\(\int x^3 \sec ^{-1}(a+b x)^2 \, dx\) [27]

Optimal result

Integrand size = 12, antiderivative size = 381 \[ \int x^3 \sec ^{-1}(a+b x)^2 \, dx=-\frac {a x}{b^3}+\frac {(a+b x)^2}{12 b^4}-\frac {(a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{3 b^4}-\frac {3 a^2 (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{b^4}+\frac {a (a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{b^4}-\frac {(a+b x)^3 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{6 b^4}-\frac {a^4 \sec ^{-1}(a+b x)^2}{4 b^4}+\frac {1}{4} x^4 \sec ^{-1}(a+b x)^2-\frac {2 i a \sec ^{-1}(a+b x) \arctan \left (e^{i \sec ^{-1}(a+b x)}\right )}{b^4}-\frac {4 i a^3 \sec ^{-1}(a+b x) \arctan \left (e^{i \sec ^{-1}(a+b x)}\right )}{b^4}+\frac {\log (a+b x)}{3 b^4}+\frac {3 a^2 \log (a+b x)}{b^4}+\frac {i a \operatorname {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^4}+\frac {2 i a^3 \operatorname {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^4}-\frac {i a \operatorname {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b^4}-\frac {2 i a^3 \operatorname {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b^4} \] Output:

-a*x/b^3+1/12*(b*x+a)^2/b^4-1/3*(b*x+a)*(1-1/(b*x+a)^2)^(1/2)*arcsec(b*x+a 
)/b^4-3*a^2*(b*x+a)*(1-1/(b*x+a)^2)^(1/2)*arcsec(b*x+a)/b^4+a*(b*x+a)^2*(1 
-1/(b*x+a)^2)^(1/2)*arcsec(b*x+a)/b^4-1/6*(b*x+a)^3*(1-1/(b*x+a)^2)^(1/2)* 
arcsec(b*x+a)/b^4-1/4*a^4*arcsec(b*x+a)^2/b^4+1/4*x^4*arcsec(b*x+a)^2-2*I* 
a*arcsec(b*x+a)*arctan(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))/b^4+I*a*polylog( 
2,-I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))/b^4+1/3*ln(b*x+a)/b^4+3*a^2*ln(b 
*x+a)/b^4+2*I*a^3*polylog(2,-I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))/b^4-I* 
a*polylog(2,I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))/b^4-2*I*a^3*polylog(2,I 
*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))/b^4-4*I*a^3*arcsec(b*x+a)*arctan(1/( 
b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))/b^4
 

Mathematica [A] (warning: unable to verify)

Time = 8.17 (sec) , antiderivative size = 667, normalized size of antiderivative = 1.75 \[ \int x^3 \sec ^{-1}(a+b x)^2 \, dx =\text {Too large to display} \] Input:

Integrate[x^3*ArcSec[a + b*x]^2,x]
 

Output:

((1 - a/(a + b*x))^3*(24*a*(2 + (1 + 2*a^2)*ArcSec[a + b*x]^2) + (2 + (-2 
+ 24*a)*ArcSec[a + b*x] + 3*(1 - 4*a + 12*a^2)*ArcSec[a + b*x]^2)/(-1 + Sq 
rt[1 - (a + b*x)^(-2)]) + 16*(1 + 9*a^2)*Log[(a + b*x)^(-1)] - 24*a*(1 + 2 
*a^2)*((Pi - 2*ArcSec[a + b*x])*(Log[1 - I/E^(I*ArcSec[a + b*x])] - Log[1 
+ I/E^(I*ArcSec[a + b*x])]) - Pi*Log[Cot[(Pi + 2*ArcSec[a + b*x])/4]] + (2 
*I)*(PolyLog[2, (-I)/E^(I*ArcSec[a + b*x])] - PolyLog[2, I/E^(I*ArcSec[a + 
 b*x])])) - (3*ArcSec[a + b*x]^2)/(Cos[ArcSec[a + b*x]/2] - Sin[ArcSec[a + 
 b*x]/2])^4 + (4*ArcSec[a + b*x]*(1 + 6*a*ArcSec[a + b*x])*Sin[ArcSec[a + 
b*x]/2])/(Cos[ArcSec[a + b*x]/2] - Sin[ArcSec[a + b*x]/2])^3 + (8*(2*ArcSe 
c[a + b*x] + 18*a^2*ArcSec[a + b*x] + 6*a^3*ArcSec[a + b*x]^2 + 3*a*(2 + A 
rcSec[a + b*x]^2))*Sin[ArcSec[a + b*x]/2])/(Cos[ArcSec[a + b*x]/2] - Sin[A 
rcSec[a + b*x]/2]) - (3*ArcSec[a + b*x]^2)/(Cos[ArcSec[a + b*x]/2] + Sin[A 
rcSec[a + b*x]/2])^4 + (4*ArcSec[a + b*x]*(1 - 6*a*ArcSec[a + b*x])*Sin[Ar 
cSec[a + b*x]/2])/(Cos[ArcSec[a + b*x]/2] + Sin[ArcSec[a + b*x]/2])^3 - (2 
 + (2 - 24*a)*ArcSec[a + b*x] + 3*(1 - 4*a + 12*a^2)*ArcSec[a + b*x]^2)/(C 
os[ArcSec[a + b*x]/2] + Sin[ArcSec[a + b*x]/2])^2 - (8*(-2*ArcSec[a + b*x] 
 - 18*a^2*ArcSec[a + b*x] + 6*a^3*ArcSec[a + b*x]^2 + 3*a*(2 + ArcSec[a + 
b*x]^2))*Sin[ArcSec[a + b*x]/2])/(Cos[ArcSec[a + b*x]/2] + Sin[ArcSec[a + 
b*x]/2])))/(48*b^4*(-1 + a/(a + b*x))^3)
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 357, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5781, 25, 4926, 3042, 4678, 2009}

