\(\int \frac {e-2 f x^2}{e^2+4 d f x^2+4 e f x^2+4 f^2 x^4} \, dx\) [524]

Optimal result

Integrand size = 37, antiderivative size = 81 \[ \int \frac {e-2 f x^2}{e^2+4 d f x^2+4 e f x^2+4 f^2 x^4} \, dx=-\frac {\log \left (e-2 \sqrt {-d} \sqrt {f} x+2 f x^2\right )}{4 \sqrt {-d} \sqrt {f}}+\frac {\log \left (e+2 \sqrt {-d} \sqrt {f} x+2 f x^2\right )}{4 \sqrt {-d} \sqrt {f}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {6, 1178, 642} \[ \int \frac {e-2 f x^2}{e^2+4 d f x^2+4 e f x^2+4 f^2 x^4} \, dx=\frac {\log \left (2 \sqrt {-d} \sqrt {f} x+e+2 f x^2\right )}{4 \sqrt {-d} \sqrt {f}}-\frac {\log \left (-2 \sqrt {-d} \sqrt {f} x+e+2 f x^2\right )}{4 \sqrt {-d} \sqrt {f}} \]

[In]

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

[Out]

-1/4*Log[e - 2*Sqrt[-d]*Sqrt[f]*x + 2*f*x^2]/(Sqrt[-d]*Sqrt[f]) + Log[e + 2*Sqrt[-d]*Sqrt[f]*x + 2*f*x^2]/(4*S
qrt[-d]*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 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1178

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e) - b/c, 2]},
Dist[e/(2*c*q), Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x
- x^2, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] &&  !GtQ[b^2
- 4*a*c, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e-2 f x^2}{e^2+4 (d+e) f x^2+4 f^2 x^4} \, dx \\ & = -\frac {\int \frac {\frac {\sqrt {-d}}{\sqrt {f}}+2 x}{-\frac {e}{2 f}-\frac {\sqrt {-d} x}{\sqrt {f}}-x^2} \, dx}{4 \sqrt {-d} \sqrt {f}}-\frac {\int \frac {\frac {\sqrt {-d}}{\sqrt {f}}-2 x}{-\frac {e}{2 f}+\frac {\sqrt {-d} x}{\sqrt {f}}-x^2} \, dx}{4 \sqrt {-d} \sqrt {f}} \\ & = -\frac {\log \left (e-2 \sqrt {-d} \sqrt {f} x+2 f x^2\right )}{4 \sqrt {-d} \sqrt {f}}+\frac {\log \left (e+2 \sqrt {-d} \sqrt {f} x+2 f x^2\right )}{4 \sqrt {-d} \sqrt {f}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(191\) vs. \(2(81)=162\).

Time = 0.09 (sec) , antiderivative size = 191, normalized size of antiderivative = 2.36 \[ \int \frac {e-2 f x^2}{e^2+4 d f x^2+4 e f x^2+4 f^2 x^4} \, dx=\frac {-\frac {\left (-d-2 e+\sqrt {d} \sqrt {d+2 e}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {f} x}{\sqrt {d+e-\sqrt {d} \sqrt {d+2 e}}}\right )}{\sqrt {d+e-\sqrt {d} \sqrt {d+2 e}}}-\frac {\left (d+2 e+\sqrt {d} \sqrt {d+2 e}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {f} x}{\sqrt {d+e+\sqrt {d} \sqrt {d+2 e}}}\right )}{\sqrt {d+e+\sqrt {d} \sqrt {d+2 e}}}}{2 \sqrt {2} \sqrt {d} \sqrt {d+2 e} \sqrt {f}} \]

[In]

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

[Out]

