\(\int (a+b \sec ^{-1}(c x))^3 \, dx\) [27]

Optimal result

Integrand size = 10, antiderivative size = 158 \[ \int \left (a+b \sec ^{-1}(c x)\right )^3 \, dx=x \left (a+b \sec ^{-1}(c x)\right )^3+\frac {6 i b \left (a+b \sec ^{-1}(c x)\right )^2 \arctan \left (e^{i \sec ^{-1}(c x)}\right )}{c}-\frac {6 i b^2 \left (a+b \sec ^{-1}(c x)\right ) \operatorname {PolyLog}\left (2,-i e^{i \sec ^{-1}(c x)}\right )}{c}+\frac {6 i b^2 \left (a+b \sec ^{-1}(c x)\right ) \operatorname {PolyLog}\left (2,i e^{i \sec ^{-1}(c x)}\right )}{c}+\frac {6 b^3 \operatorname {PolyLog}\left (3,-i e^{i \sec ^{-1}(c x)}\right )}{c}-\frac {6 b^3 \operatorname {PolyLog}\left (3,i e^{i \sec ^{-1}(c x)}\right )}{c} \] Output:

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

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.83 \[ \int \left (a+b \sec ^{-1}(c x)\right )^3 \, dx=\frac {a^3 c x+3 a^2 b c x \sec ^{-1}(c x)+3 a b^2 c x \sec ^{-1}(c x)^2+b^3 c x \sec ^{-1}(c x)^3-6 a b^2 \sec ^{-1}(c x) \log \left (1-i e^{i \sec ^{-1}(c x)}\right )-3 b^3 \sec ^{-1}(c x)^2 \log \left (1-i e^{i \sec ^{-1}(c x)}\right )+6 a b^2 \sec ^{-1}(c x) \log \left (1+i e^{i \sec ^{-1}(c x)}\right )+3 b^3 \sec ^{-1}(c x)^2 \log \left (1+i e^{i \sec ^{-1}(c x)}\right )-3 a^2 b \log \left (c \left (1+\sqrt {1-\frac {1}{c^2 x^2}}\right ) x\right )-6 i b^2 \left (a+b \sec ^{-1}(c x)\right ) \operatorname {PolyLog}\left (2,-i e^{i \sec ^{-1}(c x)}\right )+6 i b^2 \left (a+b \sec ^{-1}(c x)\right ) \operatorname {PolyLog}\left (2,i e^{i \sec ^{-1}(c x)}\right )+6 b^3 \operatorname {PolyLog}\left (3,-i e^{i \sec ^{-1}(c x)}\right )-6 b^3 \operatorname {PolyLog}\left (3,i e^{i \sec ^{-1}(c x)}\right )}{c} \] Input:

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

Output:

(a^3*c*x + 3*a^2*b*c*x*ArcSec[c*x] + 3*a*b^2*c*x*ArcSec[c*x]^2 + b^3*c*x*A 
rcSec[c*x]^3 - 6*a*b^2*ArcSec[c*x]*Log[1 - I*E^(I*ArcSec[c*x])] - 3*b^3*Ar 
cSec[c*x]^2*Log[1 - I*E^(I*ArcSec[c*x])] + 6*a*b^2*ArcSec[c*x]*Log[1 + I*E 
^(I*ArcSec[c*x])] + 3*b^3*ArcSec[c*x]^2*Log[1 + I*E^(I*ArcSec[c*x])] - 3*a 
^2*b*Log[c*(1 + Sqrt[1 - 1/(c^2*x^2)])*x] - (6*I)*b^2*(a + b*ArcSec[c*x])* 
PolyLog[2, (-I)*E^(I*ArcSec[c*x])] + (6*I)*b^2*(a + b*ArcSec[c*x])*PolyLog 
[2, I*E^(I*ArcSec[c*x])] + 6*b^3*PolyLog[3, (-I)*E^(I*ArcSec[c*x])] - 6*b^ 
3*PolyLog[3, I*E^(I*ArcSec[c*x])])/c
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.94, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {5739, 4909, 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.

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

\(\Big \downarrow \) 5739

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

\(\Big \downarrow \) 4909

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4669

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

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

Output:

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

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 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 4909

Int[((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.)*Tan[(a_.) + (b 
_.)*(x_)]^(p_.), x_Symbol] :> Simp[(c + d*x)^m*(Sec[a + b*x]^n/(b*n)), x] - 
 Simp[d*(m/(b*n))   Int[(c + d*x)^(m - 1)*Sec[a + b*x]^n, x], x] /; FreeQ[{ 
a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[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 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]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 445 vs. \(2 (216 ) = 432\).

Time = 1.15 (sec) , antiderivative size = 446, normalized size of antiderivative = 2.82

