Integrand size = 12, antiderivative size = 288 \[ \int x^2 \sec ^{-1}(a+b x)^2 \, dx=\frac {x}{3 b^2}+\frac {2 a (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{b^3}-\frac {(a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{3 b^3}+\frac {a^3 \sec ^{-1}(a+b x)^2}{3 b^3}+\frac {1}{3} x^3 \sec ^{-1}(a+b x)^2+\frac {2 i \sec ^{-1}(a+b x) \arctan \left (e^{i \sec ^{-1}(a+b x)}\right )}{3 b^3}+\frac {4 i a^2 \sec ^{-1}(a+b x) \arctan \left (e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {2 a \log (a+b x)}{b^3}-\frac {i \operatorname {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )}{3 b^3}-\frac {2 i a^2 \operatorname {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {i \operatorname {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{3 b^3}+\frac {2 i a^2 \operatorname {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b^3} \] Output:
1/3*x/b^2+2*a*(b*x+a)*(1-1/(b*x+a)^2)^(1/2)*arcsec(b*x+a)/b^3-1/3*(b*x+a)^ 2*(1-1/(b*x+a)^2)^(1/2)*arcsec(b*x+a)/b^3+1/3*a^3*arcsec(b*x+a)^2/b^3+1/3* x^3*arcsec(b*x+a)^2+2/3*I*arcsec(b*x+a)*arctan(1/(b*x+a)+I*(1-1/(b*x+a)^2) ^(1/2))/b^3+4*I*a^2*arcsec(b*x+a)*arctan(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2) )/b^3-2*a*ln(b*x+a)/b^3-1/3*I*polylog(2,-I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1 /2)))/b^3-2*I*a^2*polylog(2,-I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))/b^3+1/ 3*I*polylog(2,I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))/b^3+2*I*a^2*polylog(2 ,I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))/b^3
Time = 4.01 (sec) , antiderivative size = 473, normalized size of antiderivative = 1.64 \[ \int x^2 \sec ^{-1}(a+b x)^2 \, dx=\frac {4+2 \left (1+6 a^2\right ) \sec ^{-1}(a+b x)^2+\frac {\sec ^{-1}(a+b x) \left (2+(-1+6 a) \sec ^{-1}(a+b x)\right )}{-1+\sqrt {1-\frac {1}{(a+b x)^2}}}+24 a \log \left (\frac {1}{a+b x}\right )+2 \left (-1-6 a^2\right ) \left (\left (\pi -2 \sec ^{-1}(a+b x)\right ) \left (\log \left (1-i e^{-i \sec ^{-1}(a+b x)}\right )-\log \left (1+i e^{-i \sec ^{-1}(a+b x)}\right )\right )-\pi \log \left (\cot \left (\frac {1}{4} \left (\pi +2 \sec ^{-1}(a+b x)\right )\right )\right )+2 i \left (\operatorname {PolyLog}\left (2,-i e^{-i \sec ^{-1}(a+b x)}\right )-\operatorname {PolyLog}\left (2,i e^{-i \sec ^{-1}(a+b x)}\right )\right )\right )+\frac {2 \sec ^{-1}(a+b x)^2 \sin \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}{\left (\cos \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )-\sin \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )\right )^3}+\frac {2 \left (2+12 a \sec ^{-1}(a+b x)+\left (1+6 a^2\right ) \sec ^{-1}(a+b x)^2\right ) \sin \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}{\cos \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )-\sin \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}-\frac {2 \sec ^{-1}(a+b x)^2 \sin \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}{\left (\cos \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )+\sin \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )\right )^3}+\frac {\sec ^{-1}(a+b x) \left (2+(1-6 a) \sec ^{-1}(a+b x)\right )}{\left (\cos \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )+\sin \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )\right )^2}-\frac {2 \left (2-12 a \sec ^{-1}(a+b x)+\left (1+6 a^2\right ) \sec ^{-1}(a+b x)^2\right ) \sin \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}{\cos \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )+\sin \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}}{12 b^3} \] Input:
Integrate[x^2*ArcSec[a + b*x]^2,x]
Output:
(4 + 2*(1 + 6*a^2)*ArcSec[a + b*x]^2 + (ArcSec[a + b*x]*(2 + (-1 + 6*a)*Ar cSec[a + b*x]))/(-1 + Sqrt[1 - (a + b*x)^(-2)]) + 24*a*Log[(a + b*x)^(-1)] + 2*(-1 - 6*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])])) + (2*ArcSec[a + b*x]^2*Sin[ArcSec[a + b*x]/2])/(Cos[ ArcSec[a + b*x]/2] - Sin[ArcSec[a + b*x]/2])^3 + (2*(2 + 12*a*ArcSec[a + b *x] + (1 + 6*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]) - (2*ArcSec[a + b*x]^2*Sin[ArcSec[a + b*x]/2])/(Cos[ArcSec[a + b*x]/2] + Sin[ArcSec[a + b*x]/2])^3 + (ArcSec[a + b*x]*(2 + (1 - 6*a)*ArcSec[a + b*x]))/(Cos[ArcSec[a + b*x]/2] + Sin[ArcS ec[a + b*x]/2])^2 - (2*(2 - 12*a*ArcSec[a + b*x] + (1 + 6*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]))/(12*b^3)
Time = 0.54 (sec) , antiderivative size = 271, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {5781, 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.
