Integrand size = 10, antiderivative size = 77 \[ \int \text {sech}^{-1}\left (c e^{a+b x}\right ) \, dx=\frac {\text {sech}^{-1}\left (c e^{a+b x}\right )^2}{2 b}-\frac {\text {sech}^{-1}\left (c e^{a+b x}\right ) \log \left (1+e^{2 \text {sech}^{-1}\left (c e^{a+b x}\right )}\right )}{b}-\frac {\operatorname {PolyLog}\left (2,-e^{2 \text {sech}^{-1}\left (c e^{a+b x}\right )}\right )}{2 b} \]
1/2*arcsech(c*exp(b*x+a))^2/b-arcsech(c*exp(b*x+a))*ln(1+(1/c/exp(b*x+a)+( 1/c/exp(b*x+a)-1)^(1/2)*(1/c/exp(b*x+a)+1)^(1/2))^2)/b-1/2*polylog(2,-(1/c /exp(b*x+a)+(1/c/exp(b*x+a)-1)^(1/2)*(1/c/exp(b*x+a)+1)^(1/2))^2)/b
Leaf count is larger than twice the leaf count of optimal. \(249\) vs. \(2(77)=154\).
Time = 1.86 (sec) , antiderivative size = 249, normalized size of antiderivative = 3.23 \[ \int \text {sech}^{-1}\left (c e^{a+b x}\right ) \, dx=x \text {sech}^{-1}\left (c e^{a+b x}\right )-\frac {\sqrt {\frac {1-c e^{a+b x}}{1+c e^{a+b x}}} \sqrt {1+c e^{a+b x}} \left (\text {arctanh}\left (\sqrt {1-c^2 e^{2 (a+b x)}}\right ) \left (8 b x-4 \log \left (c^2 e^{2 (a+b x)}\right )\right )-\log ^2\left (c^2 e^{2 (a+b x)}\right )+4 \log \left (c^2 e^{2 (a+b x)}\right ) \log \left (\frac {1}{2} \left (1+\sqrt {1-c^2 e^{2 (a+b x)}}\right )\right )-2 \log ^2\left (\frac {1}{2} \left (1+\sqrt {1-c^2 e^{2 (a+b x)}}\right )\right )+4 \operatorname {PolyLog}\left (2,\frac {1}{2} \left (1-\sqrt {1-c^2 e^{2 (a+b x)}}\right )\right )\right )}{8 b \sqrt {1-c e^{a+b x}}} \]
x*ArcSech[c*E^(a + b*x)] - (Sqrt[(1 - c*E^(a + b*x))/(1 + c*E^(a + b*x))]* Sqrt[1 + c*E^(a + b*x)]*(ArcTanh[Sqrt[1 - c^2*E^(2*(a + b*x))]]*(8*b*x - 4 *Log[c^2*E^(2*(a + b*x))]) - Log[c^2*E^(2*(a + b*x))]^2 + 4*Log[c^2*E^(2*( a + b*x))]*Log[(1 + Sqrt[1 - c^2*E^(2*(a + b*x))])/2] - 2*Log[(1 + Sqrt[1 - c^2*E^(2*(a + b*x))])/2]^2 + 4*PolyLog[2, (1 - Sqrt[1 - c^2*E^(2*(a + b* x))])/2]))/(8*b*Sqrt[1 - c*E^(a + b*x)])
Result contains complex when optimal does not.
Time = 0.47 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.26, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.900, Rules used = {2720, 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.
