3.2.93 \(\int \frac {1}{(d+e x^2) (d^2-e^2 x^4)} \, dx\) [193]

Optimal. Leaf size=72 \[ \frac {x}{4 d^2 \left (d+e x^2\right )}+\frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2} \sqrt {e}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{4 d^{5/2} \sqrt {e}} \]

[Out]

1/4*x/d^2/(e*x^2+d)+1/2*arctan(x*e^(1/2)/d^(1/2))/d^(5/2)/e^(1/2)+1/4*arctanh(x*e^(1/2)/d^(1/2))/d^(5/2)/e^(1/
2)

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Rubi [A]
time = 0.03, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1164, 425, 536, 214, 211} \begin {gather*} \frac {\text {ArcTan}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2} \sqrt {e}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{4 d^{5/2} \sqrt {e}}+\frac {x}{4 d^2 \left (d+e x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x/(4*d^2*(d + e*x^2)) + ArcTan[(Sqrt[e]*x)/Sqrt[d]]/(2*d^(5/2)*Sqrt[e]) + ArcTanh[(Sqrt[e]*x)/Sqrt[d]]/(4*d^(5
/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 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 ) \left (d^2-e^2 x^4\right )} \, dx &=\int \frac {1}{\left (d-e x^2\right ) \left (d+e x^2\right )^2} \, dx\\ &=\frac {x}{4 d^2 \left (d+e x^2\right )}-\frac {\int \frac {-3 d e+e^2 x^2}{\left (d-e x^2\right ) \left (d+e x^2\right )} \, dx}{4 d^2 e}\\ &=\frac {x}{4 d^2 \left (d+e x^2\right )}+\frac {\int \frac {1}{d-e x^2} \, dx}{4 d^2}+\frac {\int \frac {1}{d+e x^2} \, dx}{2 d^2}\\ &=\frac {x}{4 d^2 \left (d+e x^2\right )}+\frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2} \sqrt {e}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{4 d^{5/2} \sqrt {e}}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 65, normalized size = 0.90 \begin {gather*} \frac {\frac {\sqrt {d} x}{d+e x^2}+\frac {2 \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}}{4 d^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[d]*x)/(d + e*x^2) + (2*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/Sqrt[e] + ArcTanh[(Sqrt[e]*x)/Sqrt[d]]/Sqrt[e])/(4*
d^(5/2))

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Maple [A]
time = 0.16, size = 54, normalized size = 0.75

method	result	size
default	\(\frac {\frac {x}{e \,x^{2}+d}+\frac {2 \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{\sqrt {d e}}}{4 d^{2}}+\frac {\arctanh \left (\frac {e x}{\sqrt {d e}}\right )}{4 d^{2} \sqrt {d e}}\)	\(54\)
risch	\(\frac {x}{4 d^{2} \left (e \,x^{2}+d \right )}-\frac {\ln \left (-e x -\sqrt {-d e}\right )}{4 \sqrt {-d e}\, d^{2}}+\frac {\ln \left (e x -\sqrt {-d e}\right )}{4 \sqrt {-d e}\, d^{2}}+\frac {\ln \left (e x +\sqrt {d e}\right )}{8 \sqrt {d e}\, d^{2}}-\frac {\ln \left (-e x +\sqrt {d e}\right )}{8 \sqrt {d e}\, d^{2}}\)	\(107\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/d^2*(x/(e*x^2+d)+2/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2)))+1/4/d^2/(d*e)^(1/2)*arctanh(e*x/(d*e)^(1/2))

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Maxima [A]
time = 0.53, size = 68, normalized size = 0.94 \begin {gather*} \frac {x}{4 \, {\left (d^{2} x^{2} e + d^{3}\right )}} + \frac {\arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {1}{2}\right )}}{2 \, d^{\frac {5}{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 )}{8 \, d^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x/(d^2*x^2*e + d^3) + 1/2*arctan(x*e^(1/2)/sqrt(d))*e^(-1/2)/d^(5/2) - 1/8*e^(-1/2)*log((x*e - sqrt(d)*e^(
1/2))/(x*e + sqrt(d)*e^(1/2)))/d^(5/2)

