Integrand size = 10, antiderivative size = 154 \[ \int x \sec ^{-1}(a+b x)^2 \, dx=-\frac {(a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{b^2}-\frac {a^2 \sec ^{-1}(a+b x)^2}{2 b^2}+\frac {1}{2} x^2 \sec ^{-1}(a+b x)^2-\frac {4 i a \sec ^{-1}(a+b x) \arctan \left (e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac {\log (a+b x)}{b^2}+\frac {2 i a \operatorname {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac {2 i a \operatorname {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b^2} \]
-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)*a rctan(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-(b*x+a)*arcsec(b*x+a)*(1-1/(b*x+a)^2)^(1/2)/ b^2
Time = 0.10 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.94 \[ \int x \sec ^{-1}(a+b x)^2 \, dx=\frac {-\left ((a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)\right )-a (a+b x) \sec ^{-1}(a+b x)^2+\frac {1}{2} (a+b x)^2 \sec ^{-1}(a+b x)^2-4 i a \sec ^{-1}(a+b x) \arctan \left (e^{i \sec ^{-1}(a+b x)}\right )+\log (a+b x)+2 i a \operatorname {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )-2 i a \operatorname {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b^2} \]
(-((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
Time = 0.42 (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 |
| \(\displaystyle \frac {-\frac {1}{2} a^2 \sec ^{-1}(a+b x)^2-4 i a \sec ^{-1}(a+b x) \arctan \left (e^{i \sec ^{-1}(a+b x)}\right )+\frac {1}{2} b^2 x^2 \sec ^{-1}(a+b x)^2+2 i a \operatorname {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )-2 i a \operatorname {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )-\log \left (\frac {1}{a+b x}\right )-(a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{b^2}\) |
(-((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
3.1.29.3.1 Defintions of rubi rules used
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]
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]
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]
Time = 0.95 (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 | \(\frac {-a \left (\operatorname {arcsec}\left (b x +a \right )^{2} \left (b x +a \right )+2 \,\operatorname {arcsec}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )-2 \,\operatorname {arcsec}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )-2 i \operatorname {dilog}\left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )+2 i \operatorname {dilog}\left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )\right )+\frac {\operatorname {arcsec}\left (b x +a \right )^{2} \left (b x +a \right )^{2}}{2}-\operatorname {arcsec}\left (b x +a \right ) \sqrt {\frac {\left (b x +a \right )^{2}-1}{\left (b x +a \right )^{2}}}\, \left (b x +a \right )-\ln \left (\frac {1}{b x +a}\right )}{b^{2}}\) | \(225\) |
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)))
\[ \int x \sec ^{-1}(a+b x)^2 \, dx=\int { x \operatorname {arcsec}\left (b x + a\right )^{2} \,d x } \]
\[ \int x \sec ^{-1}(a+b x)^2 \, dx=\int x \operatorname {asec}^{2}{\left (a + b x \right )}\, dx \]
\[ \int x \sec ^{-1}(a+b x)^2 \, dx=\int { x \operatorname {arcsec}\left (b x + a\right )^{2} \,d x } \]
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)
\[ \int x \sec ^{-1}(a+b x)^2 \, dx=\int { x \operatorname {arcsec}\left (b x + a\right )^{2} \,d x } \]
Timed out. \[ \int x \sec ^{-1}(a+b x)^2 \, dx=\int x\,{\mathrm {acos}\left (\frac {1}{a+b\,x}\right )}^2 \,d x \]