| \(\displaystyle \int x^2 \sec ^{-1}(a+b x)^2 \, dx\) |
| \(\Big \downarrow \) 5781 |
| \(\displaystyle \frac {\int b^2 x^2 (a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2d\sec ^{-1}(a+b x)}{b^3}\) |
| \(\Big \downarrow \) 4926 |
| \(\displaystyle \frac {\frac {2}{3} \int -b^3 x^3 \sec ^{-1}(a+b x)d\sec ^{-1}(a+b x)+\frac {1}{3} b^3 x^3 \sec ^{-1}(a+b x)^2}{b^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {2}{3} \int \sec ^{-1}(a+b x) \left (a-\csc \left (\sec ^{-1}(a+b x)+\frac {\pi }{2}\right )\right )^3d\sec ^{-1}(a+b x)+\frac {1}{3} b^3 x^3 \sec ^{-1}(a+b x)^2}{b^3}\) |
| \(\Big \downarrow \) 4678 |
| \(\displaystyle \frac {\frac {2}{3} \int \left (\sec ^{-1}(a+b x) a^3-3 (a+b x) \sec ^{-1}(a+b x) a^2+3 (a+b x)^2 \sec ^{-1}(a+b x) a-(a+b x)^3 \sec ^{-1}(a+b x)\right )d\sec ^{-1}(a+b x)+\frac {1}{3} b^3 x^3 \sec ^{-1}(a+b x)^2}{b^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{3} b^3 x^3 \sec ^{-1}(a+b x)^2+\frac {2}{3} \left (\frac {1}{2} a^3 \sec ^{-1}(a+b x)^2+6 i a^2 \sec ^{-1}(a+b x) \arctan \left (e^{i \sec ^{-1}(a+b x)}\right )-3 i a^2 \operatorname {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )+3 i a^2 \operatorname {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )+i \sec ^{-1}(a+b x) \arctan \left (e^{i \sec ^{-1}(a+b x)}\right )-\frac {1}{2} i \operatorname {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )+\frac {1}{2} i \operatorname {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )+\frac {1}{2} (a+b x)+3 a \log \left (\frac {1}{a+b x}\right )+3 a (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)-\frac {1}{2} (a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)\right )}{b^3}\) |
Input:
Int[x^2*ArcSec[a + b*x]^2,x]
Output:
((b^3*x^3*ArcSec[a + b*x]^2)/3 + (2*((a + b*x)/2 + 3*a*(a + b*x)*Sqrt[1 - (a + b*x)^(-2)]*ArcSec[a + b*x] - ((a + b*x)^2*Sqrt[1 - (a + b*x)^(-2)]*Ar cSec[a + b*x])/2 + (a^3*ArcSec[a + b*x]^2)/2 + I*ArcSec[a + b*x]*ArcTan[E^ (I*ArcSec[a + b*x])] + (6*I)*a^2*ArcSec[a + b*x]*ArcTan[E^(I*ArcSec[a + b* x])] + 3*a*Log[(a + b*x)^(-1)] - (I/2)*PolyLog[2, (-I)*E^(I*ArcSec[a + b*x ])] - (3*I)*a^2*PolyLog[2, (-I)*E^(I*ArcSec[a + b*x])] + (I/2)*PolyLog[2, I*E^(I*ArcSec[a + b*x])] + (3*I)*a^2*PolyLog[2, I*E^(I*ArcSec[a + b*x])])) /3)/b^3
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 = 0.76 (sec) , antiderivative size = 498, normalized size of antiderivative = 1.73
| method | result | size |
| derivativedivides | \(\frac {\operatorname {arcsec}\left (b x +a \right )^{2} a^{2} \left (b x +a \right )-\operatorname {arcsec}\left (b x +a \right )^{2} a \left (b x +a \right )^{2}+\frac {\operatorname {arcsec}\left (b x +a \right )^{2} \left (b x +a \right )^{3}}{3}+2 \,\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 )-\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 )^{2}}{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^{2}+\frac {b x}{3}+\frac {a}{3}+\frac {\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 )}{3}-\frac {\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 )}{3}+\frac {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 )}{3}-\frac {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 )}{3}-4 \ln \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right ) a +2 \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 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^{2}+2 i \operatorname {arcsec}\left (b x +a \right ) a +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^{2} \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^{2} \operatorname {arcsec}\left (b x +a \right )}{b^{3}}\) | \(498\) |
