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

Optimal. Leaf size=310 \[ -2 i \sec ^{-1}(a+b x) \text {Li}_2\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )-2 i \sec ^{-1}(a+b x) \text {Li}_2\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )+2 \text {Li}_3\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+2 \text {Li}_3\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )+\sec ^{-1}(a+b x)^2 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+\sec ^{-1}(a+b x)^2 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )+i \sec ^{-1}(a+b x) \text {Li}_2\left (-e^{2 i \sec ^{-1}(a+b x)}\right )-\frac {1}{2} \text {Li}_3\left (-e^{2 i \sec ^{-1}(a+b x)}\right )-\sec ^{-1}(a+b x)^2 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right ) \]

[Out]

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

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Rubi [A]  time = 0.49, antiderivative size = 310, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 9, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {5258, 4551, 4530, 3719, 2190, 2531, 2282, 6589, 4520} \[ -2 i \sec ^{-1}(a+b x) \text {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )-2 i \sec ^{-1}(a+b x) \text {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )+2 \text {PolyLog}\left (3,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+2 \text {PolyLog}\left (3,\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )+i \sec ^{-1}(a+b x) \text {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )-\frac {1}{2} \text {PolyLog}\left (3,-e^{2 i \sec ^{-1}(a+b x)}\right )+\sec ^{-1}(a+b x)^2 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+\sec ^{-1}(a+b x)^2 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )-\sec ^{-1}(a+b x)^2 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcSec[a + b*x]^2/x,x]

[Out]

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

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 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 3719

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*(m + 1)), x
] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*(e + f*x)))/(1 + E^(2*I*(e + f*x))), x], x] /; FreeQ[{c, d, e, f}, x] &&
IGtQ[m, 0]

Rule 4520

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)])/(Cos[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :>
Simp[(I*(e + f*x)^(m + 1))/(b*f*(m + 1)), x] + (-Dist[I, Int[((e + f*x)^m*E^(I*(c + d*x)))/(a - Rt[a^2 - b^2,
2] + b*E^(I*(c + d*x))), x], x] - Dist[I, Int[((e + f*x)^m*E^(I*(c + d*x)))/(a + Rt[a^2 - b^2, 2] + b*E^(I*(c
+ d*x))), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && PosQ[a^2 - b^2]

Rule 4530

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

Rule 4551

Int[(((e_.) + (f_.)*(x_))^(m_.)*(F_)[(c_.) + (d_.)*(x_)]^(n_.)*(G_)[(c_.) + (d_.)*(x_)]^(p_.))/((a_) + (b_.)*S
ec[(c_.) + (d_.)*(x_)]), x_Symbol] :> Int[((e + f*x)^m*Cos[c + d*x]*F[c + d*x]^n*G[c + d*x]^p)/(b + a*Cos[c +
d*x]), x] /; FreeQ[{a, b, c, d, e, f}, x] && TrigQ[F] && TrigQ[G] && IntegersQ[m, n, p]