method	result	size
derivativedivides	\(\frac {c x \left (b^{3} \operatorname {arcsec}\left (c x \right )^{3}+3 a \,b^{2} \operatorname {arcsec}\left (c x \right )^{2}+3 \,\operatorname {arcsec}\left (c x \right ) a^{2} b +a^{3}\right )-3 \ln \left (1-i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right ) b^{3} \operatorname {arcsec}\left (c x \right )^{2}+3 \ln \left (1+i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right ) b^{3} \operatorname {arcsec}\left (c x \right )^{2}-6 \ln \left (1-i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right ) \operatorname {arcsec}\left (c x \right ) a \,b^{2}+6 \,\operatorname {arcsec}\left (c x \right ) \ln \left (1+i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right ) a \,b^{2}-6 \operatorname {polylog}\left (3, i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right ) b^{3}+6 \operatorname {polylog}\left (3, -i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right ) b^{3}+6 i b^{3} \operatorname {arcsec}\left (c x \right ) \operatorname {polylog}\left (2, i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )-6 i b^{3} \operatorname {arcsec}\left (c x \right ) \operatorname {polylog}\left (2, -i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )+6 i \arctan \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right ) a^{2} b +6 i \operatorname {polylog}\left (2, i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right ) a \,b^{2}-6 i \operatorname {polylog}\left (2, -i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right ) a \,b^{2}}{c}\)	\(446\)
default	\(\frac {c x \left (b^{3} \operatorname {arcsec}\left (c x \right )^{3}+3 a \,b^{2} \operatorname {arcsec}\left (c x \right )^{2}+3 \,\operatorname {arcsec}\left (c x \right ) a^{2} b +a^{3}\right )-3 \ln \left (1-i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right ) b^{3} \operatorname {arcsec}\left (c x \right )^{2}+3 \ln \left (1+i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right ) b^{3} \operatorname {arcsec}\left (c x \right )^{2}-6 \ln \left (1-i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right ) \operatorname {arcsec}\left (c x \right ) a \,b^{2}+6 \,\operatorname {arcsec}\left (c x \right ) \ln \left (1+i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right ) a \,b^{2}-6 \operatorname {polylog}\left (3, i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right ) b^{3}+6 \operatorname {polylog}\left (3, -i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right ) b^{3}+6 i b^{3} \operatorname {arcsec}\left (c x \right ) \operatorname {polylog}\left (2, i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )-6 i b^{3} \operatorname {arcsec}\left (c x \right ) \operatorname {polylog}\left (2, -i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )+6 i \arctan \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right ) a^{2} b +6 i \operatorname {polylog}\left (2, i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right ) a \,b^{2}-6 i \operatorname {polylog}\left (2, -i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right ) a \,b^{2}}{c}\)	\(446\)

Input:

int((a+b*arcsec(c*x))^3,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [F]

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

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

Output:

integral(b^3*arcsec(c*x)^3 + 3*a*b^2*arcsec(c*x)^2 + 3*a^2*b*arcsec(c*x) + 
 a^3, x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

-3/2*a*b^2*c^2*(2*x/c^2 - log(c*x + 1)/c^3 + log(c*x - 1)/c^3)*log(c)^2 - 
12*b^3*c^2*integrate(1/4*x^2*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))/(c^2*x^2 
- 1), x)*log(c)^2 + b^3*x*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^3 - 3/4*b^3* 
x*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))*log(c^2*x^2)^2 + 12*b^3*c^2*integrat 
e(1/4*x^2*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))*log(c^2*x^2)/(c^2*x^2 - 1), 
x)*log(c) - 24*b^3*c^2*integrate(1/4*x^2*arctan(sqrt(c*x + 1)*sqrt(c*x - 1 
))*log(x)/(c^2*x^2 - 1), x)*log(c) + 12*a*b^2*c^2*integrate(1/4*x^2*log(c^ 
2*x^2)/(c^2*x^2 - 1), x)*log(c) - 24*a*b^2*c^2*integrate(1/4*x^2*log(x)/(c 
^2*x^2 - 1), x)*log(c) + 12*b^3*c^2*integrate(1/4*x^2*arctan(sqrt(c*x + 1) 
*sqrt(c*x - 1))*log(c^2*x^2)*log(x)/(c^2*x^2 - 1), x) - 12*b^3*c^2*integra 
te(1/4*x^2*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))*log(x)^2/(c^2*x^2 - 1), x) 
+ 12*a*b^2*c^2*integrate(1/4*x^2*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^2/(c^ 
2*x^2 - 1), x) + 12*b^3*c^2*integrate(1/4*x^2*arctan(sqrt(c*x + 1)*sqrt(c* 
x - 1))*log(c^2*x^2)/(c^2*x^2 - 1), x) - 3*a*b^2*c^2*integrate(1/4*x^2*log 
(c^2*x^2)^2/(c^2*x^2 - 1), x) + 12*a*b^2*c^2*integrate(1/4*x^2*log(c^2*x^2 
)*log(x)/(c^2*x^2 - 1), x) - 12*a*b^2*c^2*integrate(1/4*x^2*log(x)^2/(c^2* 
x^2 - 1), x) - 3/2*a*b^2*(log(c*x + 1)/c - log(c*x - 1)/c)*log(c)^2 + 12*b 
^3*integrate(1/4*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))/(c^2*x^2 - 1), x)*log 
(c)^2 - 12*b^3*integrate(1/4*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))*log(c^2*x 
^2)/(c^2*x^2 - 1), x)*log(c) + 24*b^3*integrate(1/4*arctan(sqrt(c*x + 1...
 

Giac [F(-2)]

Exception generated. \[ \int \left (a+b \sec ^{-1}(c x)\right )^3 \, dx=\text {Exception raised: RuntimeError} \] Input:

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

Output:

Exception raised: RuntimeError >> an error occurred running a Giac command 
:INPUT:sage2OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const ve 
cteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \left (a+b \sec ^{-1}(c x)\right )^3 \, dx=3 \left (\int \mathit {asec} \left (c x \right )d x \right ) a^{2} b +\left (\int \mathit {asec} \left (c x \right )^{3}d x \right ) b^{3}+3 \left (\int \mathit {asec} \left (c x \right )^{2}d x \right ) a \,b^{2}+a^{3} x \] Input:

int((a+b*asec(c*x))^3,x)
 

Output:

3*int(asec(c*x),x)*a**2*b + int(asec(c*x)**3,x)*b**3 + 3*int(asec(c*x)**2, 
x)*a*b**2 + a**3*x