3.1.35 \(\int \sec ^{-1}(a+b x)^3 \, dx\) [35]

3.1.35.1 Optimal result

Integrand size = 8, antiderivative size = 154 \[ \int \sec ^{-1}(a+b x)^3 \, dx=\frac {(a+b x) \sec ^{-1}(a+b x)^3}{b}+\frac {6 i \sec ^{-1}(a+b x)^2 \arctan \left (e^{i \sec ^{-1}(a+b x)}\right )}{b}-\frac {6 i \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )}{b}+\frac {6 i \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b}+\frac {6 \operatorname {PolyLog}\left (3,-i e^{i \sec ^{-1}(a+b x)}\right )}{b}-\frac {6 \operatorname {PolyLog}\left (3,i e^{i \sec ^{-1}(a+b x)}\right )}{b} \]

output

(b*x+a)*arcsec(b*x+a)^3/b+6*I*arcsec(b*x+a)^2*arctan(1/(b*x+a)+I*(1-1/(b*x 
+a)^2)^(1/2))/b-6*I*arcsec(b*x+a)*polylog(2,-I*(1/(b*x+a)+I*(1-1/(b*x+a)^2 
)^(1/2)))/b+6*I*arcsec(b*x+a)*polylog(2,I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/ 
2)))/b+6*polylog(3,-I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))/b-6*polylog(3,I 
*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))/b
 

3.1.35.2 Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.04 \[ \int \sec ^{-1}(a+b x)^3 \, dx=\frac {(a+b x) \sec ^{-1}(a+b x)^3-3 \sec ^{-1}(a+b x)^2 \left (\log \left (1-i e^{i \sec ^{-1}(a+b x)}\right )-\log \left (1+i e^{i \sec ^{-1}(a+b x)}\right )\right )-6 i \sec ^{-1}(a+b x) \left (\operatorname {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )-\operatorname {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )\right )+6 \left (\operatorname {PolyLog}\left (3,-i e^{i \sec ^{-1}(a+b x)}\right )-\operatorname {PolyLog}\left (3,i e^{i \sec ^{-1}(a+b x)}\right )\right )}{b} \]

input

Integrate[ArcSec[a + b*x]^3,x]
 

output

((a + b*x)*ArcSec[a + b*x]^3 - 3*ArcSec[a + b*x]^2*(Log[1 - I*E^(I*ArcSec[ 
a + b*x])] - Log[1 + I*E^(I*ArcSec[a + b*x])]) - (6*I)*ArcSec[a + b*x]*(Po 
lyLog[2, (-I)*E^(I*ArcSec[a + b*x])] - PolyLog[2, I*E^(I*ArcSec[a + b*x])] 
) + 6*(PolyLog[3, (-I)*E^(I*ArcSec[a + b*x])] - PolyLog[3, I*E^(I*ArcSec[a 
 + b*x])]))/b
 

3.1.35.3 Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.97, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {5775, 5739, 4244, 3042, 4669, 3011, 2720, 7143}

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.

\(\displaystyle \int \sec ^{-1}(a+b x)^3 \, dx\)

\(\Big \downarrow \) 5775

\(\displaystyle \frac {\int \sec ^{-1}(a+b x)^3d(a+b x)}{b}\)

\(\Big \downarrow \) 5739

\(\displaystyle \frac {\int (a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^3d\sec ^{-1}(a+b x)}{b}\)

\(\Big \downarrow \) 4244

\(\displaystyle \frac {(a+b x) \sec ^{-1}(a+b x)^3-3 \int (a+b x) \sec ^{-1}(a+b x)^2d\sec ^{-1}(a+b x)}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(a+b x) \sec ^{-1}(a+b x)^3-3 \int \sec ^{-1}(a+b x)^2 \csc \left (\sec ^{-1}(a+b x)+\frac {\pi }{2}\right )d\sec ^{-1}(a+b x)}{b}\)

\(\Big \downarrow \) 4669

\(\displaystyle \frac {(a+b x) \sec ^{-1}(a+b x)^3-3 \left (-2 \int \sec ^{-1}(a+b x) \log \left (1-i e^{i \sec ^{-1}(a+b x)}\right )d\sec ^{-1}(a+b x)+2 \int \sec ^{-1}(a+b x) \log \left (1+i e^{i \sec ^{-1}(a+b x)}\right )d\sec ^{-1}(a+b x)-2 i \sec ^{-1}(a+b x)^2 \arctan \left (e^{i \sec ^{-1}(a+b x)}\right )\right )}{b}\)

