\(\int \sec ^{-1}(a+b x)^2 \, dx\) [30]

Optimal result

Integrand size = 8, antiderivative size = 94 \[ \int \sec ^{-1}(a+b x)^2 \, dx=\frac {(a+b x) \sec ^{-1}(a+b x)^2}{b}+\frac {4 i \sec ^{-1}(a+b x) \arctan \left (e^{i \sec ^{-1}(a+b x)}\right )}{b}-\frac {2 i \operatorname {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )}{b}+\frac {2 i \operatorname {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b} \] Output:

(b*x+a)*arcsec(b*x+a)^2/b+4*I*arcsec(b*x+a)*arctan(1/(b*x+a)+I*(1-1/(b*x+a 
)^2)^(1/2))/b-2*I*polylog(2,-I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))/b+2*I* 
polylog(2,I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))/b
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.18 \[ \int \sec ^{-1}(a+b x)^2 \, dx=\frac {\sec ^{-1}(a+b x) \left ((a+b x) \sec ^{-1}(a+b x)-2 \log \left (1-i e^{i \sec ^{-1}(a+b x)}\right )+2 \log \left (1+i e^{i \sec ^{-1}(a+b x)}\right )\right )-2 i \operatorname {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )+2 i \operatorname {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b} \] Input:

Integrate[ArcSec[a + b*x]^2,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.95, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.875, Rules used = {5775, 5739, 4244, 3042, 4669, 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.

\(\displaystyle \int \sec ^{-1}(a+b x)^2 \, dx\)

\(\Big \downarrow \) 5775

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

\(\Big \downarrow \) 5739

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

\(\Big \downarrow \) 4244

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4669

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

Int[ArcSec[a + b*x]^2,x]
 

Output:

((a + b*x)*ArcSec[a + b*x]^2 - 2*((-2*I)*ArcSec[a + b*x]*ArcTan[E^(I*ArcSe 
c[a + b*x])] + I*PolyLog[2, (-I)*E^(I*ArcSec[a + b*x])] - I*PolyLog[2, I*E 
^(I*ArcSec[a + b*x])]))/b
 

Defintions of rubi rules used

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 4244

Int[(x_)^(m_.)*Sec[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*Tan[(a_.) + (b_.)*(x_)^( 
n_.)]^(q_.), x_Symbol] :> Simp[x^(m - n + 1)*(Sec[a + b*x^n]^p/(b*n*p)), x] 
 - Simp[(m - n + 1)/(b*n*p)   Int[x^(m - n)*Sec[a + b*x^n]^p, x], x] /; Fre 
eQ[{a, b, p}, x] && IntegerQ[n] && GeQ[m, n] && EqQ[q, 1]
 

rule 4669

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol 
] :> Simp[-2*(c + d*x)^m*(ArcTanh[E^(I*k*Pi)*E^(I*(e + f*x))]/f), x] + (-Si 
mp[d*(m/f)   Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))], x], 
 x] + Simp[d*(m/f)   Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x 
))], x], x]) /; FreeQ[{c, d, e, f}, x] && IntegerQ[2*k] && IGtQ[m, 0]
 

rule 5739

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[1/c   Subst[ 
Int[(a + b*x)^n*Sec[x]*Tan[x], x], x, ArcSec[c*x]], x] /; FreeQ[{a, b, c, n 
}, x] && IGtQ[n, 0]
 

rule 5775

Int[((a_.) + ArcSec[(c_) + (d_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[1/d 
  Subst[Int[(a + b*ArcSec[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d}, 
 x] && IGtQ[p, 0]
 

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.72

method	result	size
derivativedivides	\(\frac {\operatorname {arcsec}\left (b x +a \right )^{2} \left (b x +a \right )+2 \,\operatorname {arcsec}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )-2 \,\operatorname {arcsec}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )-2 i \operatorname {dilog}\left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )+2 i \operatorname {dilog}\left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{b}\)	\(162\)
default	\(\frac {\operatorname {arcsec}\left (b x +a \right )^{2} \left (b x +a \right )+2 \,\operatorname {arcsec}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )-2 \,\operatorname {arcsec}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )-2 i \operatorname {dilog}\left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )+2 i \operatorname {dilog}\left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{b}\)	\(162\)

Input:

int(arcsec(b*x+a)^2,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [F]

\[ \int \sec ^{-1}(a+b x)^2 \, dx=\int { \operatorname {arcsec}\left (b x + a\right )^{2} \,d x } \] Input:

integrate(arcsec(b*x+a)^2,x, algorithm="fricas")
 

Output:

integral(arcsec(b*x + a)^2, x)
 

Sympy [F]

\[ \int \sec ^{-1}(a+b x)^2 \, dx=\int \operatorname {asec}^{2}{\left (a + b x \right )}\, dx \] Input:

integrate(asec(b*x+a)**2,x)
 

Output:

Integral(asec(a + b*x)**2, x)
 

Maxima [F]

\[ \int \sec ^{-1}(a+b x)^2 \, dx=\int { \operatorname {arcsec}\left (b x + a\right )^{2} \,d x } \] Input:

integrate(arcsec(b*x+a)^2,x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

x*arctan(sqrt(b*x + a + 1)*sqrt(b*x + a - 1))^2 - 1/4*x*log(b^2*x^2 + 2*a* 
b*x + a^2)^2 - integrate((2*sqrt(b*x + a + 1)*sqrt(b*x + a - 1)*b*x*arctan 
(sqrt(b*x + a + 1)*sqrt(b*x + a - 1)) + (b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3* 
a^2 - 1)*b*x - a)*log(b*x + a)^2 - (b^3*x^3 + 2*a*b^2*x^2 + (a^2 - 1)*b*x 
+ (b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2 - 1)*b*x - a)*log(b*x + a))*log(b^ 
2*x^2 + 2*a*b*x + a^2))/(b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2 - 1)*b*x - a 
), x)
 

Giac [F]

\[ \int \sec ^{-1}(a+b x)^2 \, dx=\int { \operatorname {arcsec}\left (b x + a\right )^{2} \,d x } \] Input:

integrate(arcsec(b*x+a)^2,x, algorithm="giac")
 

Output:

integrate(arcsec(b*x + a)^2, x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \sec ^{-1}(a+b x)^2 \, dx=\int \mathit {asec} \left (b x +a \right )^{2}d x \] Input:

int(asec(b*x+a)^2,x)
 

Output:

int(asec(a + b*x)**2,x)