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

Optimal. Leaf size=89 \[ \frac {x}{8 d^2 \left (d+e x^2\right )^2}+\frac {5 x}{16 d^3 \left (d+e x^2\right )}+\frac {7 \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{16 d^{7/2} \sqrt {e}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{7/2} \sqrt {e}} \]

[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)

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Rubi [A]
time = 0.05, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1164, 425, 541, 536, 214, 211} \begin {gather*} \frac {7 \text {ArcTan}\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}}+\frac {5 x}{16 d^3 \left (d+e x^2\right )}+\frac {x}{8 d^2 \left (d+e x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[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*} \int \frac {1}{\left (d+e x^2\right )^2 \left (d^2-e^2 x^4\right )} \, dx &=\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*}

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Mathematica [A]
time = 0.04, size = 76, normalized size = 0.85 \begin {gather*} \frac {\frac {\sqrt {d} x \left (7 d+5 e x^2\right )}{\left (d+e x^2\right )^2}+\frac {7 \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}}{16 d^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[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))

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Maple [A]
time = 0.15, size = 64, normalized size = 0.72

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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))

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Maxima [A]
time = 0.52, size = 89, normalized size = 1.00 \begin {gather*} \frac {5 \, x^{3} e + 7 \, d x}{16 \, {\left (d^{3} x^{4} e^{2} + 2 \, d^{4} x^{2} e + d^{5}\right )}} + \frac {7 \, \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {1}{2}\right )}}{16 \, d^{\frac {7}{2}}} - \frac {e^{\left (-\frac {1}{2}\right )} \log \left (\frac {x e - \sqrt {d} e^{\frac {1}{2}}}{x e + \sqrt {d} e^{\frac {1}{2}}}\right )}{16 \, d^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(5*x^3*e + 7*d*x)/(d^3*x^4*e^2 + 2*d^4*x^2*e + d^5) + 7/16*arctan(x*e^(1/2)/sqrt(d))*e^(-1/2)/d^(7/2) - 1
/16*e^(-1/2)*log((x*e - sqrt(d)*e^(1/2))/(x*e + sqrt(d)*e^(1/2)))/d^(7/2)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 146 vs. \(2 (63) = 126\).
time = 0.38, size = 281, normalized size = 3.16 \begin {gather*} \left [\frac {5 \, d x^{3} e^{2} + 7 \, d^{2} x e + 7 \, {\left (x^{4} e^{2} + 2 \, d x^{2} e + d^{2}\right )} \sqrt {d} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\frac {1}{2}} + {\left (x^{4} e^{2} + 2 \, d x^{2} e + d^{2}\right )} \sqrt {d} e^{\frac {1}{2}} \log \left (\frac {x^{2} e + 2 \, \sqrt {d} x e^{\frac {1}{2}} + d}{x^{2} e - d}\right )}{16 \, {\left (d^{4} x^{4} e^{3} + 2 \, d^{5} x^{2} e^{2} + d^{6} e\right )}}, \frac {10 \, d x^{3} e^{2} + 14 \, d^{2} x e - 4 \, {\left (x^{4} e^{2} + 2 \, d x^{2} e + d^{2}\right )} \sqrt {-d e} \arctan \left (\frac {\sqrt {-d e} x}{d}\right ) - 7 \, {\left (x^{4} e^{2} + 2 \, d x^{2} e + d^{2}\right )} \sqrt {-d e} \log \left (\frac {x^{2} e - 2 \, \sqrt {-d e} x - d}{x^{2} e + d}\right )}{32 \, {\left (d^{4} x^{4} e^{3} + 2 \, d^{5} x^{2} e^{2} + d^{6} e\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(5*d*x^3*e^2 + 7*d^2*x*e + 7*(x^4*e^2 + 2*d*x^2*e + d^2)*sqrt(d)*arctan(x*e^(1/2)/sqrt(d))*e^(1/2) + (x^
4*e^2 + 2*d*x^2*e + d^2)*sqrt(d)*e^(1/2)*log((x^2*e + 2*sqrt(d)*x*e^(1/2) + d)/(x^2*e - d)))/(d^4*x^4*e^3 + 2*
d^5*x^2*e^2 + d^6*e), 1/32*(10*d*x^3*e^2 + 14*d^2*x*e - 4*(x^4*e^2 + 2*d*x^2*e + d^2)*sqrt(-d*e)*arctan(sqrt(-
d*e)*x/d) - 7*(x^4*e^2 + 2*d*x^2*e + d^2)*sqrt(-d*e)*log((x^2*e - 2*sqrt(-d*e)*x - d)/(x^2*e + d)))/(d^4*x^4*e
^3 + 2*d^5*x^2*e^2 + d^6*e)]

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 257 vs. \(2 (82) = 164\).
time = 0.27, size = 257, normalized size = 2.89 \begin {gather*} - \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]
time = 2.63, size = 67, normalized size = 0.75 \begin {gather*} \frac {7 \, \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {1}{2}\right )}}{16 \, d^{\frac {7}{2}}} - \frac {\arctan \left (\frac {x e}{\sqrt {-d e}}\right )}{8 \, \sqrt {-d e} d^{3}} + \frac {5 \, x^{3} e + 7 \, d x}{16 \, {\left (x^{2} e + d\right )}^{2} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.16, size = 96, normalized size = 1.08 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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