\(\Big \downarrow \) 3011

\(\displaystyle \frac {(a+b x) \sec ^{-1}(a+b x)^3-3 \left (2 \left (i \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )-i \int \operatorname {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )d\sec ^{-1}(a+b x)\right )-2 \left (i \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )-i \int \operatorname {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )d\sec ^{-1}(a+b x)\right )-2 i \sec ^{-1}(a+b x)^2 \arctan \left (e^{i \sec ^{-1}(a+b x)}\right )\right )}{b}\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {(a+b x) \sec ^{-1}(a+b x)^3-3 \left (2 \left (i \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )-\int e^{-i \sec ^{-1}(a+b x)} \operatorname {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )de^{i \sec ^{-1}(a+b x)}\right )-2 \left (i \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )-\int e^{-i \sec ^{-1}(a+b x)} \operatorname {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )de^{i \sec ^{-1}(a+b x)}\right )-2 i \sec ^{-1}(a+b x)^2 \arctan \left (e^{i \sec ^{-1}(a+b x)}\right )\right )}{b}\)

\(\Big \downarrow \) 7143

\(\displaystyle \frac {(a+b x) \sec ^{-1}(a+b x)^3-3 \left (-2 i \sec ^{-1}(a+b x)^2 \arctan \left (e^{i \sec ^{-1}(a+b x)}\right )+2 \left (i \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )-\operatorname {PolyLog}\left (3,-i e^{i \sec ^{-1}(a+b x)}\right )\right )-2 \left (i \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )-\operatorname {PolyLog}\left (3,i e^{i \sec ^{-1}(a+b x)}\right )\right )\right )}{b}\)

input

Int[ArcSec[a + b*x]^3,x]
 

output

((a + b*x)*ArcSec[a + b*x]^3 - 3*((-2*I)*ArcSec[a + b*x]^2*ArcTan[E^(I*Arc 
Sec[a + b*x])] + 2*(I*ArcSec[a + b*x]*PolyLog[2, (-I)*E^(I*ArcSec[a + b*x] 
)] - PolyLog[3, (-I)*E^(I*ArcSec[a + b*x])]) - 2*(I*ArcSec[a + b*x]*PolyLo 
g[2, I*E^(I*ArcSec[a + b*x])] - PolyLog[3, I*E^(I*ArcSec[a + b*x])])))/b
 

3.1.35.3.1 Defintions of rubi rules used

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 3011

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) 
*(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + 
b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F]))   Int[(f + g*x)^( 
m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e 
, f, g, n}, x] && GtQ[m, 0]
 

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]
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

3.1.35.4 Maple [F]

\[\int \operatorname {arcsec}\left (b x +a \right )^{3}d x\]

input

int(arcsec(b*x+a)^3,x)
 

output

int(arcsec(b*x+a)^3,x)
 

3.1.35.5 Fricas [F]

\[ \int \sec ^{-1}(a+b x)^3 \, dx=\int { \operatorname {arcsec}\left (b x + a\right )^{3} \,d x } \]

input

integrate(arcsec(b*x+a)^3,x, algorithm="fricas")
 

output

integral(arcsec(b*x + a)^3, x)
 

3.1.35.6 Sympy [F]

\[ \int \sec ^{-1}(a+b x)^3 \, dx=\int \operatorname {asec}^{3}{\left (a + b x \right )}\, dx \]

input

integrate(asec(b*x+a)**3,x)
 

output

Integral(asec(a + b*x)**3, x)
 

3.1.35.7 Maxima [F]

\[ \int \sec ^{-1}(a+b x)^3 \, dx=\int { \operatorname {arcsec}\left (b x + a\right )^{3} \,d x } \]

input

integrate(arcsec(b*x+a)^3,x, algorithm="maxima")
 

output

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

3.1.35.8 Giac [F]

\[ \int \sec ^{-1}(a+b x)^3 \, dx=\int { \operatorname {arcsec}\left (b x + a\right )^{3} \,d x } \]

input

integrate(arcsec(b*x+a)^3,x, algorithm="giac")
 

output

integrate(arcsec(b*x + a)^3, x)
 

3.1.35.9 Mupad [F(-1)]

Timed out. \[ \int \sec ^{-1}(a+b x)^3 \, dx=\int {\mathrm {acos}\left (\frac {1}{a+b\,x}\right )}^3 \,d x \]

input

int(acos(1/(a + b*x))^3,x)
 

output

int(acos(1/(a + b*x))^3, x)