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

Optimal. Leaf size=80 \[ \frac {2 i \text {Li}_2\left (-i e^{\text {sech}^{-1}(a+b x)}\right )}{b}-\frac {2 i \text {Li}_2\left (i e^{\text {sech}^{-1}(a+b x)}\right )}{b}+\frac {(a+b x) \text {sech}^{-1}(a+b x)^2}{b}-\frac {4 \text {sech}^{-1}(a+b x) \tan ^{-1}\left (e^{\text {sech}^{-1}(a+b x)}\right )}{b} \]

[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

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Rubi [A]  time = 0.07, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {6315, 6279, 5418, 4180, 2279, 2391} \[ \frac {2 i \text {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a+b x)}\right )}{b}-\frac {2 i \text {PolyLog}\left (2,i e^{\text {sech}^{-1}(a+b x)}\right )}{b}+\frac {(a+b x) \text {sech}^{-1}(a+b x)^2}{b}-\frac {4 \text {sech}^{-1}(a+b x) \tan ^{-1}\left (e^{\text {sech}^{-1}(a+b x)}\right )}{b} \]

Antiderivative was successfully verified.

[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 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 4180

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 5418

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 6279

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 6315

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

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Mathematica [A]  time = 0.23, size = 105, normalized size = 1.31 \[ \frac {i \left (2 \text {Li}_2\left (-i e^{-\text {sech}^{-1}(a+b x)}\right )-2 \text {Li}_2\left (i e^{-\text {sech}^{-1}(a+b x)}\right )+\text {sech}^{-1}(a+b x) \left (-i (a+b x) \text {sech}^{-1}(a+b x)+2 \log \left (1-i e^{-\text {sech}^{-1}(a+b x)}\right )-2 \log \left (1+i e^{-\text {sech}^{-1}(a+b x)}\right )\right )\right )}{b} \]

Warning: Unable to verify antiderivative.

[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

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fricas [F]  time = 0.79, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\operatorname {arsech}\left (b x + a\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsech(b*x + a)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsech(b*x + a)^2, x)

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maple [A]  time = 0.34, size = 209, normalized size = 2.61 \[ x \mathrm {arcsech}\left (b x +a \right )^{2}+\frac {\mathrm {arcsech}\left (b x +a \right )^{2} a}{b}+\frac {2 i \mathrm {arcsech}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{b}-\frac {2 i \mathrm {arcsech}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{b}+\frac {2 i \dilog \left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{b}-\frac {2 i \dilog \left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ x \log \left (\sqrt {b x + a + 1} \sqrt {-b x - a + 1} b x + \sqrt {b x + a + 1} \sqrt {-b x - a + 1} a + b x + a\right )^{2} - \int -\frac {2 \, {\left (2 \, {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} b - b\right )} x - a\right )} \sqrt {b x + a + 1} \sqrt {-b x - a + 1} \log \left (b x + a\right )^{2} + 2 \, {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} b - b\right )} x - a\right )} \log \left (b x + a\right )^{2} - {\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + {\left (a^{2} b - b\right )} x + 2 \, {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} b - b\right )} x - a\right )} \log \left (b x + a\right ) + {\left ({\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} b - b\right )} x - a\right )} \sqrt {b x + a + 1} \log \left (b x + a\right ) + {\left (2 \, b^{3} x^{3} + 4 \, a b^{2} x^{2} + {\left (2 \, a^{2} b - b\right )} x + {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} b - b\right )} x - a\right )} \log \left (b x + a\right )\right )} \sqrt {b x + a + 1}\right )} \sqrt {-b x - a + 1}\right )} \log \left (\sqrt {b x + a + 1} \sqrt {-b x - a + 1} b x + \sqrt {b x + a + 1} \sqrt {-b x - a + 1} a + b x + a\right )\right )}}{b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} b - b\right )} x - a\right )} \sqrt {b x + a + 1} \sqrt {-b x - a + 1} + {\left (3 \, a^{2} b - b\right )} x - a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {acosh}\left (\frac {1}{a+b\,x}\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acosh(1/(a + b*x))^2,x)

[Out]

int(acosh(1/(a + b*x))^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asech(a + b*x)**2, x)

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