(-(((-d - 2*e + Sqrt[d]*Sqrt[d + 2*e])*ArcTan[(Sqrt[2]*Sqrt[f]*x)/Sqrt[d + e - Sqrt[d]*Sqrt[d + 2*e]]])/Sqrt[d
+ e - Sqrt[d]*Sqrt[d + 2*e]]) - ((d + 2*e + Sqrt[d]*Sqrt[d + 2*e])*ArcTan[(Sqrt[2]*Sqrt[f]*x)/Sqrt[d + e + Sq
rt[d]*Sqrt[d + 2*e]]])/Sqrt[d + e + Sqrt[d]*Sqrt[d + 2*e]])/(2*Sqrt[2]*Sqrt[d]*Sqrt[d + 2*e]*Sqrt[f])

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.91

method	result	size
risch	\(-\frac {\ln \left (-2 \sqrt {-d f}\, f \,x^{2}-2 d f x -\sqrt {-d f}\, e \right )}{4 \sqrt {-d f}}+\frac {\ln \left (-2 \sqrt {-d f}\, f \,x^{2}+2 d f x -\sqrt {-d f}\, e \right )}{4 \sqrt {-d f}}\)	\(74\)
default	\(f^{2} \left (-\frac {\left (d f +2 e f -\sqrt {d \,f^{2} \left (d +2 e \right )}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {f x \sqrt {2}}{\sqrt {-d f -e f +\sqrt {d \,f^{2} \left (d +2 e \right )}}}\right )}{4 f^{2} \sqrt {d \,f^{2} \left (d +2 e \right )}\, \sqrt {-d f -e f +\sqrt {d \,f^{2} \left (d +2 e \right )}}}+\frac {\left (-d f -2 e f -\sqrt {d \,f^{2} \left (d +2 e \right )}\right ) \sqrt {2}\, \arctan \left (\frac {f x \sqrt {2}}{\sqrt {d f +e f +\sqrt {d \,f^{2} \left (d +2 e \right )}}}\right )}{4 f^{2} \sqrt {d \,f^{2} \left (d +2 e \right )}\, \sqrt {d f +e f +\sqrt {d \,f^{2} \left (d +2 e \right )}}}\right )\)	\(193\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.74 \[ \int \frac {e-2 f x^2}{e^2+4 d f x^2+4 e f x^2+4 f^2 x^4} \, dx=\left [-\frac {\sqrt {-d f} \log \left (\frac {4 \, f^{2} x^{4} - 4 \, {\left (d - e\right )} f x^{2} + e^{2} + 4 \, {\left (2 \, f x^{3} + e x\right )} \sqrt {-d f}}{4 \, f^{2} x^{4} + 4 \, {\left (d + e\right )} f x^{2} + e^{2}}\right )}{4 \, d f}, -\frac {\sqrt {d f} \arctan \left (\frac {\sqrt {d f} x}{d}\right ) - \sqrt {d f} \arctan \left (\frac {{\left (2 \, f x^{3} + {\left (2 \, d + e\right )} x\right )} \sqrt {d f}}{d e}\right )}{2 \, d f}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.86 \[ \int \frac {e-2 f x^2}{e^2+4 d f x^2+4 e f x^2+4 f^2 x^4} \, dx=\frac {\sqrt {- \frac {1}{d f}} \log {\left (- d x \sqrt {- \frac {1}{d f}} + \frac {e}{2 f} + x^{2} \right )}}{4} - \frac {\sqrt {- \frac {1}{d f}} \log {\left (d x \sqrt {- \frac {1}{d f}} + \frac {e}{2 f} + x^{2} \right )}}{4} \]

[In]

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

[Out]

sqrt(-1/(d*f))*log(-d*x*sqrt(-1/(d*f)) + e/(2*f) + x**2)/4 - sqrt(-1/(d*f))*log(d*x*sqrt(-1/(d*f)) + e/(2*f) +
x**2)/4

Maxima [F]

\[ \int \frac {e-2 f x^2}{e^2+4 d f x^2+4 e f x^2+4 f^2 x^4} \, dx=\int { -\frac {2 \, f x^{2} - e}{4 \, f^{2} x^{4} + 4 \, d f x^{2} + 4 \, e f x^{2} + e^{2}} \,d x } \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 266 vs. \(2 (61) = 122\).

