\(\int \frac {e^{16 x} (6 x+48 x^2)}{-2+3 e^{16 x} x^2} \, dx\) [6573]

Optimal result

Integrand size = 29, antiderivative size = 14 \[ \int \frac {e^{16 x} \left (6 x+48 x^2\right )}{-2+3 e^{16 x} x^2} \, dx=\log \left (-\frac {2}{3}+e^{16 x} x^2\right ) \]

[Out]

ln(x^2*exp(x)^16-2/3)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {1607, 6816} \[ \int \frac {e^{16 x} \left (6 x+48 x^2\right )}{-2+3 e^{16 x} x^2} \, dx=\log \left (2-3 e^{16 x} x^2\right ) \]

[In]

Int[(E^(16*x)*(6*x + 48*x^2))/(-2 + 3*E^(16*x)*x^2),x]

[Out]

Log[2 - 3*E^(16*x)*x^2]

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{16 x} x (6+48 x)}{-2+3 e^{16 x} x^2} \, dx \\ & = \log \left (2-3 e^{16 x} x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93 \[ \int \frac {e^{16 x} \left (6 x+48 x^2\right )}{-2+3 e^{16 x} x^2} \, dx=\log \left (-2+3 e^{16 x} x^2\right ) \]

[In]

Integrate[(E^(16*x)*(6*x + 48*x^2))/(-2 + 3*E^(16*x)*x^2),x]

[Out]

Log[-2 + 3*E^(16*x)*x^2]

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86

method	result	size
parallelrisch	\(\ln \left (x^{2} {\mathrm e}^{16 x}-\frac {2}{3}\right )\)	\(12\)
risch	\(2 \ln \left (x \right )+\ln \left ({\mathrm e}^{16 x}-\frac {2}{3 x^{2}}\right )\)	\(17\)

[In]

int((48*x^2+6*x)*exp(x)^16/(3*x^2*exp(x)^16-2),x,method=_RETURNVERBOSE)

[Out]

ln(x^2*exp(x)^16-2/3)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.50 \[ \int \frac {e^{16 x} \left (6 x+48 x^2\right )}{-2+3 e^{16 x} x^2} \, dx=2 \, \log \left (x\right ) + \log \left (\frac {3 \, x^{2} e^{\left (16 \, x\right )} - 2}{x^{2}}\right ) \]

[In]

integrate((48*x^2+6*x)*exp(x)^16/(3*x^2*exp(x)^16-2),x, algorithm="fricas")

[Out]

2*log(x) + log((3*x^2*e^(16*x) - 2)/x^2)

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.21 \[ \int \frac {e^{16 x} \left (6 x+48 x^2\right )}{-2+3 e^{16 x} x^2} \, dx=2 \log {\left (x \right )} + \log {\left (e^{16 x} - \frac {2}{3 x^{2}} \right )} \]

[In]

integrate((48*x**2+6*x)*exp(x)**16/(3*x**2*exp(x)**16-2),x)

[Out]

2*log(x) + log(exp(16*x) - 2/(3*x**2))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.57 \[ \int \frac {e^{16 x} \left (6 x+48 x^2\right )}{-2+3 e^{16 x} x^2} \, dx=2 \, \log \left (x\right ) + \log \left (\frac {3 \, x^{2} e^{\left (16 \, x\right )} - 2}{3 \, x^{2}}\right ) \]

[In]

integrate((48*x^2+6*x)*exp(x)^16/(3*x^2*exp(x)^16-2),x, algorithm="maxima")

[Out]

2*log(x) + log(1/3*(3*x^2*e^(16*x) - 2)/x^2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {e^{16 x} \left (6 x+48 x^2\right )}{-2+3 e^{16 x} x^2} \, dx=\log \left (3 \, x^{2} e^{\left (16 \, x\right )} - 2\right ) \]

[In]

integrate((48*x^2+6*x)*exp(x)^16/(3*x^2*exp(x)^16-2),x, algorithm="giac")

[Out]

log(3*x^2*e^(16*x) - 2)

Mupad [B] (verification not implemented)

Time = 12.54 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {e^{16 x} \left (6 x+48 x^2\right )}{-2+3 e^{16 x} x^2} \, dx=\ln \left (3\,x^2\,{\mathrm {e}}^{16\,x}-2\right ) \]

[In]

int((exp(16*x)*(6*x + 48*x^2))/(3*x^2*exp(16*x) - 2),x)

[Out]

log(3*x^2*exp(16*x) - 2)