3.539 \(\int \frac {x^2}{e^2+4 e f x^3-4 d f x^6+4 f^2 x^6} \, dx\)

Optimal. Leaf size=44 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {f} \left (e-2 x^3 (d-f)\right )}{\sqrt {d} e}\right )}{6 \sqrt {d} e \sqrt {f}} \]

[Out]

-1/6*arctanh((e-2*(d-f)*x^3)*f^(1/2)/e/d^(1/2))/e/d^(1/2)/f^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6, 1352, 618, 206} \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {f} \left (e-2 x^3 (d-f)\right )}{\sqrt {d} e}\right )}{6 \sqrt {d} e \sqrt {f}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTanh[(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 206

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

Rule 618

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 1352

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*} \int \frac {x^2}{e^2+4 e f x^3-4 d f x^6+4 f^2 x^6} \, dx &=\int \frac {x^2}{e^2+4 e f x^3+\left (-4 d f+4 f^2\right ) x^6} \, dx\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{e^2+4 e f x+\left (-4 d f+4 f^2\right ) x^2} \, dx,x,x^3\right )\\ &=-\left (\frac {2}{3} \operatorname {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 {\tanh ^{-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*}

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Mathematica [A]  time = 0.02, size = 46, normalized size = 1.05 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {f} \left (-2 d x^3+e+2 f x^3\right )}{\sqrt {d} e}\right )}{6 \sqrt {d} e \sqrt {f}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*ArcTanh[(Sqrt[f]*(e - 2*d*x^3 + 2*f*x^3))/(Sqrt[d]*e)]/(Sqrt[d]*e*Sqrt[f])

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fricas [A]  time = 0.45, size = 168, normalized size = 3.82 \[ \left [\frac {\sqrt {d f} \log \left (-\frac {4 \, {\left (d^{2} f - 2 \, d f^{2} + f^{3}\right )} x^{6} - 4 \, {\left (d e f - e f^{2}\right )} x^{3} + d e^{2} + e^{2} f + 2 \, {\left (2 \, {\left (d e - e f\right )} x^{3} - e^{2}\right )} \sqrt {d f}}{4 \, {\left (d f - f^{2}\right )} x^{6} - 4 \, e f x^{3} - e^{2}}\right )}{12 \, d e f}, \frac {\sqrt {-d f} \arctan \left (-\frac {{\left (2 \, {\left (d - f\right )} x^{3} - e\right )} \sqrt {-d f}}{d e}\right )}{6 \, d e f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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giac [A]  time = 9.22, size = 41, normalized size = 0.93 \[ -\frac {\arctan \left (\frac {2 \, d f x^{3} - 2 \, f^{2} x^{3} - f e}{\sqrt {-d f e^{2}}}\right )}{6 \, \sqrt {-d f e^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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 - f*e)/sqrt(-d*f*e^2))/sqrt(-d*f*e^2)

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maple [A]  time = 0.00, size = 42, normalized size = 0.95 \[ \frac {\arctanh \left (\frac {2 \left (4 d f -4 f^{2}\right ) x^{3}-4 e f}{4 \sqrt {d f}\, e}\right )}{6 \sqrt {d f}\, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/(d*f)^(1/2)/e*arctanh(1/4*(2*(4*d*f-4*f^2)*x^3-4*e*f)/(d*f)^(1/2)/e)

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maxima [A]  time = 1.54, size = 67, normalized size = 1.52 \[ \frac {\log \left (\frac {2 \, {\left (d f - f^{2}\right )} x^{3} - e f + \sqrt {d f} e}{2 \, {\left (d f - f^{2}\right )} x^{3} - e f - \sqrt {d f} e}\right )}{12 \, \sqrt {d f} e} \]

Verification of antiderivative is not currently implemented for this CAS.

