\(\int \text {sech}^{-1}(a+b x)^2 \, dx\) [11]

Optimal result

Integrand size = 8, antiderivative size = 80 \[ \int \text {sech}^{-1}(a+b x)^2 \, dx=\frac {(a+b x) \text {sech}^{-1}(a+b x)^2}{b}-\frac {4 \text {sech}^{-1}(a+b x) \arctan \left (e^{\text {sech}^{-1}(a+b x)}\right )}{b}+\frac {2 i \operatorname {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a+b x)}\right )}{b}-\frac {2 i \operatorname {PolyLog}\left (2,i e^{\text {sech}^{-1}(a+b x)}\right )}{b} \]

[Out]

(b*x+a)*arcsech(b*x+a)^2/b-4*arcsech(b*x+a)*arctan(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))/b+2*I*po
lylog(2,-I*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2)))/b-2*I*polylog(2,I*(1/(b*x+a)+(1/(b*x+a)-1)^(1/
2)*(1/(b*x+a)+1)^(1/2)))/b

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {6450, 6414, 5526, 4265, 2317, 2438} \[ \int \text {sech}^{-1}(a+b x)^2 \, dx=-\frac {4 \text {sech}^{-1}(a+b x) \arctan \left (e^{\text {sech}^{-1}(a+b x)}\right )}{b}+\frac {2 i \operatorname {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a+b x)}\right )}{b}-\frac {2 i \operatorname {PolyLog}\left (2,i e^{\text {sech}^{-1}(a+b x)}\right )}{b}+\frac {(a+b x) \text {sech}^{-1}(a+b x)^2}{b} \]

[In]

Int[ArcSech[a + b*x]^2,x]

[Out]

((a + b*x)*ArcSech[a + b*x]^2)/b - (4*ArcSech[a + b*x]*ArcTan[E^ArcSech[a + b*x]])/b + ((2*I)*PolyLog[2, (-I)*
E^ArcSech[a + b*x]])/b - ((2*I)*PolyLog[2, I*E^ArcSech[a + b*x]])/b

Rule 2317

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 2438

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 4265

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[-2*(c +
d*x)^m*(ArcTanh[E^((-I)*e + f*fz*x)/E^(I*k*Pi)]/(f*fz*I)), x] + (-Dist[d*(m/(f*fz*I)), Int[(c + d*x)^(m - 1)*
Log[1 - E^((-I)*e + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[d*(m/(f*fz*I)), Int[(c + d*x)^(m - 1)*Log[1 + E^((-I)*e
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 5526

Int[(x_)^(m_.)*Sech[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*Tanh[(a_.) + (b_.)*(x_)^(n_.)]^(q_.), x_Symbol] :> Simp[(-
x^(m - n + 1))*(Sech[a + b*x^n]^p/(b*n*p)), x] + Dist[(m - n + 1)/(b*n*p), Int[x^(m - n)*Sech[a + b*x^n]^p, x]
, x] /; FreeQ[{a, b, p}, x] && RationalQ[m] && IntegerQ[n] && GeQ[m - n, 0] && EqQ[q, 1]

Rule 6414

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[-c^(-1), Subst[Int[(a + b*x)^n*Sech[x]*Tanh[x]
, x], x, ArcSech[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[n, 0]

Rule 6450

Int[((a_.) + ArcSech[(c_) + (d_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcSech[x])^p, x
], x, c + d*x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \text {sech}^{-1}(x)^2 \, dx,x,a+b x\right )}{b} \\ & = -\frac {\text {Subst}\left (\int x^2 \text {sech}(x) \tanh (x) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{b} \\ & = \frac {(a+b x) \text {sech}^{-1}(a+b x)^2}{b}-\frac {2 \text {Subst}\left (\int x \text {sech}(x) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{b} \\ & = \frac {(a+b x) \text {sech}^{-1}(a+b x)^2}{b}-\frac {4 \text {sech}^{-1}(a+b x) \arctan \left (e^{\text {sech}^{-1}(a+b x)}\right )}{b}+\frac {(2 i) \text {Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{b}-\frac {(2 i) \text {Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{b} \\ & = \frac {(a+b x) \text {sech}^{-1}(a+b x)^2}{b}-\frac {4 \text {sech}^{-1}(a+b x) \arctan \left (e^{\text {sech}^{-1}(a+b x)}\right )}{b}+\frac {(2 i) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{\text {sech}^{-1}(a+b x)}\right )}{b}-\frac {(2 i) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{\text {sech}^{-1}(a+b x)}\right )}{b} \\ & = \frac {(a+b x) \text {sech}^{-1}(a+b x)^2}{b}-\frac {4 \text {sech}^{-1}(a+b x) \arctan \left (e^{\text {sech}^{-1}(a+b x)}\right )}{b}+\frac {2 i \operatorname {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a+b x)}\right )}{b}-\frac {2 i \operatorname {PolyLog}\left (2,i e^{\text {sech}^{-1}(a+b x)}\right )}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.31 \[ \int \text {sech}^{-1}(a+b x)^2 \, dx=\frac {i \left (\text {sech}^{-1}(a+b x) \left (-i (a+b x) \text {sech}^{-1}(a+b x)+2 \log \left (1-i e^{-\text {sech}^{-1}(a+b x)}\right )-2 \log \left (1+i e^{-\text {sech}^{-1}(a+b x)}\right )\right )+2 \operatorname {PolyLog}\left (2,-i e^{-\text {sech}^{-1}(a+b x)}\right )-2 \operatorname {PolyLog}\left (2,i e^{-\text {sech}^{-1}(a+b x)}\right )\right )}{b} \]

