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

Optimal. Leaf size=149 \[ -\frac {a^2 \text {sech}^{-1}(a+b x)^2}{2 b^2}-\frac {2 i a \text {Li}_2\left (-i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {2 i a \text {Li}_2\left (i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}-\frac {\log (a+b x)}{b^2}-\frac {\sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1) \text {sech}^{-1}(a+b x)}{b^2}+\frac {4 a \text {sech}^{-1}(a+b x) \tan ^{-1}\left (e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {1}{2} x^2 \text {sech}^{-1}(a+b x)^2 \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {6321, 5468, 4190, 4180, 2279, 2391, 4184, 3475} \[ -\frac {2 i a \text {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {2 i a \text {PolyLog}\left (2,i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}-\frac {a^2 \text {sech}^{-1}(a+b x)^2}{2 b^2}-\frac {\log (a+b x)}{b^2}-\frac {\sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1) \text {sech}^{-1}(a+b x)}{b^2}+\frac {4 a \text {sech}^{-1}(a+b x) \tan ^{-1}\left (e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {1}{2} x^2 \text {sech}^{-1}(a+b x)^2 \]

Antiderivative was successfully verified.

[In]

Int[x*ArcSech[a + b*x]^2,x]

[Out]

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

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 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

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 4184

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

Rule 4190

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 0]

Rule 5468

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

Rule 6321

Int[((a_.) + ArcSech[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> -Dist[(d^(m + 1)
)^(-1), Subst[Int[(a + b*x)^p*Sech[x]*Tanh[x]*(d*e - c*f + f*Sech[x])^m, x], x, ArcSech[c + d*x]], x] /; FreeQ
[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && IntegerQ[m]

Rubi steps

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

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Mathematica [A]  time = 0.43, size = 172, normalized size = 1.15 \[ \frac {-4 i a \left (\text {Li}_2\left (-i e^{-\text {sech}^{-1}(a+b x)}\right )-\text {Li}_2\left (i e^{-\text {sech}^{-1}(a+b x)}\right )\right )+2 \log \left (\frac {1}{a+b x}\right )+(a+b x)^2 \text {sech}^{-1}(a+b x)^2-2 a (a+b x) \text {sech}^{-1}(a+b x)^2-2 \sqrt {-\frac {a+b x-1}{a+b x+1}} (a+b x+1) \text {sech}^{-1}(a+b x)-4 i a \text {sech}^{-1}(a+b x) \left (\log \left (1-i e^{-\text {sech}^{-1}(a+b x)}\right )-\log \left (1+i e^{-\text {sech}^{-1}(a+b x)}\right )\right )}{2 b^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x*ArcSech[a + b*x]^2,x]

[Out]

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

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x \operatorname {arsech}\left (b x + a\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x*arcsech(b*x + a)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*arcsech(b*x + a)^2, x)

