Integrand size = 20, antiderivative size = 71 \[ \int \frac {e^{\coth ^{-1}(a x)}}{c-\frac {c}{a x}} \, dx=-\frac {2 \sqrt {1-\frac {1}{a^2 x^2}}}{c \left (a-\frac {1}{x}\right )}+\frac {\sqrt {1-\frac {1}{a^2 x^2}} x}{c}+\frac {2 \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a c} \] Output:
-2*(1-1/a^2/x^2)^(1/2)/c/(a-1/x)+(1-1/a^2/x^2)^(1/2)*x/c+2*arctanh((1-1/a^ 2/x^2)^(1/2))/a/c
Time = 0.13 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.89 \[ \int \frac {e^{\coth ^{-1}(a x)}}{c-\frac {c}{a x}} \, dx=\frac {a \sqrt {1-\frac {1}{a^2 x^2}} x (-3+a x)+2 (-1+a x) \log \left (\left (1+\sqrt {1-\frac {1}{a^2 x^2}}\right ) x\right )}{a c (-1+a x)} \] Input:
Integrate[E^ArcCoth[a*x]/(c - c/(a*x)),x]
Output:
(a*Sqrt[1 - 1/(a^2*x^2)]*x*(-3 + a*x) + 2*(-1 + a*x)*Log[(1 + Sqrt[1 - 1/( a^2*x^2)])*x])/(a*c*(-1 + a*x))
Time = 0.55 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6731, 27, 564, 27, 534, 243, 73, 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.
\(\displaystyle \int \frac {e^{\coth ^{-1}(a x)}}{c-\frac {c}{a x}} \, dx\) |
\(\Big \downarrow \) 6731 |
\(\displaystyle -c \int \frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}} x^2}{c^2 \left (a-\frac {1}{x}\right )^2}d\frac {1}{x}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {a^2 \int \frac {\sqrt {1-\frac {1}{a^2 x^2}} x^2}{\left (a-\frac {1}{x}\right )^2}d\frac {1}{x}}{c}\) |
\(\Big \downarrow \) 564 |
\(\displaystyle -\frac {a^2 \left (\frac {\int \frac {\left (a+\frac {2}{x}\right ) x^2}{a \sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}}{a^2}+\frac {2 \sqrt {1-\frac {1}{a^2 x^2}}}{a^2 \left (a-\frac {1}{x}\right )}\right )}{c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {a^2 \left (\frac {\int \frac {\left (a+\frac {2}{x}\right ) x^2}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}}{a^3}+\frac {2 \sqrt {1-\frac {1}{a^2 x^2}}}{a^2 \left (a-\frac {1}{x}\right )}\right )}{c}\) |
\(\Big \downarrow \) 534 |
\(\displaystyle -\frac {a^2 \left (\frac {2 \int \frac {x}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}-a x \sqrt {1-\frac {1}{a^2 x^2}}}{a^3}+\frac {2 \sqrt {1-\frac {1}{a^2 x^2}}}{a^2 \left (a-\frac {1}{x}\right )}\right )}{c}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle -\frac {a^2 \left (\frac {\int \frac {x}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x^2}-a x \sqrt {1-\frac {1}{a^2 x^2}}}{a^3}+\frac {2 \sqrt {1-\frac {1}{a^2 x^2}}}{a^2 \left (a-\frac {1}{x}\right )}\right )}{c}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {a^2 \left (\frac {-2 a^2 \int \frac {1}{a^2-a^2 \sqrt {1-\frac {1}{a^2 x^2}}}d\sqrt {1-\frac {1}{a^2 x^2}}-a x \sqrt {1-\frac {1}{a^2 x^2}}}{a^3}+\frac {2 \sqrt {1-\frac {1}{a^2 x^2}}}{a^2 \left (a-\frac {1}{x}\right )}\right )}{c}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {a^2 \left (\frac {2 \sqrt {1-\frac {1}{a^2 x^2}}}{a^2 \left (a-\frac {1}{x}\right )}+\frac {-2 \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )-a x \sqrt {1-\frac {1}{a^2 x^2}}}{a^3}\right )}{c}\) |