[In]

Integrate[ArcSech[a + b*x]^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 192, normalized size of antiderivative = 2.40

method	result	size
derivativedivides	\(\frac {\operatorname {arcsech}\left (b x +a \right )^{2} \left (b x +a \right )+2 i \operatorname {arcsech}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )-2 i \operatorname {arcsech}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )+2 i \operatorname {dilog}\left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )-2 i \operatorname {dilog}\left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{b}\)	\(192\)
default	\(\frac {\operatorname {arcsech}\left (b x +a \right )^{2} \left (b x +a \right )+2 i \operatorname {arcsech}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )-2 i \operatorname {arcsech}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )+2 i \operatorname {dilog}\left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )-2 i \operatorname {dilog}\left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{b}\)	\(192\)

[In]

int(arcsech(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [F]

\[ \int \text {sech}^{-1}(a+b x)^2 \, dx=\int { \operatorname {arsech}\left (b x + a\right )^{2} \,d x } \]

[In]

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

[Out]

integral(arcsech(b*x + a)^2, x)

Sympy [F]

\[ \int \text {sech}^{-1}(a+b x)^2 \, dx=\int \operatorname {asech}^{2}{\left (a + b x \right )}\, dx \]

[In]

integrate(asech(b*x+a)**2,x)

[Out]

Integral(asech(a + b*x)**2, x)

Maxima [F]

\[ \int \text {sech}^{-1}(a+b x)^2 \, dx=\int { \operatorname {arsech}\left (b x + a\right )^{2} \,d x } \]

[In]

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

[Out]

x*log(sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)*b*x + sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)*a + b*x + a)^2 - integra
te(-2*(2*(b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2*b - b)*x - a)*sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)*log(b*x + a)
^2 + 2*(b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2*b - b)*x - a)*log(b*x + a)^2 - (b^3*x^3 + 2*a*b^2*x^2 + (a^2*b -
b)*x + 2*(b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2*b - b)*x - a)*log(b*x + a) + ((b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3
*a^2*b - b)*x - a)*sqrt(b*x + a + 1)*log(b*x + a) + (2*b^3*x^3 + 4*a*b^2*x^2 + (2*a^2*b - b)*x + (b^3*x^3 + 3*
a*b^2*x^2 + a^3 + (3*a^2*b - b)*x - a)*log(b*x + a))*sqrt(b*x + a + 1))*sqrt(-b*x - a + 1))*log(sqrt(b*x + a +
1)*sqrt(-b*x - a + 1)*b*x + sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)*a + b*x + a))/(b^3*x^3 + 3*a*b^2*x^2 + a^3 +
(b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2*b - b)*x - a)*sqrt(b*x + a + 1)*sqrt(-b*x - a + 1) + (3*a^2*b - b)*x -
a), x)

Giac [F]

\[ \int \text {sech}^{-1}(a+b x)^2 \, dx=\int { \operatorname {arsech}\left (b x + a\right )^{2} \,d x } \]

[In]

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

[Out]

integrate(arcsech(b*x + a)^2, x)

Mupad [F(-1)]

Timed out. \[ \int \text {sech}^{-1}(a+b x)^2 \, dx=\int {\mathrm {acosh}\left (\frac {1}{a+b\,x}\right )}^2 \,d x \]

[In]

int(acosh(1/(a + b*x))^2,x)

[Out]

int(acosh(1/(a + b*x))^2, x)