^3} \, dx=\frac {2 \, {\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}} {\left (b x + a\right )}}{b \arcsin \left (b x + a\right )} - \frac {\operatorname {Ci}\left (4 \, \arcsin \left (b x + a\right )\right )}{b} - \frac {\operatorname {Ci}\left (2 \, \arcsin \left (b x + a\right )\right )}{b} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )}^{2}}{2 \, b \arcsin \left (b x + a\right )^{2}} \]

input

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

output

2*(-b^2*x^2 - 2*a*b*x - a^2 + 1)^(3/2)*(b*x + a)/(b*arcsin(b*x + a)) - cos 
_integral(4*arcsin(b*x + a))/b - cos_integral(2*arcsin(b*x + a))/b - 1/2*( 
b^2*x^2 + 2*a*b*x + a^2 - 1)^2/(b*arcsin(b*x + a)^2)
 

3.4.25.9 Mupad [F(-1)]

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

input

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

output

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