Integrand size = 23, antiderivative size = 55 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^2} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x}-\frac {\sqrt {e} \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{\sqrt {d}} \]
Time = 0.04 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.11 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^2} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x}+\frac {\sqrt {e} \left (\log (x)-\log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )\right )}{\sqrt {d}} \]
-(ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x) + (Sqrt[e]*(Log[x] - Log[d + Sqr t[d]*Sqrt[d + e*x^2]]))/Sqrt[d]
Time = 0.22 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {6775, 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.
\(\displaystyle \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^2} \, dx\) |
\(\Big \downarrow \) 6775 |
\(\displaystyle \sqrt {e} \int \frac {1}{x \sqrt {e x^2+d}}dx-\frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \sqrt {e} \int \frac {1}{x^2 \sqrt {e x^2+d}}dx^2-\frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\int \frac {1}{\frac {x^4}{e}-\frac {d}{e}}d\sqrt {e x^2+d}}{\sqrt {e}}-\frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x}-\frac {\sqrt {e} \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{\sqrt {d}}\) |
-(ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x) - (Sqrt[e]*ArcTanh[Sqrt[d + e*x^ 2]/Sqrt[d]])/Sqrt[d]
3.1.13.3.1 Defintions of rubi rules used
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[ArcTanh[((c_.)*(x_))/Sqrt[(a_.) + (b_.)*(x_)^2]]*((d_.)*(x_))^(m_.), x_ Symbol] :> Simp[(d*x)^(m + 1)*(ArcTanh[(c*x)/Sqrt[a + b*x^2]]/(d*(m + 1))), x] - Simp[c/(d*(m + 1)) Int[(d*x)^(m + 1)/Sqrt[a + b*x^2], x], x] /; Fre eQ[{a, b, c, d, m}, x] && EqQ[b, c^2] && NeQ[m, -1]
Time = 0.01 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.53
method | result | size |
default | \(-\frac {\operatorname {arctanh}\left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )}{x}-\frac {\sqrt {e}\, \sqrt {e \,x^{2}+d}}{d}+\frac {\sqrt {e}\, \left (\sqrt {e \,x^{2}+d}-\sqrt {d}\, \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {e \,x^{2}+d}}{x}\right )\right )}{d}\) | \(84\) |
parts | \(-\frac {\operatorname {arctanh}\left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )}{x}-\frac {\sqrt {e}\, \sqrt {e \,x^{2}+d}}{d}+\frac {\sqrt {e}\, \left (\sqrt {e \,x^{2}+d}-\sqrt {d}\, \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {e \,x^{2}+d}}{x}\right )\right )}{d}\) | \(84\) |
-arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x-e^(1/2)/d*(e*x^2+d)^(1/2)+e^(1/2)/d* ((e*x^2+d)^(1/2)-d^(1/2)*ln((2*d+2*d^(1/2)*(e*x^2+d)^(1/2))/x))
Leaf count of result is larger than twice the leaf count of optimal. 135 vs. \(2 (43) = 86\).
Time = 0.28 (sec) , antiderivative size = 273, normalized size of antiderivative = 4.96 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^2} \, dx=\left [\frac {x \sqrt {\frac {e}{d}} \log \left (-\frac {e^{2} x^{2} - 2 \, \sqrt {e x^{2} + d} d \sqrt {e} \sqrt {\frac {e}{d}} + 2 \, d e}{x^{2}}\right ) + {\left (x - 1\right )} \log \left (\frac {2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + d}{d}\right ) - x \log \left (\frac {e x + \sqrt {e x^{2} + d} \sqrt {e}}{x}\right ) + x \log \left (\frac {e x - \sqrt {e x^{2} + d} \sqrt {e}}{x}\right )}{2 \, x}, \frac {2 \, x \sqrt {-\frac {e}{d}} \arctan \left (\frac {\sqrt {e x^{2} + d} d \sqrt {e} \sqrt {-\frac {e}{d}}}{e^{2} x^{2} + d e}\right ) + {\left (x - 1\right )} \log \left (\frac {2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + d}{d}\right ) - x \log \left (\frac {e x + \sqrt {e x^{2} + d} \sqrt {e}}{x}\right ) + x \log \left (\frac {e x - \sqrt {e x^{2} + d} \sqrt {e}}{x}\right )}{2 \, x}\right ] \]
[1/2*(x*sqrt(e/d)*log(-(e^2*x^2 - 2*sqrt(e*x^2 + d)*d*sqrt(e)*sqrt(e/d) + 2*d*e)/x^2) + (x - 1)*log((2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x + d)/d) - x*log((e*x + sqrt(e*x^2 + d)*sqrt(e))/x) + x*log((e*x - sqrt(e*x^2 + d)*s qrt(e))/x))/x, 1/2*(2*x*sqrt(-e/d)*arctan(sqrt(e*x^2 + d)*d*sqrt(e)*sqrt(- e/d)/(e^2*x^2 + d*e)) + (x - 1)*log((2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x + d)/d) - x*log((e*x + sqrt(e*x^2 + d)*sqrt(e))/x) + x*log((e*x - sqrt(e* x^2 + d)*sqrt(e))/x))/x]
\[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^2} \, dx=\int \frac {\operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{x^{2}}\, dx \]
\[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^2} \, dx=\int { \frac {\operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right )}{x^{2}} \,d x } \]
d*sqrt(e)*integrate(-sqrt(e*x^2 + d)/(e^2*x^5 + d*e*x^3 - (e*x^3 + d*x)*(e *x^2 + d)), x) - 1/2*(log(sqrt(e)*x + sqrt(e*x^2 + d)) - log(-sqrt(e)*x + sqrt(e*x^2 + d)))/x
\[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^2} \, dx=\int { \frac {\operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right )}{x^{2}} \,d x } \]
Timed out. \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^2} \, dx=\int \frac {\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {e\,x^2+d}}\right )}{x^2} \,d x \]