Integrand size = 27, antiderivative size = 82 \[ \int \frac {\sqrt {d^2-e^2 x^2}}{x^3 (d+e x)} \, dx=-\frac {\sqrt {d^2-e^2 x^2}}{2 d x^2}+\frac {e \sqrt {d^2-e^2 x^2}}{d^2 x}-\frac {e^2 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^2} \]
-1/2*e^2*arctanh((-e^2*x^2+d^2)^(1/2)/d)/d^2-1/2*(-e^2*x^2+d^2)^(1/2)/d/x^ 2+e*(-e^2*x^2+d^2)^(1/2)/d^2/x
Time = 0.15 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.13 \[ \int \frac {\sqrt {d^2-e^2 x^2}}{x^3 (d+e x)} \, dx=-\frac {d (d-2 e x) \sqrt {d^2-e^2 x^2}+\sqrt {d^2} e^2 x^2 \log (x)-\sqrt {d^2} e^2 x^2 \log \left (\sqrt {d^2}-\sqrt {d^2-e^2 x^2}\right )}{2 d^3 x^2} \]
-1/2*(d*(d - 2*e*x)*Sqrt[d^2 - e^2*x^2] + Sqrt[d^2]*e^2*x^2*Log[x] - Sqrt[ d^2]*e^2*x^2*Log[Sqrt[d^2] - Sqrt[d^2 - e^2*x^2]])/(d^3*x^2)
Time = 0.24 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {566, 539, 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.
\(\displaystyle \int \frac {\sqrt {d^2-e^2 x^2}}{x^3 (d+e x)} \, dx\) |
\(\Big \downarrow \) 566 |
\(\displaystyle \int \frac {d-e x}{x^3 \sqrt {d^2-e^2 x^2}}dx\) |
\(\Big \downarrow \) 539 |
\(\displaystyle -\frac {\int \frac {d e (2 d-e x)}{x^2 \sqrt {d^2-e^2 x^2}}dx}{2 d^2}-\frac {\sqrt {d^2-e^2 x^2}}{2 d x^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {e \int \frac {2 d-e x}{x^2 \sqrt {d^2-e^2 x^2}}dx}{2 d}-\frac {\sqrt {d^2-e^2 x^2}}{2 d x^2}\) |
\(\Big \downarrow \) 534 |
\(\displaystyle -\frac {e \left (-e \int \frac {1}{x \sqrt {d^2-e^2 x^2}}dx-\frac {2 \sqrt {d^2-e^2 x^2}}{d x}\right )}{2 d}-\frac {\sqrt {d^2-e^2 x^2}}{2 d x^2}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle -\frac {e \left (-\frac {1}{2} e \int \frac {1}{x^2 \sqrt {d^2-e^2 x^2}}dx^2-\frac {2 \sqrt {d^2-e^2 x^2}}{d x}\right )}{2 d}-\frac {\sqrt {d^2-e^2 x^2}}{2 d x^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {e \left (\frac {\int \frac {1}{\frac {d^2}{e^2}-\frac {x^4}{e^2}}d\sqrt {d^2-e^2 x^2}}{e}-\frac {2 \sqrt {d^2-e^2 x^2}}{d x}\right )}{2 d}-\frac {\sqrt {d^2-e^2 x^2}}{2 d x^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {e \left (\frac {e \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d}-\frac {2 \sqrt {d^2-e^2 x^2}}{d x}\right )}{2 d}-\frac {\sqrt {d^2-e^2 x^2}}{2 d x^2}\) |
-1/2*Sqrt[d^2 - e^2*x^2]/(d*x^2) - (e*((-2*Sqrt[d^2 - e^2*x^2])/(d*x) + (e *ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/d))/(2*d)
3.1.99.3.1 Defintions of rubi rules used
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_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))), x] + Simp[1/(a*(m + 1)) Int[x^(m + 1)*(a + b*x^2)^p*(a*d*(m + 1) - b*c*(m + 2*p + 3)*x), x], x] /; FreeQ[{a, b, c, d, p}, x] && ILtQ[m, -1] && GtQ[p, -1] && IntegerQ[2*p]
Int[((x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)), x_Symbol] : > Int[x^m*(a/c + b*(x/d))*(a + b*x^2)^(p - 1), x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[p, 0]
Time = 0.40 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.91
method | result | size |