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.

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

\(\Big \downarrow \) 5781

\(\displaystyle \frac {\int b^3 x^3 (a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2d\sec ^{-1}(a+b x)}{b^4}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\int -b^3 x^3 (a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2d\sec ^{-1}(a+b x)}{b^4}\)

\(\Big \downarrow \) 4926

\(\displaystyle \frac {\frac {1}{4} b^4 x^4 \sec ^{-1}(a+b x)^2-\frac {1}{2} \int b^4 x^4 \sec ^{-1}(a+b x)d\sec ^{-1}(a+b x)}{b^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{4} b^4 x^4 \sec ^{-1}(a+b x)^2-\frac {1}{2} \int \sec ^{-1}(a+b x) \left (a-\csc \left (\sec ^{-1}(a+b x)+\frac {\pi }{2}\right )\right )^4d\sec ^{-1}(a+b x)}{b^4}\)

\(\Big \downarrow \) 4678

\(\displaystyle \frac {\frac {1}{4} b^4 x^4 \sec ^{-1}(a+b x)^2-\frac {1}{2} \int \left (\sec ^{-1}(a+b x) a^4-4 (a+b x) \sec ^{-1}(a+b x) a^3+6 (a+b x)^2 \sec ^{-1}(a+b x) a^2-4 (a+b x)^3 \sec ^{-1}(a+b x) a+(a+b x)^4 \sec ^{-1}(a+b x)\right )d\sec ^{-1}(a+b x)}{b^4}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {1}{4} b^4 x^4 \sec ^{-1}(a+b x)^2+\frac {1}{2} \left (-\frac {1}{2} a^4 \sec ^{-1}(a+b x)^2-8 i a^3 \sec ^{-1}(a+b x) \arctan \left (e^{i \sec ^{-1}(a+b x)}\right )+4 i a^3 \operatorname {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )-4 i a^3 \operatorname {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )-6 a^2 \log \left (\frac {1}{a+b x}\right )-6 a^2 (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)-4 i a \sec ^{-1}(a+b x) \arctan \left (e^{i \sec ^{-1}(a+b x)}\right )+2 i a \operatorname {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )-2 i a \operatorname {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )-2 a (a+b x)+\frac {1}{6} (a+b x)^2-\frac {2}{3} \log \left (\frac {1}{a+b x}\right )+2 a (a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)-\frac {1}{3} (a+b x)^3 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)-\frac {2}{3} (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)\right )}{b^4}\)

Input:

Int[x^3*ArcSec[a + b*x]^2,x]
 

Output:

((b^4*x^4*ArcSec[a + b*x]^2)/4 + (-2*a*(a + b*x) + (a + b*x)^2/6 - (2*(a + 
 b*x)*Sqrt[1 - (a + b*x)^(-2)]*ArcSec[a + b*x])/3 - 6*a^2*(a + b*x)*Sqrt[1 
 - (a + b*x)^(-2)]*ArcSec[a + b*x] + 2*a*(a + b*x)^2*Sqrt[1 - (a + b*x)^(- 
2)]*ArcSec[a + b*x] - ((a + b*x)^3*Sqrt[1 - (a + b*x)^(-2)]*ArcSec[a + b*x 
])/3 - (a^4*ArcSec[a + b*x]^2)/2 - (4*I)*a*ArcSec[a + b*x]*ArcTan[E^(I*Arc 
Sec[a + b*x])] - (8*I)*a^3*ArcSec[a + b*x]*ArcTan[E^(I*ArcSec[a + b*x])] - 
 (2*Log[(a + b*x)^(-1)])/3 - 6*a^2*Log[(a + b*x)^(-1)] + (2*I)*a*PolyLog[2 
, (-I)*E^(I*ArcSec[a + b*x])] + (4*I)*a^3*PolyLog[2, (-I)*E^(I*ArcSec[a + 
b*x])] - (2*I)*a*PolyLog[2, I*E^(I*ArcSec[a + b*x])] - (4*I)*a^3*PolyLog[2 
, I*E^(I*ArcSec[a + b*x])])/2)/b^4
 

