3.3.69 \(\int \frac {1}{(a+b \arcsin (c+d x))^{3/2}} \, dx\) [269]

3.3.69.1 Optimal result

Integrand size = 14, antiderivative size = 144 \[ \int \frac {1}{(a+b \arcsin (c+d x))^{3/2}} \, dx=-\frac {2 \sqrt {1-(c+d x)^2}}{b d \sqrt {a+b \arcsin (c+d x)}}-\frac {2 \sqrt {2 \pi } \cos \left (\frac {a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arcsin (c+d x)}}{\sqrt {b}}\right )}{b^{3/2} d}+\frac {2 \sqrt {2 \pi } \operatorname {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arcsin (c+d x)}}{\sqrt {b}}\right ) \sin \left (\frac {a}{b}\right )}{b^{3/2} d} \]

output

-2*cos(a/b)*FresnelS(2^(1/2)/Pi^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b^(1/2))*2 
^(1/2)*Pi^(1/2)/b^(3/2)/d+2*FresnelC(2^(1/2)/Pi^(1/2)*(a+b*arcsin(d*x+c))^ 
(1/2)/b^(1/2))*sin(a/b)*2^(1/2)*Pi^(1/2)/b^(3/2)/d-2*(1-(d*x+c)^2)^(1/2)/b 
/d/(a+b*arcsin(d*x+c))^(1/2)
 

3.3.69.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.16 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.28 \[ \int \frac {1}{(a+b \arcsin (c+d x))^{3/2}} \, dx=\frac {e^{-\frac {i (a+b \arcsin (c+d x))}{b}} \left (e^{i \arcsin (c+d x)} \sqrt {-\frac {i (a+b \arcsin (c+d x))}{b}} \Gamma \left (\frac {1}{2},-\frac {i (a+b \arcsin (c+d x))}{b}\right )+e^{\frac {i a}{b}} \left (-1-e^{2 i \arcsin (c+d x)}+e^{\frac {i (a+b \arcsin (c+d x))}{b}} \sqrt {\frac {i (a+b \arcsin (c+d x))}{b}} \Gamma \left (\frac {1}{2},\frac {i (a+b \arcsin (c+d x))}{b}\right )\right )\right )}{b d \sqrt {a+b \arcsin (c+d x)}} \]

input

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

output

(E^(I*ArcSin[c + d*x])*Sqrt[((-I)*(a + b*ArcSin[c + d*x]))/b]*Gamma[1/2, ( 
(-I)*(a + b*ArcSin[c + d*x]))/b] + E^((I*a)/b)*(-1 - E^((2*I)*ArcSin[c + d 
*x]) + E^((I*(a + b*ArcSin[c + d*x]))/b)*Sqrt[(I*(a + b*ArcSin[c + d*x]))/ 
b]*Gamma[1/2, (I*(a + b*ArcSin[c + d*x]))/b]))/(b*d*E^((I*(a + b*ArcSin[c 
+ d*x]))/b)*Sqrt[a + b*ArcSin[c + d*x]])
 

3.3.69.3 Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {5302, 5132, 5224, 25, 3042, 3787, 25, 3042, 3785, 3786, 3832, 3833}

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.

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

\(\Big \downarrow \) 5302

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

\(\Big \downarrow \) 5132

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

\(\Big \downarrow \) 5224

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3787

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3785

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

\(\Big \downarrow \) 3786

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

\(\Big \downarrow \) 3832

\(\displaystyle \frac {-\frac {2 \left (\sqrt {2 \pi } \sqrt {b} \cos \left (\frac {a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arcsin (c+d x)}}{\sqrt {b}}\right )-2 \sin \left (\frac {a}{b}\right ) \int \cos \left (\frac {a+b \arcsin (c+d x)}{b}\right )d\sqrt {a+b \arcsin (c+d x)}\right )}{b^2}-\frac {2 \sqrt {1-(c+d x)^2}}{b \sqrt {a+b \arcsin (c+d x)}}}{d}\)

