Integrand size = 19, antiderivative size = 53 \[ \int \frac {e^{\text {sech}^{-1}(c x)}}{1-c^2 x^2} \, dx=-\frac {2 \text {arctanh}\left (\frac {1}{\sqrt {1-c x} \sqrt {\frac {1}{1+c x}}}\right )}{c}+\frac {\log (x)}{c}-\frac {\log \left (1-c^2 x^2\right )}{2 c} \] Output:
-2*arctanh(1/(-c*x+1)^(1/2)/(1/(c*x+1))^(1/2))/c+ln(x)/c-1/2*ln(-c^2*x^2+1 )/c
Time = 0.20 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.38 \[ \int \frac {e^{\text {sech}^{-1}(c x)}}{1-c^2 x^2} \, dx=\frac {2 \log (x)}{c}-\frac {\log \left (1-c^2 x^2\right )}{2 c}-\frac {\log \left (1+\sqrt {\frac {1-c x}{1+c x}}+c x \sqrt {\frac {1-c x}{1+c x}}\right )}{c} \] Input:
Integrate[E^ArcSech[c*x]/(1 - c^2*x^2),x]
Output:
(2*Log[x])/c - Log[1 - c^2*x^2]/(2*c) - Log[1 + Sqrt[(1 - c*x)/(1 + c*x)] + c*x*Sqrt[(1 - c*x)/(1 + c*x)]]/c
Time = 0.56 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.36, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {6893, 243, 47, 14, 16, 2044, 103, 221}
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.
\(\displaystyle \int \frac {e^{\text {sech}^{-1}(c x)}}{1-c^2 x^2} \, dx\) |
\(\Big \downarrow \) 6893 |
\(\displaystyle \frac {\int \frac {1}{x \left (1-c^2 x^2\right )}dx}{c}+\frac {\int \frac {\sqrt {\frac {1}{c x+1}}}{x \sqrt {1-c x}}dx}{c}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\int \frac {1}{x^2 \left (1-c^2 x^2\right )}dx^2}{2 c}+\frac {\int \frac {\sqrt {\frac {1}{c x+1}}}{x \sqrt {1-c x}}dx}{c}\) |
\(\Big \downarrow \) 47 |
\(\displaystyle \frac {c^2 \int \frac {1}{1-c^2 x^2}dx^2+\int \frac {1}{x^2}dx^2}{2 c}+\frac {\int \frac {\sqrt {\frac {1}{c x+1}}}{x \sqrt {1-c x}}dx}{c}\) |
\(\Big \downarrow \) 14 |
\(\displaystyle \frac {c^2 \int \frac {1}{1-c^2 x^2}dx^2+\log \left (x^2\right )}{2 c}+\frac {\int \frac {\sqrt {\frac {1}{c x+1}}}{x \sqrt {1-c x}}dx}{c}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {\int \frac {\sqrt {\frac {1}{c x+1}}}{x \sqrt {1-c x}}dx}{c}+\frac {\log \left (x^2\right )-\log \left (1-c^2 x^2\right )}{2 c}\) |
\(\Big \downarrow \) 2044 |
\(\displaystyle \frac {\sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \int \frac {1}{x \sqrt {1-c x} \sqrt {c x+1}}dx}{c}+\frac {\log \left (x^2\right )-\log \left (1-c^2 x^2\right )}{2 c}\) |
\(\Big \downarrow \) 103 |
\(\displaystyle \frac {\log \left (x^2\right )-\log \left (1-c^2 x^2\right )}{2 c}-\sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \int \frac {1}{c-c (1-c x) (c x+1)}d\left (\sqrt {1-c x} \sqrt {c x+1}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\log \left (x^2\right )-\log \left (1-c^2 x^2\right )}{2 c}-\frac {\sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \text {arctanh}\left (\sqrt {1-c x} \sqrt {c x+1}\right )}{c}\) |
