Integrand size = 27, antiderivative size = 113 \[ \int \frac {1}{x^3 (d+e x) \sqrt {d^2-e^2 x^2}} \, dx=-\frac {3 \sqrt {d^2-e^2 x^2}}{2 d^3 x^2}+\frac {2 e \sqrt {d^2-e^2 x^2}}{d^4 x}+\frac {\sqrt {d^2-e^2 x^2}}{d^2 x^2 (d+e x)}-\frac {3 e^2 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^4} \]
-3/2*e^2*arctanh((-e^2*x^2+d^2)^(1/2)/d)/d^4-3/2*(-e^2*x^2+d^2)^(1/2)/d^3/ x^2+2*e*(-e^2*x^2+d^2)^(1/2)/d^4/x+(-e^2*x^2+d^2)^(1/2)/d^2/x^2/(e*x+d)
Time = 0.24 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.95 \[ \int \frac {1}{x^3 (d+e x) \sqrt {d^2-e^2 x^2}} \, dx=\frac {\frac {d \sqrt {d^2-e^2 x^2} \left (-d^2+d e x+4 e^2 x^2\right )}{x^2 (d+e x)}-3 \sqrt {d^2} e^2 \log (x)+3 \sqrt {d^2} e^2 \log \left (\sqrt {d^2}-\sqrt {d^2-e^2 x^2}\right )}{2 d^5} \]
((d*Sqrt[d^2 - e^2*x^2]*(-d^2 + d*e*x + 4*e^2*x^2))/(x^2*(d + e*x)) - 3*Sq rt[d^2]*e^2*Log[x] + 3*Sqrt[d^2]*e^2*Log[Sqrt[d^2] - Sqrt[d^2 - e^2*x^2]]) /(2*d^5)
Time = 0.36 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {564, 25, 2338, 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.
\(\displaystyle \int \frac {1}{x^3 (d+e x) \sqrt {d^2-e^2 x^2}} \, dx\) |
\(\Big \downarrow \) 564 |
\(\displaystyle \frac {e^2 \sqrt {d^2-e^2 x^2}}{d^4 (d+e x)}-\int -\frac {\frac {e^2 x^2}{d^3}-\frac {e x}{d^2}+\frac {1}{d}}{x^3 \sqrt {d^2-e^2 x^2}}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \frac {\frac {e^2 x^2}{d^3}-\frac {e x}{d^2}+\frac {1}{d}}{x^3 \sqrt {d^2-e^2 x^2}}dx+\frac {e^2 \sqrt {d^2-e^2 x^2}}{d^4 (d+e x)}\) |
\(\Big \downarrow \) 2338 |
\(\displaystyle -\frac {\int \frac {e (2 d-3 e x)}{d x^2 \sqrt {d^2-e^2 x^2}}dx}{2 d^2}+\frac {e^2 \sqrt {d^2-e^2 x^2}}{d^4 (d+e x)}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^3 x^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {e \int \frac {2 d-3 e x}{x^2 \sqrt {d^2-e^2 x^2}}dx}{2 d^3}+\frac {e^2 \sqrt {d^2-e^2 x^2}}{d^4 (d+e x)}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^3 x^2}\) |
\(\Big \downarrow \) 534 |
\(\displaystyle -\frac {e \left (-3 e \int \frac {1}{x \sqrt {d^2-e^2 x^2}}dx-\frac {2 \sqrt {d^2-e^2 x^2}}{d x}\right )}{2 d^3}+\frac {e^2 \sqrt {d^2-e^2 x^2}}{d^4 (d+e x)}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^3 x^2}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle -\frac {e \left (-\frac {3}{2} e \int \frac {1}{x^2 \sqrt {d^2-e^2 x^2}}dx^2-\frac {2 \sqrt {d^2-e^2 x^2}}{d x}\right )}{2 d^3}+\frac {e^2 \sqrt {d^2-e^2 x^2}}{d^4 (d+e x)}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^3 x^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {e \left (\frac {3 \int \frac {1}{\frac {d^2}{e^2}-\frac {x^4}{e^2}}d\sqrt {d^2-e^2 x^2}}{e}-\frac {2 \sqrt {d^2-e^2 x^2}}{d x}\right )}{2 d^3}+\frac {e^2 \sqrt {d^2-e^2 x^2}}{d^4 (d+e x)}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^3 x^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {e \left (\frac {3 e \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d}-\frac {2 \sqrt {d^2-e^2 x^2}}{d x}\right )}{2 d^3}+\frac {e^2 \sqrt {d^2-e^2 x^2}}{d^4 (d+e x)}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^3 x^2}\) |
-1/2*Sqrt[d^2 - e^2*x^2]/(d^3*x^2) + (e^2*Sqrt[d^2 - e^2*x^2])/(d^4*(d + e *x)) - (e*((-2*Sqrt[d^2 - e^2*x^2])/(d*x) + (3*e*ArcTanh[Sqrt[d^2 - e^2*x^ 2]/d])/d))/(2*d^3)
3.2.26.3.1 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_))^(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[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{ Q = PolynomialQuotient[Pq, c*x, x], R = PolynomialRemainder[Pq, c*x, x]}, S imp[R*(c*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] + Simp[1/(a*c*( m + 1)) Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*( m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && Lt Q[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])
