Integrand size = 19, antiderivative size = 61 \[ \int \frac {\text {sech}^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx=\frac {\text {sech}^{-1}(a+b x)^2}{2 d}-\frac {\text {sech}^{-1}(a+b x) \log \left (1+e^{2 \text {sech}^{-1}(a+b x)}\right )}{d}-\frac {\operatorname {PolyLog}\left (2,-e^{2 \text {sech}^{-1}(a+b x)}\right )}{2 d} \] Output:
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*polylog(2,-(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1 /(b*x+a)+1)^(1/2))^2)/d
Time = 0.06 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.85 \[ \int \frac {\text {sech}^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx=\frac {-\text {sech}^{-1}(a+b x) \left (\text {sech}^{-1}(a+b x)+2 \log \left (1+e^{-2 \text {sech}^{-1}(a+b x)}\right )\right )+\operatorname {PolyLog}\left (2,-e^{-2 \text {sech}^{-1}(a+b x)}\right )}{2 d} \] Input:
Integrate[ArcSech[a + b*x]/((a*d)/b + d*x),x]
Output:
(-(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)
Result contains complex when optimal does not.
Time = 0.53 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.10, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {6873, 27, 6835, 6297, 3042, 26, 4201, 2620, 2715, 2838}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\text {sech}^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx\) |
\(\Big \downarrow \) 6873 |
\(\displaystyle \frac {\int \frac {b \text {sech}^{-1}(a+b x)}{d (a+b x)}d(a+b x)}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\text {sech}^{-1}(a+b x)}{a+b x}d(a+b x)}{d}\) |
\(\Big \downarrow \) 6835 |
\(\displaystyle -\frac {\int (a+b x) \text {arccosh}\left (\frac {1}{a+b x}\right )d\frac {1}{a+b x}}{d}\) |
\(\Big \downarrow \) 6297 |
\(\displaystyle -\frac {\int (a+b x) \sqrt {\frac {\frac {1}{a+b x}-1}{1+\frac {1}{a+b x}}} \left (1+\frac {1}{a+b x}\right ) \text {arccosh}\left (\frac {1}{a+b x}\right )d\text {arccosh}\left (\frac {1}{a+b x}\right )}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int -i \text {arccosh}\left (\frac {1}{a+b x}\right ) \tan \left (i \text {arccosh}\left (\frac {1}{a+b x}\right )\right )d\text {arccosh}\left (\frac {1}{a+b x}\right )}{d}\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {i \int \text {arccosh}\left (\frac {1}{a+b x}\right ) \tan \left (i \text {arccosh}\left (\frac {1}{a+b x}\right )\right )d\text {arccosh}\left (\frac {1}{a+b x}\right )}{d}\) |
\(\Big \downarrow \) 4201 |
\(\displaystyle \frac {i \left (2 i \int \frac {e^{2 \text {arccosh}\left (\frac {1}{a+b x}\right )} \text {arccosh}\left (\frac {1}{a+b x}\right )}{1+e^{2 \text {arccosh}\left (\frac {1}{a+b x}\right )}}d\text {arccosh}\left (\frac {1}{a+b x}\right )-\frac {i}{2 (a+b x)^2}\right )}{d}\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle \frac {i \left (2 i \left (\frac {1}{2} \text {arccosh}\left (\frac {1}{a+b x}\right ) \log \left (e^{2 \text {arccosh}\left (\frac {1}{a+b x}\right )}+1\right )-\frac {1}{2} \int \log \left (1+e^{2 \text {arccosh}\left (\frac {1}{a+b x}\right )}\right )d\text {arccosh}\left (\frac {1}{a+b x}\right )\right )-\frac {i}{2 (a+b x)^2}\right )}{d}\) |
\(\Big \downarrow \) 2715 |
\(\displaystyle \frac {i \left (2 i \left (\frac {1}{2} \text {arccosh}\left (\frac {1}{a+b x}\right ) \log \left (e^{2 \text {arccosh}\left (\frac {1}{a+b x}\right )}+1\right )-\frac {1}{4} \int (a+b x) \log \left (1+e^{2 \text {arccosh}\left (\frac {1}{a+b x}\right )}\right )de^{2 \text {arccosh}\left (\frac {1}{a+b x}\right )}\right )-\frac {i}{2 (a+b x)^2}\right )}{d}\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle \frac {i \left (2 i \left (\frac {1}{2} \text {arccosh}\left (\frac {1}{a+b x}\right ) \log \left (e^{2 \text {arccosh}\left (\frac {1}{a+b x}\right )}+1\right )+\frac {1}{4} \operatorname {PolyLog}(2,-a-b x)\right )-\frac {i}{2 (a+b x)^2}\right )}{d}\) |
Input:
Int[ArcSech[a + b*x]/((a*d)/b + d*x),x]
Output:
(I*((-1/2*I)/(a + b*x)^2 + (2*I)*((ArcCosh[(a + b*x)^(-1)]*Log[1 + E^(2*Ar cCosh[(a + b*x)^(-1)])])/2 + PolyLog[2, -a - b*x]/4)))/d
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] - Si mp[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]
Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Simp[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]
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]
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] + Simp[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]
Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Simp[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]
Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> -Subst[Int[(a + b*ArcCosh[x/c])/x, x], x, 1/x] /; FreeQ[{a, b, c}, x]
Int[((a_.) + ArcSech[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^( m_.), x_Symbol] :> Simp[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]
Time = 0.36 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.82
method | result | size |
derivativedivides | \(\frac {\frac {b \operatorname {arcsech}\left (b x +a \right )^{2}}{2 d}-\frac {b \,\operatorname {arcsech}\left (b x +a \right ) \ln \left (1+\left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )^{2}\right )}{d}-\frac {b \operatorname {polylog}\left (2, -\left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )^{2}\right )}{2 d}}{b}\) | \(111\) |
default | \(\frac {\frac {b \operatorname {arcsech}\left (b x +a \right )^{2}}{2 d}-\frac {b \,\operatorname {arcsech}\left (b x +a \right ) \ln \left (1+\left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )^{2}\right )}{d}-\frac {b \operatorname {polylog}\left (2, -\left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )^{2}\right )}{2 d}}{b}\) | \(111\) |
Input:
int(arcsech(b*x+a)/(a*d/b+d*x),x,method=_RETURNVERBOSE)
Output:
1/b*(1/2/d*b*arcsech(b*x+a)^2-1/d*b*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/d*b*polylog(2,-(1/(b*x+a)+(1/(b*x +a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))^2))
\[ \int \frac {\text {sech}^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx=\int { \frac {\operatorname {arsech}\left (b x + a\right )}{d x + \frac {a d}{b}} \,d x } \] Input:
integrate(arcsech(b*x+a)/(a*d/b+d*x),x, algorithm="fricas")
Output:
integral(b*arcsech(b*x + a)/(b*d*x + a*d), x)
\[ \int \frac {\text {sech}^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx=\frac {b \int \frac {\operatorname {asech}{\left (a + b x \right )}}{a + b x}\, dx}{d} \] Input:
integrate(asech(b*x+a)/(a*d/b+d*x),x)
Output:
b*Integral(asech(a + b*x)/(a + b*x), x)/d
\[ \int \frac {\text {sech}^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx=\int { \frac {\operatorname {arsech}\left (b x + a\right )}{d x + \frac {a d}{b}} \,d x } \] Input:
integrate(arcsech(b*x+a)/(a*d/b+d*x),x, algorithm="maxima")
Output:
1/2*(2*log(sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)*b*x + sqrt(b*x + a + 1)*sq rt(-b*x - a + 1)*a + b*x + a)*log(b*x + a) - 3*log(b*x + a)^2)/d - 1/2*(lo g(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)
\[ \int \frac {\text {sech}^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx=\int { \frac {\operatorname {arsech}\left (b x + a\right )}{d x + \frac {a d}{b}} \,d x } \] Input:
integrate(arcsech(b*x+a)/(a*d/b+d*x),x, algorithm="giac")
Output:
integrate(arcsech(b*x + a)/(d*x + a*d/b), x)
Timed out. \[ \int \frac {\text {sech}^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx=\int \frac {\mathrm {acosh}\left (\frac {1}{a+b\,x}\right )}{d\,x+\frac {a\,d}{b}} \,d x \] Input:
int(acosh(1/(a + b*x))/(d*x + (a*d)/b),x)
Output:
int(acosh(1/(a + b*x))/(d*x + (a*d)/b), x)
\[ \int \frac {\text {sech}^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx=\frac {\left (\int \frac {\mathit {asech} \left (b x +a \right )}{b x +a}d x \right ) b}{d} \] Input:
int(asech(b*x+a)/(a*d/b+d*x),x)
Output:
(int(asech(a + b*x)/(a + b*x),x)*b)/d