\(\int \frac {4 e-6 x+(-3 e x+3 x^2) \log ^3(e x^2-x^3)}{(e x-x^2) \log ^3(e x^2-x^3)} \, dx\) [3475]

Optimal result

Integrand size = 58, antiderivative size = 19 \[ \int \frac {4 e-6 x+\left (-3 e x+3 x^2\right ) \log ^3\left (e x^2-x^3\right )}{\left (e x-x^2\right ) \log ^3\left (e x^2-x^3\right )} \, dx=2-3 x-\frac {1}{\log ^2\left ((e-x) x^2\right )} \]

[Out]

-3*x-1/ln((exp(1)-x)*x^2)^2+2

Rubi [F]

\[ \int \frac {4 e-6 x+\left (-3 e x+3 x^2\right ) \log ^3\left (e x^2-x^3\right )}{\left (e x-x^2\right ) \log ^3\left (e x^2-x^3\right )} \, dx=\int \frac {4 e-6 x+\left (-3 e x+3 x^2\right ) \log ^3\left (e x^2-x^3\right )}{\left (e x-x^2\right ) \log ^3\left (e x^2-x^3\right )} \, dx \]

[In]

Int[(4*E - 6*x + (-3*E*x + 3*x^2)*Log[E*x^2 - x^3]^3)/((E*x - x^2)*Log[E*x^2 - x^3]^3),x]

[Out]

-3*x + 2*Defer[Int][(2*E - 3*x)/((E - x)*x*Log[(E - x)*x^2]^3), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {4 e-6 x+\left (-3 e x+3 x^2\right ) \log ^3\left (e x^2-x^3\right )}{(e-x) x \log ^3\left (e x^2-x^3\right )} \, dx \\ & = \int \frac {4 e-6 x+\left (-3 e x+3 x^2\right ) \log ^3\left (e x^2-x^3\right )}{(e-x) x \log ^3\left ((e-x) x^2\right )} \, dx \\ & = \int \left (-3+\frac {2 (2 e-3 x)}{(e-x) x \log ^3\left ((e-x) x^2\right )}\right ) \, dx \\ & = -3 x+2 \int \frac {2 e-3 x}{(e-x) x \log ^3\left ((e-x) x^2\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {4 e-6 x+\left (-3 e x+3 x^2\right ) \log ^3\left (e x^2-x^3\right )}{\left (e x-x^2\right ) \log ^3\left (e x^2-x^3\right )} \, dx=-3 x-\frac {1}{\log ^2\left ((e-x) x^2\right )} \]

[In]

Integrate[(4*E - 6*x + (-3*E*x + 3*x^2)*Log[E*x^2 - x^3]^3)/((E*x - x^2)*Log[E*x^2 - x^3]^3),x]

[Out]

-3*x - Log[(E - x)*x^2]^(-2)

Maple [A] (verified)

Time = 1.70 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16

method	result	size
default	\(-3 x -\frac {1}{\ln \left (x^{2} {\mathrm e}-x^{3}\right )^{2}}\)	\(22\)
risch	\(-3 x -\frac {1}{\ln \left (x^{2} {\mathrm e}-x^{3}\right )^{2}}\)	\(22\)
parts	\(-3 x -\frac {1}{\ln \left (x^{2} {\mathrm e}-x^{3}\right )^{2}}\)	\(22\)
norman	\(\frac {-1-3 x \ln \left (x^{2} {\mathrm e}-x^{3}\right )^{2}}{\ln \left (x^{2} {\mathrm e}-x^{3}\right )^{2}}\)	\(37\)
parallelrisch	\(-\frac {1+6 \ln \left (\left ({\mathrm e}-x \right ) x^{2}\right )^{2} {\mathrm e}+3 \ln \left (\left ({\mathrm e}-x \right ) x^{2}\right )^{2} x}{\ln \left (\left ({\mathrm e}-x \right ) x^{2}\right )^{2}}\)	\(51\)

[In]

int(((-3*x*exp(1)+3*x^2)*ln(x^2*exp(1)-x^3)^3+4*exp(1)-6*x)/(x*exp(1)-x^2)/ln(x^2*exp(1)-x^3)^3,x,method=_RETU
RNVERBOSE)

[Out]