Time = 0.40 (sec) , antiderivative size = 266, normalized size of antiderivative = 3.28 \[ \int \frac {e-2 f x^2}{e^2+4 d f x^2+4 e f x^2+4 f^2 x^4} \, dx=\frac {\sqrt {2} {\left (d e f {\left | f \right |} - \sqrt {d^{2} + 2 \, d e} {\left (d + e\right )} f^{2} + \sqrt {d^{2} + 2 \, d e} d f^{2}\right )} \arctan \left (\frac {4 \, \sqrt {\frac {1}{2}} x}{\sqrt {\frac {4 \, d f + 4 \, e f + \sqrt {-16 \, e^{2} f^{2} + 16 \, {\left (d f + e f\right )}^{2}}}{f^{2}}}}\right )}{4 \, {\left (d^{2} + d e - \sqrt {d^{2} + 2 \, d e} d\right )} \sqrt {{\left (d + e + \sqrt {d^{2} + 2 \, d e}\right )} f} f^{2}} + \frac {\sqrt {2} {\left (d e f {\left | f \right |} + \sqrt {d^{2} + 2 \, d e} {\left (d + e\right )} f^{2} - \sqrt {d^{2} + 2 \, d e} d f^{2}\right )} \arctan \left (\frac {4 \, \sqrt {\frac {1}{2}} x}{\sqrt {\frac {4 \, d f + 4 \, e f - \sqrt {-16 \, e^{2} f^{2} + 16 \, {\left (d f + e f\right )}^{2}}}{f^{2}}}}\right )}{4 \, {\left (d^{2} + d e + \sqrt {d^{2} + 2 \, d e} d\right )} \sqrt {{\left (d + e - \sqrt {d^{2} + 2 \, d e}\right )} f} f^{2}} \]

[In]

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

[Out]

1/4*sqrt(2)*(d*e*f*abs(f) - sqrt(d^2 + 2*d*e)*(d + e)*f^2 + sqrt(d^2 + 2*d*e)*d*f^2)*arctan(4*sqrt(1/2)*x/sqrt
((4*d*f + 4*e*f + sqrt(-16*e^2*f^2 + 16*(d*f + e*f)^2))/f^2))/((d^2 + d*e - sqrt(d^2 + 2*d*e)*d)*sqrt((d + e +
sqrt(d^2 + 2*d*e))*f)*f^2) + 1/4*sqrt(2)*(d*e*f*abs(f) + sqrt(d^2 + 2*d*e)*(d + e)*f^2 - sqrt(d^2 + 2*d*e)*d*
f^2)*arctan(4*sqrt(1/2)*x/sqrt((4*d*f + 4*e*f - sqrt(-16*e^2*f^2 + 16*(d*f + e*f)^2))/f^2))/((d^2 + d*e + sqrt
(d^2 + 2*d*e)*d)*sqrt((d + e - sqrt(d^2 + 2*d*e))*f)*f^2)

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.62 \[ \int \frac {e-2 f x^2}{e^2+4 d f x^2+4 e f x^2+4 f^2 x^4} \, dx=\frac {\mathrm {atan}\left (\frac {2\,f^{3/2}\,x^3+2\,d\,\sqrt {f}\,x+e\,\sqrt {f}\,x}{\sqrt {d}\,e}\right )-\mathrm {atan}\left (\frac {\sqrt {f}\,x}{\sqrt {d}}\right )}{2\,\sqrt {d}\,\sqrt {f}} \]

[In]

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

[Out]

(atan((2*f^(3/2)*x^3 + 2*d*f^(1/2)*x + e*f^(1/2)*x)/(d^(1/2)*e)) - atan((f^(1/2)*x)/d^(1/2)))/(2*d^(1/2)*f^(1/
2))