\(\int x \sec ^{-1}(a+b x)^2 \, dx\) [29]

Optimal result

Integrand size = 10, antiderivative size = 154 \[ \int x \sec ^{-1}(a+b x)^2 \, dx=-\frac {(a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{b^2}-\frac {a^2 \sec ^{-1}(a+b x)^2}{2 b^2}+\frac {1}{2} x^2 \sec ^{-1}(a+b x)^2-\frac {4 i a \sec ^{-1}(a+b x) \arctan \left (e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac {\log (a+b x)}{b^2}+\frac {2 i a \operatorname {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac {2 i a \operatorname {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b^2} \] Output:

-(b*x+a)*(1-1/(b*x+a)^2)^(1/2)*arcsec(b*x+a)/b^2-1/2*a^2*arcsec(b*x+a)^2/b 
^2+1/2*x^2*arcsec(b*x+a)^2-4*I*a*arcsec(b*x+a)*arctan(1/(b*x+a)+I*(1-1/(b* 
x+a)^2)^(1/2))/b^2+ln(b*x+a)/b^2+2*I*a*polylog(2,-I*(1/(b*x+a)+I*(1-1/(b*x 
+a)^2)^(1/2)))/b^2-2*I*a*polylog(2,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 = 144, normalized size of antiderivative = 0.94 \[ \int x \sec ^{-1}(a+b x)^2 \, dx=\frac {-\left ((a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)\right )-a (a+b x) \sec ^{-1}(a+b x)^2+\frac {1}{2} (a+b x)^2 \sec ^{-1}(a+b x)^2-4 i a \sec ^{-1}(a+b x) \arctan \left (e^{i \sec ^{-1}(a+b x)}\right )+\log (a+b x)+2 i a \operatorname {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )-2 i a \operatorname {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b^2} \] Input:

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

Output:

(-((a + b*x)*Sqrt[1 - (a + b*x)^(-2)]*ArcSec[a + b*x]) - a*(a + b*x)*ArcSe 
c[a + b*x]^2 + ((a + b*x)^2*ArcSec[a + b*x]^2)/2 - (4*I)*a*ArcSec[a + b*x] 
*ArcTan[E^(I*ArcSec[a + b*x])] + Log[a + b*x] + (2*I)*a*PolyLog[2, (-I)*E^ 
(I*ArcSec[a + b*x])] - (2*I)*a*PolyLog[2, I*E^(I*ArcSec[a + b*x])])/b^2
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.95, 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.

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

\(\Big \downarrow \) 5781

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4926

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4678

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(-((a + b*x)*Sqrt[1 - (a + b*x)^(-2)]*ArcSec[a + b*x]) - (a^2*ArcSec[a + b 
*x]^2)/2 + (b^2*x^2*ArcSec[a + b*x]^2)/2 - (4*I)*a*ArcSec[a + b*x]*ArcTan[ 
E^(I*ArcSec[a + b*x])] - Log[(a + b*x)^(-1)] + (2*I)*a*PolyLog[2, (-I)*E^( 
I*ArcSec[a + b*x])] - (2*I)*a*PolyLog[2, I*E^(I*ArcSec[a + b*x])])/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 = 0.50 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.46

method	result	size
derivativedivides	\(\frac {-a \left (\operatorname {arcsec}\left (b x +a \right )^{2} \left (b x +a \right )+2 \,\operatorname {arcsec}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )-2 \,\operatorname {arcsec}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )-2 i \operatorname {dilog}\left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )+2 i \operatorname {dilog}\left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )\right )+\frac {\operatorname {arcsec}\left (b x +a \right )^{2} \left (b x +a \right )^{2}}{2}-\operatorname {arcsec}\left (b x +a \right ) \sqrt {\frac {\left (b x +a \right )^{2}-1}{\left (b x +a \right )^{2}}}\, \left (b x +a \right )-\ln \left (\frac {1}{b x +a}\right )}{b^{2}}\)	\(225\)
default	\(\frac {-a \left (\operatorname {arcsec}\left (b x +a \right )^{2} \left (b x +a \right )+2 \,\operatorname {arcsec}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )-2 \,\operatorname {arcsec}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )-2 i \operatorname {dilog}\left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )+2 i \operatorname {dilog}\left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )\right )+\frac {\operatorname {arcsec}\left (b x +a \right )^{2} \left (b x +a \right )^{2}}{2}-\operatorname {arcsec}\left (b x +a \right ) \sqrt {\frac {\left (b x +a \right )^{2}-1}{\left (b x +a \right )^{2}}}\, \left (b x +a \right )-\ln \left (\frac {1}{b x +a}\right )}{b^{2}}\)	\(225\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

1/2*x^2*arctan(sqrt(b*x + a + 1)*sqrt(b*x + a - 1))^2 - 1/8*x^2*log(b^2*x^ 
2 + 2*a*b*x + a^2)^2 - integrate(1/2*(2*sqrt(b*x + a + 1)*sqrt(b*x + a - 1 
)*b*x^2*arctan(sqrt(b*x + a + 1)*sqrt(b*x + a - 1)) + 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))/(b^3*x^3 + 3*a*b^2 
*x^2 + a^3 + (3*a^2 - 1)*b*x - a), x)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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