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

Optimal. Leaf size=40 \[ \frac{\tan ^{-1}\left (\frac{2 \sqrt{d} \sqrt{f} x^2}{e+2 f x^3}\right )}{2 \sqrt{d} \sqrt{f}} \]

[Out]

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

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Rubi [A]  time = 0.089305, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.05, Rules used = {2094, 205} \[ \frac{\tan ^{-1}\left (\frac{2 \sqrt{d} \sqrt{f} x^2}{e+2 f x^3}\right )}{2 \sqrt{d} \sqrt{f}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2094

Int[((x_)^(m_.)*((A_) + (B_.)*(x_)^(n_.)))/((a_) + (b_.)*(x_)^(k_.) + (c_.)*(x_)^(n_.) + (d_.)*(x_)^(n2_)), x_
Symbol] :> Dist[(A^2*(m - n + 1))/(m + 1), Subst[Int[1/(a + A^2*b*(m - n + 1)^2*x^2), x], x, x^(m + 1)/(A*(m -
n + 1) + B*(m + 1)*x^n)], x] /; FreeQ[{a, b, c, d, A, B, m, n}, x] && EqQ[n2, 2*n] && EqQ[k, 2*(m + 1)] && Eq
Q[a*B^2*(m + 1)^2 - A^2*d*(m - n + 1)^2, 0] && EqQ[B*c*(m + 1) - 2*A*d*(m - n + 1), 0]

Rule 205

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

Rubi steps

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

Mathematica [C]  time = 0.0502882, size = 86, normalized size = 2.15 \[ -\frac{\text{RootSum}\left [4 \text{$\#$1}^4 d f+4 \text{$\#$1}^3 e f+4 \text{$\#$1}^6 f^2+e^2\& ,\frac{\text{$\#$1}^3 f \log (x-\text{$\#$1})-e \log (x-\text{$\#$1})}{4 \text{$\#$1}^2 d+6 \text{$\#$1}^4 f+3 \text{$\#$1} e}\& \right ]}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-RootSum[e^2 + 4*e*f*#1^3 + 4*d*f*#1^4 + 4*f^2*#1^6 & , (-(e*Log[x - #1]) + f*Log[x - #1]*#1^3)/(3*e*#1 + 4*d*
#1^2 + 6*f*#1^4) & ]/(2*f)

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Maple [C]  time = 0.009, size = 74, normalized size = 1.9 \begin{align*} -{\frac{1}{2\,f}\sum _{{\it \_R}={\it RootOf} \left ( 4\,{f}^{2}{{\it \_Z}}^{6}+4\,df{{\it \_Z}}^{4}+4\,fe{{\it \_Z}}^{3}+{e}^{2} \right ) }{\frac{ \left ({{\it \_R}}^{4}f-{\it \_R}\,e \right ) \ln \left ( x-{\it \_R} \right ) }{6\,f{{\it \_R}}^{5}+4\,d{{\it \_R}}^{3}+3\,e{{\it \_R}}^{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/f*sum((_R^4*f-_R*e)/(6*_R^5*f+4*_R^3*d+3*_R^2*e)*ln(x-_R),_R=RootOf(4*_Z^6*f^2+4*_Z^4*d*f+4*_Z^3*e*f+e^2)
)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -2 \, \int \frac{{\left (f x^{3} - e\right )} x}{4 \, f^{2} x^{6} + 4 \, d f x^{4} + 4 \, e f x^{3} + e^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.33453, size = 347, normalized size = 8.68 \begin{align*} \left [-\frac{\sqrt{-d f} \log \left (\frac{4 \, f^{2} x^{6} - 4 \, d f x^{4} + 4 \, e f x^{3} + e^{2} + 4 \,{\left (2 \, f x^{5} + e x^{2}\right )} \sqrt{-d f}}{4 \, f^{2} x^{6} + 4 \, d f x^{4} + 4 \, e f x^{3} + 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^{4} + 2 \, d x^{2} + e x\right )} \sqrt{d f}}{d e}\right )}{2 \, d f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*sqrt(-d*f)*log((4*f^2*x^6 - 4*d*f*x^4 + 4*e*f*x^3 + e^2 + 4*(2*f*x^5 + e*x^2)*sqrt(-d*f))/(4*f^2*x^6 + 4
*d*f*x^4 + 4*e*f*x^3 + e^2))/(d*f), -1/2*(sqrt(d*f)*arctan(sqrt(d*f)*x/d) - sqrt(d*f)*arctan((2*f*x^4 + 2*d*x^
2 + e*x)*sqrt(d*f)/(d*e)))/(d*f)]

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Sympy [B]  time = 1.15566, size = 73, normalized size = 1.82 \begin{align*} \frac{\sqrt{- \frac{1}{d f}} \log{\left (- d x^{2} \sqrt{- \frac{1}{d f}} + \frac{e}{2 f} + x^{3} \right )}}{4} - \frac{\sqrt{- \frac{1}{d f}} \log{\left (d x^{2} \sqrt{- \frac{1}{d f}} + \frac{e}{2 f} + x^{3} \right )}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(-1/(d*f))*log(-d*x**2*sqrt(-1/(d*f)) + e/(2*f) + x**3)/4 - sqrt(-1/(d*f))*log(d*x**2*sqrt(-1/(d*f)) + e/(
2*f) + x**3)/4

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{2 \,{\left (f x^{3} - e\right )} x}{4 \, f^{2} x^{6} + 4 \, d f x^{4} + 4 \, e f x^{3} + e^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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