\(\int \frac {e^{3 x} (-1-x+3 x^2)+3 e^{3 x} x \log (2 x)}{x^3+2 x^2 \log (2 x)+x \log ^2(2 x)} \, dx\) [7330]

Optimal result

Integrand size = 53, antiderivative size = 23 \[ \int \frac {e^{3 x} \left (-1-x+3 x^2\right )+3 e^{3 x} x \log (2 x)}{x^3+2 x^2 \log (2 x)+x \log ^2(2 x)} \, dx=-\frac {1}{5}+\log \left (4 e^{\frac {e^{3 x}}{x+\log (2 x)}}\right ) \]

[Out]

ln(4*exp(exp(3*x)/(ln(2*x)+x)))-1/5

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {6820, 2326} \[ \int \frac {e^{3 x} \left (-1-x+3 x^2\right )+3 e^{3 x} x \log (2 x)}{x^3+2 x^2 \log (2 x)+x \log ^2(2 x)} \, dx=\frac {e^{3 x} \left (x^2+x \log (2 x)\right )}{x (x+\log (2 x))^2} \]

[In]

Int[(E^(3*x)*(-1 - x + 3*x^2) + 3*E^(3*x)*x*Log[2*x])/(x^3 + 2*x^2*Log[2*x] + x*Log[2*x]^2),x]

[Out]

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

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6820

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

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.61 \[ \int \frac {e^{3 x} \left (-1-x+3 x^2\right )+3 e^{3 x} x \log (2 x)}{x^3+2 x^2 \log (2 x)+x \log ^2(2 x)} \, dx=\frac {e^{3 x}}{x+\log (2 x)} \]

[In]

Integrate[(E^(3*x)*(-1 - x + 3*x^2) + 3*E^(3*x)*x*Log[2*x])/(x^3 + 2*x^2*Log[2*x] + x*Log[2*x]^2),x]

[Out]

E^(3*x)/(x + Log[2*x])

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.61

method	result	size
risch	\(\frac {{\mathrm e}^{3 x}}{\ln \left (2 x \right )+x}\)	\(14\)
parallelrisch	\(\frac {{\mathrm e}^{3 x}}{\ln \left (2 x \right )+x}\)	\(14\)

[In]

int((3*x*exp(3*x)*ln(2*x)+(3*x^2-x-1)*exp(3*x))/(x*ln(2*x)^2+2*x^2*ln(2*x)+x^3),x,method=_RETURNVERBOSE)

[Out]

exp(3*x)/(ln(2*x)+x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.57 \[ \int \frac {e^{3 x} \left (-1-x+3 x^2\right )+3 e^{3 x} x \log (2 x)}{x^3+2 x^2 \log (2 x)+x \log ^2(2 x)} \, dx=\frac {e^{\left (3 \, x\right )}}{x + \log \left (2 \, x\right )} \]

[In]

integrate((3*x*exp(3*x)*log(2*x)+(3*x^2-x-1)*exp(3*x))/(x*log(2*x)^2+2*x^2*log(2*x)+x^3),x, algorithm="fricas"
)

[Out]

e^(3*x)/(x + log(2*x))

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.43 \[ \int \frac {e^{3 x} \left (-1-x+3 x^2\right )+3 e^{3 x} x \log (2 x)}{x^3+2 x^2 \log (2 x)+x \log ^2(2 x)} \, dx=\frac {e^{3 x}}{x + \log {\left (2 x \right )}} \]

[In]

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

[Out]

exp(3*x)/(x + log(2*x))

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.57 \[ \int \frac {e^{3 x} \left (-1-x+3 x^2\right )+3 e^{3 x} x \log (2 x)}{x^3+2 x^2 \log (2 x)+x \log ^2(2 x)} \, dx=\frac {e^{\left (3 \, x\right )}}{x + \log \left (2\right ) + \log \left (x\right )} \]

[In]

integrate((3*x*exp(3*x)*log(2*x)+(3*x^2-x-1)*exp(3*x))/(x*log(2*x)^2+2*x^2*log(2*x)+x^3),x, algorithm="maxima"
)

[Out]

e^(3*x)/(x + log(2) + log(x))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.57 \[ \int \frac {e^{3 x} \left (-1-x+3 x^2\right )+3 e^{3 x} x \log (2 x)}{x^3+2 x^2 \log (2 x)+x \log ^2(2 x)} \, dx=\frac {e^{\left (3 \, x\right )}}{x + \log \left (2 \, x\right )} \]

[In]

integrate((3*x*exp(3*x)*log(2*x)+(3*x^2-x-1)*exp(3*x))/(x*log(2*x)^2+2*x^2*log(2*x)+x^3),x, algorithm="giac")

[Out]

e^(3*x)/(x + log(2*x))

Mupad [B] (verification not implemented)

Time = 14.57 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.57 \[ \int \frac {e^{3 x} \left (-1-x+3 x^2\right )+3 e^{3 x} x \log (2 x)}{x^3+2 x^2 \log (2 x)+x \log ^2(2 x)} \, dx=\frac {{\mathrm {e}}^{3\,x}}{x+\ln \left (2\,x\right )} \]

[In]

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

[Out]

exp(3*x)/(x + log(2*x))