\(\int \frac {\sec ^{-1}(a+b x)}{x} \, dx\) [23]

Optimal result

Integrand size = 10, antiderivative size = 200 \[ \int \frac {\sec ^{-1}(a+b x)}{x} \, dx=\sec ^{-1}(a+b x) \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+\sec ^{-1}(a+b x) \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )-i \operatorname {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )-i \operatorname {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )+\frac {1}{2} i \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.42 \[ \int \frac {\sec ^{-1}(a+b x)}{x} \, dx=-4 i \arcsin \left (\frac {\sqrt {\frac {-1+a}{a}}}{\sqrt {2}}\right ) \arctan \left (\frac {(1+a) \tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt {1-a^2}}\right )+\left (\sec ^{-1}(a+b x)-2 \arcsin \left (\frac {\sqrt {\frac {-1+a}{a}}}{\sqrt {2}}\right )\right ) \log \left (1+\frac {\left (-1+\sqrt {1-a^2}\right ) e^{i \sec ^{-1}(a+b x)}}{a}\right )+\left (\sec ^{-1}(a+b x)+2 \arcsin \left (\frac {\sqrt {\frac {-1+a}{a}}}{\sqrt {2}}\right )\right ) \log \left (1-\frac {\left (1+\sqrt {1-a^2}\right ) e^{i \sec ^{-1}(a+b x)}}{a}\right )-\sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )-i \left (\operatorname {PolyLog}\left (2,-\frac {\left (-1+\sqrt {1-a^2}\right ) e^{i \sec ^{-1}(a+b x)}}{a}\right )+\operatorname {PolyLog}\left (2,\frac {\left (1+\sqrt {1-a^2}\right ) e^{i \sec ^{-1}(a+b x)}}{a}\right )\right )+\frac {1}{2} i \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.10 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.32, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.400, Rules used = {5781, 25, 5062, 5041, 25, 3042, 4202, 2620, 2715, 2838, 5031, 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.

\(\displaystyle \int \frac {\sec ^{-1}(a+b x)}{x} \, dx\)

\(\Big \downarrow \) 5781

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 5062

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

\(\Big \downarrow \) 5041

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4202

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

\(\Big \downarrow \) 5031

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

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

Output:

(I/2)*ArcSec[a + b*x]^2 + a*(((-1/2*I)*ArcSec[a + b*x]^2)/a - I*((I*ArcSec 
[a + b*x]*Log[1 - (a*E^(I*ArcSec[a + b*x]))/(1 - Sqrt[1 - a^2])])/a + Poly 
Log[2, (a*E^(I*ArcSec[a + b*x]))/(1 - Sqrt[1 - a^2])]/a) - I*((I*ArcSec[a 
+ b*x]*Log[1 - (a*E^(I*ArcSec[a + b*x]))/(1 + Sqrt[1 - a^2])])/a + PolyLog 
[2, (a*E^(I*ArcSec[a + b*x]))/(1 + Sqrt[1 - a^2])]/a)) - (2*I)*((-1/2*I)*A 
rcSec[a + b*x]*Log[1 + E^((2*I)*ArcSec[a + b*x])] - PolyLog[2, -E^((2*I)*A 
rcSec[a + b*x])]/4)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 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 5031

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] + (-Simp[I   Int[(e + f*x)^m*(E^(I*(c + d*x))/(a - Rt[a^2 - b^2, 2] + 
b*E^(I*(c + d*x)))), x], x] - Simp[I   Int[(e + f*x)^m*(E^(I*(c + d*x))/(a 
+ Rt[a^2 - b^2, 2] + b*E^(I*(c + d*x)))), x], x]) /; FreeQ[{a, b, c, d, e, 
f}, x] && IGtQ[m, 0] && PosQ[a^2 - b^2]
 