[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]

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

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mupad [B]  time = 3.53, size = 923, normalized size = 20.98 \[ \frac {\mathrm {atan}\left (\frac {\frac {\left (x^3\,\left (32\,d^3\,f^3-96\,d^2\,f^4+96\,d\,f^5-32\,f^6\right )+\frac {x^3\,\left (-64\,e\,d^3\,f^4+192\,e\,d^2\,f^5-192\,e\,d\,f^6+64\,e\,f^7\right )+16\,e^2\,f^6-48\,d\,e^2\,f^5+48\,d^2\,e^2\,f^4-16\,d^3\,e^2\,f^3-\frac {\frac {x^3\,\left (-384\,d^3\,e^2\,f^5+1152\,d^2\,e^2\,f^6-1152\,d\,e^2\,f^7+384\,e^2\,f^8\right )}{12}+16\,e^3\,f^7-48\,d\,e^3\,f^6+48\,d^2\,e^3\,f^5-16\,d^3\,e^3\,f^4}{\sqrt {d}\,e\,\sqrt {f}}}{\sqrt {d}\,e\,\sqrt {f}}\right )\,1{}\mathrm {i}}{\sqrt {d}\,e\,\sqrt {f}}+\frac {\left (x^3\,\left (32\,d^3\,f^3-96\,d^2\,f^4+96\,d\,f^5-32\,f^6\right )-\frac {x^3\,\left (-64\,e\,d^3\,f^4+192\,e\,d^2\,f^5-192\,e\,d\,f^6+64\,e\,f^7\right )+16\,e^2\,f^6-48\,d\,e^2\,f^5+48\,d^2\,e^2\,f^4-16\,d^3\,e^2\,f^3+\frac {\frac {x^3\,\left (-384\,d^3\,e^2\,f^5+1152\,d^2\,e^2\,f^6-1152\,d\,e^2\,f^7+384\,e^2\,f^8\right )}{12}+16\,e^3\,f^7-48\,d\,e^3\,f^6+48\,d^2\,e^3\,f^5-16\,d^3\,e^3\,f^4}{\sqrt {d}\,e\,\sqrt {f}}}{\sqrt {d}\,e\,\sqrt {f}}\right )\,1{}\mathrm {i}}{\sqrt {d}\,e\,\sqrt {f}}}{\frac {x^3\,\left (32\,d^3\,f^3-96\,d^2\,f^4+96\,d\,f^5-32\,f^6\right )+\frac {x^3\,\left (-64\,e\,d^3\,f^4+192\,e\,d^2\,f^5-192\,e\,d\,f^6+64\,e\,f^7\right )+16\,e^2\,f^6-48\,d\,e^2\,f^5+48\,d^2\,e^2\,f^4-16\,d^3\,e^2\,f^3-\frac {\frac {x^3\,\left (-384\,d^3\,e^2\,f^5+1152\,d^2\,e^2\,f^6-1152\,d\,e^2\,f^7+384\,e^2\,f^8\right )}{12}+16\,e^3\,f^7-48\,d\,e^3\,f^6+48\,d^2\,e^3\,f^5-16\,d^3\,e^3\,f^4}{\sqrt {d}\,e\,\sqrt {f}}}{\sqrt {d}\,e\,\sqrt {f}}}{\sqrt {d}\,e\,\sqrt {f}}-\frac {x^3\,\left (32\,d^3\,f^3-96\,d^2\,f^4+96\,d\,f^5-32\,f^6\right )-\frac {x^3\,\left (-64\,e\,d^3\,f^4+192\,e\,d^2\,f^5-192\,e\,d\,f^6+64\,e\,f^7\right )+16\,e^2\,f^6-48\,d\,e^2\,f^5+48\,d^2\,e^2\,f^4-16\,d^3\,e^2\,f^3+\frac {\frac {x^3\,\left (-384\,d^3\,e^2\,f^5+1152\,d^2\,e^2\,f^6-1152\,d\,e^2\,f^7+384\,e^2\,f^8\right )}{12}+16\,e^3\,f^7-48\,d\,e^3\,f^6+48\,d^2\,e^3\,f^5-16\,d^3\,e^3\,f^4}{\sqrt {d}\,e\,\sqrt {f}}}{\sqrt {d}\,e\,\sqrt {f}}}{\sqrt {d}\,e\,\sqrt {f}}}\right )\,1{}\mathrm {i}}{6\,\sqrt {d}\,e\,\sqrt {f}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan((((x^3*(96*d*f^5 - 32*f^6 - 96*d^2*f^4 + 32*d^3*f^3) + (x^3*(64*e*f^7 + 192*d^2*e*f^5 - 64*d^3*e*f^4 - 1
