\(\int \frac {1+e^{1+\frac {1}{5} e^{15+2 x^2}-x} (2-x+\frac {1}{5} e^{15+2 x^2} (-8 x+4 x^2))}{-2+x} \, dx\) [8270]

Optimal result

Integrand size = 56, antiderivative size = 30 \[ \int \frac {1+e^{1+\frac {1}{5} e^{15+2 x^2}-x} \left (2-x+\frac {1}{5} e^{15+2 x^2} \left (-8 x+4 x^2\right )\right )}{-2+x} \, dx=4+e^{1+\frac {1}{5} e^{15+2 x^2}-x}+\log (4 (2-x)) \]

[Out]

ln(-4*x+8)+4+exp(exp(-ln(5)+2*x^2+15)-x+1)

Rubi [F]

\[ \int \frac {1+e^{1+\frac {1}{5} e^{15+2 x^2}-x} \left (2-x+\frac {1}{5} e^{15+2 x^2} \left (-8 x+4 x^2\right )\right )}{-2+x} \, dx=\int \frac {1+e^{1+\frac {1}{5} e^{15+2 x^2}-x} \left (2-x+\frac {1}{5} e^{15+2 x^2} \left (-8 x+4 x^2\right )\right )}{-2+x} \, dx \]

[In]

Int[(1 + E^(1 + E^(15 + 2*x^2)/5 - x)*(2 - x + (E^(15 + 2*x^2)*(-8*x + 4*x^2))/5))/(-2 + x),x]

[Out]

Log[2 - x] - Defer[Int][E^((5 + E^(15 + 2*x^2) - 5*x)/5), x] + (4*Defer[Int][E^((80 + E^(15 + 2*x^2) - 5*x + 1
0*x^2)/5)*x, x])/5

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {4}{5} e^{\frac {1}{5} \left (80+e^{15+2 x^2}-5 x+10 x^2\right )} x+\frac {e^{-x} \left (2 e^{1+\frac {1}{5} e^{15+2 x^2}}+e^x-e^{1+\frac {1}{5} e^{15+2 x^2}} x\right )}{-2+x}\right ) \, dx \\ & = \frac {4}{5} \int e^{\frac {1}{5} \left (80+e^{15+2 x^2}-5 x+10 x^2\right )} x \, dx+\int \frac {e^{-x} \left (2 e^{1+\frac {1}{5} e^{15+2 x^2}}+e^x-e^{1+\frac {1}{5} e^{15+2 x^2}} x\right )}{-2+x} \, dx \\ & = \frac {4}{5} \int e^{\frac {1}{5} \left (80+e^{15+2 x^2}-5 x+10 x^2\right )} x \, dx+\int \left (-e^{\frac {1}{5} \left (5+e^{15+2 x^2}-5 x\right )}+\frac {1}{-2+x}\right ) \, dx \\ & = \log (2-x)+\frac {4}{5} \int e^{\frac {1}{5} \left (80+e^{15+2 x^2}-5 x+10 x^2\right )} x \, dx-\int e^{\frac {1}{5} \left (5+e^{15+2 x^2}-5 x\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {1+e^{1+\frac {1}{5} e^{15+2 x^2}-x} \left (2-x+\frac {1}{5} e^{15+2 x^2} \left (-8 x+4 x^2\right )\right )}{-2+x} \, dx=e^{1+\frac {1}{5} e^{15+2 x^2}-x}+\log (-2+x) \]

[In]

Integrate[(1 + E^(1 + E^(15 + 2*x^2)/5 - x)*(2 - x + (E^(15 + 2*x^2)*(-8*x + 4*x^2))/5))/(-2 + x),x]

[Out]

E^(1 + E^(15 + 2*x^2)/5 - x) + Log[-2 + x]

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73

method	result	size
risch	\({\mathrm e}^{\frac {{\mathrm e}^{2 x^{2}+15}}{5}-x +1}+\ln \left (-2+x \right )\)	\(22\)
norman	\({\mathrm e}^{{\mathrm e}^{-\ln \left (5\right )+2 x^{2}+15}-x +1}+\ln \left (-2+x \right )\)	\(24\)
parallelrisch	\({\mathrm e}^{{\mathrm e}^{-\ln \left (5\right )+2 x^{2}+15}-x +1}+\ln \left (-2+x \right )\)	\(24\)
parts	\({\mathrm e}^{{\mathrm e}^{-\ln \left (5\right )+2 x^{2}+15}-x +1}+\ln \left (-2+x \right )\)	\(24\)

