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

Optimal. Leaf size=78 \[ x \left (a+b \text {sech}^{-1}(c x)\right )^2-\frac {4 b \tan ^{-1}\left (e^{\text {sech}^{-1}(c x)}\right ) \left (a+b \text {sech}^{-1}(c x)\right )}{c}+\frac {2 i b^2 \text {Li}_2\left (-i e^{\text {sech}^{-1}(c x)}\right )}{c}-\frac {2 i b^2 \text {Li}_2\left (i e^{\text {sech}^{-1}(c x)}\right )}{c} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6279, 5451, 4180, 2279, 2391} \[ \frac {2 i b^2 \text {PolyLog}\left (2,-i e^{\text {sech}^{-1}(c x)}\right )}{c}-\frac {2 i b^2 \text {PolyLog}\left (2,i e^{\text {sech}^{-1}(c x)}\right )}{c}+x \left (a+b \text {sech}^{-1}(c x)\right )^2-\frac {4 b \tan ^{-1}\left (e^{\text {sech}^{-1}(c x)}\right ) \left (a+b \text {sech}^{-1}(c x)\right )}{c} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSech[c*x])^2,x]

[Out]

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

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 5451

Int[((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.)*Tanh[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> -Si
mp[((c + d*x)^m*Sech[a + b*x]^n)/(b*n), x] + Dist[(d*m)/(b*n), Int[(c + d*x)^(m - 1)*Sech[a + b*x]^n, x], x] /
; FreeQ[{a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]

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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.27, size = 126, normalized size = 1.62 \[ a^2 x+\frac {2 a b \left (c x \text {sech}^{-1}(c x)-2 \tan ^{-1}\left (\tanh \left (\frac {1}{2} \text {sech}^{-1}(c x)\right )\right )\right )}{c}+\frac {i b^2 \left (2 \text {Li}_2\left (-i e^{-\text {sech}^{-1}(c x)}\right )-2 \text {Li}_2\left (i e^{-\text {sech}^{-1}(c x)}\right )+\text {sech}^{-1}(c x) \left (-i c x \text {sech}^{-1}(c x)+2 \log \left (1-i e^{-\text {sech}^{-1}(c x)}\right )-2 \log \left (1+i e^{-\text {sech}^{-1}(c x)}\right )\right )\right )}{c} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSech[c*x])^2,x]

[Out]

a^2*x + (2*a*b*(c*x*ArcSech[c*x] - 2*ArcTan[Tanh[ArcSech[c*x]/2]]))/c + (I*b^2*(ArcSech[c*x]*((-I)*c*x*ArcSech
[c*x] + 2*Log[1 - I/E^ArcSech[c*x]] - 2*Log[1 + I/E^ArcSech[c*x]]) + 2*PolyLog[2, (-I)/E^ArcSech[c*x]] - 2*Pol
yLog[2, I/E^ArcSech[c*x]]))/c

________________________________________________________________________________________

fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b^{2} \operatorname {arsech}\left (c x\right )^{2} + 2 \, a b \operatorname {arsech}\left (c x\right ) + a^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(b^2*arcsech(c*x)^2 + 2*a*b*arcsech(c*x) + a^2, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arcsech(c*x) + a)^2, x)

________________________________________________________________________________________

maple [A]  time = 0.30, size = 250, normalized size = 3.21 \[ x \,b^{2} \mathrm {arcsech}\left (c x \right )^{2}-\frac {2 i \mathrm {arcsech}\left (c x \right ) \ln \left (1-i \left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )\right ) b^{2}}{c}+\frac {2 i \mathrm {arcsech}\left (c x \right ) \ln \left (1+i \left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )\right ) b^{2}}{c}+2 x a b \,\mathrm {arcsech}\left (c x \right )+\frac {2 i \dilog \left (1+i \left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )\right ) b^{2}}{c}-\frac {2 i \dilog \left (1-i \left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )\right ) b^{2}}{c}+a^{2} x -\frac {2 \arctan \left (\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right ) a b}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*log(sqrt(c*x + 1)*sqrt(-c*x + 1) + 1)^2 - integrate(-(c^2*x^2*log(c)^2 + (c^2*x^2 - 1)*log(x)^2 + (c^2*x^2*
log(c)^2 + (c^2*x^2 - 1)*log(x)^2 - log(c)^2 + 2*(c^2*x^2*log(c) - log(c))*log(x))*sqrt(c*x + 1)*sqrt(-c*x + 1
) - 2*(c^2*x^2*log(c) + (c^2*x^2*(log(c) + 1) + (c^2*x^2 - 1)*log(x) - log(c))*sqrt(c*x + 1)*sqrt(-c*x + 1) +
(c^2*x^2 - 1)*log(x) - log(c))*log(sqrt(c*x + 1)*sqrt(-c*x + 1) + 1) - log(c)^2 + 2*(c^2*x^2*log(c) - log(c))*
log(x))/(c^2*x^2 + (c^2*x^2 - 1)*sqrt(c*x + 1)*sqrt(-c*x + 1) - 1), x))*b^2 + a^2*x + 2*(c*x*arcsech(c*x) - ar
ctan(sqrt(1/(c^2*x^2) - 1)))*a*b/c

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acosh(1/(c*x)))^2,x)

[Out]

int((a + b*acosh(1/(c*x)))^2, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asech(c*x))**2, x)

________________________________________________________________________________________