risch | \(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, \left (-2 e x +d \right )}{2 d^{2} x^{2}}-\frac {e^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 d \sqrt {d^{2}}}\) | \(75\) |
default | \(\frac {-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{2 d^{2} x^{2}}-\frac {e^{2} \left (\sqrt {-e^{2} x^{2}+d^{2}}-\frac {d^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}}\right )}{2 d^{2}}}{d}+\frac {e^{2} \left (\sqrt {-e^{2} x^{2}+d^{2}}-\frac {d^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}}\right )}{d^{3}}-\frac {e \left (-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{d^{2} x}-\frac {2 e^{2} \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )}{d^{2}}\right )}{d^{2}}-\frac {e^{2} \left (\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}+\frac {d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {e^{2}}}\right )}{d^{3}}\) | \(326\) |
-1/2*(-e^2*x^2+d^2)^(1/2)*(-2*e*x+d)/d^2/x^2-1/2/d*e^2/(d^2)^(1/2)*ln((2*d ^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)
Time = 0.27 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.77 \[ \int \frac {\sqrt {d^2-e^2 x^2}}{x^3 (d+e x)} \, dx=\frac {e^{2} x^{2} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + \sqrt {-e^{2} x^{2} + d^{2}} {\left (2 \, e x - d\right )}}{2 \, d^{2} x^{2}} \]
1/2*(e^2*x^2*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + sqrt(-e^2*x^2 + d^2)*(2* e*x - d))/(d^2*x^2)
\[ \int \frac {\sqrt {d^2-e^2 x^2}}{x^3 (d+e x)} \, dx=\int \frac {\sqrt {- \left (- d + e x\right ) \left (d + e x\right )}}{x^{3} \left (d + e x\right )}\, dx \]
\[ \int \frac {\sqrt {d^2-e^2 x^2}}{x^3 (d+e x)} \, dx=\int { \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{{\left (e x + d\right )} x^{3}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 193 vs. \(2 (72) = 144\).
Time = 0.28 (sec) , antiderivative size = 193, normalized size of antiderivative = 2.35 \[ \int \frac {\sqrt {d^2-e^2 x^2}}{x^3 (d+e x)} \, dx=\frac {{\left (e^{3} - \frac {4 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} e}{x}\right )} e^{4} x^{2}}{8 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2} d^{2} {\left | e \right |}} - \frac {e^{3} \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |} \right |}}{2 \, e^{2} {\left | x \right |}}\right )}{2 \, d^{2} {\left | e \right |}} + \frac {\frac {4 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} d^{2} e {\left | e \right |}}{x} - \frac {{\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2} d^{2} {\left | e \right |}}{e x^{2}}}{8 \, d^{4} e^{2}} \]
1/8*(e^3 - 4*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))*e/x)*e^4*x^2/((d*e + sqrt (-e^2*x^2 + d^2)*abs(e))^2*d^2*abs(e)) - 1/2*e^3*log(1/2*abs(-2*d*e - 2*sq rt(-e^2*x^2 + d^2)*abs(e))/(e^2*abs(x)))/(d^2*abs(e)) + 1/8*(4*(d*e + sqrt (-e^2*x^2 + d^2)*abs(e))*d^2*e*abs(e)/x - (d*e + sqrt(-e^2*x^2 + d^2)*abs( e))^2*d^2*abs(e)/(e*x^2))/(d^4*e^2)
Timed out. \[ \int \frac {\sqrt {d^2-e^2 x^2}}{x^3 (d+e x)} \, dx=\int \frac {\sqrt {d^2-e^2\,x^2}}{x^3\,\left (d+e\,x\right )} \,d x \]