\(\displaystyle \int \text {sech}^{-1}\left (c e^{a+b x}\right ) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int e^{-a-b x} \text {sech}^{-1}\left (c e^{a+b x}\right )de^{a+b x}}{b}\) |
\(\Big \downarrow \) 6835 |
\(\displaystyle -\frac {\int e^{-a-b x} \text {arccosh}\left (\frac {e^{-a-b x}}{c}\right )de^{-a-b x}}{b}\) |
\(\Big \downarrow \) 6297 |
\(\displaystyle -\frac {\int c e^{a+b x} \sqrt {\frac {\frac {e^{-a-b x}}{c}-1}{1+\frac {e^{-a-b x}}{c}}} \left (1+\frac {e^{-a-b x}}{c}\right ) \text {arccosh}\left (\frac {e^{-a-b x}}{c}\right )d\text {arccosh}\left (\frac {e^{-a-b x}}{c}\right )}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int -i \text {arccosh}\left (\frac {e^{-a-b x}}{c}\right ) \tan \left (i \text {arccosh}\left (\frac {e^{-a-b x}}{c}\right )\right )d\text {arccosh}\left (\frac {e^{-a-b x}}{c}\right )}{b}\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {i \int \text {arccosh}\left (\frac {e^{-a-b x}}{c}\right ) \tan \left (i \text {arccosh}\left (\frac {e^{-a-b x}}{c}\right )\right )d\text {arccosh}\left (\frac {e^{-a-b x}}{c}\right )}{b}\) |
\(\Big \downarrow \) 4201 |
\(\displaystyle \frac {i \left (2 i \int \frac {e^{a+b x+2 \text {arccosh}\left (\frac {e^{-a-b x}}{c}\right )}}{1+e^{2 \text {arccosh}\left (\frac {e^{-a-b x}}{c}\right )}}d\text {arccosh}\left (\frac {e^{-a-b x}}{c}\right )-\frac {1}{2} i e^{2 a+2 b x}\right )}{b}\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle \frac {i \left (2 i \left (\frac {1}{2} \text {arccosh}\left (\frac {e^{-a-b x}}{c}\right ) \log \left (e^{2 \text {arccosh}\left (\frac {e^{-a-b x}}{c}\right )}+1\right )-\frac {1}{2} \int \log \left (1+e^{2 \text {arccosh}\left (\frac {e^{-a-b x}}{c}\right )}\right )d\text {arccosh}\left (\frac {e^{-a-b x}}{c}\right )\right )-\frac {1}{2} i e^{2 a+2 b x}\right )}{b}\) |
\(\Big \downarrow \) 2715 |
\(\displaystyle \frac {i \left (2 i \left (\frac {1}{2} \text {arccosh}\left (\frac {e^{-a-b x}}{c}\right ) \log \left (e^{2 \text {arccosh}\left (\frac {e^{-a-b x}}{c}\right )}+1\right )-\frac {1}{4} \int e^{2 \text {arccosh}\left (\frac {e^{-a-b x}}{c}\right )} \log \left (1+e^{2 \text {arccosh}\left (\frac {e^{-a-b x}}{c}\right )}\right )de^{2 \text {arccosh}\left (\frac {e^{-a-b x}}{c}\right )}\right )-\frac {1}{2} i e^{2 a+2 b x}\right )}{b}\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle \frac {i \left (2 i \left (\frac {1}{4} \operatorname {PolyLog}\left (2,-e^{2 \text {arccosh}\left (\frac {e^{-a-b x}}{c}\right )}\right )+\frac {1}{2} \text {arccosh}\left (\frac {e^{-a-b x}}{c}\right ) \log \left (e^{2 \text {arccosh}\left (\frac {e^{-a-b x}}{c}\right )}+1\right )\right )-\frac {1}{2} i e^{2 a+2 b x}\right )}{b}\) |
