\(\int \frac {e-4 f x^3}{e^2-4 d f x^2+4 e f x^3+4 f^2 x^6} \, dx\) [527]

Optimal result

Integrand size = 37, antiderivative size = 38 \[ \int \frac {e-4 f x^3}{e^2-4 d f x^2+4 e f x^3+4 f^2 x^6} \, dx=\frac {\text {arctanh}\left (\frac {2 \sqrt {d} \sqrt {f} x}{e+2 f x^3}\right )}{2 \sqrt {d} \sqrt {f}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {2118, 214} \[ \int \frac {e-4 f x^3}{e^2-4 d f x^2+4 e f x^3+4 f^2 x^6} \, dx=\frac {\text {arctanh}\left (\frac {2 \sqrt {d} \sqrt {f} x}{e+2 f x^3}\right )}{2 \sqrt {d} \sqrt {f}} \]

[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 214

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

Rule 2118

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*} \text {integral}& = \left (2 e^2\right ) \text {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*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 0.04 (sec) , antiderivative size = 87, normalized size of antiderivative = 2.29 \[ \int \frac {e-4 f x^3}{e^2-4 d f x^2+4 e f x^3+4 f^2 x^6} \, dx=-\frac {\text {RootSum}\left [e^2-4 d f \text {$\#$1}^2+4 e f \text {$\#$1}^3+4 f^2 \text {$\#$1}^6\&,\frac {-e \log (x-\text {$\#$1})+4 f \log (x-\text {$\#$1}) \text {$\#$1}^3}{-2 d \text {$\#$1}+3 e \text {$\#$1}^2+6 f \text {$\#$1}^5}\&\right ]}{4 f} \]

[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

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.08 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.89

method	result	size
default	\(\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (4 f^{2} \textit {\_Z}^{6}+4 e f \,\textit {\_Z}^{3}-4 d f \,\textit {\_Z}^{2}+e^{2}\right )}{\sum }\frac {\left (4 \textit {\_R}^{3} f -e \right ) \ln \left (x -\textit {\_R} \right )}{-6 f \,\textit {\_R}^{5}-3 e \,\textit {\_R}^{2}+2 d \textit {\_R}}}{4 f}\)	\(72\)
risch	\(\frac {\ln \left (\left (-32 f \left (d f \right )^{\frac {3}{2}} d +54 e^{2} f^{2} d \right ) x^{3}+\left (54 e^{2} \left (d f \right )^{\frac {3}{2}}-32 d^{3} f^{2}\right ) x -16 e \left (d f \right )^{\frac {3}{2}} d +27 e^{3} f d \right )}{4 \sqrt {d f}}-\frac {\ln \left (\left (-32 f \left (d f \right )^{\frac {3}{2}} d -54 e^{2} f^{2} d \right ) x^{3}+\left (54 e^{2} \left (d f \right )^{\frac {3}{2}}+32 d^{3} f^{2}\right ) x -16 e \left (d f \right )^{\frac {3}{2}} d -27 e^{3} f d \right )}{4 \sqrt {d f}}\)	\(140\)

[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,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (28) = 56\).

Time = 0.32 (sec) , antiderivative size = 155, normalized size of antiderivative = 4.08 \[ \int \frac {e-4 f x^3}{e^2-4 d f x^2+4 e f x^3+4 f^2 x^6} \, dx=\left [\frac {\sqrt {d f} \log \left (\frac {4 \, f^{2} x^{6} + 4 \, e f x^{3} + 4 \, d f x^{2} + e^{2} + 4 \, {\left (2 \, f x^{4} + e x\right )} \sqrt {d f}}{4 \, f^{2} x^{6} + 4 \, e f x^{3} - 4 \, d f x^{2} + e^{2}}\right )}{4 \, d f}, -\frac {\sqrt {-d f} \arctan \left (\frac {\sqrt {-d f} x^{2}}{d}\right ) - \sqrt {-d f} \arctan \left (\frac {{\left (2 \, f x^{5} + e x^{2} - 2 \, d x\right )} \sqrt {-d f}}{d e}\right )}{2 \, d f}\right ] \]

[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)]

Sympy [A] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.66 \[ \int \frac {e-4 f x^3}{e^2-4 d f x^2+4 e f x^3+4 f^2 x^6} \, dx=- \frac {\sqrt {\frac {1}{d f}} \log {\left (- d x \sqrt {\frac {1}{d f}} + \frac {e}{2 f} + x^{3} \right )}}{4} + \frac {\sqrt {\frac {1}{d f}} \log {\left (d x \sqrt {\frac {1}{d f}} + \frac {e}{2 f} + x^{3} \right )}}{4} \]

[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

Maxima [F]

\[ \int \frac {e-4 f x^3}{e^2-4 d f x^2+4 e f x^3+4 f^2 x^6} \, dx=\int { -\frac {4 \, f x^{3} - e}{4 \, f^{2} x^{6} + 4 \, e f x^{3} - 4 \, d f x^{2} + e^{2}} \,d x } \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (28) = 56\).

Time = 0.33 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.66 \[ \int \frac {e-4 f x^3}{e^2-4 d f x^2+4 e f x^3+4 f^2 x^6} \, dx=\frac {\sqrt {d f} \log \left ({\left | 2 \, f x^{3} + 2 \, \sqrt {d f} x + e \right |}\right )}{4 \, d f} - \frac {\sqrt {d f} \log \left ({\left | 2 \, f x^{3} - 2 \, \sqrt {d f} x + e \right |}\right )}{4 \, d f} \]

[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]

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

Mupad [B] (verification not implemented)

Time = 17.55 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.76 \[ \int \frac {e-4 f x^3}{e^2-4 d f x^2+4 e f x^3+4 f^2 x^6} \, dx=-\frac {\mathrm {atanh}\left (\frac {-32\,f\,d^2\,x+27\,e^3+54\,f\,e^2\,x^3}{16\,d^{3/2}\,e\,\sqrt {f}+32\,d^{3/2}\,f^{3/2}\,x^3-54\,\sqrt {d}\,e^2\,\sqrt {f}\,x}\right )}{2\,\sqrt {d}\,\sqrt {f}} \]

[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))