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

Optimal. Leaf size=30 \[ 4+e^{1+\frac {1}{5} e^{15+2 x^2}-x}+\log (4 (2-x)) \]

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Rubi [F]  time = 0.68, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 25, normalized size = 0.83 \begin {gather*} e^{1+\frac {1}{5} e^{15+2 x^2}-x}+\log (-2+x) \end {gather*}

Antiderivative was successfully verified.

[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]

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fricas [A]  time = 0.60, size = 23, normalized size = 0.77 \begin {gather*} e^{\left (-x + e^{\left (2 \, x^{2} - \log \relax (5) + 15\right )} + 1\right )} + \log \left (x - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)/(x-2),x, algorithm="frica
s")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (4 \, {\left (x^{2} - 2 \, x\right )} e^{\left (2 \, x^{2} - \log \relax (5) + 15\right )} - x + 2\right )} e^{\left (-x + e^{\left (2 \, x^{2} - \log \relax (5) + 15\right )} + 1\right )} + 1}{x - 2}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)/(x-2),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)

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maple [A]  time = 0.33, size = 22, normalized size = 0.73

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

Verification of antiderivative is not currently implemented for this CAS.

[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)/(x-2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.62, size = 21, normalized size = 0.70 \begin {gather*} e^{\left (-x + \frac {1}{5} \, e^{\left (2 \, x^{2} + 15\right )} + 1\right )} + \log \left (x - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)/(x-2),x, algorithm="maxim
a")

[Out]

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

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mupad [B]  time = 0.15, size = 21, normalized size = 0.70 \begin {gather*} \ln \left (x-2\right )+{\mathrm {e}}^{\frac {{\mathrm {e}}^{15}\,{\mathrm {e}}^{2\,x^2}}{5}-x+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.37, size = 19, normalized size = 0.63 \begin {gather*} e^{- x + \frac {e^{2 x^{2} + 15}}{5} + 1} + \log {\left (x - 2 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)/(x-2),x)

[Out]

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

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