\(\int \frac {3 e^{1+16 x^2}-2 x^2-64 x^4}{e^{1+16 x^2} x-2 x^3} \, dx\) [9093]

Optimal result

Integrand size = 42, antiderivative size = 26 \[ \int \frac {3 e^{1+16 x^2}-2 x^2-64 x^4}{e^{1+16 x^2} x-2 x^3} \, dx=\log \left (\frac {2 x^3}{3 \left (2-4 e^{-1-16 x^2} x^2\right )}\right ) \]

[Out]

ln(2/3*x^3/(2-4*x^2/exp(16*x^2+1)))

Rubi [F]

\[ \int \frac {3 e^{1+16 x^2}-2 x^2-64 x^4}{e^{1+16 x^2} x-2 x^3} \, dx=\int \frac {3 e^{1+16 x^2}-2 x^2-64 x^4}{e^{1+16 x^2} x-2 x^3} \, dx \]

[In]

Int[(3*E^(1 + 16*x^2) - 2*x^2 - 64*x^4)/(E^(1 + 16*x^2)*x - 2*x^3),x]

[Out]

3*Log[x] + 2*Defer[Subst][Defer[Int][(E^(1 + 16*x) - 2*x)^(-1), x], x, x^2] - 32*Defer[Subst][Defer[Int][x/(E^
(1 + 16*x) - 2*x), x], x, x^2]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {3}{x}-\frac {4 x \left (-1+16 x^2\right )}{e^{1+16 x^2}-2 x^2}\right ) \, dx \\ & = 3 \log (x)-4 \int \frac {x \left (-1+16 x^2\right )}{e^{1+16 x^2}-2 x^2} \, dx \\ & = 3 \log (x)-2 \text {Subst}\left (\int \frac {-1+16 x}{e^{1+16 x}-2 x} \, dx,x,x^2\right ) \\ & = 3 \log (x)-2 \text {Subst}\left (\int \left (-\frac {1}{e^{1+16 x}-2 x}+\frac {16 x}{e^{1+16 x}-2 x}\right ) \, dx,x,x^2\right ) \\ & = 3 \log (x)+2 \text {Subst}\left (\int \frac {1}{e^{1+16 x}-2 x} \, dx,x,x^2\right )-32 \text {Subst}\left (\int \frac {x}{e^{1+16 x}-2 x} \, dx,x,x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 2.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {3 e^{1+16 x^2}-2 x^2-64 x^4}{e^{1+16 x^2} x-2 x^3} \, dx=16 x^2+3 \log (x)-\log \left (e^{1+16 x^2}-2 x^2\right ) \]

[In]

Integrate[(3*E^(1 + 16*x^2) - 2*x^2 - 64*x^4)/(E^(1 + 16*x^2)*x - 2*x^3),x]

[Out]

16*x^2 + 3*Log[x] - Log[E^(1 + 16*x^2) - 2*x^2]

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08

method	result	size
parallelrisch	\(16 x^{2}+3 \ln \left (x \right )-\ln \left (x^{2}-\frac {{\mathrm e}^{16 x^{2}+1}}{2}\right )\)	\(28\)
risch	\(3 \ln \left (x \right )+16 x^{2}+1-\ln \left (-2 x^{2}+{\mathrm e}^{16 x^{2}+1}\right )\)	\(29\)
norman	\(16 x^{2}+3 \ln \left (x \right )-\ln \left (2 x^{2}-{\mathrm e}^{16 x^{2}+1}\right )\)	\(30\)

[In]

int((3*exp(16*x^2+1)-64*x^4-2*x^2)/(x*exp(16*x^2+1)-2*x^3),x,method=_RETURNVERBOSE)

[Out]

16*x^2+3*ln(x)-ln(x^2-1/2*exp(16*x^2+1))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {3 e^{1+16 x^2}-2 x^2-64 x^4}{e^{1+16 x^2} x-2 x^3} \, dx=16 \, x^{2} - \log \left (-2 \, x^{2} + e^{\left (16 \, x^{2} + 1\right )}\right ) + 3 \, \log \left (x\right ) \]

[In]

integrate((3*exp(16*x^2+1)-64*x^4-2*x^2)/(x*exp(16*x^2+1)-2*x^3),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {3 e^{1+16 x^2}-2 x^2-64 x^4}{e^{1+16 x^2} x-2 x^3} \, dx=16 x^{2} + 3 \log {\left (x \right )} - \log {\left (- 2 x^{2} + e^{16 x^{2} + 1} \right )} \]

[In]

integrate((3*exp(16*x**2+1)-64*x**4-2*x**2)/(x*exp(16*x**2+1)-2*x**3),x)

[Out]

16*x**2 + 3*log(x) - log(-2*x**2 + exp(16*x**2 + 1))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27 \[ \int \frac {3 e^{1+16 x^2}-2 x^2-64 x^4}{e^{1+16 x^2} x-2 x^3} \, dx=16 \, x^{2} - \log \left (-{\left (2 \, x^{2} - e^{\left (16 \, x^{2} + 1\right )}\right )} e^{\left (-1\right )}\right ) + 3 \, \log \left (x\right ) \]

[In]

integrate((3*exp(16*x^2+1)-64*x^4-2*x^2)/(x*exp(16*x^2+1)-2*x^3),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int \frac {3 e^{1+16 x^2}-2 x^2-64 x^4}{e^{1+16 x^2} x-2 x^3} \, dx=16 \, x^{2} + \frac {3}{2} \, \log \left (16 \, x^{2}\right ) - \log \left (16 \, x^{2} - 8 \, e^{\left (16 \, x^{2} + 1\right )}\right ) + 1 \]

[In]

integrate((3*exp(16*x^2+1)-64*x^4-2*x^2)/(x*exp(16*x^2+1)-2*x^3),x, algorithm="giac")

[Out]

16*x^2 + 3/2*log(16*x^2) - log(16*x^2 - 8*e^(16*x^2 + 1)) + 1

Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {3 e^{1+16 x^2}-2 x^2-64 x^4}{e^{1+16 x^2} x-2 x^3} \, dx=3\,\ln \left (x\right )-\ln \left (2\,x^2-\mathrm {e}\,{\mathrm {e}}^{16\,x^2}\right )+16\,x^2 \]

[In]

int(-(2*x^2 - 3*exp(16*x^2 + 1) + 64*x^4)/(x*exp(16*x^2 + 1) - 2*x^3),x)

[Out]

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