92*d*e*f^6) + 16*e^2*f^6 - 48*d*e^2*f^5 + 48*d^2*e^2*f^4 - 16*d^3*e^2*f^3 - ((x^3*(384*e^2*f^8 - 1152*d*e^2*f^
7 + 1152*d^2*e^2*f^6 - 384*d^3*e^2*f^5))/12 + 16*e^3*f^7 - 48*d*e^3*f^6 + 48*d^2*e^3*f^5 - 16*d^3*e^3*f^4)/(d^
(1/2)*e*f^(1/2)))/(d^(1/2)*e*f^(1/2)))*1i)/(d^(1/2)*e*f^(1/2)) + ((x^3*(96*d*f^5 - 32*f^6 - 96*d^2*f^4 + 32*d^
3*f^3) - (x^3*(64*e*f^7 + 192*d^2*e*f^5 - 64*d^3*e*f^4 - 192*d*e*f^6) + 16*e^2*f^6 - 48*d*e^2*f^5 + 48*d^2*e^2
*f^4 - 16*d^3*e^2*f^3 + ((x^3*(384*e^2*f^8 - 1152*d*e^2*f^7 + 1152*d^2*e^2*f^6 - 384*d^3*e^2*f^5))/12 + 16*e^3
*f^7 - 48*d*e^3*f^6 + 48*d^2*e^3*f^5 - 16*d^3*e^3*f^4)/(d^(1/2)*e*f^(1/2)))/(d^(1/2)*e*f^(1/2)))*1i)/(d^(1/2)*
e*f^(1/2)))/((x^3*(96*d*f^5 - 32*f^6 - 96*d^2*f^4 + 32*d^3*f^3) + (x^3*(64*e*f^7 + 192*d^2*e*f^5 - 64*d^3*e*f^
4 - 192*d*e*f^6) + 16*e^2*f^6 - 48*d*e^2*f^5 + 48*d^2*e^2*f^4 - 16*d^3*e^2*f^3 - ((x^3*(384*e^2*f^8 - 1152*d*e
^2*f^7 + 1152*d^2*e^2*f^6 - 384*d^3*e^2*f^5))/12 + 16*e^3*f^7 - 48*d*e^3*f^6 + 48*d^2*e^3*f^5 - 16*d^3*e^3*f^4
)/(d^(1/2)*e*f^(1/2)))/(d^(1/2)*e*f^(1/2)))/(d^(1/2)*e*f^(1/2)) - (x^3*(96*d*f^5 - 32*f^6 - 96*d^2*f^4 + 32*d^
3*f^3) - (x^3*(64*e*f^7 + 192*d^2*e*f^5 - 64*d^3*e*f^4 - 192*d*e*f^6) + 16*e^2*f^6 - 48*d*e^2*f^5 + 48*d^2*e^2
*f^4 - 16*d^3*e^2*f^3 + ((x^3*(384*e^2*f^8 - 1152*d*e^2*f^7 + 1152*d^2*e^2*f^6 - 384*d^3*e^2*f^5))/12 + 16*e^3
*f^7 - 48*d*e^3*f^6 + 48*d^2*e^3*f^5 - 16*d^3*e^3*f^4)/(d^(1/2)*e*f^(1/2)))/(d^(1/2)*e*f^(1/2)))/(d^(1/2)*e*f^
(1/2))))*1i)/(6*d^(1/2)*e*f^(1/2))

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sympy [A]  time = 0.85, size = 75, normalized size = 1.70 \[ - \frac {\frac {\sqrt {\frac {1}{d f}} \log {\left (x^{3} + \frac {- d e \sqrt {\frac {1}{d f}} - e}{2 d - 2 f} \right )}}{12} - \frac {\sqrt {\frac {1}{d f}} \log {\left (x^{3} + \frac {d e \sqrt {\frac {1}{d f}} - e}{2 d - 2 f} \right )}}{12}}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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