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Fricas [A] Leaf count of result is larger than twice the leaf count of optimal. 105 vs. \(2 (47) = 94\).
time = 0.36, size = 200, normalized size = 2.78 \begin {gather*} \left [\frac {4 \, {\left (x^{2} e + d\right )} \sqrt {d} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\frac {1}{2}} + {\left (x^{2} e + d\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 ) + 2 \, d x e}{8 \, {\left (d^{3} x^{2} e^{2} + d^{4} e\right )}}, \frac {d x e - {\left (x^{2} e + d\right )} \sqrt {-d e} \arctan \left (\frac {\sqrt {-d e} x}{d}\right ) - {\left (x^{2} e + d\right )} \sqrt {-d e} \log \left (\frac {x^{2} e - 2 \, \sqrt {-d e} x - d}{x^{2} e + d}\right )}{4 \, {\left (d^{3} x^{2} e^{2} + d^{4} e\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(4*(x^2*e + d)*sqrt(d)*arctan(x*e^(1/2)/sqrt(d))*e^(1/2) + (x^2*e + d)*sqrt(d)*e^(1/2)*log((x^2*e + 2*sqr
t(d)*x*e^(1/2) + d)/(x^2*e - d)) + 2*d*x*e)/(d^3*x^2*e^2 + d^4*e), 1/4*(d*x*e - (x^2*e + d)*sqrt(-d*e)*arctan(
sqrt(-d*e)*x/d) - (x^2*e + d)*sqrt(-d*e)*log((x^2*e - 2*sqrt(-d*e)*x - d)/(x^2*e + d)))/(d^3*x^2*e^2 + d^4*e)]

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 226 vs. \(2 (63) = 126\).
time = 0.20, size = 226, normalized size = 3.14 \begin {gather*} \frac {x}{4 d^{3} + 4 d^{2} e x^{2}} - \frac {\sqrt {\frac {1}{d^{5} e}} \log {\left (- \frac {d^{8} e \left (\frac {1}{d^{5} e}\right )^{\frac {3}{2}}}{10} - \frac {9 d^{3} \sqrt {\frac {1}{d^{5} e}}}{10} + x \right )}}{8} + \frac {\sqrt {\frac {1}{d^{5} e}} \log {\left (\frac {d^{8} e \left (\frac {1}{d^{5} e}\right )^{\frac {3}{2}}}{10} + \frac {9 d^{3} \sqrt {\frac {1}{d^{5} e}}}{10} + x \right )}}{8} - \frac {\sqrt {- \frac {1}{d^{5} e}} \log {\left (- \frac {4 d^{8} e \left (- \frac {1}{d^{5} e}\right )^{\frac {3}{2}}}{5} - \frac {9 d^{3} \sqrt {- \frac {1}{d^{5} e}}}{5} + x \right )}}{4} + \frac {\sqrt {- \frac {1}{d^{5} e}} \log {\left (\frac {4 d^{8} e \left (- \frac {1}{d^{5} e}\right )^{\frac {3}{2}}}{5} + \frac {9 d^{3} \sqrt {- \frac {1}{d^{5} e}}}{5} + x \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(4*d**3 + 4*d**2*e*x**2) - sqrt(1/(d**5*e))*log(-d**8*e*(1/(d**5*e))**(3/2)/10 - 9*d**3*sqrt(1/(d**5*e))/10
+ x)/8 + sqrt(1/(d**5*e))*log(d**8*e*(1/(d**5*e))**(3/2)/10 + 9*d**3*sqrt(1/(d**5*e))/10 + x)/8 - sqrt(-1/(d**
5*e))*log(-4*d**8*e*(-1/(d**5*e))**(3/2)/5 - 9*d**3*sqrt(-1/(d**5*e))/5 + x)/4 + sqrt(-1/(d**5*e))*log(4*d**8*
e*(-1/(d**5*e))**(3/2)/5 + 9*d**3*sqrt(-1/(d**5*e))/5 + x)/4

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Giac [A]
time = 3.03, size = 56, normalized size = 0.78 \begin {gather*} \frac {\arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {1}{2}\right )}}{2 \, d^{\frac {5}{2}}} - \frac {\arctan \left (\frac {x e}{\sqrt {-d e}}\right )}{4 \, \sqrt {-d e} d^{2}} + \frac {x}{4 \, {\left (x^{2} e + d\right )} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.16, size = 74, normalized size = 1.03 \begin {gather*} \frac {x}{4\,d^2\,\left (e\,x^2+d\right )}+\frac {\mathrm {atanh}\left (\frac {x\,\sqrt {d^5\,e}}{d^3}\right )\,\sqrt {d^5\,e}}{4\,d^5\,e}-\frac {\mathrm {atanh}\left (\frac {x\,\sqrt {-d^5\,e}}{d^3}\right )\,\sqrt {-d^5\,e}}{2\,d^5\,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)),x)

[Out]

x/(4*d^2*(d + e*x^2)) + (atanh((x*(d^5*e)^(1/2))/d^3)*(d^5*e)^(1/2))/(4*d^5*e) - (atanh((x*(-d^5*e)^(1/2))/d^3
)*(-d^5*e)^(1/2))/(2*d^5*e)

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