Integrand size = 26, antiderivative size = 80 \[ \int \frac {1}{\left (d+e x^2\right )^{3/2} \left (d^2-e^2 x^4\right )} \, dx=\frac {x}{6 d^2 \left (d+e x^2\right )^{3/2}}+\frac {7 x}{12 d^3 \sqrt {d+e x^2}}+\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt {e} x}{\sqrt {d+e x^2}}\right )}{4 \sqrt {2} d^3 \sqrt {e}} \]
1/6*x/d^2/(e*x^2+d)^(3/2)+1/8*arctanh(x*2^(1/2)*e^(1/2)/(e*x^2+d)^(1/2))/d ^3*2^(1/2)/e^(1/2)+7/12*x/d^3/(e*x^2+d)^(1/2)
Time = 0.18 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (d+e x^2\right )^{3/2} \left (d^2-e^2 x^4\right )} \, dx=\frac {\frac {2 \left (9 d x+7 e x^3\right )}{\left (d+e x^2\right )^{3/2}}+\frac {3 \sqrt {2} \text {arctanh}\left (\frac {d-e x^2+\sqrt {e} x \sqrt {d+e x^2}}{\sqrt {2} d}\right )}{\sqrt {e}}}{24 d^3} \]
((2*(9*d*x + 7*e*x^3))/(d + e*x^2)^(3/2) + (3*Sqrt[2]*ArcTanh[(d - e*x^2 + Sqrt[e]*x*Sqrt[d + e*x^2])/(Sqrt[2]*d)])/Sqrt[e])/(24*d^3)
Time = 0.23 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.10, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {1388, 316, 25, 27, 402, 27, 291, 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.
| \(\displaystyle \int \frac {1}{\left (d+e x^2\right )^{3/2} \left (d^2-e^2 x^4\right )} \, dx\) |
| \(\Big \downarrow \) 1388 |
| \(\displaystyle \int \frac {1}{\left (d-e x^2\right ) \left (d+e x^2\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle \frac {x}{6 d^2 \left (d+e x^2\right )^{3/2}}-\frac {\int -\frac {e \left (5 d-2 e x^2\right )}{\left (d-e x^2\right ) \left (e x^2+d\right )^{3/2}}dx}{6 d^2 e}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {e \left (5 d-2 e x^2\right )}{\left (d-e x^2\right ) \left (e x^2+d\right )^{3/2}}dx}{6 d^2 e}+\frac {x}{6 d^2 \left (d+e x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {5 d-2 e x^2}{\left (d-e x^2\right ) \left (e x^2+d\right )^{3/2}}dx}{6 d^2}+\frac {x}{6 d^2 \left (d+e x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {7 x}{2 d \sqrt {d+e x^2}}-\frac {\int -\frac {3 d^2 e}{\left (d-e x^2\right ) \sqrt {e x^2+d}}dx}{2 d^2 e}}{6 d^2}+\frac {x}{6 d^2 \left (d+e x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {3}{2} \int \frac {1}{\left (d-e x^2\right ) \sqrt {e x^2+d}}dx+\frac {7 x}{2 d \sqrt {d+e x^2}}}{6 d^2}+\frac {x}{6 d^2 \left (d+e x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {3}{2} \int \frac {1}{d-\frac {2 d e x^2}{e x^2+d}}d\frac {x}{\sqrt {e x^2+d}}+\frac {7 x}{2 d \sqrt {d+e x^2}}}{6 d^2}+\frac {x}{6 d^2 \left (d+e x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {3 \text {arctanh}\left (\frac {\sqrt {2} \sqrt {e} x}{\sqrt {d+e x^2}}\right )}{2 \sqrt {2} d \sqrt {e}}+\frac {7 x}{2 d \sqrt {d+e x^2}}}{6 d^2}+\frac {x}{6 d^2 \left (d+e x^2\right )^{3/2}}\) |
x/(6*d^2*(d + e*x^2)^(3/2)) + ((7*x)/(2*d*Sqrt[d + e*x^2]) + (3*ArcTanh[(S qrt[2]*Sqrt[e]*x)/Sqrt[d + e*x^2]])/(2*Sqrt[2]*d*Sqrt[e]))/(6*d^2)
3.2.98.3.1 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_)^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[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d)) Int[(a + b*x^2)^(p + 1)*(c + d*x ^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x ], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ! ( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, p, q, x]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -1]
