Integrand size = 24, antiderivative size = 89 \[ \int \frac {1}{\left (d+e x^2\right )^2 \left (d^2-e^2 x^4\right )} \, dx=\frac {x}{8 d^2 \left (d+e x^2\right )^2}+\frac {5 x}{16 d^3 \left (d+e x^2\right )}+\frac {7 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{16 d^{7/2} \sqrt {e}}+\frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{7/2} \sqrt {e}} \] Output:
1/8*x/d^2/(e*x^2+d)^2+5/16*x/d^3/(e*x^2+d)+7/16*arctan(e^(1/2)*x/d^(1/2))/ d^(7/2)/e^(1/2)+1/8*arctanh(e^(1/2)*x/d^(1/2))/d^(7/2)/e^(1/2)
Time = 0.06 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\left (d+e x^2\right )^2 \left (d^2-e^2 x^4\right )} \, dx=\frac {\frac {\sqrt {d} x \left (7 d+5 e x^2\right )}{\left (d+e x^2\right )^2}+\frac {7 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}}{16 d^{7/2}} \] Input:
Integrate[1/((d + e*x^2)^2*(d^2 - e^2*x^4)),x]
Output:
((Sqrt[d]*x*(7*d + 5*e*x^2))/(d + e*x^2)^2 + (7*ArcTan[(Sqrt[e]*x)/Sqrt[d] ])/Sqrt[e] + (2*ArcTanh[(Sqrt[e]*x)/Sqrt[d]])/Sqrt[e])/(16*d^(7/2))
Time = 0.41 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.13, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {1388, 316, 25, 27, 402, 27, 397, 218, 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.
\(\displaystyle \int \frac {1}{\left (d+e x^2\right )^2 \left (d^2-e^2 x^4\right )} \, dx\) |
\(\Big \downarrow \) 1388 |
\(\displaystyle \int \frac {1}{\left (d-e x^2\right ) \left (d+e x^2\right )^3}dx\) |
\(\Big \downarrow \) 316 |
\(\displaystyle \frac {x}{8 d^2 \left (d+e x^2\right )^2}-\frac {\int -\frac {e \left (7 d-3 e x^2\right )}{\left (d-e x^2\right ) \left (e x^2+d\right )^2}dx}{8 d^2 e}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {e \left (7 d-3 e x^2\right )}{\left (d-e x^2\right ) \left (e x^2+d\right )^2}dx}{8 d^2 e}+\frac {x}{8 d^2 \left (d+e x^2\right )^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {7 d-3 e x^2}{\left (d-e x^2\right ) \left (e x^2+d\right )^2}dx}{8 d^2}+\frac {x}{8 d^2 \left (d+e x^2\right )^2}\) |
\(\Big \downarrow \) 402 |
\(\displaystyle \frac {\frac {5 x}{2 d \left (d+e x^2\right )}-\frac {\int -\frac {2 d e \left (9 d-5 e x^2\right )}{\left (d-e x^2\right ) \left (e x^2+d\right )}dx}{4 d^2 e}}{8 d^2}+\frac {x}{8 d^2 \left (d+e x^2\right )^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int \frac {9 d-5 e x^2}{\left (d-e x^2\right ) \left (e x^2+d\right )}dx}{2 d}+\frac {5 x}{2 d \left (d+e x^2\right )}}{8 d^2}+\frac {x}{8 d^2 \left (d+e x^2\right )^2}\) |
\(\Big \downarrow \) 397 |
\(\displaystyle \frac {\frac {2 \int \frac {1}{d-e x^2}dx+7 \int \frac {1}{e x^2+d}dx}{2 d}+\frac {5 x}{2 d \left (d+e x^2\right )}}{8 d^2}+\frac {x}{8 d^2 \left (d+e x^2\right )^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {2 \int \frac {1}{d-e x^2}dx+\frac {7 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} \sqrt {e}}}{2 d}+\frac {5 x}{2 d \left (d+e x^2\right )}}{8 d^2}+\frac {x}{8 d^2 \left (d+e x^2\right )^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {\frac {7 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} \sqrt {e}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} \sqrt {e}}}{2 d}+\frac {5 x}{2 d \left (d+e x^2\right )}}{8 d^2}+\frac {x}{8 d^2 \left (d+e x^2\right )^2}\) |
