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

Optimal. Leaf size=244 \[ -\frac{2 b \text{PolyLog}\left (2,\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}+\frac{2 b \text{PolyLog}\left (2,\frac{a e^{i \sec ^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )}{a \sqrt{1-a^2}}-\frac{2 i b \sec ^{-1}(a+b x) \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}+\frac{2 i b \sec ^{-1}(a+b x) \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )}{a \sqrt{1-a^2}}-\frac{b \sec ^{-1}(a+b x)^2}{a}-\frac{\sec ^{-1}(a+b x)^2}{x} \]

[Out]

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

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Rubi [A]  time = 0.394866, antiderivative size = 244, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {5258, 4426, 4191, 3321, 2264, 2190, 2279, 2391} \[ -\frac{2 b \text{PolyLog}\left (2,\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}+\frac{2 b \text{PolyLog}\left (2,\frac{a e^{i \sec ^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )}{a \sqrt{1-a^2}}-\frac{2 i b \sec ^{-1}(a+b x) \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}+\frac{2 i b \sec ^{-1}(a+b x) \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )}{a \sqrt{1-a^2}}-\frac{b \sec ^{-1}(a+b x)^2}{a}-\frac{\sec ^{-1}(a+b x)^2}{x} \]

Antiderivative was successfully verified.

[In]

Int[ArcSec[a + b*x]^2/x^2,x]

[Out]

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

Rule 5258

Int[((a_.) + ArcSec[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[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]

Rule 4426

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] - Dist[(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] && I
GtQ[m, 0] && NeQ[n, -1]

Rule 4191

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

Rule 3321

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)]), x_Symbol] :> Dist[2, Int[((c
+ d*x)^m*E^(I*Pi*(k - 1/2))*E^(I*(e + f*x)))/(b + 2*a*E^(I*Pi*(k - 1/2))*E^(I*(e + f*x)) - b*E^(2*I*k*Pi)*E^(
2*I*(e + f*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[2*k] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[m, 0]

Rule 2190

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

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \frac{\sec ^{-1}(a+b x)^2}{x^2} \, dx &=b \operatorname{Subst}\left (\int \frac{x^2 \sec (x) \tan (x)}{(-a+\sec (x))^2} \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=-\frac{\sec ^{-1}(a+b x)^2}{x}+(2 b) \operatorname{Subst}\left (\int \frac{x}{-a+\sec (x)} \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=-\frac{\sec ^{-1}(a+b x)^2}{x}+(2 b) \operatorname{Subst}\left (\int \left (-\frac{x}{a}+\frac{x}{a (1-a \cos (x))}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=-\frac{b \sec ^{-1}(a+b x)^2}{a}-\frac{\sec ^{-1}(a+b x)^2}{x}+\frac{(2 b) \operatorname{Subst}\left (\int \frac{x}{1-a \cos (x)} \, dx,x,\sec ^{-1}(a+b x)\right )}{a}\\ &=-\frac{b \sec ^{-1}(a+b x)^2}{a}-\frac{\sec ^{-1}(a+b x)^2}{x}+\frac{(4 b) \operatorname{Subst}\left (\int \frac{e^{i x} x}{-a+2 e^{i x}-a e^{2 i x}} \, dx,x,\sec ^{-1}(a+b x)\right )}{a}\\ &=-\frac{b \sec ^{-1}(a+b x)^2}{a}-\frac{\sec ^{-1}(a+b x)^2}{x}-\frac{(4 b) \operatorname{Subst}\left (\int \frac{e^{i x} x}{2-2 \sqrt{1-a^2}-2 a e^{i x}} \, dx,x,\sec ^{-1}(a+b x)\right )}{\sqrt{1-a^2}}+\frac{(4 b) \operatorname{Subst}\left (\int \frac{e^{i x} x}{2+2 \sqrt{1-a^2}-2 a e^{i x}} \, dx,x,\sec ^{-1}(a+b x)\right )}{\sqrt{1-a^2}}\\ &=-\frac{b \sec ^{-1}(a+b x)^2}{a}-\frac{\sec ^{-1}(a+b x)^2}{x}-\frac{2 i b \sec ^{-1}(a+b x) \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}+\frac{2 i b \sec ^{-1}(a+b x) \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}+\frac{(2 i b) \operatorname{Subst}\left (\int \log \left (1-\frac{2 a e^{i x}}{2-2 \sqrt{1-a^2}}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{a \sqrt{1-a^2}}-\frac{(2 i b) \operatorname{Subst}\left (\int \log \left (1-\frac{2 a e^{i x}}{2+2 \sqrt{1-a^2}}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{a \sqrt{1-a^2}}\\ &=-\frac{b \sec ^{-1}(a+b x)^2}{a}-\frac{\sec ^{-1}(a+b x)^2}{x}-\frac{2 i b \sec ^{-1}(a+b x) \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}+\frac{2 i b \sec ^{-1}(a+b x) \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}+\frac{(2 b) \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{2 a x}{2-2 \sqrt{1-a^2}}\right )}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )}{a \sqrt{1-a^2}}-\frac{(2 b) \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{2 a x}{2+2 \sqrt{1-a^2}}\right )}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )}{a \sqrt{1-a^2}}\\ &=-\frac{b \sec ^{-1}(a+b x)^2}{a}-\frac{\sec ^{-1}(a+b x)^2}{x}-\frac{2 i b \sec ^{-1}(a+b x) \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}+\frac{2 i b \sec ^{-1}(a+b x) \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}-\frac{2 b \text{Li}_2\left (\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}+\frac{2 b \text{Li}_2\left (\frac{a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}\\ \end{align*}