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x] /; FreeQ[{a, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && (Integer Q[p] || (GtQ[a, 0] && GtQ[d, 0]))
Time = 0.31 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.86
| method | result | size |
| pseudoelliptic | \(\frac {14 e^{\frac {3}{2}} x^{3}+3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {e \,x^{2}+d}\, \sqrt {2}}{2 x \sqrt {e}}\right ) \left (e \,x^{2}+d \right )^{\frac {3}{2}}+18 \sqrt {e}\, d x}{24 \sqrt {e}\, \left (e \,x^{2}+d \right )^{\frac {3}{2}} d^{3}}\) | \(69\) |
| default | \(-\frac {e \left (\frac {1}{2 d \sqrt {\left (x -\frac {\sqrt {e d}}{e}\right )^{2} e +2 \sqrt {e d}\, \left (x -\frac {\sqrt {e d}}{e}\right )+2 d}}-\frac {\sqrt {e d}\, \left (2 e \left (x -\frac {\sqrt {e d}}{e}\right )+2 \sqrt {e d}\right )}{4 d^{2} e \sqrt {\left (x -\frac {\sqrt {e d}}{e}\right )^{2} e +2 \sqrt {e d}\, \left (x -\frac {\sqrt {e d}}{e}\right )+2 d}}-\frac {\sqrt {2}\, \ln \left (\frac {4 d +2 \sqrt {e d}\, \left (x -\frac {\sqrt {e d}}{e}\right )+2 \sqrt {2}\, \sqrt {d}\, \sqrt {\left (x -\frac {\sqrt {e d}}{e}\right )^{2} e +2 \sqrt {e d}\, \left (x -\frac {\sqrt {e d}}{e}\right )+2 d}}{x -\frac {\sqrt {e d}}{e}}\right )}{4 d^{\frac {3}{2}}}\right )}{2 \left (\sqrt {e d}-\sqrt {-e d}\right ) \left (\sqrt {e d}+\sqrt {-e d}\right ) \sqrt {e d}}+\frac {e \left (\frac {1}{2 d \sqrt {\left (x +\frac {\sqrt {e d}}{e}\right )^{2} e -2 \sqrt {e d}\, \left (x +\frac {\sqrt {e d}}{e}\right )+2 d}}+\frac {\sqrt {e d}\, \left (2 e \left (x +\frac {\sqrt {e d}}{e}\right )-2 \sqrt {e d}\right )}{4 d^{2} e \sqrt {\left (x +\frac {\sqrt {e d}}{e}\right )^{2} e -2 \sqrt {e d}\, \left (x +\frac {\sqrt {e d}}{e}\right )+2 d}}-\frac {\sqrt {2}\, \ln \left (\frac {4 d -2 \sqrt {e d}\, \left (x +\frac {\sqrt {e d}}{e}\right )+2 \sqrt {2}\, \sqrt {d}\, \sqrt {\left (x +\frac {\sqrt {e d}}{e}\right )^{2} e -2 \sqrt {e d}\, \left (x +\frac {\sqrt {e d}}{e}\right )+2 d}}{x +\frac {\sqrt {e d}}{e}}\right )}{4 d^{\frac {3}{2}}}\right )}{2 \left (\sqrt {e d}-\sqrt {-e d}\right ) \left (\sqrt {e d}+\sqrt {-e d}\right ) \sqrt {e d}}-\frac {e \left (\frac {1}{3 \sqrt {-e d}\, \left (x +\frac {\sqrt {-e d}}{e}\right ) \sqrt {\left (x +\frac {\sqrt {-e d}}{e}\right )^{2} e -2 \sqrt {-e d}\, \left (x +\frac {\sqrt {-e d}}{e}\right )}}+\frac {2 e \left (x +\frac {\sqrt {-e d}}{e}\right )-2 \sqrt {-e d}}{3 \sqrt {-e d}\, d \sqrt {\left (x +\frac {\sqrt {-e d}}{e}\right )^{2} e -2 \sqrt {-e d}\, \left (x +\frac {\sqrt {-e d}}{e}\right )}}\right )}{2 \sqrt {-e d}\, \left (\sqrt {e d}-\sqrt {-e d}\right ) \left (\sqrt {e d}+\sqrt {-e d}\right )}+\frac {e \left (-\frac {1}{3 \sqrt {-e d}\, \left (x -\frac {\sqrt {-e d}}{e}\right ) \sqrt {\left (x -\frac {\sqrt {-e d}}{e}\right )^{2} e +2 \sqrt {-e d}\, \left (x -\frac {\sqrt {-e d}}{e}\right )}}-\frac {2 e \left (x -\frac {\sqrt {-e d}}{e}\right )+2 \sqrt {-e d}}{3 \sqrt {-e d}\, d \sqrt {\left (x -\frac {\sqrt {-e d}}{e}\right )^{2} e +2 \sqrt {-e d}\, \left (x -\frac {\sqrt {-e d}}{e}\right )}}\right )}{2 \sqrt {-e d}\, \left (\sqrt {e d}-\sqrt {-e d}\right ) \left (\sqrt {e d}+\sqrt {-e d}\right )}\) | \(865\) |
1/24*(14*e^(3/2)*x^3+3*2^(1/2)*arctanh(1/2*(e*x^2+d)^(1/2)/x*2^(1/2)/e^(1/ 2))*(e*x^2+d)^(3/2)+18*e^(1/2)*d*x)/e^(1/2)/(e*x^2+d)^(3/2)/d^3
Leaf count of result is larger than twice the leaf count of optimal. 127 vs. \(2 (60) = 120\).