Input:
Int[E^ArcCoth[a*x]/(c - c/(a*x)),x]
Output:
-((a^2*((2*Sqrt[1 - 1/(a^2*x^2)])/(a^2*(a - x^(-1))) + (-(a*Sqrt[1 - 1/(a^ 2*x^2)]*x) - 2*ArcTanh[Sqrt[1 - 1/(a^2*x^2)]])/a^3))/c)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-c)*x^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1))), x] + Simp[d Int[ x^(m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[m, 0] && GtQ[p, -1] && EqQ[m + 2*p + 3, 0]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[(-(-c)^(m - n - 2))*d^(2*n - m + 3)*(Sqrt[a + b*x^2]/(2^(n + 1)*b ^(n + 2)*(c + d*x))), x] - Simp[d^(2*n + 2)/b^(n + 1) Int[(x^m/Sqrt[a + b *x^2])*ExpandToSum[((2^(-n - 1)*(-c)^(m - n - 1))/(d^m*x^m) - (-c + d*x)^(- n - 1))/(c + d*x), x], x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^ 2, 0] && ILtQ[m, 0] && ILtQ[n, 0] && EqQ[n + p, -3/2]
Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.), x_Symbol] :> S imp[-c^n Subst[Int[(c + d*x)^(p - n)*((1 - x^2/a^2)^(n/2)/x^2), x], x, 1/ x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[c + a*d, 0] && IntegerQ[(n - 1)/2] && IntegerQ[2*p]
Leaf count of result is larger than twice the leaf count of optimal. \(142\) vs. \(2(65)=130\).
Time = 0.11 (sec) , antiderivative size = 143, normalized size of antiderivative = 2.01
method | result | size |
risch | \(\frac {a x -1}{a c \sqrt {\frac {a x -1}{a x +1}}}+\frac {\left (\frac {2 \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}-1}\right )}{a \sqrt {a^{2}}}-\frac {2 \sqrt {\left (x -\frac {1}{a}\right )^{2} a^{2}+2 a \left (x -\frac {1}{a}\right )}}{a^{3} \left (x -\frac {1}{a}\right )}\right ) a \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{c \sqrt {\frac {a x -1}{a x +1}}\, \left (a x +1\right )}\) | \(143\) |
default | \(-\frac {-2 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}\, a^{2} x^{2}-2 \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a^{3} x^{2}+\left (\left (a x -1\right ) \left (a x +1\right )\right )^{\frac {3}{2}} \sqrt {a^{2}}+4 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}\, a x +4 \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a^{2} x -2 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}-2 a \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right )}{a \sqrt {a^{2}}\, \left (a x -1\right ) c \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {\frac {a x -1}{a x +1}}}\) | \(250\) |
Input:
int(1/((a*x-1)/(a*x+1))^(1/2)/(c-c/a/x),x,method=_RETURNVERBOSE)
Output:
1/a*(a*x-1)/c/((a*x-1)/(a*x+1))^(1/2)+(2/a*ln(a^2*x/(a^2)^(1/2)+(a^2*x^2-1 )^(1/2))/(a^2)^(1/2)-2/a^3/(x-1/a)*((x-1/a)^2*a^2+2*a*(x-1/a))^(1/2))*a/c/ ((a*x-1)/(a*x+1))^(1/2)*((a*x-1)*(a*x+1))^(1/2)/(a*x+1)
