\(\int x \sec ^{-1}(a+b x)^3 \, dx\) [34]

Optimal result

Integrand size = 10, antiderivative size = 278 \[ \int x \sec ^{-1}(a+b x)^3 \, dx=\frac {3 i \sec ^{-1}(a+b x)^2}{2 b^2}-\frac {3 (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{2 b^2}-\frac {a^2 \sec ^{-1}(a+b x)^3}{2 b^2}+\frac {1}{2} x^2 \sec ^{-1}(a+b x)^3-\frac {6 i a \sec ^{-1}(a+b x)^2 \arctan \left (e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac {3 \sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac {6 i a \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac {6 i a \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac {3 i \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )}{2 b^2}-\frac {6 a \operatorname {PolyLog}\left (3,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac {6 a \operatorname {PolyLog}\left (3,i e^{i \sec ^{-1}(a+b x)}\right )}{b^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 257, normalized size of antiderivative = 0.92 \[ \int x \sec ^{-1}(a+b x)^3 \, dx=\frac {\frac {3}{2} i \sec ^{-1}(a+b x)^2-\frac {3}{2} (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2-a (a+b x) \sec ^{-1}(a+b x)^3+\frac {1}{2} (a+b x)^2 \sec ^{-1}(a+b x)^3-6 i a \sec ^{-1}(a+b x)^2 \arctan \left (e^{i \sec ^{-1}(a+b x)}\right )-3 \sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )+6 i a \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )-6 i a \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )+\frac {3}{2} i \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )-6 a \operatorname {PolyLog}\left (3,-i e^{i \sec ^{-1}(a+b x)}\right )+6 a \operatorname {PolyLog}\left (3,i e^{i \sec ^{-1}(a+b x)}\right )}{b^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 253, normalized size of antiderivative = 0.91, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, 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.

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

\(\Big \downarrow \) 5781

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4926

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4678

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

((b^2*x^2*ArcSec[a + b*x]^3)/2 - (3*((-I)*ArcSec[a + b*x]^2 + (a + b*x)*Sq 
rt[1 - (a + b*x)^(-2)]*ArcSec[a + b*x]^2 + (a^2*ArcSec[a + b*x]^3)/3 + (4* 
I)*a*ArcSec[a + b*x]^2*ArcTan[E^(I*ArcSec[a + b*x])] + 2*ArcSec[a + b*x]*L 
og[1 + E^((2*I)*ArcSec[a + b*x])] - (4*I)*a*ArcSec[a + b*x]*PolyLog[2, (-I 
)*E^(I*ArcSec[a + b*x])] + (4*I)*a*ArcSec[a + b*x]*PolyLog[2, I*E^(I*ArcSe 
c[a + b*x])] - I*PolyLog[2, -E^((2*I)*ArcSec[a + b*x])] + 4*a*PolyLog[3, ( 
-I)*E^(I*ArcSec[a + b*x])] - 4*a*PolyLog[3, I*E^(I*ArcSec[a + b*x])]))/2)/ 
b^2
 

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 = 0.73 (sec) , antiderivative size = 379, normalized size of antiderivative = 1.36

