\(\int \frac {\arccos (a+b x)}{x} \, dx\) [5]

Optimal result

Integrand size = 10, antiderivative size = 177 \[ \int \frac {\arccos (a+b x)}{x} \, dx=-\frac {1}{2} i \arccos (a+b x)^2+\arccos (a+b x) \log \left (1-\frac {e^{i \arccos (a+b x)}}{a-i \sqrt {1-a^2}}\right )+\arccos (a+b x) \log \left (1-\frac {e^{i \arccos (a+b x)}}{a+i \sqrt {1-a^2}}\right )-i \operatorname {PolyLog}\left (2,\frac {e^{i \arccos (a+b x)}}{a-i \sqrt {1-a^2}}\right )-i \operatorname {PolyLog}\left (2,\frac {e^{i \arccos (a+b x)}}{a+i \sqrt {1-a^2}}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.29 \[ \int \frac {\arccos (a+b x)}{x} \, dx=-\frac {1}{2} i \arccos (a+b x)^2-4 i \arcsin \left (\frac {\sqrt {1-a}}{\sqrt {2}}\right ) \arctan \left (\frac {(1+a) \tan \left (\frac {1}{2} \arccos (a+b x)\right )}{\sqrt {-1+a^2}}\right )+\left (\arccos (a+b x)-2 \arcsin \left (\frac {\sqrt {1-a}}{\sqrt {2}}\right )\right ) \log \left (1+\left (-a+\sqrt {-1+a^2}\right ) e^{i \arccos (a+b x)}\right )+\left (\arccos (a+b x)+2 \arcsin \left (\frac {\sqrt {1-a}}{\sqrt {2}}\right )\right ) \log \left (1-\left (a+\sqrt {-1+a^2}\right ) e^{i \arccos (a+b x)}\right )-i \left (\operatorname {PolyLog}\left (2,\left (a-\sqrt {-1+a^2}\right ) e^{i \arccos (a+b x)}\right )+\operatorname {PolyLog}\left (2,\left (a+\sqrt {-1+a^2}\right ) e^{i \arccos (a+b x)}\right )\right ) \] Input:

Integrate[ArcCos[a + b*x]/x,x]
 

Output:

(-1/2*I)*ArcCos[a + b*x]^2 - (4*I)*ArcSin[Sqrt[1 - a]/Sqrt[2]]*ArcTan[((1 
+ a)*Tan[ArcCos[a + b*x]/2])/Sqrt[-1 + a^2]] + (ArcCos[a + b*x] - 2*ArcSin 
[Sqrt[1 - a]/Sqrt[2]])*Log[1 + (-a + Sqrt[-1 + a^2])*E^(I*ArcCos[a + b*x]) 
] + (ArcCos[a + b*x] + 2*ArcSin[Sqrt[1 - a]/Sqrt[2]])*Log[1 - (a + Sqrt[-1 
 + a^2])*E^(I*ArcCos[a + b*x])] - I*(PolyLog[2, (a - Sqrt[-1 + a^2])*E^(I* 
ArcCos[a + b*x])] + PolyLog[2, (a + Sqrt[-1 + a^2])*E^(I*ArcCos[a + b*x])] 
)
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {5305, 25, 27, 5241, 5033, 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.

\(\displaystyle \int \frac {\arccos (a+b x)}{x} \, dx\)

\(\Big \downarrow \) 5305

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

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\int -\frac {\arccos (a+b x)}{x}d(a+b x)}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle -\int -\frac {\arccos (a+b x)}{b x}d(a+b x)\)

\(\Big \downarrow \) 5241

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

\(\Big \downarrow \) 5033

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

Int[ArcCos[a + b*x]/x,x]
 

Output:

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

Defintions of rubi rules used

rule 25

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 5033

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)])/(Cos[(c_.) + (d_.) 
*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[I*((e + f*x)^(m + 1)/(b*f*(m + 1))) 
, x] + (Int[(e + f*x)^m*(E^(I*(c + d*x))/(I*a - Rt[-a^2 + b^2, 2] + I*b*E^( 
I*(c + d*x)))), x] + Int[(e + f*x)^m*(E^(I*(c + d*x))/(I*a + Rt[-a^2 + b^2, 
 2] + I*b*E^(I*(c + d*x)))), x]) /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 
 0] && NegQ[a^2 - b^2]
 

rule 5241

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

rule 5305

Int[((a_.) + ArcCos[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m 
_.), x_Symbol] :> Simp[1/d   Subst[Int[((d*e - c*f)/d + f*(x/d))^m*(a + b*A 
rcCos[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
 