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 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int \frac {\sec ^{-1}(a+b x)^2}{x} \, dx &=\operatorname {Subst}\left (\int \frac {x^2 \sec (x) \tan (x)}{-a+\sec (x)} \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=\operatorname {Subst}\left (\int \frac {x^2 \tan (x)}{1-a \cos (x)} \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=a \operatorname {Subst}\left (\int \frac {x^2 \sin (x)}{1-a \cos (x)} \, dx,x,\sec ^{-1}(a+b x)\right )+\operatorname {Subst}\left (\int x^2 \tan (x) \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=-\left (2 i \operatorname {Subst}\left (\int \frac {e^{2 i x} x^2}{1+e^{2 i x}} \, dx,x,\sec ^{-1}(a+b x)\right )\right )-(i a) \operatorname {Subst}\left (\int \frac {e^{i x} x^2}{1-\sqrt {1-a^2}-a e^{i x}} \, dx,x,\sec ^{-1}(a+b x)\right )-(i a) \operatorname {Subst}\left (\int \frac {e^{i x} x^2}{1+\sqrt {1-a^2}-a e^{i x}} \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=\sec ^{-1}(a+b x)^2 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+\sec ^{-1}(a+b x)^2 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\sec ^{-1}(a+b x)^2 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )-2 \operatorname {Subst}\left (\int x \log \left (1-\frac {a e^{i x}}{1-\sqrt {1-a^2}}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )-2 \operatorname {Subst}\left (\int x \log \left (1-\frac {a e^{i x}}{1+\sqrt {1-a^2}}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )+2 \operatorname {Subst}\left (\int x \log \left (1+e^{2 i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=\sec ^{-1}(a+b x)^2 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+\sec ^{-1}(a+b x)^2 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\sec ^{-1}(a+b x)^2 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )-2 i \sec ^{-1}(a+b x) \text {Li}_2\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )-2 i \sec ^{-1}(a+b x) \text {Li}_2\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )+i \sec ^{-1}(a+b x) \text {Li}_2\left (-e^{2 i \sec ^{-1}(a+b x)}\right )-i \operatorname {Subst}\left (\int \text {Li}_2\left (-e^{2 i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )+2 i \operatorname {Subst}\left (\int \text {Li}_2\left (\frac {a e^{i x}}{1-\sqrt {1-a^2}}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )+2 i \operatorname {Subst}\left (\int \text {Li}_2\left (\frac {a e^{i x}}{1+\sqrt {1-a^2}}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=\sec ^{-1}(a+b x)^2 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+\sec ^{-1}(a+b x)^2 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\sec ^{-1}(a+b x)^2 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )-2 i \sec ^{-1}(a+b x) \text {Li}_2\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )-2 i \sec ^{-1}(a+b x) \text {Li}_2\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )+i \sec ^{-1}(a+b x) \text {Li}_2\left (-e^{2 i \sec ^{-1}(a+b x)}\right )-\frac {1}{2} \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{2 i \sec ^{-1}(a+b x)}\right )+2 \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {a x}{1-\sqrt {1-a^2}}\right )}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )+2 \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {a x}{1+\sqrt {1-a^2}}\right )}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )\\ &=\sec ^{-1}(a+b x)^2 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+\sec ^{-1}(a+b x)^2 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\sec ^{-1}(a+b x)^2 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )-2 i \sec ^{-1}(a+b x) \text {Li}_2\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )-2 i \sec ^{-1}(a+b x) \text {Li}_2\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )+i \sec ^{-1}(a+b x) \text {Li}_2\left (-e^{2 i \sec ^{-1}(a+b x)}\right )+2 \text {Li}_3\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+2 \text {Li}_3\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\frac {1}{2} \text {Li}_3\left (-e^{2 i \sec ^{-1}(a+b x)}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 2.38, size = 813, normalized size = 2.62 \[ \log \left (\frac {e^{i \sec ^{-1}(a+b x)} a}{\sqrt {1-a^2}-1}+1\right ) \sec ^{-1}(a+b x)^2+\log \left (\frac {e^{i \sec ^{-1}(a+b x)} \left (\sqrt {1-a^2}-1\right )}{a}+1\right ) \sec ^{-1}(a+b x)^2+\log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right ) \sec ^{-1}(a+b x)^2+\log \left (1-\frac {\left (\sqrt {1-a^2}+1\right ) e^{i \sec ^{-1}(a+b x)}}{a}\right ) \sec ^{-1}(a+b x)^2-2 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right ) \sec ^{-1}(a+b x)^2+\log \left (\frac {2 \left (i \sqrt {1-\frac {1}{(a+b x)^2}}+\frac {1}{a+b x}\right )}{a+b x}\right ) \sec ^{-1}(a+b x)^2-\log \left (\frac {\left (\sqrt {1-a^2}-1\right ) \left (i \sqrt {1-\frac {1}{(a+b x)^2}}+\frac {1}{a+b x}\right )}{a}+1\right ) \sec ^{-1}(a+b x)^2-\log \left (1-\frac {\left (\sqrt {1-a^2}+1\right ) \left (i \sqrt {1-\frac {1}{(a+b x)^2}}+\frac {1}{a+b x}\right )}{a}\right ) \sec ^{-1}(a+b x)^2-4 \sin ^{-1}\left (\frac {\sqrt {\frac {a-1}{a}}}{\sqrt {2}}\right ) \log \left (\frac {e^{i \sec ^{-1}(a+b x)} \left (\sqrt {1-a^2}-1\right )}{a}+1\right ) \sec ^{-1}(a+b x)+4 \sin ^{-1}\left (\frac {\sqrt {\frac {a-1}{a}}}{\sqrt {2}}\right ) \log \left (1-\frac {\left (\sqrt {1-a^2}+1\right ) e^{i \sec ^{-1}(a+b x)}}{a}\right ) \sec ^{-1}(a+b x)+4 \sin ^{-1}\left (\frac {\sqrt {\frac {a-1}{a}}}{\sqrt {2}}\right ) \log \left (\frac {\left (\sqrt {1-a^2}-1\right ) \left (i \sqrt {1-\frac {1}{(a+b x)^2}}+\frac {1}{a+b x}\right )}{a}+1\right ) \sec ^{-1}(a+b x)-4 \sin ^{-1}\left (\frac {\sqrt {\frac {a-1}{a}}}{\sqrt {2}}\right ) \log \left (1-\frac {\left (\sqrt {1-a^2}+1\right ) \left (i \sqrt {1-\frac {1}{(a+b x)^2}}+\frac {1}{a+b x}\right )}{a}\right ) \sec ^{-1}(a+b x)-2 i \text {Li}_2\left (-\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}-1}\right ) \sec ^{-1}(a+b x)-2 i \text {Li}_2\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right ) \sec ^{-1}(a+b x)+i \text {Li}_2\left (-e^{2 i \sec ^{-1}(a+b x)}\right ) \sec ^{-1}(a+b x)+2 \text {Li}_3\left (-\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}-1}\right )+2 \text {Li}_3\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )-\frac {1}{2} \text {Li}_3\left (-e^{2 i \sec ^{-1}(a+b x)}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcSec[a + b*x]^2/x,x]