(I*((-1/2*I)*E^(2*a + 2*b*x) + (2*I)*((ArcCosh[E^(-a - b*x)/c]*Log[1 + E^( 2*ArcCosh[E^(-a - b*x)/c])])/2 + PolyLog[2, -E^(2*ArcCosh[E^(-a - b*x)/c]) ]/4)))/b
3.1.31.3.1 Defintions of rubi rules used
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[(((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[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
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]
Time = 1.10 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.75
method | result | size |
derivativedivides | \(\frac {\frac {\operatorname {arcsech}\left ({\mathrm e}^{b x +a} c \right )^{2}}{2}-\operatorname {arcsech}\left ({\mathrm e}^{b x +a} c \right ) \ln \left (1+\left (\frac {{\mathrm e}^{-b x -a}}{c}+\sqrt {\frac {{\mathrm e}^{-b x -a}}{c}-1}\, \sqrt {\frac {{\mathrm e}^{-b x -a}}{c}+1}\right )^{2}\right )-\frac {\operatorname {polylog}\left (2, -\left (\frac {{\mathrm e}^{-b x -a}}{c}+\sqrt {\frac {{\mathrm e}^{-b x -a}}{c}-1}\, \sqrt {\frac {{\mathrm e}^{-b x -a}}{c}+1}\right )^{2}\right )}{2}}{b}\) | \(135\) |
default | \(\frac {\frac {\operatorname {arcsech}\left ({\mathrm e}^{b x +a} c \right )^{2}}{2}-\operatorname {arcsech}\left ({\mathrm e}^{b x +a} c \right ) \ln \left (1+\left (\frac {{\mathrm e}^{-b x -a}}{c}+\sqrt {\frac {{\mathrm e}^{-b x -a}}{c}-1}\, \sqrt {\frac {{\mathrm e}^{-b x -a}}{c}+1}\right )^{2}\right )-\frac {\operatorname {polylog}\left (2, -\left (\frac {{\mathrm e}^{-b x -a}}{c}+\sqrt {\frac {{\mathrm e}^{-b x -a}}{c}-1}\, \sqrt {\frac {{\mathrm e}^{-b x -a}}{c}+1}\right )^{2}\right )}{2}}{b}\) | \(135\) |
1/b*(1/2*arcsech(exp(b*x+a)*c)^2-arcsech(exp(b*x+a)*c)*ln(1+(1/c/exp(b*x+a )+(1/c/exp(b*x+a)-1)^(1/2)*(1/c/exp(b*x+a)+1)^(1/2))^2)-1/2*polylog(2,-(1/ c/exp(b*x+a)+(1/c/exp(b*x+a)-1)^(1/2)*(1/c/exp(b*x+a)+1)^(1/2))^2))
Exception generated. \[ \int \text {sech}^{-1}\left (c e^{a+b x}\right ) \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
\[ \int \text {sech}^{-1}\left (c e^{a+b x}\right ) \, dx=\int \operatorname {asech}{\left (c e^{a + b x} \right )}\, dx \]
\[ \int \text {sech}^{-1}\left (c e^{a+b x}\right ) \, dx=\int { \operatorname {arsech}\left (c e^{\left (b x + a\right )}\right ) \,d x } \]
b*c^2*integrate(x*e^(2*b*x + 2*a)/(c^2*e^(2*b*x + 2*a) + (c^2*e^(2*b*x + 2 *a) - 1)*e^(1/2*log(c*e^(b*x + a) + 1) + 1/2*log(-c*e^(b*x + a) + 1)) - 1) , x) - 1/2*b*x^2 - (a + log(c))*x + x*log(sqrt(c*e^(b*x + a) + 1)*sqrt(-c* e^(b*x + a) + 1) + 1) - 1/2*(b*x*log(c*e^(b*x + a) + 1) + dilog(-c*e^(b*x + a)))/b - 1/2*(b*x*log(-c*e^(b*x + a) + 1) + dilog(c*e^(b*x + a)))/b
\[ \int \text {sech}^{-1}\left (c e^{a+b x}\right ) \, dx=\int { \operatorname {arsech}\left (c e^{\left (b x + a\right )}\right ) \,d x } \]
Timed out. \[ \int \text {sech}^{-1}\left (c e^{a+b x}\right ) \, dx=\int \mathrm {acosh}\left (\frac {{\mathrm {e}}^{-a-b\,x}}{c}\right ) \,d x \]