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

Optimal. Leaf size=38 \[ \frac {\tanh ^{-1}\left (\frac {2 \sqrt {d} \sqrt {f} x}{e+2 f x^3}\right )}{2 \sqrt {d} \sqrt {f}} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {2093, 208} \[ \frac {\tanh ^{-1}\left (\frac {2 \sqrt {d} \sqrt {f} x}{e+2 f x^3}\right )}{2 \sqrt {d} \sqrt {f}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 208

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

Rule 2093

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

Rubi steps

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

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Mathematica [C]  time = 0.06, size = 87, normalized size = 2.29 \[ -\frac {\text {RootSum}\left [4 \text {$\#$1}^6 f^2+4 \text {$\#$1}^3 e f-4 \text {$\#$1}^2 d f+e^2\& ,\frac {4 \text {$\#$1}^3 f \log (x-\text {$\#$1})-e \log (x-\text {$\#$1})}{6 \text {$\#$1}^5 f+3 \text {$\#$1}^2 e-2 \text {$\#$1} d}\& \right ]}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*RootSum[e^2 - 4*d*f*#1^2 + 4*e*f*#1^3 + 4*f^2*#1^6 & , (-(e*Log[x - #1]) + 4*f*Log[x - #1]*#1^3)/(-2*d*#1
+ 3*e*#1^2 + 6*f*#1^5) & ]/f

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fricas [B]  time = 0.71, size = 155, normalized size = 4.08 \[ \left [\frac {\sqrt {d f} \log \left (\frac {4 \, f^{2} x^{6} + 4 \, e f x^{3} + 4 \, d f x^{2} + e^{2} + 4 \, {\left (2 \, f x^{4} + e x\right )} \sqrt {d f}}{4 \, f^{2} x^{6} + 4 \, e f x^{3} - 4 \, d f x^{2} + e^{2}}\right )}{4 \, d f}, -\frac {\sqrt {-d f} \arctan \left (\frac {\sqrt {-d f} x^{2}}{d}\right ) - \sqrt {-d f} \arctan \left (\frac {{\left (2 \, f x^{5} + e x^{2} - 2 \, d x\right )} \sqrt {-d f}}{d e}\right )}{2 \, d f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.01, size = 70, normalized size = 1.84 \[ \frac {\left (-4 \RootOf \left (4 f^{2} \textit {\_Z}^{6}+4 e f \,\textit {\_Z}^{3}-4 d f \,\textit {\_Z}^{2}+e^{2}\right )^{3} f +e \right ) \ln \left (-\RootOf \left (4 f^{2} \textit {\_Z}^{6}+4 e f \,\textit {\_Z}^{3}-4 d f \,\textit {\_Z}^{2}+e^{2}\right )+x \right )}{4 f \left (6 f \RootOf \left (4 f^{2} \textit {\_Z}^{6}+4 e f \,\textit {\_Z}^{3}-4 d f \,\textit {\_Z}^{2}+e^{2}\right )^{5}+3 e \RootOf \left (4 f^{2} \textit {\_Z}^{6}+4 e f \,\textit {\_Z}^{3}-4 d f \,\textit {\_Z}^{2}+e^{2}\right )^{2}-2 d \RootOf \left (4 f^{2} \textit {\_Z}^{6}+4 e f \,\textit {\_Z}^{3}-4 d f \,\textit {\_Z}^{2}+e^{2}\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.12, size = 67, normalized size = 1.76 \[ -\frac {\mathrm {atanh}\left (\frac {-32\,f\,d^2\,x+27\,e^3+54\,f\,e^2\,x^3}{16\,d^{3/2}\,e\,\sqrt {f}+32\,d^{3/2}\,f^{3/2}\,x^3-54\,\sqrt {d}\,e^2\,\sqrt {f}\,x}\right )}{2\,\sqrt {d}\,\sqrt {f}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-atanh((27*e^3 - 32*d^2*f*x + 54*e^2*f*x^3)/(16*d^(3/2)*e*f^(1/2) + 32*d^(3/2)*f^(3/2)*x^3 - 54*d^(1/2)*e^2*f^
(1/2)*x))/(2*d^(1/2)*f^(1/2))

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sympy [A]  time = 0.74, size = 63, normalized size = 1.66 \[ - \frac {\sqrt {\frac {1}{d f}} \log {\left (- d x \sqrt {\frac {1}{d f}} + \frac {e}{2 f} + x^{3} \right )}}{4} + \frac {\sqrt {\frac {1}{d f}} \log {\left (d x \sqrt {\frac {1}{d f}} + \frac {e}{2 f} + x^{3} \right )}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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