Time = 0.41 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.04
method | result | size |
risch | \(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, \left (-2 e x +d \right )}{2 d^{4} x^{2}}-\frac {3 e^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 d^{3} \sqrt {d^{2}}}+\frac {e \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{d^{4} \left (x +\frac {d}{e}\right )}\) | \(117\) |
default | \(\frac {-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{2 d^{2} x^{2}}-\frac {e^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 d^{2} \sqrt {d^{2}}}}{d}-\frac {e^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{3} \sqrt {d^{2}}}+\frac {e \sqrt {-e^{2} x^{2}+d^{2}}}{d^{4} x}+\frac {e \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{d^{4} \left (x +\frac {d}{e}\right )}\) | \(183\) |
-1/2*(-e^2*x^2+d^2)^(1/2)*(-2*e*x+d)/d^4/x^2-3/2*e^2/d^3/(d^2)^(1/2)*ln((2 *d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)+e/d^4/(x+d/e)*(-(x+d/e)^2*e^2+ 2*d*e*(x+d/e))^(1/2)
Time = 0.26 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^3 (d+e x) \sqrt {d^2-e^2 x^2}} \, dx=\frac {2 \, e^{3} x^{3} + 2 \, d e^{2} x^{2} + 3 \, {\left (e^{3} x^{3} + d e^{2} x^{2}\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + {\left (4 \, e^{2} x^{2} + d e x - d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{2 \, {\left (d^{4} e x^{3} + d^{5} x^{2}\right )}} \]
1/2*(2*e^3*x^3 + 2*d*e^2*x^2 + 3*(e^3*x^3 + d*e^2*x^2)*log(-(d - sqrt(-e^2 *x^2 + d^2))/x) + (4*e^2*x^2 + d*e*x - d^2)*sqrt(-e^2*x^2 + d^2))/(d^4*e*x ^3 + d^5*x^2)
\[ \int \frac {1}{x^3 (d+e x) \sqrt {d^2-e^2 x^2}} \, dx=\int \frac {1}{x^{3} \sqrt {- \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )}\, dx \]
\[ \int \frac {1}{x^3 (d+e x) \sqrt {d^2-e^2 x^2}} \, dx=\int { \frac {1}{\sqrt {-e^{2} x^{2} + d^{2}} {\left (e x + d\right )} x^{3}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 256 vs. \(2 (101) = 202\).
Time = 0.29 (sec) , antiderivative size = 256, normalized size of antiderivative = 2.27 \[ \int \frac {1}{x^3 (d+e x) \sqrt {d^2-e^2 x^2}} \, dx=\frac {{\left (e^{3} - \frac {3 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} e}{x} - \frac {20 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2}}{e x^{2}}\right )} e^{4} x^{2}}{8 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2} d^{4} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} + 1\right )} {\left | e \right |}} - \frac {3 \, e^{3} \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |} \right |}}{2 \, e^{2} {\left | x \right |}}\right )}{2 \, d^{4} {\left | e \right |}} + \frac {\frac {4 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} d^{4} e {\left | e \right |}}{x} - \frac {{\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2} d^{4} {\left | e \right |}}{e x^{2}}}{8 \, d^{8} e^{2}} \]
1/8*(e^3 - 3*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))*e/x - 20*(d*e + sqrt(-e^2 *x^2 + d^2)*abs(e))^2/(e*x^2))*e^4*x^2/((d*e + sqrt(-e^2*x^2 + d^2)*abs(e) )^2*d^4*((d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) + 1)*abs(e)) - 3/2*e^ 3*log(1/2*abs(-2*d*e - 2*sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*abs(x)))/(d^4*a bs(e)) + 1/8*(4*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))*d^4*e*abs(e)/x - (d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^2*d^4*abs(e)/(e*x^2))/(d^8*e^2)
Timed out. \[ \int \frac {1}{x^3 (d+e x) \sqrt {d^2-e^2 x^2}} \, dx=\int \frac {1}{x^3\,\sqrt {d^2-e^2\,x^2}\,\left (d+e\,x\right )} \,d x \]