\(\int \frac {a+b \arccos (\frac {\sqrt {1-c x}}{\sqrt {1+c x}})}{1-c^2 x^2} \, dx\) [81]

Optimal result

Integrand size = 38, antiderivative size = 141 \[ \int \frac {a+b \arccos \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{1-c^2 x^2} \, dx=\frac {i \left (a+b \arccos \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2}{2 b c}-\frac {\left (a+b \arccos \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \log \left (1+e^{2 i \arccos \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{c}+\frac {i b \operatorname {PolyLog}\left (2,-e^{2 i \arccos \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{2 c} \] Output:

1/2*I*(a+b*arccos((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^2/b/c-(a+b*arccos((-c*x+1 
)^(1/2)/(c*x+1)^(1/2)))*ln(1+((-c*x+1)^(1/2)/(c*x+1)^(1/2)+I*(1-(-c*x+1)/( 
c*x+1))^(1/2))^2)/c+1/2*I*b*polylog(2,-((-c*x+1)^(1/2)/(c*x+1)^(1/2)+I*(1- 
(-c*x+1)/(c*x+1))^(1/2))^2)/c
                                                                                    
                                                                                    
 

Mathematica [F]

\[ \int \frac {a+b \arccos \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{1-c^2 x^2} \, dx=\int \frac {a+b \arccos \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{1-c^2 x^2} \, dx \] Input:

Integrate[(a + b*ArcCos[Sqrt[1 - c*x]/Sqrt[1 + c*x]])/(1 - c^2*x^2),x]
 

Output:

Integrate[(a + b*ArcCos[Sqrt[1 - c*x]/Sqrt[1 + c*x]])/(1 - c^2*x^2), x]
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.184, Rules used = {7232, 5137, 3042, 4202, 2620, 2715, 2838}

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.

\(\displaystyle \int \frac {a+b \arccos \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}{1-c^2 x^2} \, dx\)

\(\Big \downarrow \) 7232

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

\(\Big \downarrow \) 5137

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4202

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

\(\Big \downarrow \) 2620

\(\displaystyle \frac {\frac {i \left (a+b \arccos \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2}{2 b}-2 i \left (\frac {1}{2} i b \int \log \left (1+e^{2 i \arccos \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right )d\arccos \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )-\frac {1}{2} i \log \left (1+e^{2 i \arccos \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \arccos \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )\right )}{c}\)

\(\Big \downarrow \) 2715

\(\displaystyle \frac {\frac {i \left (a+b \arccos \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2}{2 b}-2 i \left (\frac {1}{4} b \int \frac {\sqrt {c x+1} \log \left (1+e^{2 i \arccos \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right )}{\sqrt {1-c x}}de^{2 i \arccos \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}-\frac {1}{2} i \log \left (1+e^{2 i \arccos \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \arccos \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )\right )}{c}\)

\(\Big \downarrow \) 2838

\(\displaystyle \frac {\frac {i \left (a+b \arccos \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2}{2 b}-2 i \left (-\frac {1}{2} i \log \left (1+e^{2 i \arccos \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \arccos \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )-\frac {1}{4} b \operatorname {PolyLog}\left (2,-e^{2 i \arccos \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right )\right )}{c}\)

Input:

Int[(a + b*ArcCos[Sqrt[1 - c*x]/Sqrt[1 + c*x]])/(1 - c^2*x^2),x]
 

Output:

(((I/2)*(a + b*ArcCos[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^2)/b - (2*I)*((-1/2*I) 
*(a + b*ArcCos[Sqrt[1 - c*x]/Sqrt[1 + c*x]])*Log[1 + E^((2*I)*ArcCos[Sqrt[ 
1 - c*x]/Sqrt[1 + c*x]])] - (b*PolyLog[2, -E^((2*I)*ArcCos[Sqrt[1 - c*x]/S 
qrt[1 + c*x]])])/4))/c
 

Defintions of rubi rules used

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
 

rule 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

rule 3042

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

rule 4202

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I 
*((c + d*x)^(m + 1)/(d*(m + 1))), x] - Simp[2*I   Int[(c + d*x)^m*(E^(2*I*( 
e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] && IGt 
Q[m, 0]
 

rule 5137

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> -Subst[Int[ 
(a + b*x)^n*Tan[x], x], x, ArcCos[c*x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0 
]
 

rule 7232

Int[((a_.) + (b_.)*(F_)[((c_.)*Sqrt[(d_.) + (e_.)*(x_)])/Sqrt[(f_.) + (g_.) 
*(x_)]])^(n_.)/((A_.) + (C_.)*(x_)^2), x_Symbol] :> Simp[2*e*(g/(C*(e*f - d 
*g)))   Subst[Int[(a + b*F[c*x])^n/x, x], x, Sqrt[d + e*x]/Sqrt[f + g*x]], 
x] /; FreeQ[{a, b, c, d, e, f, g, A, C, F}, x] && EqQ[C*d*f - A*e*g, 0] && 
EqQ[e*f + d*g, 0] && IGtQ[n, 0]
 

