\(\int \frac {-x+3 e^{x^3} x^3+(-e^{x^3}+x) \log (-e^{x^3}+x)}{e^{x^3} x^2-x^3} \, dx\) [3818]

Optimal result

Integrand size = 52, antiderivative size = 14 \[ \int \frac {-x+3 e^{x^3} x^3+\left (-e^{x^3}+x\right ) \log \left (-e^{x^3}+x\right )}{e^{x^3} x^2-x^3} \, dx=\frac {\log \left (-e^{x^3}+x\right )}{x} \]

[Out]

ln(-exp(x^3)+x)/x

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.058, Rules used = {6874, 14, 2631} \[ \int \frac {-x+3 e^{x^3} x^3+\left (-e^{x^3}+x\right ) \log \left (-e^{x^3}+x\right )}{e^{x^3} x^2-x^3} \, dx=\frac {\log \left (x-e^{x^3}\right )}{x} \]

[In]

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

[Out]

Log[-E^x^3 + x]/x

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2631

Int[Log[u_]*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[(a + b*x)^(m + 1)*(Log[u]/(b*(m + 1))), x] - Dist[1/
(b*(m + 1)), Int[SimplifyIntegrand[(a + b*x)^(m + 1)*(D[u, x]/u), x], x], x] /; FreeQ[{a, b, m}, x] && Inverse
FunctionFreeQ[u, x] && NeQ[m, -1]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {-1+3 x^3}{\left (e^{x^3}-x\right ) x}+\frac {3 x^3-\log \left (-e^{x^3}+x\right )}{x^2}\right ) \, dx \\ & = \int \frac {-1+3 x^3}{\left (e^{x^3}-x\right ) x} \, dx+\int \frac {3 x^3-\log \left (-e^{x^3}+x\right )}{x^2} \, dx \\ & = \int \left (-\frac {1}{\left (e^{x^3}-x\right ) x}+\frac {3 x^2}{e^{x^3}-x}\right ) \, dx+\int \left (3 x-\frac {\log \left (-e^{x^3}+x\right )}{x^2}\right ) \, dx \\ & = \frac {3 x^2}{2}+3 \int \frac {x^2}{e^{x^3}-x} \, dx-\int \frac {1}{\left (e^{x^3}-x\right ) x} \, dx-\int \frac {\log \left (-e^{x^3}+x\right )}{x^2} \, dx \\ & = \frac {3 x^2}{2}+\frac {\log \left (-e^{x^3}+x\right )}{x}+3 \int \frac {x^2}{e^{x^3}-x} \, dx-\int \frac {1}{\left (e^{x^3}-x\right ) x} \, dx-\int \frac {1-3 e^{x^3} x^2}{x \left (-e^{x^3}+x\right )} \, dx \\ & = \frac {3 x^2}{2}+\frac {\log \left (-e^{x^3}+x\right )}{x}+3 \int \frac {x^2}{e^{x^3}-x} \, dx-\int \frac {1}{\left (e^{x^3}-x\right ) x} \, dx-\int \left (3 x+\frac {-1+3 x^3}{\left (e^{x^3}-x\right ) x}\right ) \, dx \\ & = \frac {\log \left (-e^{x^3}+x\right )}{x}+3 \int \frac {x^2}{e^{x^3}-x} \, dx-\int \frac {1}{\left (e^{x^3}-x\right ) x} \, dx-\int \frac {-1+3 x^3}{\left (e^{x^3}-x\right ) x} \, dx \\ & = \frac {\log \left (-e^{x^3}+x\right )}{x}+3 \int \frac {x^2}{e^{x^3}-x} \, dx-\int \frac {1}{\left (e^{x^3}-x\right ) x} \, dx-\int \left (-\frac {1}{\left (e^{x^3}-x\right ) x}+\frac {3 x^2}{e^{x^3}-x}\right ) \, dx \\ & = \frac {\log \left (-e^{x^3}+x\right )}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {-x+3 e^{x^3} x^3+\left (-e^{x^3}+x\right ) \log \left (-e^{x^3}+x\right )}{e^{x^3} x^2-x^3} \, dx=\frac {\log \left (-e^{x^3}+x\right )}{x} \]

[In]

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

[Out]

Log[-E^x^3 + x]/x

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00

method	result	size
norman	\(\frac {\ln \left (-{\mathrm e}^{x^{3}}+x \right )}{x}\)	\(14\)
risch	\(\frac {\ln \left (-{\mathrm e}^{x^{3}}+x \right )}{x}\)	\(14\)
parallelrisch	\(\frac {\ln \left (-{\mathrm e}^{x^{3}}+x \right )}{x}\)	\(14\)

[In]

int(((-exp(x^3)+x)*ln(-exp(x^3)+x)+3*x^3*exp(x^3)-x)/(x^2*exp(x^3)-x^3),x,method=_RETURNVERBOSE)

[Out]

ln(-exp(x^3)+x)/x

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93 \[ \int \frac {-x+3 e^{x^3} x^3+\left (-e^{x^3}+x\right ) \log \left (-e^{x^3}+x\right )}{e^{x^3} x^2-x^3} \, dx=\frac {\log \left (x - e^{\left (x^{3}\right )}\right )}{x} \]

[In]

integrate(((-exp(x^3)+x)*log(-exp(x^3)+x)+3*x^3*exp(x^3)-x)/(x^2*exp(x^3)-x^3),x, algorithm="fricas")

[Out]

log(x - e^(x^3))/x

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.57 \[ \int \frac {-x+3 e^{x^3} x^3+\left (-e^{x^3}+x\right ) \log \left (-e^{x^3}+x\right )}{e^{x^3} x^2-x^3} \, dx=\frac {\log {\left (x - e^{x^{3}} \right )}}{x} \]

[In]

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

[Out]

log(x - exp(x**3))/x

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93 \[ \int \frac {-x+3 e^{x^3} x^3+\left (-e^{x^3}+x\right ) \log \left (-e^{x^3}+x\right )}{e^{x^3} x^2-x^3} \, dx=\frac {\log \left (x - e^{\left (x^{3}\right )}\right )}{x} \]

[In]

integrate(((-exp(x^3)+x)*log(-exp(x^3)+x)+3*x^3*exp(x^3)-x)/(x^2*exp(x^3)-x^3),x, algorithm="maxima")

[Out]

log(x - e^(x^3))/x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93 \[ \int \frac {-x+3 e^{x^3} x^3+\left (-e^{x^3}+x\right ) \log \left (-e^{x^3}+x\right )}{e^{x^3} x^2-x^3} \, dx=\frac {\log \left (x - e^{\left (x^{3}\right )}\right )}{x} \]

[In]

integrate(((-exp(x^3)+x)*log(-exp(x^3)+x)+3*x^3*exp(x^3)-x)/(x^2*exp(x^3)-x^3),x, algorithm="giac")

[Out]

log(x - e^(x^3))/x

Mupad [B] (verification not implemented)

Time = 9.60 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93 \[ \int \frac {-x+3 e^{x^3} x^3+\left (-e^{x^3}+x\right ) \log \left (-e^{x^3}+x\right )}{e^{x^3} x^2-x^3} \, dx=\frac {\ln \left (x-{\mathrm {e}}^{x^3}\right )}{x} \]

[In]

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

[Out]

log(x - exp(x^3))/x