\(\int \frac {1}{\sqrt {a+b \arccos (c+d x)}} \, dx\) [20]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [F(-2)]
Sympy [F]
Maxima [F]
Giac [C] (verification not implemented)
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 14, antiderivative size = 106 \[ \int \frac {1}{\sqrt {a+b \arccos (c+d x)}} \, dx=-\frac {\sqrt {2 \pi } \cos \left (\frac {a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arccos (c+d x)}}{\sqrt {b}}\right )}{\sqrt {b} d}+\frac {\sqrt {2 \pi } \operatorname {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arccos (c+d x)}}{\sqrt {b}}\right ) \sin \left (\frac {a}{b}\right )}{\sqrt {b} d} \] Output:

-2^(1/2)*Pi^(1/2)*cos(a/b)*FresnelS(2^(1/2)/Pi^(1/2)*(a+b*arccos(d*x+c))^( 
1/2)/b^(1/2))/b^(1/2)/d+2^(1/2)*Pi^(1/2)*FresnelC(2^(1/2)/Pi^(1/2)*(a+b*ar 
ccos(d*x+c))^(1/2)/b^(1/2))*sin(a/b)/b^(1/2)/d
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.08 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.21 \[ \int \frac {1}{\sqrt {a+b \arccos (c+d x)}} \, dx=\frac {e^{-\frac {i a}{b}} \left (\sqrt {-\frac {i (a+b \arccos (c+d x))}{b}} \Gamma \left (\frac {1}{2},-\frac {i (a+b \arccos (c+d x))}{b}\right )+e^{\frac {2 i a}{b}} \sqrt {\frac {i (a+b \arccos (c+d x))}{b}} \Gamma \left (\frac {1}{2},\frac {i (a+b \arccos (c+d x))}{b}\right )\right )}{2 d \sqrt {a+b \arccos (c+d x)}} \] Input:

Integrate[1/Sqrt[a + b*ArcCos[c + d*x]],x]
 

Output:

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

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.02, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.786, Rules used = {5303, 5135, 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.

\(\displaystyle \int \frac {1}{\sqrt {a+b \arccos (c+d x)}} \, dx\)

\(\Big \downarrow \) 5303

\(\displaystyle \frac {\int \frac {1}{\sqrt {a+b \arccos (c+d x)}}d(c+d x)}{d}\)

\(\Big \downarrow \) 5135

\(\displaystyle -\frac {\int -\frac {\sin \left (\frac {a}{b}-\frac {a+b \arccos (c+d x)}{b}\right )}{\sqrt {a+b \arccos (c+d x)}}d(a+b \arccos (c+d x))}{b d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {\sin \left (\frac {a}{b}-\frac {a+b \arccos (c+d x)}{b}\right )}{\sqrt {a+b \arccos (c+d x)}}d(a+b \arccos (c+d x))}{b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\sin \left (\frac {a}{b}-\frac {a+b \arccos (c+d x)}{b}\right )}{\sqrt {a+b \arccos (c+d x)}}d(a+b \arccos (c+d x))}{b d}\)

\(\Big \downarrow \) 3787

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3785

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

\(\Big \downarrow \) 3786

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

\(\Big \downarrow \) 3832

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

\(\Big \downarrow \) 3833

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

Input:

Int[1/Sqrt[a + b*ArcCos[c + d*x]],x]
 

Output:

-((Sqrt[b]*Sqrt[2*Pi]*Cos[a/b]*FresnelS[(Sqrt[2/Pi]*Sqrt[a + b*ArcCos[c + 
d*x]])/Sqrt[b]] - Sqrt[b]*Sqrt[2*Pi]*FresnelC[(Sqrt[2/Pi]*Sqrt[a + b*ArcCo 
s[c + d*x]])/Sqrt[b]]*Sin[a/b])/(b*d))
 

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 5135
Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[-(b*c)^(-1) 
  Subst[Int[x^n*Sin[-a/b + x/b], x], x, a + b*ArcCos[c*x]], x] /; FreeQ[{a, 
 b, c, n}, x]
 

rule 5303
Int[((a_.) + ArcCos[(c_) + (d_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[1/d 
  Subst[Int[(a + b*ArcCos[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, 
n}, x]
 
Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.88

method result size
default \(\frac {\sqrt {2}\, \sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, \left (\cos \left (\frac {a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, \sqrt {a +b \arccos \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right )+\sin \left (\frac {a}{b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {2}\, \sqrt {a +b \arccos \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right )\right )}{d}\) \(93\)

Input:

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

Output:

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

Fricas [F(-2)]

Exception generated. \[ \int \frac {1}{\sqrt {a+b \arccos (c+d x)}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {1}{\sqrt {a+b \arccos (c+d x)}} \, dx=\int \frac {1}{\sqrt {a + b \operatorname {acos}{\left (c + d x \right )}}}\, dx \] Input:

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

Output:

Integral(1/sqrt(a + b*acos(c + d*x)), x)
 

Maxima [F]

\[ \int \frac {1}{\sqrt {a+b \arccos (c+d x)}} \, dx=\int { \frac {1}{\sqrt {b \arccos \left (d x + c\right ) + a}} \,d x } \] Input:

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

Output:

integrate(1/sqrt(b*arccos(d*x + c) + a), x)
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.58 \[ \int \frac {1}{\sqrt {a+b \arccos (c+d x)}} \, dx=\frac {i \, \sqrt {\pi } \operatorname {erf}\left (-\frac {i \, \sqrt {2} \sqrt {b \arccos \left (d x + c\right ) + a}}{2 \, \sqrt {{\left | b \right |}}} - \frac {\sqrt {2} \sqrt {b \arccos \left (d x + c\right ) + a} \sqrt {{\left | b \right |}}}{2 \, b}\right ) e^{\left (\frac {i \, a}{b}\right )}}{d {\left (\frac {i \, \sqrt {2} b}{\sqrt {{\left | b \right |}}} + \sqrt {2} \sqrt {{\left | b \right |}}\right )}} - \frac {i \, \sqrt {\pi } \operatorname {erf}\left (\frac {i \, \sqrt {2} \sqrt {b \arccos \left (d x + c\right ) + a}}{2 \, \sqrt {{\left | b \right |}}} - \frac {\sqrt {2} \sqrt {b \arccos \left (d x + c\right ) + a} \sqrt {{\left | b \right |}}}{2 \, b}\right ) e^{\left (-\frac {i \, a}{b}\right )}}{d {\left (-\frac {i \, \sqrt {2} b}{\sqrt {{\left | b \right |}}} + \sqrt {2} \sqrt {{\left | b \right |}}\right )}} \] Input:

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

Output:

I*sqrt(pi)*erf(-1/2*I*sqrt(2)*sqrt(b*arccos(d*x + c) + a)/sqrt(abs(b)) - 1 
/2*sqrt(2)*sqrt(b*arccos(d*x + c) + a)*sqrt(abs(b))/b)*e^(I*a/b)/(d*(I*sqr 
t(2)*b/sqrt(abs(b)) + sqrt(2)*sqrt(abs(b)))) - I*sqrt(pi)*erf(1/2*I*sqrt(2 
)*sqrt(b*arccos(d*x + c) + a)/sqrt(abs(b)) - 1/2*sqrt(2)*sqrt(b*arccos(d*x 
 + c) + a)*sqrt(abs(b))/b)*e^(-I*a/b)/(d*(-I*sqrt(2)*b/sqrt(abs(b)) + sqrt 
(2)*sqrt(abs(b))))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {a+b \arccos (c+d x)}} \, dx=\int \frac {1}{\sqrt {a+b\,\mathrm {acos}\left (c+d\,x\right )}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\sqrt {a+b \arccos (c+d x)}} \, dx=\int \frac {\sqrt {\mathit {acos} \left (d x +c \right ) b +a}}{\mathit {acos} \left (d x +c \right ) b +a}d x \] Input:

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

Output:

int(sqrt(acos(c + d*x)*b + a)/(acos(c + d*x)*b + a),x)