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

Optimal. Leaf size=200 \[ -i \text {Li}_2\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )-i \text {Li}_2\left (\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )+\sec ^{-1}(a+b x) \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+\sec ^{-1}(a+b x) \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )+\frac {1}{2} i \text {Li}_2\left (-e^{2 i \sec ^{-1}(a+b x)}\right )-\sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.31, antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {5258, 4551, 4530, 3719, 2190, 2279, 2391, 4520} \[ -i \text {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )-i \text {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )+\frac {1}{2} i \text {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )+\sec ^{-1}(a+b x) \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+\sec ^{-1}(a+b x) \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )-\sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcSec[a + b*x]/x,x]

[Out]

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

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]

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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.32, size = 284, normalized size = 1.42 \[ -i \left (\text {Li}_2\left (-\frac {\left (\sqrt {1-a^2}-1\right ) e^{i \sec ^{-1}(a+b x)}}{a}\right )+\text {Li}_2\left (\frac {\left (\sqrt {1-a^2}+1\right ) e^{i \sec ^{-1}(a+b x)}}{a}\right )\right )+\log \left (1+\frac {\left (\sqrt {1-a^2}-1\right ) e^{i \sec ^{-1}(a+b x)}}{a}\right ) \left (\sec ^{-1}(a+b x)-2 \sin ^{-1}\left (\frac {\sqrt {\frac {a-1}{a}}}{\sqrt {2}}\right )\right )+\log \left (1-\frac {\left (\sqrt {1-a^2}+1\right ) e^{i \sec ^{-1}(a+b x)}}{a}\right ) \left (\sec ^{-1}(a+b x)+2 \sin ^{-1}\left (\frac {\sqrt {\frac {a-1}{a}}}{\sqrt {2}}\right )\right )-4 i \sin ^{-1}\left (\frac {\sqrt {\frac {a-1}{a}}}{\sqrt {2}}\right ) \tan ^{-1}\left (\frac {(a+1) \tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt {1-a^2}}\right )+\frac {1}{2} i \text {Li}_2\left (-e^{2 i \sec ^{-1}(a+b x)}\right )-\sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcSec[a + b*x]/x,x]

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.97, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arcsec}\left (b x + a\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsec(b*x + a)/x, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsec(b*x + a)/x, x)

________________________________________________________________________________________

maple [A]  time = 0.84, size = 374, normalized size = 1.87 \[ -\mathrm {arcsec}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )-\mathrm {arcsec}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )+\mathrm {arcsec}\left (b x +a \right ) \ln \left (\frac {-a \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )+\sqrt {-a^{2}+1}+1}{1+\sqrt {-a^{2}+1}}\right )+\mathrm {arcsec}\left (b x +a \right ) \ln \left (\frac {a \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )+\sqrt {-a^{2}+1}-1}{-1+\sqrt {-a^{2}+1}}\right )-i \dilog \left (\frac {-a \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )+\sqrt {-a^{2}+1}+1}{1+\sqrt {-a^{2}+1}}\right )-i \dilog \left (\frac {a \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )+\sqrt {-a^{2}+1}-1}{-1+\sqrt {-a^{2}+1}}\right )+i \dilog \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )+i \dilog \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsec(b*x + a)/x, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acos(1/(a + b*x))/x,x)

[Out]

int(acos(1/(a + b*x))/x, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asec(a + b*x)/x, x)

________________________________________________________________________________________