Input:
Int[1/((d + e*x^2)^2*(d^2 - e^2*x^4)),x]
Output:
x/(8*d^2*(d + e*x^2)^2) + ((5*x)/(2*d*(d + e*x^2)) + ((7*ArcTan[(Sqrt[e]*x )/Sqrt[d]])/(Sqrt[d]*Sqrt[e]) + (2*ArcTanh[(Sqrt[e]*x)/Sqrt[d]])/(Sqrt[d]* Sqrt[e]))/(2*d))/(8*d^2)
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)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
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[((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[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, 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 = 0.30 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.72
method | result | size |
default | \(\frac {\frac {\frac {5}{2} e \,x^{3}+\frac {7}{2} d x}{\left (e \,x^{2}+d \right )^{2}}+\frac {7 \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 \sqrt {d e}}}{8 d^{3}}+\frac {\operatorname {arctanh}\left (\frac {e x}{\sqrt {d e}}\right )}{8 d^{3} \sqrt {d e}}\) | \(64\) |
risch | \(\frac {\frac {5 e \,x^{3}}{16 d^{3}}+\frac {7 x}{16 d^{2}}}{\left (e \,x^{2}+d \right )^{2}}+\frac {\ln \left (e x +\sqrt {d e}\right )}{16 \sqrt {d e}\, d^{3}}-\frac {\ln \left (-e x +\sqrt {d e}\right )}{16 \sqrt {d e}\, d^{3}}-\frac {7 \ln \left (-e x -\sqrt {-d e}\right )}{32 \sqrt {-d e}\, d^{3}}+\frac {7 \ln \left (e x -\sqrt {-d e}\right )}{32 \sqrt {-d e}\, d^{3}}\) | \(118\) |
Input:
int(1/(e*x^2+d)^2/(-e^2*x^4+d^2),x,method=_RETURNVERBOSE)
Output:
1/8/d^3*((5/2*e*x^3+7/2*d*x)/(e*x^2+d)^2+7/2/(d*e)^(1/2)*arctan(e*x/(d*e)^ (1/2)))+1/8/d^3/(d*e)^(1/2)*arctanh(e*x/(d*e)^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (65) = 130\).
Time = 0.09 (sec) , antiderivative size = 278, normalized size of antiderivative = 3.12 \[ \int \frac {1}{\left (d+e x^2\right )^2 \left (d^2-e^2 x^4\right )} \, dx=\left [\frac {5 \, d e^{2} x^{3} + 7 \, d^{2} e x + 7 \, {\left (e^{2} x^{4} + 2 \, d e x^{2} + d^{2}\right )} \sqrt {d e} \arctan \left (\frac {\sqrt {d e} x}{d}\right ) + {\left (e^{2} x^{4} + 2 \, d e x^{2} + d^{2}\right )} \sqrt {d e} \log \left (\frac {e x^{2} + 2 \, \sqrt {d e} x + d}{e x^{2} - d}\right )}{16 \, {\left (d^{4} e^{3} x^{4} + 2 \, d^{5} e^{2} x^{2} + d^{6} e\right )}}, \frac {10 \, d e^{2} x^{3} + 14 \, d^{2} e x - 4 \, {\left (e^{2} x^{4} + 2 \, d e x^{2} + d^{2}\right )} \sqrt {-d e} \arctan \left (\frac {\sqrt {-d e} x}{d}\right ) - 7 \, {\left (e^{2} x^{4} + 2 \, d e x^{2} + d^{2}\right )} \sqrt {-d e} \log \left (\frac {e x^{2} - 2 \, \sqrt {-d e} x - d}{e x^{2} + d}\right )}{32 \, {\left (d^{4} e^{3} x^{4} + 2 \, d^{5} e^{2} x^{2} + d^{6} e\right )}}\right ] \] Input:
integrate(1/(e*x^2+d)^2/(-e^2*x^4+d^2),x, algorithm="fricas")
Output:
[1/16*(5*d*e^2*x^3 + 7*d^2*e*x + 7*(e^2*x^4 + 2*d*e*x^2 + d^2)*sqrt(d*e)*a rctan(sqrt(d*e)*x/d) + (e^2*x^4 + 2*d*e*x^2 + d^2)*sqrt(d*e)*log((e*x^2 + 2*sqrt(d*e)*x + d)/(e*x^2 - d)))/(d^4*e^3*x^4 + 2*d^5*e^2*x^2 + d^6*e), 1/ 32*(10*d*e^2*x^3 + 14*d^2*e*x - 4*(e^2*x^4 + 2*d*e*x^2 + d^2)*sqrt(-d*e)*a rctan(sqrt(-d*e)*x/d) - 7*(e^2*x^4 + 2*d*e*x^2 + d^2)*sqrt(-d*e)*log((e*x^ 2 - 2*sqrt(-d*e)*x - d)/(e*x^2 + d)))/(d^4*e^3*x^4 + 2*d^5*e^2*x^2 + d^6*e )]
Leaf count of result is larger than twice the leaf count of optimal. 257 vs. \(2 (82) = 164\).
