3.1.98 \(\int \frac {\text {sech}^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx\) [98]

Optimal. Leaf size=61 \[ \frac {\text {sech}^{-1}(a+b x)^2}{2 d}-\frac {\text {sech}^{-1}(a+b x) \log \left (1+e^{2 \text {sech}^{-1}(a+b x)}\right )}{d}-\frac {\text {PolyLog}\left (2,-e^{2 \text {sech}^{-1}(a+b x)}\right )}{2 d} \]

[Out]

1/2*arcsech(b*x+a)^2/d-arcsech(b*x+a)*ln(1+(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))^2)/d-1/2*polylo
g(2,-(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))^2)/d

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Rubi [A]
time = 0.07, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {6454, 12, 6416, 5882, 3799, 2221, 2317, 2438} \begin {gather*} -\frac {\text {Li}_2\left (-e^{2 \text {sech}^{-1}(a+b x)}\right )}{2 d}+\frac {\text {sech}^{-1}(a+b x)^2}{2 d}-\frac {\text {sech}^{-1}(a+b x) \log \left (e^{2 \text {sech}^{-1}(a+b x)}+1\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcSech[a + b*x]/((a*d)/b + d*x),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 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 3799

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (Complex[0, fz_])*(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*fz*x))/(1 + E^(2*((-I)*e + f*fz*x)))), x]
, x] /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]

Rule 5882

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

Rule 6416

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> -Subst[Int[(a + b*ArcCosh[x/c])/x, x], x, 1/x] /; F
reeQ[{a, b, c}, x]

Rule 6454

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

Rubi steps

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

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Mathematica [A]
time = 0.04, size = 52, normalized size = 0.85 \begin {gather*} \frac {-\text {sech}^{-1}(a+b x) \left (\text {sech}^{-1}(a+b x)+2 \log \left (1+e^{-2 \text {sech}^{-1}(a+b x)}\right )\right )+\text {PolyLog}\left (2,-e^{-2 \text {sech}^{-1}(a+b x)}\right )}{2 d} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcSech[a + b*x]/((a*d)/b + d*x),x]

[Out]

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

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Maple [A]
time = 0.49, size = 111, normalized size = 1.82

method	result	size
derivativedivides	\(\frac {\frac {b \mathrm {arcsech}\left (b x +a \right )^{2}}{2 d}-\frac {b \,\mathrm {arcsech}\left (b x +a \right ) \ln \left (1+\left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )^{2}\right )}{d}-\frac {b \polylog \left (2, -\left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )^{2}\right )}{2 d}}{b}\)	\(111\)
default	\(\frac {\frac {b \mathrm {arcsech}\left (b x +a \right )^{2}}{2 d}-\frac {b \,\mathrm {arcsech}\left (b x +a \right ) \ln \left (1+\left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )^{2}\right )}{d}-\frac {b \polylog \left (2, -\left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )^{2}\right )}{2 d}}{b}\)	\(111\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsech(b*x+a)/(a*d/b+d*x),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(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)*log(b*
x + a) - 3*log(b*x + a)^2)/d - 1/2*(log(b*x + a + 1)*log(b*x + a) + dilog(-b*x - a))/d - 1/2*(log(b*x + a)*log
(-b*x - a + 1) + dilog(b*x + a))/d + integrate((b^2*x + a*b)*log(b*x + a)/(b^2*d*x^2 + 2*a*b*d*x + a^2*d + (b^
2*d*x^2 + 2*a*b*d*x + a^2*d - d)*sqrt(b*x + a + 1)*sqrt(-b*x - a + 1) - d), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(b*arcsech(b*x + a)/(b*d*x + a*d), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {b \int \frac {\operatorname {asech}{\left (a + b x \right )}}{a + b x}\, dx}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asech(b*x+a)/(a*d/b+d*x),x)

[Out]

b*Integral(asech(a + b*x)/(a + b*x), x)/d

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsech(b*x + a)/(d*x + a*d/b), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\mathrm {acosh}\left (\frac {1}{a+b\,x}\right )}{d\,x+\frac {a\,d}{b}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acosh(1/(a + b*x))/(d*x + (a*d)/b),x)

[Out]

int(acosh(1/(a + b*x))/(d*x + (a*d)/b), x)

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