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

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

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Rubi [A]  time = 0.06, 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 = {1150, 414, 522, 208, 205} \begin {gather*} \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 {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 205

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

Rule 208

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

Rule 414

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]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 522

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 1150

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[(d + e*x^2)^(p + q)*(a/d + (c*x^
2)/e)^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 = 0.04, 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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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (d+e x^2\right ) \left (d^2-e^2 x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.88, size = 189, normalized size = 2.62 \begin {gather*} \left [\frac {2 \, d e x + 4 \, {\left (e x^{2} + d\right )} \sqrt {d e} \arctan \left (\frac {\sqrt {d e} x}{d}\right ) + {\left (e x^{2} + d\right )} \sqrt {d e} \log \left (\frac {e x^{2} + 2 \, \sqrt {d e} x + d}{e x^{2} - d}\right )}{8 \, {\left (d^{3} e^{2} x^{2} + d^{4} e\right )}}, \frac {d e x - {\left (e x^{2} + d\right )} \sqrt {-d e} \arctan \left (\frac {\sqrt {-d e} x}{d}\right ) - {\left (e x^{2} + d\right )} \sqrt {-d e} \log \left (\frac {e x^{2} - 2 \, \sqrt {-d e} x - d}{e x^{2} + d}\right )}{4 \, {\left (d^{3} e^{2} x^{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*(2*d*e*x + 4*(e*x^2 + d)*sqrt(d*e)*arctan(sqrt(d*e)*x/d) + (e*x^2 + d)*sqrt(d*e)*log((e*x^2 + 2*sqrt(d*e)
*x + d)/(e*x^2 - d)))/(d^3*e^2*x^2 + d^4*e), 1/4*(d*e*x - (e*x^2 + d)*sqrt(-d*e)*arctan(sqrt(-d*e)*x/d) - (e*x
^2 + d)*sqrt(-d*e)*log((e*x^2 - 2*sqrt(-d*e)*x - d)/(e*x^2 + d)))/(d^3*e^2*x^2 + d^4*e)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \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]

Exception raised: NotImplementedError >> Unable to parse Giac output: -((d^2*exp(2)^3)^(1/4)*abs(d)*exp(1)^2-d
*exp(2)*(d^2*exp(2)^3)^(1/4))/(4*d^4*exp(2)*exp(1)^2-4*d^4*exp(2)^2)*ln(abs(x-(d^2/exp(2))^(1/4)))+((d^2*exp(2
)^3)^(1/4))^3/(4*d^4*exp(2)^2*exp(1)-4*d^4*exp(1)*exp(2)^2)*ln(abs(x+(d^2/exp(2))^(1/4)))-((d^2*exp(2)^3)^(1/4
)*abs(d)*exp(1)^2+d*exp(2)*(d^2*exp(2)^3)^(1/4))/(2*d^4*exp(2)*exp(1)^2-2*d^4*exp(2)^2)*atan(x/(d^2/exp(2))^(1
/4))-2*exp(1)^2*1/2/(exp(2)*d^2-d^2*exp(1)^2)/sqrt(d*exp(1))*atan(x*exp(1)/sqrt(d*exp(1)))

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maple [A]  time = 0.01, size = 55, normalized size = 0.76 \begin {gather*} \frac {x}{4 \left (e \,x^{2}+d \right ) d^{2}}+\frac {\arctanh \left (\frac {e x}{\sqrt {d e}}\right )}{4 \sqrt {d e}\, d^{2}}+\frac {\arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 \sqrt {d e}\, d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.44, size = 71, normalized size = 0.99 \begin {gather*} \frac {x}{4 \, {\left (d^{2} e x^{2} + d^{3}\right )}} + \frac {\arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 \, \sqrt {d e} d^{2}} - \frac {\log \left (\frac {e x - \sqrt {d e}}{e x + \sqrt {d e}}\right )}{8 \, \sqrt {d e} 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="maxima")

[Out]

1/4*x/(d^2*e*x^2 + d^3) + 1/2*arctan(e*x/sqrt(d*e))/(sqrt(d*e)*d^2) - 1/8*log((e*x - sqrt(d*e))/(e*x + sqrt(d*
e)))/(sqrt(d*e)*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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sympy [B]  time = 0.45, 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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