Mathematica [B]  time = 2.22178, size = 686, normalized size = 2.81 \[ -\frac{\frac{(a+b x) \sec ^{-1}(a+b x)^2}{x}+\frac{2 b \left (i \left (\text{PolyLog}\left (2,\frac{\left (1+i \sqrt{a^2-1}\right ) \left (\sqrt{a^2-1} \tan \left (\frac{1}{2} \sec ^{-1}(a+b x)\right )-a+1\right )}{a \left (\sqrt{a^2-1} \tan \left (\frac{1}{2} \sec ^{-1}(a+b x)\right )+a-1\right )}\right )-\text{PolyLog}\left (2,\frac{\left (1-i \sqrt{a^2-1}\right ) \left (\sqrt{a^2-1} \tan \left (\frac{1}{2} \sec ^{-1}(a+b x)\right )-a+1\right )}{a \left (\sqrt{a^2-1} \tan \left (\frac{1}{2} \sec ^{-1}(a+b x)\right )+a-1\right )}\right )\right )+2 \sec ^{-1}(a+b x) \tanh ^{-1}\left (\frac{(a-1) \cot \left (\frac{1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt{a^2-1}}\right )-2 \cos ^{-1}\left (\frac{1}{a}\right ) \tanh ^{-1}\left (\frac{(a+1) \tan \left (\frac{1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt{a^2-1}}\right )-\log \left (\frac{(a-1) \left (\sqrt{a^2-1}+i a+i\right ) \left (\tan \left (\frac{1}{2} \sec ^{-1}(a+b x)\right )-i\right )}{a \left (\sqrt{a^2-1} \tan \left (\frac{1}{2} \sec ^{-1}(a+b x)\right )+a-1\right )}\right ) \left (\cos ^{-1}\left (\frac{1}{a}\right )-2 i \tanh ^{-1}\left (\frac{(a+1) \tan \left (\frac{1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt{a^2-1}}\right )\right )-\log \left (\frac{(a-1) \left (\sqrt{a^2-1}-i a-i\right ) \left (\tan \left (\frac{1}{2} \sec ^{-1}(a+b x)\right )+i\right )}{a \left (\sqrt{a^2-1} \tan \left (\frac{1}{2} \sec ^{-1}(a+b x)\right )+a-1\right )}\right ) \left (\cos ^{-1}\left (\frac{1}{a}\right )+2 i \tanh ^{-1}\left (\frac{(a+1) \tan \left (\frac{1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt{a^2-1}}\right )\right )+\log \left (\frac{\sqrt{a^2-1} e^{-\frac{1}{2} i \sec ^{-1}(a+b x)}}{\sqrt{2} \sqrt{a} \sqrt{-\frac{b x}{a+b x}}}\right ) \left (-2 i \tanh ^{-1}\left (\frac{(a-1) \cot \left (\frac{1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt{a^2-1}}\right )+2 i \tanh ^{-1}\left (\frac{(a+1) \tan \left (\frac{1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt{a^2-1}}\right )+\cos ^{-1}\left (\frac{1}{a}\right )\right )+\log \left (\frac{\sqrt{a^2-1} e^{\frac{1}{2} i \sec ^{-1}(a+b x)}}{\sqrt{2} \sqrt{a} \sqrt{-\frac{b x}{a+b x}}}\right ) \left (\cos ^{-1}\left (\frac{1}{a}\right )+2 i \left (\tanh ^{-1}\left (\frac{(a-1) \cot \left (\frac{1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt{a^2-1}}\right )-\tanh ^{-1}\left (\frac{(a+1) \tan \left (\frac{1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt{a^2-1}}\right )\right )\right )\right )}{\sqrt{a^2-1}}}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSec[a + b*x]^2/x^2,x]

[Out]

