3.6.31 \(\int \frac {x}{e^2+4 e f x^2-4 d f x^4+4 f^2 x^4} \, dx\) [531]

Optimal. Leaf size=44 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {f} \left (e-2 (d-f) x^2\right )}{\sqrt {d} e}\right )}{4 \sqrt {d} e \sqrt {f}} \]

[Out]

-1/4*arctanh((e-2*(d-f)*x^2)*f^(1/2)/e/d^(1/2))/e/d^(1/2)/f^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {6, 1121, 632, 212} \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {\sqrt {f} \left (e-2 x^2 (d-f)\right )}{\sqrt {d} e}\right )}{4 \sqrt {d} e \sqrt {f}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*ArcTanh[(Sqrt[f]*(e - 2*(d - f)*x^2))/(Sqrt[d]*e)]/(Sqrt[d]*e*Sqrt[f])

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1121

Int[(x_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(a + b*x + c*x^2)^p, x],
x, x^2], x] /; FreeQ[{a, b, c, p}, x]

Rubi steps

\begin {align*} \int \frac {x}{e^2+4 e f x^2-4 d f x^4+4 f^2 x^4} \, dx &=\int \frac {x}{e^2+4 e f x^2+\left (-4 d f+4 f^2\right ) x^4} \, dx\\ &=\frac {1}{2} \text {Subst}\left (\int \frac {1}{e^2+4 e f x+\left (-4 d f+4 f^2\right ) x^2} \, dx,x,x^2\right )\\ &=-\text {Subst}\left (\int \frac {1}{16 d e^2 f-x^2} \, dx,x,4 f \left (e-2 (d-f) x^2\right )\right )\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {f} \left (e+2 (-d+f) x^2\right )}{\sqrt {d} e}\right )}{4 \sqrt {d} e \sqrt {f}}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 46, normalized size = 1.05 \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {\sqrt {f} \left (e-2 d x^2+2 f x^2\right )}{\sqrt {d} e}\right )}{4 \sqrt {d} e \sqrt {f}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.02, size = 42, normalized size = 0.95

method	result	size
default	\(\frac {\arctanh \left (\frac {2 \left (4 d f -4 f^{2}\right ) x^{2}-4 e f}{4 \sqrt {d f}\, e}\right )}{4 \sqrt {d f}\, e}\)	\(42\)
risch	\(\frac {\ln \left (\left (-2 \sqrt {d f}-2 f \right ) x^{2}-e \right )}{8 \sqrt {d f}\, e}-\frac {\ln \left (\left (-2 \sqrt {d f}+2 f \right ) x^{2}+e \right )}{8 \sqrt {d f}\, e}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/(d*f)^(1/2)/e*arctanh(1/4*(2*(4*d*f-4*f^2)*x^2-4*e*f)/(d*f)^(1/2)/e)

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Maxima [A]
time = 0.51, size = 70, normalized size = 1.59 \begin {gather*} \frac {e^{\left (-1\right )} \log \left (\frac {2 \, {\left (d f - f^{2}\right )} x^{2} - f e + \sqrt {d f} e}{2 \, {\left (d f - f^{2}\right )} x^{2} - f e - \sqrt {d f} e}\right )}{8 \, \sqrt {d f}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*e^(-1)*log((2*(d*f - f^2)*x^2 - f*e + sqrt(d*f)*e)/(2*(d*f - f^2)*x^2 - f*e - sqrt(d*f)*e))/sqrt(d*f)

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Fricas [A]
time = 0.36, size = 160, normalized size = 3.64 \begin {gather*} \left [\frac {\sqrt {d f} e^{\left (-1\right )} \log \left (-\frac {4 \, {\left (d^{2} f - 2 \, d f^{2} + f^{3}\right )} x^{4} - 4 \, {\left (d f - f^{2}\right )} x^{2} e + {\left (d + f\right )} e^{2} + 2 \, {\left (2 \, {\left (d - f\right )} x^{2} e - e^{2}\right )} \sqrt {d f}}{4 \, {\left (d f - f^{2}\right )} x^{4} - 4 \, f x^{2} e - e^{2}}\right )}{8 \, d f}, \frac {\sqrt {-d f} \arctan \left (-\frac {{\left (2 \, {\left (d - f\right )} x^{2} - e\right )} \sqrt {-d f} e^{\left (-1\right )}}{d}\right ) e^{\left (-1\right )}}{4 \, d f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.38, size = 75, normalized size = 1.70 \begin {gather*} - \frac {\frac {\sqrt {\frac {1}{d f}} \log {\left (x^{2} + \frac {- d e \sqrt {\frac {1}{d f}} - e}{2 d - 2 f} \right )}}{8} - \frac {\sqrt {\frac {1}{d f}} \log {\left (x^{2} + \frac {d e \sqrt {\frac {1}{d f}} - e}{2 d - 2 f} \right )}}{8}}{e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(sqrt(1/(d*f))*log(x**2 + (-d*e*sqrt(1/(d*f)) - e)/(2*d - 2*f))/8 - sqrt(1/(d*f))*log(x**2 + (d*e*sqrt(1/(d*f
)) - e)/(2*d - 2*f))/8)/e

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Giac [A]
time = 6.92, size = 41, normalized size = 0.93 \begin {gather*} -\frac {\arctan \left (\frac {{\left (2 \, d f x^{2} - 2 \, f^{2} x^{2} - f e\right )} e^{\left (-1\right )}}{\sqrt {-d f}}\right ) e^{\left (-1\right )}}{4 \, \sqrt {-d f}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 3.11, size = 199, normalized size = 4.52 \begin {gather*} \frac {\mathrm {atanh}\left (\frac {16\,d^{3/2}\,f^{3/2}\,x^2}{\frac {8\,e\,f^3}{d}-32\,f^3\,x^2-16\,e\,f^2+16\,d\,f^2\,x^2+8\,d\,e\,f+\frac {16\,f^4\,x^2}{d}}-\frac {32\,\sqrt {d}\,f^{5/2}\,x^2}{\frac {8\,e\,f^3}{d}-32\,f^3\,x^2-16\,e\,f^2+16\,d\,f^2\,x^2+8\,d\,e\,f+\frac {16\,f^4\,x^2}{d}}+\frac {16\,f^{7/2}\,x^2}{\sqrt {d}\,\left (\frac {8\,e\,f^3}{d}-32\,f^3\,x^2-16\,e\,f^2+16\,d\,f^2\,x^2+8\,d\,e\,f+\frac {16\,f^4\,x^2}{d}\right )}\right )}{4\,\sqrt {d}\,e\,\sqrt {f}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atanh((16*d^(3/2)*f^(3/2)*x^2)/((8*e*f^3)/d - 32*f^3*x^2 - 16*e*f^2 + 16*d*f^2*x^2 + 8*d*e*f + (16*f^4*x^2)/d)
- (32*d^(1/2)*f^(5/2)*x^2)/((8*e*f^3)/d - 32*f^3*x^2 - 16*e*f^2 + 16*d*f^2*x^2 + 8*d*e*f + (16*f^4*x^2)/d) +
(16*f^(7/2)*x^2)/(d^(1/2)*((8*e*f^3)/d - 32*f^3*x^2 - 16*e*f^2 + 16*d*f^2*x^2 + 8*d*e*f + (16*f^4*x^2)/d)))/(4
*d^(1/2)*e*f^(1/2))

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