\(\Big \downarrow \) 3833

\(\displaystyle \frac {-\frac {2 \left (\sqrt {2 \pi } \sqrt {b} \cos \left (\frac {a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arcsin (c+d x)}}{\sqrt {b}}\right )-\sqrt {2 \pi } \sqrt {b} \sin \left (\frac {a}{b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arcsin (c+d x)}}{\sqrt {b}}\right )\right )}{b^2}-\frac {2 \sqrt {1-(c+d x)^2}}{b \sqrt {a+b \arcsin (c+d x)}}}{d}\)

input

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

output

((-2*Sqrt[1 - (c + d*x)^2])/(b*Sqrt[a + b*ArcSin[c + d*x]]) - (2*(Sqrt[b]* 
Sqrt[2*Pi]*Cos[a/b]*FresnelS[(Sqrt[2/Pi]*Sqrt[a + b*ArcSin[c + d*x]])/Sqrt 
[b]] - Sqrt[b]*Sqrt[2*Pi]*FresnelC[(Sqrt[2/Pi]*Sqrt[a + b*ArcSin[c + d*x]] 
)/Sqrt[b]]*Sin[a/b]))/b^2)/d
 

3.3.69.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3042

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

rule 3785

Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> S 
imp[2/d   Subst[Int[Cos[f*(x^2/d)], x], x, Sqrt[c + d*x]], x] /; FreeQ[{c, 
d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]
 

rule 3786

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

rule 3787

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Cos 
[(d*e - c*f)/d]   Int[Sin[c*(f/d) + f*x]/Sqrt[c + d*x], x], x] + Simp[Sin[( 
d*e - c*f)/d]   Int[Cos[c*(f/d) + f*x]/Sqrt[c + d*x], x], x] /; FreeQ[{c, d 
, e, f}, x] && ComplexFreeQ[f] && NeQ[d*e - c*f, 0]
 

rule 3832

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[ 
d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]
 

rule 3833

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[ 
d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]
 

rule 5132

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

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 5302

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[1/d 
  Subst[Int[(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, 
n}, x]
 

3.3.69.4 Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.18

method	result	size
default	\(-\frac {2 \left (-\sqrt {a +b \arcsin \left (d x +c \right )}\, \cos \left (\frac {a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right ) \sqrt {-\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {2}-\sqrt {a +b \arcsin \left (d x +c \right )}\, \sin \left (\frac {a}{b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right ) \sqrt {-\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {2}+\cos \left (-\frac {a +b \arcsin \left (d x +c \right )}{b}+\frac {a}{b}\right )\right )}{d b \sqrt {a +b \arcsin \left (d x +c \right )}}\)	\(170\)

input

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

output

-2/d/b*(-(a+b*arcsin(d*x+c))^(1/2)*cos(a/b)*FresnelS(2^(1/2)/Pi^(1/2)/(-1/ 
b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*(-1/b)^(1/2)*Pi^(1/2)*2^(1/2)-(a+b*a 
rcsin(d*x+c))^(1/2)*sin(a/b)*FresnelC(2^(1/2)/Pi^(1/2)/(-1/b)^(1/2)*(a+b*a 
rcsin(d*x+c))^(1/2)/b)*(-1/b)^(1/2)*Pi^(1/2)*2^(1/2)+cos(-(a+b*arcsin(d*x+ 
c))/b+a/b))/(a+b*arcsin(d*x+c))^(1/2)
 

3.3.69.5 Fricas [F(-2)]

Exception generated. \[ \int \frac {1}{(a+b \arcsin (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \]

input

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

output

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (constant residues)
 

3.3.69.6 Sympy [F]

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

input

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

output

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

3.3.69.7 Maxima [F]

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

input

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

output

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

3.3.69.8 Giac [F]

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

input

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

output

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

3.3.69.9 Mupad [F(-1)]

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

input

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

output

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