-((((a + b*x)*ArcSec[a + b*x]^2)/x + (2*b*(2*ArcSec[a + b*x]*ArcTanh[((-1 + a)*Cot[ArcSec[a + b*x]/2])/Sqrt[-1
+ a^2]] - 2*ArcCos[a^(-1)]*ArcTanh[((1 + a)*Tan[ArcSec[a + b*x]/2])/Sqrt[-1 + a^2]] + (ArcCos[a^(-1)] - (2*I)
*ArcTanh[((-1 + a)*Cot[ArcSec[a + b*x]/2])/Sqrt[-1 + a^2]] + (2*I)*ArcTanh[((1 + a)*Tan[ArcSec[a + b*x]/2])/Sq
rt[-1 + a^2]])*Log[Sqrt[-1 + a^2]/(Sqrt[2]*Sqrt[a]*E^((I/2)*ArcSec[a + b*x])*Sqrt[-((b*x)/(a + b*x))])] + (Arc
Cos[a^(-1)] + (2*I)*(ArcTanh[((-1 + a)*Cot[ArcSec[a + b*x]/2])/Sqrt[-1 + a^2]] - ArcTanh[((1 + a)*Tan[ArcSec[a
+ b*x]/2])/Sqrt[-1 + a^2]]))*Log[(Sqrt[-1 + a^2]*E^((I/2)*ArcSec[a + b*x]))/(Sqrt[2]*Sqrt[a]*Sqrt[-((b*x)/(a
+ b*x))])] - (ArcCos[a^(-1)] - (2*I)*ArcTanh[((1 + a)*Tan[ArcSec[a + b*x]/2])/Sqrt[-1 + a^2]])*Log[((-1 + a)*(
I + I*a + Sqrt[-1 + a^2])*(-I + Tan[ArcSec[a + b*x]/2]))/(a*(-1 + a + Sqrt[-1 + a^2]*Tan[ArcSec[a + b*x]/2]))]
- (ArcCos[a^(-1)] + (2*I)*ArcTanh[((1 + a)*Tan[ArcSec[a + b*x]/2])/Sqrt[-1 + a^2]])*Log[((-1 + a)*(-I - I*a +
Sqrt[-1 + a^2])*(I + Tan[ArcSec[a + b*x]/2]))/(a*(-1 + a + Sqrt[-1 + a^2]*Tan[ArcSec[a + b*x]/2]))] + I*(-Pol
yLog[2, ((1 - I*Sqrt[-1 + a^2])*(1 - a + Sqrt[-1 + a^2]*Tan[ArcSec[a + b*x]/2]))/(a*(-1 + a + Sqrt[-1 + a^2]*T
an[ArcSec[a + b*x]/2]))] + PolyLog[2, ((1 + I*Sqrt[-1 + a^2])*(1 - a + Sqrt[-1 + a^2]*Tan[ArcSec[a + b*x]/2]))
/(a*(-1 + a + Sqrt[-1 + a^2]*Tan[ArcSec[a + b*x]/2]))])))/Sqrt[-1 + a^2])/a)

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Maple [A]  time = 0.404, size = 341, normalized size = 1.4 \begin{align*} -{\frac{b \left ({\rm arcsec} \left (bx+a\right ) \right ) ^{2}}{a}}-{\frac{ \left ({\rm arcsec} \left (bx+a\right ) \right ) ^{2}}{x}}-{\frac{2\,ib{\rm arcsec} \left (bx+a\right )}{a \left ({a}^{2}-1 \right ) }\sqrt{-{a}^{2}+1}\ln \left ({ \left ( -a \left ( \left ( bx+a \right ) ^{-1}+i\sqrt{1- \left ( bx+a \right ) ^{-2}} \right ) +\sqrt{-{a}^{2}+1}+1 \right ) \left ( 1+\sqrt{-{a}^{2}+1} \right ) ^{-1}} \right ) }+{\frac{2\,ib{\rm arcsec} \left (bx+a\right )}{a \left ({a}^{2}-1 \right ) }\sqrt{-{a}^{2}+1}\ln \left ({ \left ( a \left ( \left ( bx+a \right ) ^{-1}+i\sqrt{1- \left ( bx+a \right ) ^{-2}} \right ) +\sqrt{-{a}^{2}+1}-1 \right ) \left ( -1+\sqrt{-{a}^{2}+1} \right ) ^{-1}} \right ) }-2\,{\frac{b\sqrt{-{a}^{2}+1}}{a \left ({a}^{2}-1 \right ) }{\it dilog} \left ({\frac{1}{1+\sqrt{-{a}^{2}+1}} \left ( -a \left ( \left ( bx+a \right ) ^{-1}+i\sqrt{1- \left ( bx+a \right ) ^{-2}} \right ) +\sqrt{-{a}^{2}+1}+1 \right ) } \right ) }+2\,{\frac{b\sqrt{-{a}^{2}+1}}{a \left ({a}^{2}-1 \right ) }{\it dilog} \left ({\frac{1}{-1+\sqrt{-{a}^{2}+1}} \left ( a \left ( \left ( bx+a \right ) ^{-1}+i\sqrt{1- \left ( bx+a \right ) ^{-2}} \right ) +\sqrt{-{a}^{2}+1}-1 \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\operatorname{arcsec}\left (b x + a\right )^{2}}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsec(b*x + a)^2/x^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{asec}^{2}{\left (a + b x \right )}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asec(a + b*x)**2/x**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arcsec}\left (b x + a\right )^{2}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsec(b*x + a)^2/x^2, x)