| default | \(\frac {\operatorname {arcsec}\left (b x +a \right )^{2} a^{2} \left (b x +a \right )-\operatorname {arcsec}\left (b x +a \right )^{2} a \left (b x +a \right )^{2}+\frac {\operatorname {arcsec}\left (b x +a \right )^{2} \left (b x +a \right )^{3}}{3}+2 \,\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 )-\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 )^{2}}{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^{2}+\frac {b x}{3}+\frac {a}{3}+\frac {\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 )}{3}-\frac {\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 )}{3}+\frac {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 )}{3}-\frac {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 )}{3}-4 \ln \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right ) a +2 \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 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^{2}+2 i \operatorname {arcsec}\left (b x +a \right ) a +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^{2} \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^{2} \operatorname {arcsec}\left (b x +a \right )}{b^{3}}\) | \(498\) |
Input:
int(x^2*arcsec(b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
1/b^3*(arcsec(b*x+a)^2*a^2*(b*x+a)-arcsec(b*x+a)^2*a*(b*x+a)^2+1/3*arcsec( b*x+a)^2*(b*x+a)^3+2*arcsec(b*x+a)*(((b*x+a)^2-1)/(b*x+a)^2)^(1/2)*a*(b*x+ a)-1/3*arcsec(b*x+a)*(((b*x+a)^2-1)/(b*x+a)^2)^(1/2)*(b*x+a)^2+2*I*dilog(1 -I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))*a^2+1/3*b*x+1/3*a+1/3*arcsec(b*x+a )*ln(1+I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))-1/3*arcsec(b*x+a)*ln(1-I*(1/ (b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))+1/3*I*dilog(1-I*(1/(b*x+a)+I*(1-1/(b*x+a )^2)^(1/2)))-1/3*I*dilog(1+I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))-4*ln(1/( b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))*a+2*ln(1+(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2 ))^2)*a-2*I*dilog(1+I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))*a^2+2*I*arcsec( b*x+a)*a+2*ln(1+I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))*a^2*arcsec(b*x+a)-2 *ln(1-I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))*a^2*arcsec(b*x+a))
\[ \int x^2 \sec ^{-1}(a+b x)^2 \, dx=\int { x^{2} \operatorname {arcsec}\left (b x + a\right )^{2} \,d x } \] Input:
integrate(x^2*arcsec(b*x+a)^2,x, algorithm="fricas")
Output:
integral(x^2*arcsec(b*x + a)^2, x)
\[ \int x^2 \sec ^{-1}(a+b x)^2 \, dx=\int x^{2} \operatorname {asec}^{2}{\left (a + b x \right )}\, dx \] Input:
integrate(x**2*asec(b*x+a)**2,x)
Output:
Integral(x**2*asec(a + b*x)**2, x)
\[ \int x^2 \sec ^{-1}(a+b x)^2 \, dx=\int { x^{2} \operatorname {arcsec}\left (b x + a\right )^{2} \,d x } \] Input:
integrate(x^2*arcsec(b*x+a)^2,x, algorithm="maxima")
Output:
1/3*x^3*arctan(sqrt(b*x + a + 1)*sqrt(b*x + a - 1))^2 - 1/12*x^3*log(b^2*x ^2 + 2*a*b*x + a^2)^2 - integrate(1/3*(2*sqrt(b*x + a + 1)*sqrt(b*x + a - 1)*b*x^3*arctan(sqrt(b*x + a + 1)*sqrt(b*x + a - 1)) + 3*(b^3*x^5 + 3*a*b^ 2*x^4 + (3*a^2 - 1)*b*x^3 + (a^3 - a)*x^2)*log(b*x + a)^2 - (b^3*x^5 + 2*a *b^2*x^4 + (a^2 - 1)*b*x^3 + 3*(b^3*x^5 + 3*a*b^2*x^4 + (3*a^2 - 1)*b*x^3 + (a^3 - a)*x^2)*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)
\[ \int x^2 \sec ^{-1}(a+b x)^2 \, dx=\int { x^{2} \operatorname {arcsec}\left (b x + a\right )^{2} \,d x } \] Input:
integrate(x^2*arcsec(b*x+a)^2,x, algorithm="giac")
Output:
integrate(x^2*arcsec(b*x + a)^2, x)
Timed out. \[ \int x^2 \sec ^{-1}(a+b x)^2 \, dx=\int x^2\,{\mathrm {acos}\left (\frac {1}{a+b\,x}\right )}^2 \,d x \] Input:
int(x^2*acos(1/(a + b*x))^2,x)
Output:
int(x^2*acos(1/(a + b*x))^2, x)
\[ \int x^2 \sec ^{-1}(a+b x)^2 \, dx=\int \mathit {asec} \left (b x +a \right )^{2} x^{2}d x \] Input:
int(x^2*asec(b*x+a)^2,x)
Output:
int(asec(a + b*x)**2*x**2,x)