Time = 0.29 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.89 \[ \int \frac {1}{\left (d+e x^2\right )^2 \left (d^2-e^2 x^4\right )} \, dx=- \frac {\sqrt {\frac {1}{d^{7} e}} \log {\left (- \frac {20 d^{11} e \left (\frac {1}{d^{7} e}\right )^{\frac {3}{2}}}{371} - \frac {351 d^{4} \sqrt {\frac {1}{d^{7} e}}}{371} + x \right )}}{16} + \frac {\sqrt {\frac {1}{d^{7} e}} \log {\left (\frac {20 d^{11} e \left (\frac {1}{d^{7} e}\right )^{\frac {3}{2}}}{371} + \frac {351 d^{4} \sqrt {\frac {1}{d^{7} e}}}{371} + x \right )}}{16} - \frac {7 \sqrt {- \frac {1}{d^{7} e}} \log {\left (- \frac {245 d^{11} e \left (- \frac {1}{d^{7} e}\right )^{\frac {3}{2}}}{106} - \frac {351 d^{4} \sqrt {- \frac {1}{d^{7} e}}}{106} + x \right )}}{32} + \frac {7 \sqrt {- \frac {1}{d^{7} e}} \log {\left (\frac {245 d^{11} e \left (- \frac {1}{d^{7} e}\right )^{\frac {3}{2}}}{106} + \frac {351 d^{4} \sqrt {- \frac {1}{d^{7} e}}}{106} + x \right )}}{32} - \frac {- 7 d x - 5 e x^{3}}{16 d^{5} + 32 d^{4} e x^{2} + 16 d^{3} e^{2} x^{4}} \] Input:
integrate(1/(e*x**2+d)**2/(-e**2*x**4+d**2),x)
Output:
-sqrt(1/(d**7*e))*log(-20*d**11*e*(1/(d**7*e))**(3/2)/371 - 351*d**4*sqrt( 1/(d**7*e))/371 + x)/16 + sqrt(1/(d**7*e))*log(20*d**11*e*(1/(d**7*e))**(3 /2)/371 + 351*d**4*sqrt(1/(d**7*e))/371 + x)/16 - 7*sqrt(-1/(d**7*e))*log( -245*d**11*e*(-1/(d**7*e))**(3/2)/106 - 351*d**4*sqrt(-1/(d**7*e))/106 + x )/32 + 7*sqrt(-1/(d**7*e))*log(245*d**11*e*(-1/(d**7*e))**(3/2)/106 + 351* d**4*sqrt(-1/(d**7*e))/106 + x)/32 - (-7*d*x - 5*e*x**3)/(16*d**5 + 32*d** 4*e*x**2 + 16*d**3*e**2*x**4)
Exception generated. \[ \int \frac {1}{\left (d+e x^2\right )^2 \left (d^2-e^2 x^4\right )} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/(e*x^2+d)^2/(-e^2*x^4+d^2),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
Time = 0.10 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.74 \[ \int \frac {1}{\left (d+e x^2\right )^2 \left (d^2-e^2 x^4\right )} \, dx=\frac {7 \, \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{16 \, \sqrt {d e} d^{3}} - \frac {\arctan \left (\frac {e x}{\sqrt {-d e}}\right )}{8 \, \sqrt {-d e} d^{3}} + \frac {5 \, e x^{3} + 7 \, d x}{16 \, {\left (e x^{2} + d\right )}^{2} d^{3}} \] Input:
integrate(1/(e*x^2+d)^2/(-e^2*x^4+d^2),x, algorithm="giac")
Output:
7/16*arctan(e*x/sqrt(d*e))/(sqrt(d*e)*d^3) - 1/8*arctan(e*x/sqrt(-d*e))/(s qrt(-d*e)*d^3) + 1/16*(5*e*x^3 + 7*d*x)/((e*x^2 + d)^2*d^3)