Time = 0.28 (sec) , antiderivative size = 279, normalized size of antiderivative = 3.49 \[ \int \frac {1}{\left (d+e x^2\right )^{3/2} \left (d^2-e^2 x^4\right )} \, dx=\left [\frac {3 \, \sqrt {2} {\left (e^{2} x^{4} + 2 \, d e x^{2} + d^{2}\right )} \sqrt {e} \log \left (\frac {17 \, e^{2} x^{4} + 14 \, d e x^{2} + 4 \, \sqrt {2} {\left (3 \, e x^{3} + d x\right )} \sqrt {e x^{2} + d} \sqrt {e} + d^{2}}{e^{2} x^{4} - 2 \, d e x^{2} + d^{2}}\right ) + 8 \, {\left (7 \, e^{2} x^{3} + 9 \, d e x\right )} \sqrt {e x^{2} + d}}{96 \, {\left (d^{3} e^{3} x^{4} + 2 \, d^{4} e^{2} x^{2} + d^{5} e\right )}}, -\frac {3 \, \sqrt {2} {\left (e^{2} x^{4} + 2 \, d e x^{2} + d^{2}\right )} \sqrt {-e} \arctan \left (\frac {\sqrt {2} {\left (3 \, e x^{2} + d\right )} \sqrt {e x^{2} + d} \sqrt {-e}}{4 \, {\left (e^{2} x^{3} + d e x\right )}}\right ) - 4 \, {\left (7 \, e^{2} x^{3} + 9 \, d e x\right )} \sqrt {e x^{2} + d}}{48 \, {\left (d^{3} e^{3} x^{4} + 2 \, d^{4} e^{2} x^{2} + d^{5} e\right )}}\right ] \]
[1/96*(3*sqrt(2)*(e^2*x^4 + 2*d*e*x^2 + d^2)*sqrt(e)*log((17*e^2*x^4 + 14* d*e*x^2 + 4*sqrt(2)*(3*e*x^3 + d*x)*sqrt(e*x^2 + d)*sqrt(e) + d^2)/(e^2*x^ 4 - 2*d*e*x^2 + d^2)) + 8*(7*e^2*x^3 + 9*d*e*x)*sqrt(e*x^2 + d))/(d^3*e^3* x^4 + 2*d^4*e^2*x^2 + d^5*e), -1/48*(3*sqrt(2)*(e^2*x^4 + 2*d*e*x^2 + d^2) *sqrt(-e)*arctan(1/4*sqrt(2)*(3*e*x^2 + d)*sqrt(e*x^2 + d)*sqrt(-e)/(e^2*x ^3 + d*e*x)) - 4*(7*e^2*x^3 + 9*d*e*x)*sqrt(e*x^2 + d))/(d^3*e^3*x^4 + 2*d ^4*e^2*x^2 + d^5*e)]
\[ \int \frac {1}{\left (d+e x^2\right )^{3/2} \left (d^2-e^2 x^4\right )} \, dx=- \int \frac {1}{- d^{3} \sqrt {d + e x^{2}} - d^{2} e x^{2} \sqrt {d + e x^{2}} + d e^{2} x^{4} \sqrt {d + e x^{2}} + e^{3} x^{6} \sqrt {d + e x^{2}}}\, dx \]
-Integral(1/(-d**3*sqrt(d + e*x**2) - d**2*e*x**2*sqrt(d + e*x**2) + d*e** 2*x**4*sqrt(d + e*x**2) + e**3*x**6*sqrt(d + e*x**2)), x)
\[ \int \frac {1}{\left (d+e x^2\right )^{3/2} \left (d^2-e^2 x^4\right )} \, dx=\int { -\frac {1}{{\left (e^{2} x^{4} - d^{2}\right )} {\left (e x^{2} + d\right )}^{\frac {3}{2}}} \,d x } \]
Time = 0.32 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.41 \[ \int \frac {1}{\left (d+e x^2\right )^{3/2} \left (d^2-e^2 x^4\right )} \, dx=\frac {x {\left (\frac {7 \, e x^{2}}{d^{3}} + \frac {9}{d^{2}}\right )}}{12 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}}} + \frac {\sqrt {2} \log \left (\frac {{\left | 2 \, {\left (\sqrt {e} x - \sqrt {e x^{2} + d}\right )}^{2} - 4 \, \sqrt {2} {\left | d \right |} - 6 \, d \right |}}{{\left | 2 \, {\left (\sqrt {e} x - \sqrt {e x^{2} + d}\right )}^{2} + 4 \, \sqrt {2} {\left | d \right |} - 6 \, d \right |}}\right )}{16 \, d^{2} \sqrt {e} {\left | d \right |}} \]
1/12*x*(7*e*x^2/d^3 + 9/d^2)/(e*x^2 + d)^(3/2) + 1/16*sqrt(2)*log(abs(2*(s qrt(e)*x - sqrt(e*x^2 + d))^2 - 4*sqrt(2)*abs(d) - 6*d)/abs(2*(sqrt(e)*x - sqrt(e*x^2 + d))^2 + 4*sqrt(2)*abs(d) - 6*d))/(d^2*sqrt(e)*abs(d))
Timed out. \[ \int \frac {1}{\left (d+e x^2\right )^{3/2} \left (d^2-e^2 x^4\right )} \, dx=\int \frac {1}{\left (d^2-e^2\,x^4\right )\,{\left (e\,x^2+d\right )}^{3/2}} \,d x \]