[In]

int((((4*x^2-8*x)*exp(-ln(5)+2*x^2+15)+2-x)*exp(exp(-ln(5)+2*x^2+15)-x+1)+1)/(-2+x),x,method=_RETURNVERBOSE)

[Out]

exp(1/5*exp(2*x^2+15)-x+1)+ln(-2+x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int \frac {1+e^{1+\frac {1}{5} e^{15+2 x^2}-x} \left (2-x+\frac {1}{5} e^{15+2 x^2} \left (-8 x+4 x^2\right )\right )}{-2+x} \, dx=e^{\left (-x + e^{\left (2 \, x^{2} - \log \left (5\right ) + 15\right )} + 1\right )} + \log \left (x - 2\right ) \]

[In]

integrate((((4*x^2-8*x)*exp(-log(5)+2*x^2+15)+2-x)*exp(exp(-log(5)+2*x^2+15)-x+1)+1)/(-2+x),x, algorithm="fric
as")

[Out]

e^(-x + e^(2*x^2 - log(5) + 15) + 1) + log(x - 2)

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.63 \[ \int \frac {1+e^{1+\frac {1}{5} e^{15+2 x^2}-x} \left (2-x+\frac {1}{5} e^{15+2 x^2} \left (-8 x+4 x^2\right )\right )}{-2+x} \, dx=e^{- x + \frac {e^{2 x^{2} + 15}}{5} + 1} + \log {\left (x - 2 \right )} \]

[In]

integrate((((4*x**2-8*x)*exp(-ln(5)+2*x**2+15)+2-x)*exp(exp(-ln(5)+2*x**2+15)-x+1)+1)/(-2+x),x)

[Out]

exp(-x + exp(2*x**2 + 15)/5 + 1) + log(x - 2)

Maxima [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.70 \[ \int \frac {1+e^{1+\frac {1}{5} e^{15+2 x^2}-x} \left (2-x+\frac {1}{5} e^{15+2 x^2} \left (-8 x+4 x^2\right )\right )}{-2+x} \, dx=e^{\left (-x + \frac {1}{5} \, e^{\left (2 \, x^{2} + 15\right )} + 1\right )} + \log \left (x - 2\right ) \]

[In]

integrate((((4*x^2-8*x)*exp(-log(5)+2*x^2+15)+2-x)*exp(exp(-log(5)+2*x^2+15)-x+1)+1)/(-2+x),x, algorithm="maxi
ma")

[Out]

e^(-x + 1/5*e^(2*x^2 + 15) + 1) + log(x - 2)

Giac [F]

\[ \int \frac {1+e^{1+\frac {1}{5} e^{15+2 x^2}-x} \left (2-x+\frac {1}{5} e^{15+2 x^2} \left (-8 x+4 x^2\right )\right )}{-2+x} \, dx=\int { \frac {{\left (4 \, {\left (x^{2} - 2 \, x\right )} e^{\left (2 \, x^{2} - \log \left (5\right ) + 15\right )} - x + 2\right )} e^{\left (-x + e^{\left (2 \, x^{2} - \log \left (5\right ) + 15\right )} + 1\right )} + 1}{x - 2} \,d x } \]

[In]

integrate((((4*x^2-8*x)*exp(-log(5)+2*x^2+15)+2-x)*exp(exp(-log(5)+2*x^2+15)-x+1)+1)/(-2+x),x, algorithm="giac
")

[Out]

integrate(((4*(x^2 - 2*x)*e^(2*x^2 - log(5) + 15) - x + 2)*e^(-x + e^(2*x^2 - log(5) + 15) + 1) + 1)/(x - 2),
x)

Mupad [B] (verification not implemented)

Time = 12.53 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.70 \[ \int \frac {1+e^{1+\frac {1}{5} e^{15+2 x^2}-x} \left (2-x+\frac {1}{5} e^{15+2 x^2} \left (-8 x+4 x^2\right )\right )}{-2+x} \, dx=\ln \left (x-2\right )+{\mathrm {e}}^{\frac {{\mathrm {e}}^{15}\,{\mathrm {e}}^{2\,x^2}}{5}-x+1} \]

[In]

int(-(exp(exp(2*x^2 - log(5) + 15) - x + 1)*(x + exp(2*x^2 - log(5) + 15)*(8*x - 4*x^2) - 2) - 1)/(x - 2),x)

[Out]

log(x - 2) + exp((exp(15)*exp(2*x^2))/5 - x + 1)