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

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

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Rubi [A]  time = 0.08, 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 = {1150, 414, 527, 522, 208, 205} \begin {gather*} \frac {5 x}{16 d^3 \left (d+e x^2\right )}+\frac {x}{8 d^2 \left (d+e x^2\right )^2}+\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 {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 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 527

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 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 )^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 = 0.06, 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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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (d+e x^2\right )^2 \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)^2*(d^2 - e^2*x^4)),x]

[Out]

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

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fricas [B]  time = 1.24, size = 278, normalized size = 3.12 \begin {gather*} \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 ] \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*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)]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: -(-2*(d^2*exp(2)^3)^(1/4)*abs(d)*exp(1)^
2+d*(d^2*exp(2)^3)^(1/4)*exp(1)^2+d*exp(2)*(d^2*exp(2)^3)^(1/4))/(4*d^5*exp(1)^4-8*d^5*exp(2)*exp(1)^2+4*d^5*e
xp(2)^2)*ln(abs(x-(d^2/exp(2))^(1/4)))+exp(2)*(d^2*exp(2)^3)^(1/4)/(4*d^4*exp(2)*exp(1)^2-8*d^4*exp(1)*exp(2)*
exp(1)+4*d^4*exp(2)^2)*ln(abs(x+(d^2/exp(2))^(1/4)))-(-2*(d^2*exp(2)^3)^(1/4)*abs(d)*exp(1)^2-d*(d^2*exp(2)^3)
^(1/4)*exp(1)^2-d*exp(2)*(d^2*exp(2)^3)^(1/4))/(2*d^5*exp(1)^4-4*d^5*exp(2)*exp(1)^2+2*d^5*exp(2)^2)*atan(x/(d
^2/exp(2))^(1/4))-(-5*exp(2)*exp(1)^2+exp(1)^4)*1/2/(-exp(2)^2*d^3+2*exp(2)*d^3*exp(1)^2-d^3*exp(1)^4)/sqrt(d*
exp(1))*atan(x*exp(1)/sqrt(d*exp(1)))+x*exp(1)^2/(-2*exp(2)*d^3+2*d^3*exp(1)^2)/(x^2*exp(1)+d)

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maple [A]  time = 0.01, size = 73, normalized size = 0.82 \begin {gather*} \frac {5 e \,x^{3}}{16 \left (e \,x^{2}+d \right )^{2} d^{3}}+\frac {7 x}{16 \left (e \,x^{2}+d \right )^{2} d^{2}}+\frac {\arctanh \left (\frac {e x}{\sqrt {d e}}\right )}{8 \sqrt {d e}\, d^{3}}+\frac {7 \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{16 \sqrt {d e}\, d^{3}} \end {gather*}

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)

[Out]

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

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maxima [A]  time = 2.49, size = 92, normalized size = 1.03 \begin {gather*} \frac {5 \, e x^{3} + 7 \, d x}{16 \, {\left (d^{3} e^{2} x^{4} + 2 \, d^{4} e x^{2} + d^{5}\right )}} + \frac {7 \, \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{16 \, \sqrt {d e} d^{3}} - \frac {\log \left (\frac {e x - \sqrt {d e}}{e x + \sqrt {d e}}\right )}{16 \, \sqrt {d e} 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="maxima")

[Out]

1/16*(5*e*x^3 + 7*d*x)/(d^3*e^2*x^4 + 2*d^4*e*x^2 + d^5) + 7/16*arctan(e*x/sqrt(d*e))/(sqrt(d*e)*d^3) - 1/16*l
og((e*x - sqrt(d*e))/(e*x + sqrt(d*e)))/(sqrt(d*e)*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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sympy [B]  time = 0.72, 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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