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.21

method	result	size
default	\(-\frac {a \ln \left (c x -1\right )}{2 c}+\frac {a \ln \left (c x +1\right )}{2 c}+\frac {i b \arccos \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2}}{2 c}-\frac {b \arccos \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \ln \left (1+\left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+i \sqrt {1-\frac {-c x +1}{c x +1}}\right )^{2}\right )}{c}+\frac {i b \operatorname {polylog}\left (2, -\left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+i \sqrt {1-\frac {-c x +1}{c x +1}}\right )^{2}\right )}{2 c}\)	\(171\)
parts	\(-\frac {a \ln \left (c x -1\right )}{2 c}+\frac {a \ln \left (c x +1\right )}{2 c}+\frac {i b \arccos \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2}}{2 c}-\frac {b \arccos \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \ln \left (1+\left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+i \sqrt {1-\frac {-c x +1}{c x +1}}\right )^{2}\right )}{c}+\frac {i b \operatorname {polylog}\left (2, -\left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+i \sqrt {1-\frac {-c x +1}{c x +1}}\right )^{2}\right )}{2 c}\)	\(171\)

Input:

int((a+b*arccos((-c*x+1)^(1/2)/(c*x+1)^(1/2)))/(-c^2*x^2+1),x,method=_RETU 
RNVERBOSE)
 

Output:

-1/2*a/c*ln(c*x-1)+1/2*a/c*ln(c*x+1)+1/2*I*b/c*arccos((-c*x+1)^(1/2)/(c*x+ 
1)^(1/2))^2-b/c*arccos((-c*x+1)^(1/2)/(c*x+1)^(1/2))*ln(1+((-c*x+1)^(1/2)/ 
(c*x+1)^(1/2)+I*(1-(-c*x+1)/(c*x+1))^(1/2))^2)+1/2*I*b*polylog(2,-((-c*x+1 
)^(1/2)/(c*x+1)^(1/2)+I*(1-(-c*x+1)/(c*x+1))^(1/2))^2)/c
 

Fricas [F]

\[ \int \frac {a+b \arccos \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{1-c^2 x^2} \, dx=\int { -\frac {b \arccos \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a}{c^{2} x^{2} - 1} \,d x } \] Input:

integrate((a+b*arccos((-c*x+1)^(1/2)/(c*x+1)^(1/2)))/(-c^2*x^2+1),x, algor 
ithm="fricas")
 

Output:

integral(-(b*arccos(sqrt(-c*x + 1)/sqrt(c*x + 1)) + a)/(c^2*x^2 - 1), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {a+b \arccos \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{1-c^2 x^2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

\[ \int \frac {a+b \arccos \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{1-c^2 x^2} \, dx=\int { -\frac {b \arccos \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a}{c^{2} x^{2} - 1} \,d x } \] Input:

integrate((a+b*arccos((-c*x+1)^(1/2)/(c*x+1)^(1/2)))/(-c^2*x^2+1),x, algor 
ithm="maxima")
                                                                                    
                                                                                    
 

Output:

1/2*a*(log(c*x + 1)/c - log(c*x - 1)/c) - b*integrate(arctan2(sqrt(2)*sqrt 
(c)*sqrt(x), sqrt(-c*x + 1))/(c^2*x^2 - 1), x)
 

Giac [F]

\[ \int \frac {a+b \arccos \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{1-c^2 x^2} \, dx=\int { -\frac {b \arccos \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a}{c^{2} x^{2} - 1} \,d x } \] Input:

integrate((a+b*arccos((-c*x+1)^(1/2)/(c*x+1)^(1/2)))/(-c^2*x^2+1),x, algor 
ithm="giac")
 

Output:

integrate(-(b*arccos(sqrt(-c*x + 1)/sqrt(c*x + 1)) + a)/(c^2*x^2 - 1), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \arccos \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{1-c^2 x^2} \, dx=\int -\frac {a+b\,\mathrm {acos}\left (\frac {\sqrt {1-c\,x}}{\sqrt {c\,x+1}}\right )}{c^2\,x^2-1} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {a+b \arccos \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{1-c^2 x^2} \, dx=\frac {-2 \left (\int \frac {\mathit {acos} \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )}{c^{2} x^{2}-1}d x \right ) b c -\mathrm {log}\left (c^{2} x -c \right ) a +\mathrm {log}\left (c^{2} x +c \right ) a}{2 c} \] Input:

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

Output:

( - 2*int(acos(sqrt( - c*x + 1)/sqrt(c*x + 1))/(c**2*x**2 - 1),x)*b*c - lo 
g(c**2*x - c)*a + log(c**2*x + c)*a)/(2*c)