method	result	size
derivativedivides	\(\frac {-\frac {\operatorname {arcsec}\left (b x +a \right )^{2} \left (2 \,\operatorname {arcsec}\left (b x +a \right ) a \left (b x +a \right )-\operatorname {arcsec}\left (b x +a \right ) \left (b x +a \right )^{2}+3 \sqrt {\frac {\left (b x +a \right )^{2}-1}{\left (b x +a \right )^{2}}}\, \left (b x +a \right )+3 i\right )}{2}+6 i \operatorname {polylog}\left (2, -i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right ) a \,\operatorname {arcsec}\left (b x +a \right )-3 \ln \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right ) a \operatorname {arcsec}\left (b x +a \right )^{2}-6 \operatorname {polylog}\left (3, -i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right ) a -6 i \operatorname {polylog}\left (2, i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right ) a \,\operatorname {arcsec}\left (b x +a \right )+3 \ln \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right ) a \operatorname {arcsec}\left (b x +a \right )^{2}+6 \operatorname {polylog}\left (3, i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right ) a +3 i \operatorname {arcsec}\left (b x +a \right )^{2}-3 \,\operatorname {arcsec}\left (b x +a \right ) \ln \left (1+{\left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}^{2}\right )+\frac {3 i \operatorname {polylog}\left (2, -{\left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}^{2}\right )}{2}}{b^{2}}\)	\(379\)
default	\(\frac {-\frac {\operatorname {arcsec}\left (b x +a \right )^{2} \left (2 \,\operatorname {arcsec}\left (b x +a \right ) a \left (b x +a \right )-\operatorname {arcsec}\left (b x +a \right ) \left (b x +a \right )^{2}+3 \sqrt {\frac {\left (b x +a \right )^{2}-1}{\left (b x +a \right )^{2}}}\, \left (b x +a \right )+3 i\right )}{2}+6 i \operatorname {polylog}\left (2, -i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right ) a \,\operatorname {arcsec}\left (b x +a \right )-3 \ln \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right ) a \operatorname {arcsec}\left (b x +a \right )^{2}-6 \operatorname {polylog}\left (3, -i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right ) a -6 i \operatorname {polylog}\left (2, i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right ) a \,\operatorname {arcsec}\left (b x +a \right )+3 \ln \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right ) a \operatorname {arcsec}\left (b x +a \right )^{2}+6 \operatorname {polylog}\left (3, i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right ) a +3 i \operatorname {arcsec}\left (b x +a \right )^{2}-3 \,\operatorname {arcsec}\left (b x +a \right ) \ln \left (1+{\left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}^{2}\right )+\frac {3 i \operatorname {polylog}\left (2, -{\left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}^{2}\right )}{2}}{b^{2}}\)	\(379\)

Input:

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

Output:

1/b^2*(-1/2*arcsec(b*x+a)^2*(2*arcsec(b*x+a)*a*(b*x+a)-arcsec(b*x+a)*(b*x+ 
a)^2+3*(((b*x+a)^2-1)/(b*x+a)^2)^(1/2)*(b*x+a)+3*I)+6*I*polylog(2,-I*(1/(b 
*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))*a*arcsec(b*x+a)-3*ln(1+I*(1/(b*x+a)+I*(1-1 
/(b*x+a)^2)^(1/2)))*a*arcsec(b*x+a)^2-6*polylog(3,-I*(1/(b*x+a)+I*(1-1/(b* 
x+a)^2)^(1/2)))*a-6*I*polylog(2,I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))*a*a 
rcsec(b*x+a)+3*ln(1-I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))*a*arcsec(b*x+a) 
^2+6*polylog(3,I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))*a+3*I*arcsec(b*x+a)^ 
2-3*arcsec(b*x+a)*ln(1+(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))^2)+3/2*I*polylo 
g(2,-(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))^2))
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

1/2*x^2*arctan(sqrt(b*x + a + 1)*sqrt(b*x + a - 1))^3 - 3/8*x^2*arctan(sqr 
t(b*x + a + 1)*sqrt(b*x + a - 1))*log(b^2*x^2 + 2*a*b*x + a^2)^2 - integra 
te(3/8*((4*b*x^2*arctan(sqrt(b*x + a + 1)*sqrt(b*x + a - 1))^2 - b*x^2*log 
(b^2*x^2 + 2*a*b*x + a^2)^2)*sqrt(b*x + a + 1)*sqrt(b*x + a - 1) + 4*(2*(b 
^3*x^4 + 3*a*b^2*x^3 + (3*a^2 - 1)*b*x^2 + (a^3 - a)*x)*log(b*x + a)^2 - ( 
b^3*x^4 + 2*a*b^2*x^3 + (a^2 - 1)*b*x^2 + 2*(b^3*x^4 + 3*a*b^2*x^3 + (3*a^ 
2 - 1)*b*x^2 + (a^3 - a)*x)*log(b*x + a))*log(b^2*x^2 + 2*a*b*x + a^2))*ar 
ctan(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)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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