Input:
Int[E^ArcSech[c*x]/(1 - c^2*x^2),x]
Output:
-((Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcTanh[Sqrt[1 - c*x]*Sqrt[1 + c*x]] )/c) + (Log[x^2] - Log[1 - c^2*x^2])/(2*c)
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[b/(b*c - a*d) Int[1/(a + b*x), x], x] - Simp[d/(b*c - a*d) Int[1/(c + d*x), x ], x] /; FreeQ[{a, b, c, d}, x]
Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_ ))), x_] :> Simp[b*f Subst[Int[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sq rt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[2*b*d *e - f*(b*c + a*d), 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[(u_.)*((c_.)*((a_.) + (b_.)*(x_)^(n_.))^(q_))^(p_), x_Symbol] :> Simp[S imp[(c*(a + b*x^n)^q)^p/(a + b*x^n)^(p*q)] Int[u*(a + b*x^n)^(p*q), x], x ] /; FreeQ[{a, b, c, n, p, q}, x] && GeQ[a, 0]
Int[E^ArcSech[(c_.)*(x_)]/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[1/(a*c) Int[Sqrt[1/(1 + c*x)]/(x*Sqrt[1 - c*x]), x], x] + Simp[1/c Int[1/(x*(a + b*x^2)), x], x] /; FreeQ[{a, b, c}, x] && EqQ[b + a*c^2, 0]
Time = 0.26 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.55
method | result | size |
default | \(-\frac {\sqrt {-\frac {x c -1}{x c}}\, x \sqrt {\frac {x c +1}{x c}}\, \operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )}{\sqrt {-c^{2} x^{2}+1}}+\frac {-\frac {\ln \left (x c +1\right )}{2}-\frac {\ln \left (x c -1\right )}{2}+\ln \left (x \right )}{c}\) | \(82\) |
Input:
int((1/c/x+(1/c/x-1)^(1/2)*(1/c/x+1)^(1/2))/(-c^2*x^2+1),x,method=_RETURNV ERBOSE)
Output:
-(-(c*x-1)/x/c)^(1/2)*x*((c*x+1)/x/c)^(1/2)*arctanh(1/(-c^2*x^2+1)^(1/2))/ (-c^2*x^2+1)^(1/2)+1/c*(-1/2*ln(c*x+1)-1/2*ln(c*x-1)+ln(x))
Leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (45) = 90\).
Time = 0.09 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.74 \[ \int \frac {e^{\text {sech}^{-1}(c x)}}{1-c^2 x^2} \, dx=-\frac {\log \left (c^{2} x^{2} - 1\right ) + \log \left (c x \sqrt {\frac {c x + 1}{c x}} \sqrt {-\frac {c x - 1}{c x}} + 1\right ) - \log \left (c x \sqrt {\frac {c x + 1}{c x}} \sqrt {-\frac {c x - 1}{c x}} - 1\right ) - 2 \, \log \left (x\right )}{2 \, c} \] Input:
integrate((1/c/x+(1/c/x-1)^(1/2)*(1/c/x+1)^(1/2))/(-c^2*x^2+1),x, algorith m="fricas")
Output:
-1/2*(log(c^2*x^2 - 1) + log(c*x*sqrt((c*x + 1)/(c*x))*sqrt(-(c*x - 1)/(c* x)) + 1) - log(c*x*sqrt((c*x + 1)/(c*x))*sqrt(-(c*x - 1)/(c*x)) - 1) - 2*l og(x))/c
\[ \int \frac {e^{\text {sech}^{-1}(c x)}}{1-c^2 x^2} \, dx=- \frac {\int \frac {c x \sqrt {-1 + \frac {1}{c x}} \sqrt {1 + \frac {1}{c x}}}{c^{2} x^{3} - x}\, dx + \int \frac {1}{c^{2} x^{3} - x}\, dx}{c} \] Input:
integrate((1/c/x+(1/c/x-1)**(1/2)*(1/c/x+1)**(1/2))/(-c**2*x**2+1),x)
Output:
-(Integral(c*x*sqrt(-1 + 1/(c*x))*sqrt(1 + 1/(c*x))/(c**2*x**3 - x), x) + Integral(1/(c**2*x**3 - x), x))/c
\[ \int \frac {e^{\text {sech}^{-1}(c x)}}{1-c^2 x^2} \, dx=\int { -\frac {\sqrt {\frac {1}{c x} + 1} \sqrt {\frac {1}{c x} - 1} + \frac {1}{c x}}{c^{2} x^{2} - 1} \,d x } \] Input:
integrate((1/c/x+(1/c/x-1)^(1/2)*(1/c/x+1)^(1/2))/(-c^2*x^2+1),x, algorith m="maxima")
Output:
integrate(1/x, x)/c - 1/2*log(c*x + 1)/c - 1/2*log(c*x - 1)/c - integrate( sqrt(c*x + 1)*sqrt(-c*x + 1)/(c^3*x^3 - c*x), x)
\[ \int \frac {e^{\text {sech}^{-1}(c x)}}{1-c^2 x^2} \, dx=\int { -\frac {\sqrt {\frac {1}{c x} + 1} \sqrt {\frac {1}{c x} - 1} + \frac {1}{c x}}{c^{2} x^{2} - 1} \,d x } \] Input:
integrate((1/c/x+(1/c/x-1)^(1/2)*(1/c/x+1)^(1/2))/(-c^2*x^2+1),x, algorith m="giac")
Output:
integrate(-(sqrt(1/(c*x) + 1)*sqrt(1/(c*x) - 1) + 1/(c*x))/(c^2*x^2 - 1), x)
Time = 25.41 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.11 \[ \int \frac {e^{\text {sech}^{-1}(c x)}}{1-c^2 x^2} \, dx=\frac {\ln \left (x\right )}{c}-\frac {4\,\mathrm {atanh}\left (\frac {\sqrt {\frac {1}{c\,x}-1}-\mathrm {i}}{\sqrt {\frac {1}{c\,x}+1}-1}\right )}{c}-\frac {\ln \left (3\,c^2\,x^2-3\right )}{2\,c} \] Input:
int(-((1/(c*x) - 1)^(1/2)*(1/(c*x) + 1)^(1/2) + 1/(c*x))/(c^2*x^2 - 1),x)
Output:
log(x)/c - (4*atanh(((1/(c*x) - 1)^(1/2) - 1i)/((1/(c*x) + 1)^(1/2) - 1))) /c - log(3*c^2*x^2 - 3)/(2*c)
Time = 0.16 (sec) , antiderivative size = 128, normalized size of antiderivative = 2.42 \[ \int \frac {e^{\text {sech}^{-1}(c x)}}{1-c^2 x^2} \, dx=\frac {-2 \,\mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-c x +1}}{\sqrt {2}}\right )}{2}\right )-1\right )+2 \,\mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-c x +1}}{\sqrt {2}}\right )}{2}\right )+1\right )-2 \,\mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-c x +1}}{\sqrt {2}}\right )}{2}\right )-1\right )+2 \,\mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-c x +1}}{\sqrt {2}}\right )}{2}\right )+1\right )-\mathrm {log}\left (c^{2} x -c \right )-\mathrm {log}\left (c^{2} x +c \right )+2 \,\mathrm {log}\left (x \right )}{2 c} \] Input:
int((1/c/x+(1/c/x-1)^(1/2)*(1/c/x+1)^(1/2))/(-c^2*x^2+1),x)
Output:
( - 2*log( - sqrt(2) + tan(asin(sqrt( - c*x + 1)/sqrt(2))/2) - 1) + 2*log( - sqrt(2) + tan(asin(sqrt( - c*x + 1)/sqrt(2))/2) + 1) - 2*log(sqrt(2) + tan(asin(sqrt( - c*x + 1)/sqrt(2))/2) - 1) + 2*log(sqrt(2) + tan(asin(sqrt ( - c*x + 1)/sqrt(2))/2) + 1) - log(c**2*x - c) - log(c**2*x + c) + 2*log( x))/(2*c)