Defintions of rubi rules used

rule 25

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

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4678

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

rule 4926

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

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 = 0.85 (sec) , antiderivative size = 673, normalized size of antiderivative = 1.77

method	result	size
derivativedivides	\(\frac {-\operatorname {arcsec}\left (b x +a \right )^{2} a^{3} \left (b x +a \right )+\frac {3 \operatorname {arcsec}\left (b x +a \right )^{2} a^{2} \left (b x +a \right )^{2}}{2}-\operatorname {arcsec}\left (b x +a \right )^{2} a \left (b x +a \right )^{3}+\frac {\operatorname {arcsec}\left (b x +a \right )^{2} \left (b x +a \right )^{4}}{4}-3 \,\operatorname {arcsec}\left (b x +a \right ) \sqrt {\frac {\left (b x +a \right )^{2}-1}{\left (b x +a \right )^{2}}}\, a^{2} \left (b x +a \right )+\operatorname {arcsec}\left (b x +a \right ) \sqrt {\frac {\left (b x +a \right )^{2}-1}{\left (b x +a \right )^{2}}}\, a \left (b x +a \right )^{2}-\frac {\operatorname {arcsec}\left (b x +a \right ) \sqrt {\frac {\left (b x +a \right )^{2}-1}{\left (b x +a \right )^{2}}}\, \left (b x +a \right )^{3}}{6}+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 ) a -\frac {\operatorname {arcsec}\left (b x +a \right ) \sqrt {\frac {\left (b x +a \right )^{2}-1}{\left (b x +a \right )^{2}}}\, \left (b x +a \right )}{3}-3 i a^{2} \operatorname {arcsec}\left (b x +a \right )-\left (b x +a \right ) a +\frac {\left (b x +a \right )^{2}}{12}+\frac {2 \ln \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{3}-\frac {\ln \left (1+{\left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}^{2}\right )}{3}+6 \ln \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right ) a^{2}-3 \ln \left (1+{\left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}^{2}\right ) a^{2}-2 \ln \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right ) a^{3} \operatorname {arcsec}\left (b x +a \right )+2 \ln \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right ) a^{3} \operatorname {arcsec}\left (b x +a \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 ) a^{3}-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 ) a -\ln \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right ) a \,\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 ) a \,\operatorname {arcsec}\left (b x +a \right )-\frac {i \operatorname {arcsec}\left (b x +a \right )}{3}-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 ) a^{3}}{b^{4}}\)	\(673\)
default	\(\frac {-\operatorname {arcsec}\left (b x +a \right )^{2} a^{3} \left (b x +a \right )+\frac {3 \operatorname {arcsec}\left (b x +a \right )^{2} a^{2} \left (b x +a \right )^{2}}{2}-\operatorname {arcsec}\left (b x +a \right )^{2} a \left (b x +a \right )^{3}+\frac {\operatorname {arcsec}\left (b x +a \right )^{2} \left (b x +a \right )^{4}}{4}-3 \,\operatorname {arcsec}\left (b x +a \right ) \sqrt {\frac {\left (b x +a \right )^{2}-1}{\left (b x +a \right )^{2}}}\, a^{2} \left (b x +a \right )+\operatorname {arcsec}\left (b x +a \right ) \sqrt {\frac {\left (b x +a \right )^{2}-1}{\left (b x +a \right )^{2}}}\, a \left (b x +a \right )^{2}-\frac {\operatorname {arcsec}\left (b x +a \right ) \sqrt {\frac {\left (b x +a \right )^{2}-1}{\left (b x +a \right )^{2}}}\, \left (b x +a \right )^{3}}{6}+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 ) a -\frac {\operatorname {arcsec}\left (b x +a \right ) \sqrt {\frac {\left (b x +a \right )^{2}-1}{\left (b x +a \right )^{2}}}\, \left (b x +a \right )}{3}-3 i a^{2} \operatorname {arcsec}\left (b x +a \right )-\left (b x +a \right ) a +\frac {\left (b x +a \right )^{2}}{12}+\frac {2 \ln \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{3}-\frac {\ln \left (1+{\left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}^{2}\right )}{3}+6 \ln \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right ) a^{2}-3 \ln \left (1+{\left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}^{2}\right ) a^{2}-2 \ln \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right ) a^{3} \operatorname {arcsec}\left (b x +a \right )+2 \ln \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right ) a^{3} \operatorname {arcsec}\left (b x +a \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 ) a^{3}-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 ) a -\ln \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right ) a \,\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 ) a \,\operatorname {arcsec}\left (b x +a \right )-\frac {i \operatorname {arcsec}\left (b x +a \right )}{3}-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 ) a^{3}}{b^{4}}\)	\(673\)