-3*x-1/ln(x^2*exp(1)-x^3)^2

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.95 \[ \int \frac {4 e-6 x+\left (-3 e x+3 x^2\right ) \log ^3\left (e x^2-x^3\right )}{\left (e x-x^2\right ) \log ^3\left (e x^2-x^3\right )} \, dx=-\frac {3 \, x \log \left (-x^{3} + x^{2} e\right )^{2} + 1}{\log \left (-x^{3} + x^{2} e\right )^{2}} \]

[In]

integrate(((-3*x*exp(1)+3*x^2)*log(x^2*exp(1)-x^3)^3+4*exp(1)-6*x)/(x*exp(1)-x^2)/log(x^2*exp(1)-x^3)^3,x, alg
orithm="fricas")

[Out]

-(3*x*log(-x^3 + x^2*e)^2 + 1)/log(-x^3 + x^2*e)^2

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {4 e-6 x+\left (-3 e x+3 x^2\right ) \log ^3\left (e x^2-x^3\right )}{\left (e x-x^2\right ) \log ^3\left (e x^2-x^3\right )} \, dx=- 3 x - \frac {1}{\log {\left (- x^{3} + e x^{2} \right )}^{2}} \]

[In]

integrate(((-3*x*exp(1)+3*x**2)*ln(x**2*exp(1)-x**3)**3+4*exp(1)-6*x)/(x*exp(1)-x**2)/ln(x**2*exp(1)-x**3)**3,
x)

[Out]

-3*x - 1/log(-x**3 + E*x**2)**2

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (21) = 42\).

Time = 0.23 (sec) , antiderivative size = 64, normalized size of antiderivative = 3.37 \[ \int \frac {4 e-6 x+\left (-3 e x+3 x^2\right ) \log ^3\left (e x^2-x^3\right )}{\left (e x-x^2\right ) \log ^3\left (e x^2-x^3\right )} \, dx=-\frac {12 \, x \log \left (x\right )^{2} + 12 \, x \log \left (x\right ) \log \left (-x + e\right ) + 3 \, x \log \left (-x + e\right )^{2} + 1}{4 \, \log \left (x\right )^{2} + 4 \, \log \left (x\right ) \log \left (-x + e\right ) + \log \left (-x + e\right )^{2}} \]

[In]

integrate(((-3*x*exp(1)+3*x^2)*log(x^2*exp(1)-x^3)^3+4*exp(1)-6*x)/(x*exp(1)-x^2)/log(x^2*exp(1)-x^3)^3,x, alg
orithm="maxima")

[Out]

-(12*x*log(x)^2 + 12*x*log(x)*log(-x + e) + 3*x*log(-x + e)^2 + 1)/(4*log(x)^2 + 4*log(x)*log(-x + e) + log(-x
+ e)^2)

Giac [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.95 \[ \int \frac {4 e-6 x+\left (-3 e x+3 x^2\right ) \log ^3\left (e x^2-x^3\right )}{\left (e x-x^2\right ) \log ^3\left (e x^2-x^3\right )} \, dx=-\frac {3 \, x \log \left (-x^{3} + x^{2} e\right )^{2} + 1}{\log \left (-x^{3} + x^{2} e\right )^{2}} \]

[In]

integrate(((-3*x*exp(1)+3*x^2)*log(x^2*exp(1)-x^3)^3+4*exp(1)-6*x)/(x*exp(1)-x^2)/log(x^2*exp(1)-x^3)^3,x, alg
orithm="giac")

[Out]

-(3*x*log(-x^3 + x^2*e)^2 + 1)/log(-x^3 + x^2*e)^2

Mupad [B] (verification not implemented)

Time = 9.14 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {4 e-6 x+\left (-3 e x+3 x^2\right ) \log ^3\left (e x^2-x^3\right )}{\left (e x-x^2\right ) \log ^3\left (e x^2-x^3\right )} \, dx=-3\,x-\frac {1}{{\ln \left (x^2\,\mathrm {e}-x^3\right )}^2} \]

[In]

int(-(6*x - 4*exp(1) + log(x^2*exp(1) - x^3)^3*(3*x*exp(1) - 3*x^2))/(log(x^2*exp(1) - x^3)^3*(x*exp(1) - x^2)
),x)

[Out]

- 3*x - 1/log(x^2*exp(1) - x^3)^2