[Out]

ArcSec[a + b*x]^2*Log[1 + (a*E^(I*ArcSec[a + b*x]))/(-1 + Sqrt[1 - a^2])] + ArcSec[a + b*x]^2*Log[1 + ((-1 + S
qrt[1 - a^2])*E^(I*ArcSec[a + b*x]))/a] - 4*ArcSec[a + b*x]*ArcSin[Sqrt[(-1 + a)/a]/Sqrt[2]]*Log[1 + ((-1 + Sq
rt[1 - a^2])*E^(I*ArcSec[a + b*x]))/a] + ArcSec[a + b*x]^2*Log[1 - (a*E^(I*ArcSec[a + b*x]))/(1 + Sqrt[1 - a^2
])] + ArcSec[a + b*x]^2*Log[1 - ((1 + Sqrt[1 - a^2])*E^(I*ArcSec[a + b*x]))/a] + 4*ArcSec[a + b*x]*ArcSin[Sqrt
[(-1 + a)/a]/Sqrt[2]]*Log[1 - ((1 + Sqrt[1 - a^2])*E^(I*ArcSec[a + b*x]))/a] - 2*ArcSec[a + b*x]^2*Log[1 + E^(
(2*I)*ArcSec[a + b*x])] + ArcSec[a + b*x]^2*Log[(2*((a + b*x)^(-1) + I*Sqrt[1 - (a + b*x)^(-2)]))/(a + b*x)] -
ArcSec[a + b*x]^2*Log[1 + ((-1 + Sqrt[1 - a^2])*((a + b*x)^(-1) + I*Sqrt[1 - (a + b*x)^(-2)]))/a] + 4*ArcSec[
a + b*x]*ArcSin[Sqrt[(-1 + a)/a]/Sqrt[2]]*Log[1 + ((-1 + Sqrt[1 - a^2])*((a + b*x)^(-1) + I*Sqrt[1 - (a + b*x)
^(-2)]))/a] - ArcSec[a + b*x]^2*Log[1 - ((1 + Sqrt[1 - a^2])*((a + b*x)^(-1) + I*Sqrt[1 - (a + b*x)^(-2)]))/a]
- 4*ArcSec[a + b*x]*ArcSin[Sqrt[(-1 + a)/a]/Sqrt[2]]*Log[1 - ((1 + Sqrt[1 - a^2])*((a + b*x)^(-1) + I*Sqrt[1
- (a + b*x)^(-2)]))/a] - (2*I)*ArcSec[a + b*x]*PolyLog[2, -((a*E^(I*ArcSec[a + b*x]))/(-1 + Sqrt[1 - a^2]))] -
(2*I)*ArcSec[a + b*x]*PolyLog[2, (a*E^(I*ArcSec[a + b*x]))/(1 + Sqrt[1 - a^2])] + I*ArcSec[a + b*x]*PolyLog[2
, -E^((2*I)*ArcSec[a + b*x])] + 2*PolyLog[3, -((a*E^(I*ArcSec[a + b*x]))/(-1 + Sqrt[1 - a^2]))] + 2*PolyLog[3,
(a*E^(I*ArcSec[a + b*x]))/(1 + Sqrt[1 - a^2])] - PolyLog[3, -E^((2*I)*ArcSec[a + b*x])]/2

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fricas [F]  time = 3.07, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arcsec}\left (b x + a\right )^{2}}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsec(b*x + a)^2/x, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arcsec}\left (b x + a\right )^{2}}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsec(b*x + a)^2/x, x)

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maple [F]  time = 1.45, size = 0, normalized size = 0.00 \[ \int \frac {\mathrm {arcsec}\left (b x +a \right )^{2}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arcsec}\left (b x + a\right )^{2}}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsec(b*x + a)^2/x, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {acos}\left (\frac {1}{a+b\,x}\right )}^2}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acos(1/(a + b*x))^2/x,x)

[Out]

int(acos(1/(a + b*x))^2/x, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asec}^{2}{\left (a + b x \right )}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asec(a + b*x)**2/x, x)

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