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

Optimal. Leaf size=61 \[ \frac {x}{2 d^2 \sqrt {d+e x^2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {e} x}{\sqrt {d+e x^2}}\right )}{2 \sqrt {2} d^2 \sqrt {e}} \]

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Rubi [A]  time = 0.04, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1150, 382, 377, 208} \begin {gather*} \frac {x}{2 d^2 \sqrt {d+e x^2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {e} x}{\sqrt {d+e x^2}}\right )}{2 \sqrt {2} d^2 \sqrt {e}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 382

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[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a*d
)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && EqQ[
n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[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}{\sqrt {d+e x^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 )^{3/2}} \, dx\\ &=\frac {x}{2 d^2 \sqrt {d+e x^2}}+\frac {\int \frac {1}{\left (d-e x^2\right ) \sqrt {d+e x^2}} \, dx}{2 d}\\ &=\frac {x}{2 d^2 \sqrt {d+e x^2}}+\frac {\operatorname {Subst}\left (\int \frac {1}{d-2 d e x^2} \, dx,x,\frac {x}{\sqrt {d+e x^2}}\right )}{2 d}\\ &=\frac {x}{2 d^2 \sqrt {d+e x^2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {e} x}{\sqrt {d+e x^2}}\right )}{2 \sqrt {2} d^2 \sqrt {e}}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 108, normalized size = 1.77 \begin {gather*} \frac {\frac {4 x}{\sqrt {d+e x^2}}-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {d}-\sqrt {e} x}{\sqrt {2} \sqrt {d+e x^2}}\right )}{\sqrt {e}}+\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {d}+\sqrt {e} x}{\sqrt {2} \sqrt {d+e x^2}}\right )}{\sqrt {e}}}{8 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((4*x)/Sqrt[d + e*x^2] - (Sqrt[2]*ArcTanh[(Sqrt[d] - Sqrt[e]*x)/(Sqrt[2]*Sqrt[d + e*x^2])])/Sqrt[e] + (Sqrt[2]
*ArcTanh[(Sqrt[d] + Sqrt[e]*x)/(Sqrt[2]*Sqrt[d + e*x^2])])/Sqrt[e])/(8*d^2)

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IntegrateAlgebraic [A]  time = 0.15, size = 84, normalized size = 1.38 \begin {gather*} \frac {x}{2 d^2 \sqrt {d+e x^2}}+\frac {\tanh ^{-1}\left (-\frac {e x^2}{\sqrt {2} d}+\frac {\sqrt {e} x \sqrt {d+e x^2}}{\sqrt {2} d}+\frac {1}{\sqrt {2}}\right )}{2 \sqrt {2} d^2 \sqrt {e}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x/(2*d^2*Sqrt[d + e*x^2]) + ArcTanh[1/Sqrt[2] - (e*x^2)/(Sqrt[2]*d) + (Sqrt[e]*x*Sqrt[d + e*x^2])/(Sqrt[2]*d)]
/(2*Sqrt[2]*d^2*Sqrt[e])

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fricas [B]  time = 1.08, size = 209, normalized size = 3.43 \begin {gather*} \left [\frac {\sqrt {2} {\left (e x^{2} + d\right )} \sqrt {e} \log \left (\frac {17 \, e^{2} x^{4} + 14 \, d e x^{2} + 4 \, \sqrt {2} {\left (3 \, e x^{3} + d x\right )} \sqrt {e x^{2} + d} \sqrt {e} + d^{2}}{e^{2} x^{4} - 2 \, d e x^{2} + d^{2}}\right ) + 8 \, \sqrt {e x^{2} + d} e x}{16 \, {\left (d^{2} e^{2} x^{2} + d^{3} e\right )}}, -\frac {\sqrt {2} {\left (e x^{2} + d\right )} \sqrt {-e} \arctan \left (\frac {\sqrt {2} {\left (3 \, e x^{2} + d\right )} \sqrt {e x^{2} + d} \sqrt {-e}}{4 \, {\left (e^{2} x^{3} + d e x\right )}}\right ) - 4 \, \sqrt {e x^{2} + d} e x}{8 \, {\left (d^{2} e^{2} x^{2} + d^{3} e\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(sqrt(2)*(e*x^2 + d)*sqrt(e)*log((17*e^2*x^4 + 14*d*e*x^2 + 4*sqrt(2)*(3*e*x^3 + d*x)*sqrt(e*x^2 + d)*sq
rt(e) + d^2)/(e^2*x^4 - 2*d*e*x^2 + d^2)) + 8*sqrt(e*x^2 + d)*e*x)/(d^2*e^2*x^2 + d^3*e), -1/8*(sqrt(2)*(e*x^2
+ d)*sqrt(-e)*arctan(1/4*sqrt(2)*(3*e*x^2 + d)*sqrt(e*x^2 + d)*sqrt(-e)/(e^2*x^3 + d*e*x)) - 4*sqrt(e*x^2 + d
)*e*x)/(d^2*e^2*x^2 + d^3*e)]

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giac [A]  time = 0.33, size = 1, normalized size = 0.02 \begin {gather*} +\infty \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

+Infinity

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maple [B]  time = 0.02, size = 441, normalized size = 7.23 \begin {gather*} -\frac {\sqrt {2}\, e \ln \left (\frac {4 d +2 \sqrt {2}\, \sqrt {2 d +\left (x -\frac {\sqrt {d e}}{e}\right )^{2} e +2 \sqrt {d e}\, \left (x -\frac {\sqrt {d e}}{e}\right )}\, \sqrt {d}+2 \sqrt {d e}\, \left (x -\frac {\sqrt {d e}}{e}\right )}{x -\frac {\sqrt {d e}}{e}}\right )}{4 \sqrt {d e}\, \left (\sqrt {-d e}+\sqrt {d e}\right ) \left (\sqrt {-d e}-\sqrt {d e}\right ) \sqrt {d}}+\frac {\sqrt {2}\, e \ln \left (\frac {4 d +2 \sqrt {2}\, \sqrt {2 d +\left (x +\frac {\sqrt {d e}}{e}\right )^{2} e -2 \sqrt {d e}\, \left (x +\frac {\sqrt {d e}}{e}\right )}\, \sqrt {d}-2 \sqrt {d e}\, \left (x +\frac {\sqrt {d e}}{e}\right )}{x +\frac {\sqrt {d e}}{e}}\right )}{4 \sqrt {d e}\, \left (\sqrt {-d e}+\sqrt {d e}\right ) \left (\sqrt {-d e}-\sqrt {d e}\right ) \sqrt {d}}-\frac {\sqrt {\left (x -\frac {\sqrt {-d e}}{e}\right )^{2} e +2 \sqrt {-d e}\, \left (x -\frac {\sqrt {-d e}}{e}\right )}}{2 \left (\sqrt {-d e}+\sqrt {d e}\right ) \left (\sqrt {-d e}-\sqrt {d e}\right ) \left (x -\frac {\sqrt {-d e}}{e}\right ) d}-\frac {\sqrt {\left (x +\frac {\sqrt {-d e}}{e}\right )^{2} e -2 \sqrt {-d e}\, \left (x +\frac {\sqrt {-d e}}{e}\right )}}{2 \left (\sqrt {-d e}+\sqrt {d e}\right ) \left (\sqrt {-d e}-\sqrt {d e}\right ) \left (x +\frac {\sqrt {-d e}}{e}\right ) d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{\left (d^2-e^2\,x^4\right )\,\sqrt {e\,x^2+d}} \,d x \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)^(1/2)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {1}{- d^{2} \sqrt {d + e x^{2}} + e^{2} x^{4} \sqrt {d + e x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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