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.12

method	result	size
derivativedivides	\(-\frac {i \arccos \left (b x +a \right )^{2}}{2}+\arccos \left (b x +a \right ) \ln \left (\frac {\sqrt {a^{2}-1}-b x -i \sqrt {1-\left (b x +a \right )^{2}}}{a +\sqrt {a^{2}-1}}\right )+\arccos \left (b x +a \right ) \ln \left (\frac {\sqrt {a^{2}-1}+b x +i \sqrt {1-\left (b x +a \right )^{2}}}{-a +\sqrt {a^{2}-1}}\right )-i \operatorname {dilog}\left (\frac {\sqrt {a^{2}-1}-b x -i \sqrt {1-\left (b x +a \right )^{2}}}{a +\sqrt {a^{2}-1}}\right )-i \operatorname {dilog}\left (\frac {\sqrt {a^{2}-1}+b x +i \sqrt {1-\left (b x +a \right )^{2}}}{-a +\sqrt {a^{2}-1}}\right )\)	\(199\)
default	\(-\frac {i \arccos \left (b x +a \right )^{2}}{2}+\arccos \left (b x +a \right ) \ln \left (\frac {\sqrt {a^{2}-1}-b x -i \sqrt {1-\left (b x +a \right )^{2}}}{a +\sqrt {a^{2}-1}}\right )+\arccos \left (b x +a \right ) \ln \left (\frac {\sqrt {a^{2}-1}+b x +i \sqrt {1-\left (b x +a \right )^{2}}}{-a +\sqrt {a^{2}-1}}\right )-i \operatorname {dilog}\left (\frac {\sqrt {a^{2}-1}-b x -i \sqrt {1-\left (b x +a \right )^{2}}}{a +\sqrt {a^{2}-1}}\right )-i \operatorname {dilog}\left (\frac {\sqrt {a^{2}-1}+b x +i \sqrt {1-\left (b x +a \right )^{2}}}{-a +\sqrt {a^{2}-1}}\right )\)	\(199\)

Input:

int(arccos(b*x+a)/x,x,method=_RETURNVERBOSE)
 

Output:

-1/2*I*arccos(b*x+a)^2+arccos(b*x+a)*ln(((a^2-1)^(1/2)-b*x-I*(1-(b*x+a)^2) 
^(1/2))/(a+(a^2-1)^(1/2)))+arccos(b*x+a)*ln(((a^2-1)^(1/2)+b*x+I*(1-(b*x+a 
)^2)^(1/2))/(-a+(a^2-1)^(1/2)))-I*dilog(((a^2-1)^(1/2)-b*x-I*(1-(b*x+a)^2) 
^(1/2))/(a+(a^2-1)^(1/2)))-I*dilog(((a^2-1)^(1/2)+b*x+I*(1-(b*x+a)^2)^(1/2 
))/(-a+(a^2-1)^(1/2)))
 

Fricas [F]

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

integrate(arccos(b*x+a)/x,x, algorithm="fricas")
 

Output:

integral(arccos(b*x + a)/x, x)
 

Sympy [F]

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

integrate(acos(b*x+a)/x,x)
 

Output:

Integral(acos(a + b*x)/x, x)
 

Maxima [F]

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

integrate(arccos(b*x+a)/x,x, algorithm="maxima")
 

Output:

integrate(arccos(b*x + a)/x, x)
 

Giac [F]

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

integrate(arccos(b*x+a)/x,x, algorithm="giac")
 

Output:

integrate(arccos(b*x + a)/x, x)
 

Mupad [F(-1)]

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

int(acos(a + b*x)/x,x)
                                                                                    
                                                                                    
 

Output:

int(acos(a + b*x)/x, x)
 

Reduce [F]

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

int(acos(b*x+a)/x,x)
 

Output:

int(acos(a + b*x)/x,x)