\(\int \frac {x^2}{e^2+4 e f x^3+4 d f x^6+4 f^2 x^6} \, dx\) [538]

Optimal result

Integrand size = 32, antiderivative size = 42 \[ \int \frac {x^2}{e^2+4 e f x^3+4 d f x^6+4 f^2 x^6} \, dx=\frac {\arctan \left (\frac {\sqrt {f} \left (e+2 (d+f) x^3\right )}{\sqrt {d} e}\right )}{6 \sqrt {d} e \sqrt {f}} \]

[Out]

1/6*arctan((e+2*(d+f)*x^3)*f^(1/2)/e/d^(1/2))/e/d^(1/2)/f^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6, 1366, 632, 210} \[ \int \frac {x^2}{e^2+4 e f x^3+4 d f x^6+4 f^2 x^6} \, dx=\frac {\arctan \left (\frac {\sqrt {f} \left (2 x^3 (d+f)+e\right )}{\sqrt {d} e}\right )}{6 \sqrt {d} e \sqrt {f}} \]

[In]

Int[x^2/(e^2 + 4*e*f*x^3 + 4*d*f*x^6 + 4*f^2*x^6),x]

[Out]

ArcTan[(Sqrt[f]*(e + 2*(d + f)*x^3))/(Sqrt[d]*e)]/(6*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 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[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 1366

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[(a + b*x +
c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[Simplify[m - n + 1], 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2}{e^2+4 e f x^3+4 \left (d f+f^2\right ) x^6} \, dx \\ & = \frac {1}{3} \text {Subst}\left (\int \frac {1}{e^2+4 e f x+4 \left (d f+f^2\right ) x^2} \, dx,x,x^3\right ) \\ & = -\left (\frac {2}{3} \text {Subst}\left (\int \frac {1}{-16 d e^2 f-x^2} \, dx,x,4 f \left (e+2 (d+f) x^3\right )\right )\right ) \\ & = \frac {\tan ^{-1}\left (\frac {\sqrt {f} \left (e+2 (d+f) x^3\right )}{\sqrt {d} e}\right )}{6 \sqrt {d} e \sqrt {f}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00 \[ \int \frac {x^2}{e^2+4 e f x^3+4 d f x^6+4 f^2 x^6} \, dx=\frac {\arctan \left (\frac {\sqrt {f} \left (e+2 (d+f) x^3\right )}{\sqrt {d} e}\right )}{6 \sqrt {d} e \sqrt {f}} \]

[In]

Integrate[x^2/(e^2 + 4*e*f*x^3 + 4*d*f*x^6 + 4*f^2*x^6),x]

[Out]

ArcTan[(Sqrt[f]*(e + 2*(d + f)*x^3))/(Sqrt[d]*e)]/(6*Sqrt[d]*e*Sqrt[f])

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00

method	result	size
default	\(\frac {\arctan \left (\frac {2 \left (4 d f +4 f^{2}\right ) x^{3}+4 e f}{4 \sqrt {d f}\, e}\right )}{6 \sqrt {d f}\, e}\)	\(42\)
risch	\(-\frac {\ln \left (\left (2 \sqrt {-d f}-2 f \right ) x^{3}-e \right )}{12 \sqrt {-d f}\, e}+\frac {\ln \left (\left (2 \sqrt {-d f}+2 f \right ) x^{3}+e \right )}{12 \sqrt {-d f}\, e}\)	\(64\)

[In]

int(x^2/(4*d*f*x^6+4*f^2*x^6+4*e*f*x^3+e^2),x,method=_RETURNVERBOSE)

[Out]

1/6/(d*f)^(1/2)/e*arctan(1/4*(2*(4*d*f+4*f^2)*x^3+4*e*f)/(d*f)^(1/2)/e)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 155, normalized size of antiderivative = 3.69 \[ \int \frac {x^2}{e^2+4 e f x^3+4 d f x^6+4 f^2 x^6} \, dx=\left [-\frac {\sqrt {-d f} \log \left (\frac {4 \, {\left (d^{2} f + 2 \, d f^{2} + f^{3}\right )} x^{6} + 4 \, {\left (d e f + e f^{2}\right )} x^{3} - d e^{2} + e^{2} f - 2 \, {\left (2 \, {\left (d e + e f\right )} x^{3} + e^{2}\right )} \sqrt {-d f}}{4 \, {\left (d f + f^{2}\right )} x^{6} + 4 \, e f x^{3} + e^{2}}\right )}{12 \, d e f}, \frac {\sqrt {d f} \arctan \left (\frac {{\left (2 \, {\left (d + f\right )} x^{3} + e\right )} \sqrt {d f}}{d e}\right )}{6 \, d e f}\right ] \]

