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]

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

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Rubi [A]  time = 0.09, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, 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 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]

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]

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*}

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Mathematica [C]  time = 0.05, size = 86, normalized size = 2.15 \[ -\frac {\text {RootSum}\left [4 \text {$\#$1}^6 f^2+4 \text {$\#$1}^4 d f+4 \text {$\#$1}^3 e f+e^2\& ,\frac {\text {$\#$1}^3 f \log (x-\text {$\#$1})-e \log (x-\text {$\#$1})}{6 \text {$\#$1}^4 f+4 \text {$\#$1}^2 d+3 \text {$\#$1} e}\& \right ]}{2 f} \]

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]

-1/2*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) & ]/f

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fricas [B]  time = 0.43, size = 153, normalized size = 3.82 \[ \left [-\frac {\sqrt {-d f} \log \left (\frac {4 \, f^{2} x^{6} - 4 \, d f x^{4} + 4 \, e f x^{3} + e^{2} + 4 \, {\left (2 \, f x^{5} + e x^{2}\right )} \sqrt {-d f}}{4 \, f^{2} x^{6} + 4 \, d f x^{4} + 4 \, e f x^{3} + e^{2}}\right )}{4 \, d f}, -\frac {\sqrt {d f} \arctan \left (\frac {\sqrt {d f} x}{d}\right ) - \sqrt {d f} \arctan \left (\frac {{\left (2 \, f x^{4} + 2 \, d x^{2} + e x\right )} \sqrt {d f}}{d e}\right )}{2 \, d f}\right ] \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {2 \, {\left (f x^{3} - e\right )} x}{4 \, f^{2} x^{6} + 4 \, d f x^{4} + 4 \, e f x^{3} + e^{2}}\,{d x} \]

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)

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maple [C]  time = 0.01, size = 74, normalized size = 1.85 \[ -\frac {\left (\RootOf \left (4 f^{2} \textit {\_Z}^{6}+4 d f \,\textit {\_Z}^{4}+4 e f \,\textit {\_Z}^{3}+e^{2}\right )^{4} f -\RootOf \left (4 f^{2} \textit {\_Z}^{6}+4 d f \,\textit {\_Z}^{4}+4 e f \,\textit {\_Z}^{3}+e^{2}\right ) e \right ) \ln \left (-\RootOf \left (4 f^{2} \textit {\_Z}^{6}+4 d f \,\textit {\_Z}^{4}+4 e f \,\textit {\_Z}^{3}+e^{2}\right )+x \right )}{2 f \left (6 f \RootOf \left (4 f^{2} \textit {\_Z}^{6}+4 d f \,\textit {\_Z}^{4}+4 e f \,\textit {\_Z}^{3}+e^{2}\right )^{5}+4 d \RootOf \left (4 f^{2} \textit {\_Z}^{6}+4 d f \,\textit {\_Z}^{4}+4 e f \,\textit {\_Z}^{3}+e^{2}\right )^{3}+3 e \RootOf \left (4 f^{2} \textit {\_Z}^{6}+4 d f \,\textit {\_Z}^{4}+4 e f \,\textit {\_Z}^{3}+e^{2}\right )^{2}\right )} \]

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(-_R+x),_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 = 0.00, size = 0, normalized size = 0.00 \[ -2 \, \int \frac {{\left (f x^{3} - e\right )} x}{4 \, f^{2} x^{6} + 4 \, d f x^{4} + 4 \, e f x^{3} + e^{2}}\,{d x} \]

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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mupad [B]  time = 3.49, size = 233, normalized size = 5.82 \[ \frac {\mathrm {atan}\left (\frac {128\,d^{7/2}\,\sqrt {f}\,x^2}{64\,d^3\,e+729\,f\,e^3}-\frac {216\,d^{3/2}\,e^2\,\sqrt {f}}{64\,d^3\,e+729\,f\,e^3}+\frac {128\,d^{5/2}\,f^{3/2}\,x^4}{64\,d^3\,e+729\,f\,e^3}+\frac {216\,d^{3/2}\,e\,\sqrt {f}}{64\,d^3+729\,f\,e^2}+\frac {729\,d^{3/2}\,e^2\,f^{3/2}\,x}{64\,d^5+729\,f\,d^2\,e^2}+\frac {1458\,d^{3/2}\,e\,f^{5/2}\,x^4}{64\,d^5+729\,f\,d^2\,e^2}+\frac {64\,d^{5/2}\,e\,\sqrt {f}\,x}{64\,d^3\,e+729\,f\,e^3}+\frac {1458\,d^{3/2}\,e\,f^{3/2}\,x^2}{64\,d^4+729\,f\,d\,e^2}\right )-\mathrm {atan}\left (\frac {\sqrt {f}\,x}{\sqrt {d}}\right )}{2\,\sqrt {d}\,\sqrt {f}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan((128*d^(7/2)*f^(1/2)*x^2)/(64*d^3*e + 729*e^3*f) - (216*d^(3/2)*e^2*f^(1/2))/(64*d^3*e + 729*e^3*f) + (1
28*d^(5/2)*f^(3/2)*x^4)/(64*d^3*e + 729*e^3*f) + (216*d^(3/2)*e*f^(1/2))/(729*e^2*f + 64*d^3) + (729*d^(3/2)*e
^2*f^(3/2)*x)/(64*d^5 + 729*d^2*e^2*f) + (1458*d^(3/2)*e*f^(5/2)*x^4)/(64*d^5 + 729*d^2*e^2*f) + (64*d^(5/2)*e
*f^(1/2)*x)/(64*d^3*e + 729*e^3*f) + (1458*d^(3/2)*e*f^(3/2)*x^2)/(64*d^4 + 729*d*e^2*f)) - atan((f^(1/2)*x)/d
^(1/2)))/(2*d^(1/2)*f^(1/2))

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sympy [B]  time = 1.12, size = 73, normalized size = 1.82 \[ \frac {\sqrt {- \frac {1}{d f}} \log {\left (- d x^{2} \sqrt {- \frac {1}{d f}} + \frac {e}{2 f} + x^{3} \right )}}{4} - \frac {\sqrt {- \frac {1}{d f}} \log {\left (d x^{2} \sqrt {- \frac {1}{d f}} + \frac {e}{2 f} + x^{3} \right )}}{4} \]

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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