\(\int \frac {e^{12 x-3 x \log (\frac {-2+5 x}{x})} (-30+60 x+(6-15 x) \log (\frac {-2+5 x}{x}))}{-2+5 x} \, dx\) [1244]

Optimal result

Integrand size = 48, antiderivative size = 23 \[ \int \frac {e^{12 x-3 x \log \left (\frac {-2+5 x}{x}\right )} \left (-30+60 x+(6-15 x) \log \left (\frac {-2+5 x}{x}\right )\right )}{-2+5 x} \, dx=e^{3 x \left (4-\log \left (5+x-\frac {2+x^2}{x}\right )\right )} \]

[Out]

exp(3*(4-ln(5-(x^2+2)/x+x))*x)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.021, Rules used = {6838} \[ \int \frac {e^{12 x-3 x \log \left (\frac {-2+5 x}{x}\right )} \left (-30+60 x+(6-15 x) \log \left (\frac {-2+5 x}{x}\right )\right )}{-2+5 x} \, dx=e^{12 x} \left (-\frac {2-5 x}{x}\right )^{-3 x} \]

[In]

Int[(E^(12*x - 3*x*Log[(-2 + 5*x)/x])*(-30 + 60*x + (6 - 15*x)*Log[(-2 + 5*x)/x]))/(-2 + 5*x),x]

[Out]

E^(12*x)/(-((2 - 5*x)/x))^(3*x)

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = e^{12 x} \left (-\frac {2-5 x}{x}\right )^{-3 x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {e^{12 x-3 x \log \left (\frac {-2+5 x}{x}\right )} \left (-30+60 x+(6-15 x) \log \left (\frac {-2+5 x}{x}\right )\right )}{-2+5 x} \, dx=e^{12 x} \left (5-\frac {2}{x}\right )^{-3 x} \]

[In]

Integrate[(E^(12*x - 3*x*Log[(-2 + 5*x)/x])*(-30 + 60*x + (6 - 15*x)*Log[(-2 + 5*x)/x]))/(-2 + 5*x),x]

[Out]

E^(12*x)/(5 - 2/x)^(3*x)

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83

method	result	size
norman	\({\mathrm e}^{-3 x \ln \left (\frac {5 x -2}{x}\right )+12 x}\)	\(19\)
risch	\(\left (\frac {5 x -2}{x}\right )^{-3 x} {\mathrm e}^{12 x}\)	\(19\)
parallelrisch	\({\mathrm e}^{-3 x \ln \left (\frac {5 x -2}{x}\right )+12 x}\)	\(19\)

[In]

int(((-15*x+6)*ln((5*x-2)/x)+60*x-30)*exp(-3*x*ln((5*x-2)/x)+12*x)/(5*x-2),x,method=_RETURNVERBOSE)

[Out]

exp(-3*x*ln((5*x-2)/x)+12*x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {e^{12 x-3 x \log \left (\frac {-2+5 x}{x}\right )} \left (-30+60 x+(6-15 x) \log \left (\frac {-2+5 x}{x}\right )\right )}{-2+5 x} \, dx=e^{\left (-3 \, x \log \left (\frac {5 \, x - 2}{x}\right ) + 12 \, x\right )} \]

[In]

integrate(((-15*x+6)*log((5*x-2)/x)+60*x-30)*exp(-3*x*log((5*x-2)/x)+12*x)/(5*x-2),x, algorithm="fricas")

[Out]

e^(-3*x*log((5*x - 2)/x) + 12*x)

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int \frac {e^{12 x-3 x \log \left (\frac {-2+5 x}{x}\right )} \left (-30+60 x+(6-15 x) \log \left (\frac {-2+5 x}{x}\right )\right )}{-2+5 x} \, dx=e^{- 3 x \log {\left (\frac {5 x - 2}{x} \right )} + 12 x} \]

[In]

integrate(((-15*x+6)*ln((5*x-2)/x)+60*x-30)*exp(-3*x*ln((5*x-2)/x)+12*x)/(5*x-2),x)

[Out]

exp(-3*x*log((5*x - 2)/x) + 12*x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {e^{12 x-3 x \log \left (\frac {-2+5 x}{x}\right )} \left (-30+60 x+(6-15 x) \log \left (\frac {-2+5 x}{x}\right )\right )}{-2+5 x} \, dx=e^{\left (-3 \, x \log \left (5 \, x - 2\right ) + 3 \, x \log \left (x\right ) + 12 \, x\right )} \]

[In]

integrate(((-15*x+6)*log((5*x-2)/x)+60*x-30)*exp(-3*x*log((5*x-2)/x)+12*x)/(5*x-2),x, algorithm="maxima")

[Out]

e^(-3*x*log(5*x - 2) + 3*x*log(x) + 12*x)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.70 \[ \int \frac {e^{12 x-3 x \log \left (\frac {-2+5 x}{x}\right )} \left (-30+60 x+(6-15 x) \log \left (\frac {-2+5 x}{x}\right )\right )}{-2+5 x} \, dx=e^{\left (-3 \, x \log \left (-\frac {2}{x} + 5\right ) + 12 \, x\right )} \]

[In]

integrate(((-15*x+6)*log((5*x-2)/x)+60*x-30)*exp(-3*x*log((5*x-2)/x)+12*x)/(5*x-2),x, algorithm="giac")

[Out]

e^(-3*x*log(-2/x + 5) + 12*x)

Mupad [B] (verification not implemented)

Time = 8.35 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {e^{12 x-3 x \log \left (\frac {-2+5 x}{x}\right )} \left (-30+60 x+(6-15 x) \log \left (\frac {-2+5 x}{x}\right )\right )}{-2+5 x} \, dx=\frac {{\mathrm {e}}^{12\,x}}{{\left (5-\frac {2}{x}\right )}^{3\,x}} \]

[In]

int(-(exp(12*x - 3*x*log((5*x - 2)/x))*(log((5*x - 2)/x)*(15*x - 6) - 60*x + 30))/(5*x - 2),x)

[Out]

exp(12*x)/(5 - 2/x)^(3*x)