Input:

int(x^3*arcsec(b*x+a)^2,x,method=_RETURNVERBOSE)
 

Output:

1/b^4*(-arcsec(b*x+a)^2*a^3*(b*x+a)+3/2*arcsec(b*x+a)^2*a^2*(b*x+a)^2-arcs 
ec(b*x+a)^2*a*(b*x+a)^3+1/4*arcsec(b*x+a)^2*(b*x+a)^4-3*arcsec(b*x+a)*(((b 
*x+a)^2-1)/(b*x+a)^2)^(1/2)*a^2*(b*x+a)+arcsec(b*x+a)*(((b*x+a)^2-1)/(b*x+ 
a)^2)^(1/2)*a*(b*x+a)^2-1/6*arcsec(b*x+a)*(((b*x+a)^2-1)/(b*x+a)^2)^(1/2)* 
(b*x+a)^3+I*dilog(1+I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))*a-1/3*arcsec(b* 
x+a)*(((b*x+a)^2-1)/(b*x+a)^2)^(1/2)*(b*x+a)-3*I*a^2*arcsec(b*x+a)-(b*x+a) 
*a+1/12*(b*x+a)^2+2/3*ln(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))-1/3*ln(1+(1/(b 
*x+a)+I*(1-1/(b*x+a)^2)^(1/2))^2)+6*ln(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))* 
a^2-3*ln(1+(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))^2)*a^2-2*ln(1+I*(1/(b*x+a)+ 
I*(1-1/(b*x+a)^2)^(1/2)))*a^3*arcsec(b*x+a)+2*ln(1-I*(1/(b*x+a)+I*(1-1/(b* 
x+a)^2)^(1/2)))*a^3*arcsec(b*x+a)+2*I*dilog(1+I*(1/(b*x+a)+I*(1-1/(b*x+a)^ 
2)^(1/2)))*a^3-I*dilog(1-I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))*a-ln(1+I*( 
1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))*a*arcsec(b*x+a)+ln(1-I*(1/(b*x+a)+I*(1 
-1/(b*x+a)^2)^(1/2)))*a*arcsec(b*x+a)-1/3*I*arcsec(b*x+a)-2*I*dilog(1-I*(1 
/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))*a^3)
 

Fricas [F]

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

integrate(x^3*arcsec(b*x+a)^2,x, algorithm="fricas")
                                                                                    
                                                                                    
 

Output:

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

Sympy [F]

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

integrate(x**3*asec(b*x+a)**2,x)
 

Output:

Integral(x**3*asec(a + b*x)**2, x)
 

Maxima [F]

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

integrate(x^3*arcsec(b*x+a)^2,x, algorithm="maxima")
 

Output:

1/4*x^4*arctan(sqrt(b*x + a + 1)*sqrt(b*x + a - 1))^2 - 1/16*x^4*log(b^2*x 
^2 + 2*a*b*x + a^2)^2 - integrate(1/4*(2*sqrt(b*x + a + 1)*sqrt(b*x + a - 
1)*b*x^4*arctan(sqrt(b*x + a + 1)*sqrt(b*x + a - 1)) + 4*(b^3*x^6 + 3*a*b^ 
2*x^5 + (3*a^2 - 1)*b*x^4 + (a^3 - a)*x^3)*log(b*x + a)^2 - (b^3*x^6 + 2*a 
*b^2*x^5 + (a^2 - 1)*b*x^4 + 4*(b^3*x^6 + 3*a*b^2*x^5 + (3*a^2 - 1)*b*x^4 
+ (a^3 - a)*x^3)*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(-2)]

Exception generated. \[ \int x^3 \sec ^{-1}(a+b x)^2 \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(x^3*arcsec(b*x+a)^2,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. \[ \int x^3 \sec ^{-1}(a+b x)^2 \, dx=\int x^3\,{\mathrm {acos}\left (\frac {1}{a+b\,x}\right )}^2 \,d x \] Input:

int(x^3*acos(1/(a + b*x))^2,x)
 

Output:

int(x^3*acos(1/(a + b*x))^2, x)
 

Reduce [F]

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

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

Output:

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