[In]

integrate(x^2/(4*d*f*x^6+4*f^2*x^6+4*e*f*x^3+e^2),x, algorithm="fricas")

[Out]

[-1/12*sqrt(-d*f)*log((4*(d^2*f + 2*d*f^2 + f^3)*x^6 + 4*(d*e*f + e*f^2)*x^3 - d*e^2 + e^2*f - 2*(2*(d*e + e*f
)*x^3 + e^2)*sqrt(-d*f))/(4*(d*f + f^2)*x^6 + 4*e*f*x^3 + e^2))/(d*e*f), 1/6*sqrt(d*f)*arctan((2*(d + f)*x^3 +
e)*sqrt(d*f)/(d*e))/(d*e*f)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (37) = 74\).

Time = 0.45 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.86 \[ \int \frac {x^2}{e^2+4 e f x^3+4 d f x^6+4 f^2 x^6} \, dx=\frac {- \frac {\sqrt {- \frac {1}{d f}} \log {\left (x^{3} + \frac {- d e \sqrt {- \frac {1}{d f}} + e}{2 d + 2 f} \right )}}{12} + \frac {\sqrt {- \frac {1}{d f}} \log {\left (x^{3} + \frac {d e \sqrt {- \frac {1}{d f}} + e}{2 d + 2 f} \right )}}{12}}{e} \]

[In]

integrate(x**2/(4*d*f*x**6+4*f**2*x**6+4*e*f*x**3+e**2),x)

[Out]

(-sqrt(-1/(d*f))*log(x**3 + (-d*e*sqrt(-1/(d*f)) + e)/(2*d + 2*f))/12 + sqrt(-1/(d*f))*log(x**3 + (d*e*sqrt(-1
/(d*f)) + e)/(2*d + 2*f))/12)/e

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.86 \[ \int \frac {x^2}{e^2+4 e f x^3+4 d f x^6+4 f^2 x^6} \, dx=\frac {\arctan \left (\frac {2 \, {\left (d f + f^{2}\right )} x^{3} + e f}{\sqrt {d f} e}\right )}{6 \, \sqrt {d f} e} \]

[In]

integrate(x^2/(4*d*f*x^6+4*f^2*x^6+4*e*f*x^3+e^2),x, algorithm="maxima")

[Out]

1/6*arctan((2*(d*f + f^2)*x^3 + e*f)/(sqrt(d*f)*e))/(sqrt(d*f)*e)

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.93 \[ \int \frac {x^2}{e^2+4 e f x^3+4 d f x^6+4 f^2 x^6} \, dx=\frac {\arctan \left (\frac {2 \, d f x^{3} + 2 \, f^{2} x^{3} + e f}{\sqrt {d f} e}\right )}{6 \, \sqrt {d f} e} \]

[In]

integrate(x^2/(4*d*f*x^6+4*f^2*x^6+4*e*f*x^3+e^2),x, algorithm="giac")

[Out]

1/6*arctan((2*d*f*x^3 + 2*f^2*x^3 + e*f)/(sqrt(d*f)*e))/(sqrt(d*f)*e)

Mupad [B] (verification not implemented)

Time = 17.47 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00 \[ \int \frac {x^2}{e^2+4 e f x^3+4 d f x^6+4 f^2 x^6} \, dx=\frac {\mathrm {atan}\left (\frac {e\,\sqrt {f}+2\,f^{3/2}\,x^3+2\,d\,\sqrt {f}\,x^3}{\sqrt {d}\,e}\right )}{6\,\sqrt {d}\,e\,\sqrt {f}} \]

[In]

int(x^2/(e^2 + 4*f^2*x^6 + 4*d*f*x^6 + 4*e*f*x^3),x)

[Out]

atan((e*f^(1/2) + 2*f^(3/2)*x^3 + 2*d*f^(1/2)*x^3)/(d^(1/2)*e))/(6*d^(1/2)*e*f^(1/2))