Time = 0.15 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.32 \[ \int \frac {e^{\coth ^{-1}(a x)}}{c-\frac {c}{a x}} \, dx=\frac {2 \, {\left (a x - 1\right )} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 2 \, {\left (a x - 1\right )} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right ) + {\left (a^{2} x^{2} - 2 \, a x - 3\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{a^{2} c x - a c} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(1/2)/(c-c/a/x),x, algorithm="fricas")
Output:
(2*(a*x - 1)*log(sqrt((a*x - 1)/(a*x + 1)) + 1) - 2*(a*x - 1)*log(sqrt((a* x - 1)/(a*x + 1)) - 1) + (a^2*x^2 - 2*a*x - 3)*sqrt((a*x - 1)/(a*x + 1)))/ (a^2*c*x - a*c)
\[ \int \frac {e^{\coth ^{-1}(a x)}}{c-\frac {c}{a x}} \, dx=\frac {a \int \frac {x}{a x \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}} - \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}\, dx}{c} \] Input:
integrate(1/((a*x-1)/(a*x+1))**(1/2)/(c-c/a/x),x)
Output:
a*Integral(x/(a*x*sqrt(a*x/(a*x + 1) - 1/(a*x + 1)) - sqrt(a*x/(a*x + 1) - 1/(a*x + 1))), x)/c
Time = 0.04 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.63 \[ \int \frac {e^{\coth ^{-1}(a x)}}{c-\frac {c}{a x}} \, dx=-2 \, a {\left (\frac {\frac {2 \, {\left (a x - 1\right )}}{a x + 1} - 1}{a^{2} c \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} - a^{2} c \sqrt {\frac {a x - 1}{a x + 1}}} - \frac {\log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{2} c} + \frac {\log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{2} c}\right )} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(1/2)/(c-c/a/x),x, algorithm="maxima")
Output:
-2*a*((2*(a*x - 1)/(a*x + 1) - 1)/(a^2*c*((a*x - 1)/(a*x + 1))^(3/2) - a^2 *c*sqrt((a*x - 1)/(a*x + 1))) - log(sqrt((a*x - 1)/(a*x + 1)) + 1)/(a^2*c) + log(sqrt((a*x - 1)/(a*x + 1)) - 1)/(a^2*c))
\[ \int \frac {e^{\coth ^{-1}(a x)}}{c-\frac {c}{a x}} \, dx=\int { \frac {1}{{\left (c - \frac {c}{a x}\right )} \sqrt {\frac {a x - 1}{a x + 1}}} \,d x } \] Input:
integrate(1/((a*x-1)/(a*x+1))^(1/2)/(c-c/a/x),x, algorithm="giac")
Output:
undef
Time = 0.05 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.87 \[ \int \frac {e^{\coth ^{-1}(a x)}}{c-\frac {c}{a x}} \, dx=\frac {2\,a\,x+8\,\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )\,\sqrt {\frac {a\,x-1}{a\,x+1}}-6}{2\,a\,c\,\sqrt {\frac {a\,x-1}{a\,x+1}}} \] Input:
int(1/((c - c/(a*x))*((a*x - 1)/(a*x + 1))^(1/2)),x)
Output:
(2*a*x + 8*atanh(((a*x - 1)/(a*x + 1))^(1/2))*((a*x - 1)/(a*x + 1))^(1/2) - 6)/(2*a*c*((a*x - 1)/(a*x + 1))^(1/2))
Time = 0.17 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.99 \[ \int \frac {e^{\coth ^{-1}(a x)}}{c-\frac {c}{a x}} \, dx=\frac {8 \sqrt {a x -1}\, \mathrm {log}\left (\frac {\sqrt {a x -1}+\sqrt {a x +1}}{\sqrt {2}}\right )-5 \sqrt {a x -1}+2 \sqrt {a x +1}\, a x -6 \sqrt {a x +1}}{2 \sqrt {a x -1}\, a c} \] Input:
int(1/((a*x-1)/(a*x+1))^(1/2)/(c-c/a/x),x)
Output:
(8*sqrt(a*x - 1)*log((sqrt(a*x - 1) + sqrt(a*x + 1))/sqrt(2)) - 5*sqrt(a*x - 1) + 2*sqrt(a*x + 1)*a*x - 6*sqrt(a*x + 1))/(2*sqrt(a*x - 1)*a*c)