rule 5041

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

rule 5062

Int[(((e_.) + (f_.)*(x_))^(m_.)*(F_)[(c_.) + (d_.)*(x_)]^(n_.)*(G_)[(c_.) + 
 (d_.)*(x_)]^(p_.))/((a_) + (b_.)*Sec[(c_.) + (d_.)*(x_)]), x_Symbol] :> In 
t[(e + f*x)^m*Cos[c + d*x]*F[c + d*x]^n*(G[c + d*x]^p/(b + a*Cos[c + d*x])) 
, x] /; FreeQ[{a, b, c, d, e, f}, x] && TrigQ[F] && TrigQ[G] && IntegersQ[m 
, n, p]
 

rule 5781

Int[((a_.) + ArcSec[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m 
_.), x_Symbol] :> Simp[1/d^(m + 1)   Subst[Int[(a + b*x)^p*Sec[x]*Tan[x]*(d 
*e - c*f + f*Sec[x])^m, x], x, ArcSec[c + d*x]], x] /; FreeQ[{a, b, c, d, e 
, f}, x] && IGtQ[p, 0] && IntegerQ[m]
 

Maple [A] (verified)

Time = 0.90 (sec) , antiderivative size = 374, normalized size of antiderivative = 1.87

method	result	size
derivativedivides	\(\operatorname {arcsec}\left (b x +a \right ) \ln \left (\frac {-a \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )+\sqrt {-a^{2}+1}+1}{1+\sqrt {-a^{2}+1}}\right )+\operatorname {arcsec}\left (b x +a \right ) \ln \left (\frac {a \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )+\sqrt {-a^{2}+1}-1}{-1+\sqrt {-a^{2}+1}}\right )-\operatorname {arcsec}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )-\operatorname {arcsec}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )+i \operatorname {dilog}\left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )+i \operatorname {dilog}\left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )-i \operatorname {dilog}\left (\frac {-a \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )+\sqrt {-a^{2}+1}+1}{1+\sqrt {-a^{2}+1}}\right )-i \operatorname {dilog}\left (\frac {a \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )+\sqrt {-a^{2}+1}-1}{-1+\sqrt {-a^{2}+1}}\right )\)	\(374\)
default	\(\operatorname {arcsec}\left (b x +a \right ) \ln \left (\frac {-a \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )+\sqrt {-a^{2}+1}+1}{1+\sqrt {-a^{2}+1}}\right )+\operatorname {arcsec}\left (b x +a \right ) \ln \left (\frac {a \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )+\sqrt {-a^{2}+1}-1}{-1+\sqrt {-a^{2}+1}}\right )-\operatorname {arcsec}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )-\operatorname {arcsec}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )+i \operatorname {dilog}\left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )+i \operatorname {dilog}\left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )-i \operatorname {dilog}\left (\frac {-a \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )+\sqrt {-a^{2}+1}+1}{1+\sqrt {-a^{2}+1}}\right )-i \operatorname {dilog}\left (\frac {a \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )+\sqrt {-a^{2}+1}-1}{-1+\sqrt {-a^{2}+1}}\right )\)	\(374\)

Input:

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

Output:

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

Fricas [F]

\[ \int \frac {\sec ^{-1}(a+b x)}{x} \, dx=\int { \frac {\operatorname {arcsec}\left (b x + a\right )}{x} \,d x } \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {\sec ^{-1}(a+b x)}{x} \, dx=\int \frac {\operatorname {asec}{\left (a + b x \right )}}{x}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {\sec ^{-1}(a+b x)}{x} \, dx=\int { \frac {\operatorname {arcsec}\left (b x + a\right )}{x} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {\sec ^{-1}(a+b x)}{x} \, dx=\int { \frac {\operatorname {arcsec}\left (b x + a\right )}{x} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

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

int(acos(1/(a + b*x))/x,x)
 

Output:

int(acos(1/(a + b*x))/x, x)
 

Reduce [F]

\[ \int \frac {\sec ^{-1}(a+b x)}{x} \, dx=\int \frac {\mathit {asec} \left (b x +a \right )}{x}d x \] Input:

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

Output:

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