Time = 0.18 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.08 \[ \int \frac {1}{\left (d+e x^2\right )^2 \left (d^2-e^2 x^4\right )} \, dx=\frac {\frac {7\,x}{16\,d^2}+\frac {5\,e\,x^3}{16\,d^3}}{d^2+2\,d\,e\,x^2+e^2\,x^4}+\frac {\mathrm {atanh}\left (\frac {x\,\sqrt {d^7\,e}}{d^4}\right )\,\sqrt {d^7\,e}}{8\,d^7\,e}-\frac {7\,\mathrm {atanh}\left (\frac {x\,\sqrt {-d^7\,e}}{d^4}\right )\,\sqrt {-d^7\,e}}{16\,d^7\,e} \] Input:
int(1/((d^2 - e^2*x^4)*(d + e*x^2)^2),x)
Output:
((7*x)/(16*d^2) + (5*e*x^3)/(16*d^3))/(d^2 + e^2*x^4 + 2*d*e*x^2) + (atanh ((x*(d^7*e)^(1/2))/d^4)*(d^7*e)^(1/2))/(8*d^7*e) - (7*atanh((x*(-d^7*e)^(1 /2))/d^4)*(-d^7*e)^(1/2))/(16*d^7*e)
Time = 0.18 (sec) , antiderivative size = 244, normalized size of antiderivative = 2.74 \[ \int \frac {1}{\left (d+e x^2\right )^2 \left (d^2-e^2 x^4\right )} \, dx=\frac {7 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) d^{2}+14 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) d e \,x^{2}+7 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) e^{2} x^{4}+\sqrt {e}\, \sqrt {d}\, \mathrm {log}\left (-\sqrt {e}\, \sqrt {d}-e x \right ) d^{2}+2 \sqrt {e}\, \sqrt {d}\, \mathrm {log}\left (-\sqrt {e}\, \sqrt {d}-e x \right ) d e \,x^{2}+\sqrt {e}\, \sqrt {d}\, \mathrm {log}\left (-\sqrt {e}\, \sqrt {d}-e x \right ) e^{2} x^{4}-\sqrt {e}\, \sqrt {d}\, \mathrm {log}\left (\sqrt {e}\, \sqrt {d}-e x \right ) d^{2}-2 \sqrt {e}\, \sqrt {d}\, \mathrm {log}\left (\sqrt {e}\, \sqrt {d}-e x \right ) d e \,x^{2}-\sqrt {e}\, \sqrt {d}\, \mathrm {log}\left (\sqrt {e}\, \sqrt {d}-e x \right ) e^{2} x^{4}+7 d^{2} e x +5 d \,e^{2} x^{3}}{16 d^{4} e \left (e^{2} x^{4}+2 d e \,x^{2}+d^{2}\right )} \] Input:
int(1/(e*x^2+d)^2/(-e^2*x^4+d^2),x)
Output:
(7*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*d**2 + 14*sqrt(e)*sqrt(d) *atan((e*x)/(sqrt(e)*sqrt(d)))*d*e*x**2 + 7*sqrt(e)*sqrt(d)*atan((e*x)/(sq rt(e)*sqrt(d)))*e**2*x**4 + sqrt(e)*sqrt(d)*log( - sqrt(e)*sqrt(d) - e*x)* d**2 + 2*sqrt(e)*sqrt(d)*log( - sqrt(e)*sqrt(d) - e*x)*d*e*x**2 + sqrt(e)* sqrt(d)*log( - sqrt(e)*sqrt(d) - e*x)*e**2*x**4 - sqrt(e)*sqrt(d)*log(sqrt (e)*sqrt(d) - e*x)*d**2 - 2*sqrt(e)*sqrt(d)*log(sqrt(e)*sqrt(d) - e*x)*d*e *x**2 - sqrt(e)*sqrt(d)*log(sqrt(e)*sqrt(d) - e*x)*e**2*x**4 + 7*d**2*e*x + 5*d*e**2*x**3)/(16*d**4*e*(d**2 + 2*d*e*x**2 + e**2*x**4))