\(\int \frac {1}{(d+e x^2)^2 (d^2-e^2 x^4)} \, dx\) [194]

Optimal result

Integrand size = 24, antiderivative size = 89 \[ \int \frac {1}{\left (d+e x^2\right )^2 \left (d^2-e^2 x^4\right )} \, dx=\frac {x}{8 d^2 \left (d+e x^2\right )^2}+\frac {5 x}{16 d^3 \left (d+e x^2\right )}+\frac {7 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{16 d^{7/2} \sqrt {e}}+\frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{7/2} \sqrt {e}} \]

[Out]

1/8*x/d^2/(e*x^2+d)^2+5/16*x/d^3/(e*x^2+d)+7/16*arctan(x*e^(1/2)/d^(1/2))/d^(7/2)/e^(1/2)+1/8*arctanh(x*e^(1/2
)/d^(1/2))/d^(7/2)/e^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1164, 425, 541, 536, 214, 211} \[ \int \frac {1}{\left (d+e x^2\right )^2 \left (d^2-e^2 x^4\right )} \, dx=\frac {7 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{16 d^{7/2} \sqrt {e}}+\frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{7/2} \sqrt {e}}+\frac {5 x}{16 d^3 \left (d+e x^2\right )}+\frac {x}{8 d^2 \left (d+e x^2\right )^2} \]

[In]

Int[1/((d + e*x^2)^2*(d^2 - e^2*x^4)),x]

[Out]

x/(8*d^2*(d + e*x^2)^2) + (5*x)/(16*d^3*(d + e*x^2)) + (7*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(16*d^(7/2)*Sqrt[e]) +
ArcTanh[(Sqrt[e]*x)/Sqrt[d]]/(8*d^(7/2)*Sqrt[e])

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 214

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]

Rule 425

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*x*(a + b*x^n)^(p + 1)*
((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - a*d))), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1
)*(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c,
d, n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomi
alQ[a, b, c, d, n, p, q, x]

Rule 536

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 541

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[(
-(b*e - a*f))*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*n*(b*c - a*d)*(p + 1))), x] + Dist[1/(a*n*(b*c - a
*d)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*
f)*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 1164

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[(d + e*x^2)^(p + q)*(a/d + (c/e)
*x^2)^p, x] /; FreeQ[{a, c, d, e, q}, x] && EqQ[c*d^2 + a*e^2, 0] && IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{\left (d-e x^2\right ) \left (d+e x^2\right )^3} \, dx \\ & = \frac {x}{8 d^2 \left (d+e x^2\right )^2}-\frac {\int \frac {-7 d e+3 e^2 x^2}{\left (d-e x^2\right ) \left (d+e x^2\right )^2} \, dx}{8 d^2 e} \\ & = \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 {\int \frac {18 d^2 e^2-10 d e^3 x^2}{\left (d-e x^2\right ) \left (d+e x^2\right )} \, dx}{32 d^4 e^2} \\ & = \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 {\int \frac {1}{d-e x^2} \, dx}{8 d^3}+\frac {7 \int \frac {1}{d+e x^2} \, dx}{16 d^3} \\ & = \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 \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{16 d^{7/2} \sqrt {e}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{7/2} \sqrt {e}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\left (d+e x^2\right )^2 \left (d^2-e^2 x^4\right )} \, dx=\frac {\frac {\sqrt {d} x \left (7 d+5 e x^2\right )}{\left (d+e x^2\right )^2}+\frac {7 \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}}{16 d^{7/2}} \]

[In]

Integrate[1/((d + e*x^2)^2*(d^2 - e^2*x^4)),x]

[Out]

((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))

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.72

method	result	size
default	\(\frac {\operatorname {arctanh}\left (\frac {e x}{\sqrt {e d}}\right )}{8 d^{3} \sqrt {e d}}+\frac {\frac {\frac {5}{2} e \,x^{3}+\frac {7}{2} d x}{\left (e \,x^{2}+d \right )^{2}}+\frac {7 \arctan \left (\frac {e x}{\sqrt {e d}}\right )}{2 \sqrt {e d}}}{8 d^{3}}\)	\(64\)
risch	\(\frac {\frac {5 e \,x^{3}}{16 d^{3}}+\frac {7 x}{16 d^{2}}}{\left (e \,x^{2}+d \right )^{2}}-\frac {7 \ln \left (-e x -\sqrt {-e d}\right )}{32 \sqrt {-e d}\, d^{3}}+\frac {7 \ln \left (e x -\sqrt {-e d}\right )}{32 \sqrt {-e d}\, d^{3}}+\frac {\ln \left (e x +\sqrt {e d}\right )}{16 \sqrt {e d}\, d^{3}}-\frac {\ln \left (-e x +\sqrt {e d}\right )}{16 \sqrt {e d}\, d^{3}}\)	\(118\)

[In]

int(1/(e*x^2+d)^2/(-e^2*x^4+d^2),x,method=_RETURNVERBOSE)

[Out]

1/8/d^3/(e*d)^(1/2)*arctanh(e*x/(e*d)^(1/2))+1/8/d^3*((5/2*e*x^3+7/2*d*x)/(e*x^2+d)^2+7/2/(e*d)^(1/2)*arctan(e
*x/(e*d)^(1/2)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (65) = 130\).