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maple [A]  time = 1.31, size = 396, normalized size = 2.66 \[ \frac {x^{2} \mathrm {arcsech}\left (b x +a \right )^{2}}{2}-\frac {\sqrt {-\frac {b x +a -1}{b x +a}}\, \sqrt {\frac {b x +a +1}{b x +a}}\, \mathrm {arcsech}\left (b x +a \right ) x}{b}-\frac {\sqrt {-\frac {b x +a -1}{b x +a}}\, \sqrt {\frac {b x +a +1}{b x +a}}\, \mathrm {arcsech}\left (b x +a \right ) a}{b^{2}}-\frac {a^{2} \mathrm {arcsech}\left (b x +a \right )^{2}}{2 b^{2}}+\frac {\mathrm {arcsech}\left (b x +a \right )}{b^{2}}-\frac {2 \ln \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )}{b^{2}}+\frac {\ln \left (1+\left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )^{2}\right )}{b^{2}}-\frac {2 i a \,\mathrm {arcsech}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{b^{2}}+\frac {2 i a \,\mathrm {arcsech}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{b^{2}}-\frac {2 i a \dilog \left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{b^{2}}+\frac {2 i a \dilog \left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*arcsech(b*x+a)^2-1/b*(-(b*x+a-1)/(b*x+a))^(1/2)*((b*x+a+1)/(b*x+a))^(1/2)*arcsech(b*x+a)*x-1/b^2*(-(b*
x+a-1)/(b*x+a))^(1/2)*((b*x+a+1)/(b*x+a))^(1/2)*arcsech(b*x+a)*a-1/2*a^2*arcsech(b*x+a)^2/b^2+1/b^2*arcsech(b*
x+a)-2/b^2*ln(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))+1/b^2*ln(1+(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/
(b*x+a)+1)^(1/2))^2)-2*I/b^2*a*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^2*a*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^2*a*dilog(1+I*(1/(b*x+a
)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2)))+2*I/b^2*a*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 \[ \frac {1}{2} \, x^{2} \log \left (\sqrt {b x + a + 1} \sqrt {-b x - a + 1} b x + \sqrt {b x + a + 1} \sqrt {-b x - a + 1} a + b x + a\right )^{2} - \int -\frac {4 \, {\left (b^{3} x^{4} + 3 \, a b^{2} x^{3} + {\left (3 \, a^{2} b - b\right )} x^{2} + {\left (a^{3} - a\right )} x\right )} \sqrt {b x + a + 1} \sqrt {-b x - a + 1} \log \left (b x + a\right )^{2} + 4 \, {\left (b^{3} x^{4} + 3 \, a b^{2} x^{3} + {\left (3 \, a^{2} b - b\right )} x^{2} + {\left (a^{3} - a\right )} x\right )} \log \left (b x + a\right )^{2} - {\left (b^{3} x^{4} + 2 \, a b^{2} x^{3} + {\left (a^{2} b - b\right )} x^{2} + 4 \, {\left (b^{3} x^{4} + 3 \, a b^{2} x^{3} + {\left (3 \, a^{2} b - b\right )} x^{2} + {\left (a^{3} - a\right )} x\right )} \log \left (b x + a\right ) + {\left (2 \, {\left (b^{3} x^{4} + 3 \, a b^{2} x^{3} + {\left (3 \, a^{2} b - b\right )} x^{2} + {\left (a^{3} - a\right )} x\right )} \sqrt {b x + a + 1} \log \left (b x + a\right ) + {\left (2 \, b^{3} x^{4} + 4 \, a b^{2} x^{3} + {\left (2 \, a^{2} b - b\right )} x^{2} + 2 \, {\left (b^{3} x^{4} + 3 \, a b^{2} x^{3} + {\left (3 \, a^{2} b - b\right )} x^{2} + {\left (a^{3} - a\right )} x\right )} \log \left (b x + a\right )\right )} \sqrt {b x + a + 1}\right )} \sqrt {-b x - a + 1}\right )} \log \left (\sqrt {b x + a + 1} \sqrt {-b x - a + 1} b x + \sqrt {b x + a + 1} \sqrt {-b x - a + 1} a + b x + a\right )}{b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} b - b\right )} x - a\right )} \sqrt {b x + a + 1} \sqrt {-b x - a + 1} + {\left (3 \, a^{2} b - b\right )} x - a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*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 - i
ntegrate(-(4*(b^3*x^4 + 3*a*b^2*x^3 + (3*a^2*b - b)*x^2 + (a^3 - a)*x)*sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)*lo
g(b*x + a)^2 + 4*(b^3*x^4 + 3*a*b^2*x^3 + (3*a^2*b - b)*x^2 + (a^3 - a)*x)*log(b*x + a)^2 - (b^3*x^4 + 2*a*b^2
*x^3 + (a^2*b - b)*x^2 + 4*(b^3*x^4 + 3*a*b^2*x^3 + (3*a^2*b - b)*x^2 + (a^3 - a)*x)*log(b*x + a) + (2*(b^3*x^
4 + 3*a*b^2*x^3 + (3*a^2*b - b)*x^2 + (a^3 - a)*x)*sqrt(b*x + a + 1)*log(b*x + a) + (2*b^3*x^4 + 4*a*b^2*x^3 +
(2*a^2*b - b)*x^2 + 2*(b^3*x^4 + 3*a*b^2*x^3 + (3*a^2*b - b)*x^2 + (a^3 - a)*x)*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 \[ \int x\,{\mathrm {acosh}\left (\frac {1}{a+b\,x}\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*acosh(1/(a + b*x))^2,x)

[Out]

int(x*acosh(1/(a + b*x))^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*asech(a + b*x)**2, x)

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