Time = 0.27 (sec) , antiderivative size = 278, normalized size of antiderivative = 3.12 \[ \int \frac {1}{\left (d+e x^2\right )^2 \left (d^2-e^2 x^4\right )} \, dx=\left [\frac {5 \, d e^{2} x^{3} + 7 \, d^{2} e x + 7 \, {\left (e^{2} x^{4} + 2 \, d e x^{2} + d^{2}\right )} \sqrt {d e} \arctan \left (\frac {\sqrt {d e} x}{d}\right ) + {\left (e^{2} x^{4} + 2 \, d e x^{2} + d^{2}\right )} \sqrt {d e} \log \left (\frac {e x^{2} + 2 \, \sqrt {d e} x + d}{e x^{2} - d}\right )}{16 \, {\left (d^{4} e^{3} x^{4} + 2 \, d^{5} e^{2} x^{2} + d^{6} e\right )}}, \frac {10 \, d e^{2} x^{3} + 14 \, d^{2} e x - 4 \, {\left (e^{2} x^{4} + 2 \, d e x^{2} + d^{2}\right )} \sqrt {-d e} \arctan \left (\frac {\sqrt {-d e} x}{d}\right ) - 7 \, {\left (e^{2} x^{4} + 2 \, d e x^{2} + d^{2}\right )} \sqrt {-d e} \log \left (\frac {e x^{2} - 2 \, \sqrt {-d e} x - d}{e x^{2} + d}\right )}{32 \, {\left (d^{4} e^{3} x^{4} + 2 \, d^{5} e^{2} x^{2} + d^{6} e\right )}}\right ] \]

[In]

integrate(1/(e*x^2+d)^2/(-e^2*x^4+d^2),x, algorithm="fricas")

[Out]

[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)*arctan(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)*arctan(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)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 257 vs. \(2 (82) = 164\).

Time = 0.27 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.89 \[ \int \frac {1}{\left (d+e x^2\right )^2 \left (d^2-e^2 x^4\right )} \, dx=- \frac {\sqrt {\frac {1}{d^{7} e}} \log {\left (- \frac {20 d^{11} e \left (\frac {1}{d^{7} e}\right )^{\frac {3}{2}}}{371} - \frac {351 d^{4} \sqrt {\frac {1}{d^{7} e}}}{371} + x \right )}}{16} + \frac {\sqrt {\frac {1}{d^{7} e}} \log {\left (\frac {20 d^{11} e \left (\frac {1}{d^{7} e}\right )^{\frac {3}{2}}}{371} + \frac {351 d^{4} \sqrt {\frac {1}{d^{7} e}}}{371} + x \right )}}{16} - \frac {7 \sqrt {- \frac {1}{d^{7} e}} \log {\left (- \frac {245 d^{11} e \left (- \frac {1}{d^{7} e}\right )^{\frac {3}{2}}}{106} - \frac {351 d^{4} \sqrt {- \frac {1}{d^{7} e}}}{106} + x \right )}}{32} + \frac {7 \sqrt {- \frac {1}{d^{7} e}} \log {\left (\frac {245 d^{11} e \left (- \frac {1}{d^{7} e}\right )^{\frac {3}{2}}}{106} + \frac {351 d^{4} \sqrt {- \frac {1}{d^{7} e}}}{106} + x \right )}}{32} - \frac {- 7 d x - 5 e x^{3}}{16 d^{5} + 32 d^{4} e x^{2} + 16 d^{3} e^{2} x^{4}} \]

[In]

integrate(1/(e*x**2+d)**2/(-e**2*x**4+d**2),x)

[Out]

-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))*lo
g(-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)

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\left (d+e x^2\right )^2 \left (d^2-e^2 x^4\right )} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/(e*x^2+d)^2/(-e^2*x^4+d^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.74 \[ \int \frac {1}{\left (d+e x^2\right )^2 \left (d^2-e^2 x^4\right )} \, dx=\frac {7 \, \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{16 \, \sqrt {d e} d^{3}} - \frac {\arctan \left (\frac {e x}{\sqrt {-d e}}\right )}{8 \, \sqrt {-d e} d^{3}} + \frac {5 \, e x^{3} + 7 \, d x}{16 \, {\left (e x^{2} + d\right )}^{2} d^{3}} \]

[In]

integrate(1/(e*x^2+d)^2/(-e^2*x^4+d^2),x, algorithm="giac")

[Out]

7/16*arctan(e*x/sqrt(d*e))/(sqrt(d*e)*d^3) - 1/8*arctan(e*x/sqrt(-d*e))/(sqrt(-d*e)*d^3) + 1/16*(5*e*x^3 + 7*d
*x)/((e*x^2 + d)^2*d^3)

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.08 \[ \int \frac {1}{\left (d+e x^2\right )^2 \left (d^2-e^2 x^4\right )} \, dx=\frac {\frac {7\,x}{16\,d^2}+\frac {5\,e\,x^3}{16\,d^3}}{d^2+2\,d\,e\,x^2+e^2\,x^4}+\frac {\mathrm {atanh}\left (\frac {x\,\sqrt {d^7\,e}}{d^4}\right )\,\sqrt {d^7\,e}}{8\,d^7\,e}-\frac {7\,\mathrm {atanh}\left (\frac {x\,\sqrt {-d^7\,e}}{d^4}\right )\,\sqrt {-d^7\,e}}{16\,d^7\,e} \]

[In]

int(1/((d^2 - e^2*x^4)*(d + e*x^2)^2),x)

[Out]

((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)