3.56.63 \(\int (-5+e^6+e^{-5-x+e^2 x} (-5+5 x-5 e^2 x)-5 \log (x)) \, dx\)

Optimal. Leaf size=32 \[ x \left (e^6+5 \left (-e^{-5-x+e^2 x}+\frac {12}{x}-\log (x)\right )\right ) \]

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Rubi [B]  time = 0.06, antiderivative size = 81, normalized size of antiderivative = 2.53, number of steps used = 6, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {6, 2187, 2176, 2194, 2295} \begin {gather*} -\left (5-e^6\right ) x+5 x+\frac {5 e^{-\left (\left (1-e^2\right ) x\right )-5} \left (1-\left (1-e^2\right ) x\right )}{1-e^2}-\frac {5 e^{-\left (\left (1-e^2\right ) x\right )-5}}{1-e^2}-5 x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-5 + E^6 + E^(-5 - x + E^2*x)*(-5 + 5*x - 5*E^2*x) - 5*Log[x],x]

[Out]

(-5*E^(-5 - (1 - E^2)*x))/(1 - E^2) + 5*x - (5 - E^6)*x + (5*E^(-5 - (1 - E^2)*x)*(1 - (1 - E^2)*x))/(1 - E^2)
- 5*x*Log[x]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2187

Int[((a_.) + (b_.)*((F_)^((g_.)*(v_)))^(n_.))^(p_.)*(u_)^(m_.), x_Symbol] :> Int[NormalizePowerOfLinear[u, x]^
m*(a + b*(F^(g*ExpandToSum[v, x]))^n)^p, x] /; FreeQ[{F, a, b, g, n, p}, x] && LinearQ[v, x] && PowerOfLinearQ
[u, x] &&  !(LinearMatchQ[v, x] && PowerOfLinearMatchQ[u, x]) && IntegerQ[m]

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left (\left (5-e^6\right ) x\right )-5 \int \log (x) \, dx+\int e^{-5-x+e^2 x} \left (-5+5 x-5 e^2 x\right ) \, dx\\ &=5 x-\left (5-e^6\right ) x-5 x \log (x)+\int e^{-5-x+e^2 x} \left (-5+\left (5-5 e^2\right ) x\right ) \, dx\\ &=5 x-\left (5-e^6\right ) x-5 x \log (x)+\int e^{-5-\left (1-e^2\right ) x} \left (-5+5 \left (1-e^2\right ) x\right ) \, dx\\ &=5 x-\left (5-e^6\right ) x+\frac {5 e^{-5-\left (1-e^2\right ) x} \left (1-\left (1-e^2\right ) x\right )}{1-e^2}-5 x \log (x)+5 \int e^{-5+\left (-1+e^2\right ) x} \, dx\\ &=-\frac {5 e^{-5-\left (1-e^2\right ) x}}{1-e^2}+5 x-\left (5-e^6\right ) x+\frac {5 e^{-5-\left (1-e^2\right ) x} \left (1-\left (1-e^2\right ) x\right )}{1-e^2}-5 x \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 25, normalized size = 0.78 \begin {gather*} e^6 x-5 e^{-5+\left (-1+e^2\right ) x} x-5 x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-5 + E^6 + E^(-5 - x + E^2*x)*(-5 + 5*x - 5*E^2*x) - 5*Log[x],x]

[Out]

E^6*x - 5*E^(-5 + (-1 + E^2)*x)*x - 5*x*Log[x]

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fricas [A]  time = 0.59, size = 23, normalized size = 0.72 \begin {gather*} x e^{6} - 5 \, x e^{\left (x e^{2} - x - 5\right )} - 5 \, x \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5*log(x)+(-5*exp(2)*x+5*x-5)*exp(exp(2)*x-x-5)+exp(6)-5,x, algorithm="fricas")

[Out]

x*e^6 - 5*x*e^(x*e^2 - x - 5) - 5*x*log(x)

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giac [B]  time = 0.13, size = 75, normalized size = 2.34 \begin {gather*} x e^{6} - 5 \, x \log \relax (x) - \frac {5 \, {\left (x e^{2} - x - 1\right )} e^{\left (x e^{2} - x - 3\right )}}{e^{4} - 2 \, e^{2} + 1} + \frac {5 \, {\left (x e^{2} - x - e^{2}\right )} e^{\left (x e^{2} - x - 5\right )}}{e^{4} - 2 \, e^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5*log(x)+(-5*exp(2)*x+5*x-5)*exp(exp(2)*x-x-5)+exp(6)-5,x, algorithm="giac")

[Out]

x*e^6 - 5*x*log(x) - 5*(x*e^2 - x - 1)*e^(x*e^2 - x - 3)/(e^4 - 2*e^2 + 1) + 5*(x*e^2 - x - e^2)*e^(x*e^2 - x
- 5)/(e^4 - 2*e^2 + 1)

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maple [A]  time = 0.06, size = 24, normalized size = 0.75

method	result	size
norman	\(x \,{\mathrm e}^{6}-5 x \ln \relax (x )-5 \,{\mathrm e}^{{\mathrm e}^{2} x -x -5} x\)	\(24\)
risch	\(x \,{\mathrm e}^{6}-5 x \ln \relax (x )-5 \,{\mathrm e}^{{\mathrm e}^{2} x -x -5} x\)	\(24\)
default	\(\frac {\frac {5 \left ({\mathrm e}^{-5+\left ({\mathrm e}^{2}-1\right ) x} \left (-5+\left ({\mathrm e}^{2}-1\right ) x \right )-{\mathrm e}^{-5+\left ({\mathrm e}^{2}-1\right ) x}\right )}{{\mathrm e}^{2}-1}+\frac {25 \,{\mathrm e}^{-5+\left ({\mathrm e}^{2}-1\right ) x}}{{\mathrm e}^{2}-1}-\frac {25 \,{\mathrm e}^{-5+\left ({\mathrm e}^{2}-1\right ) x} {\mathrm e}^{2}}{{\mathrm e}^{2}-1}-\frac {5 \,{\mathrm e}^{2} \left ({\mathrm e}^{-5+\left ({\mathrm e}^{2}-1\right ) x} \left (-5+\left ({\mathrm e}^{2}-1\right ) x \right )-{\mathrm e}^{-5+\left ({\mathrm e}^{2}-1\right ) x}\right )}{{\mathrm e}^{2}-1}-5 \,{\mathrm e}^{-5+\left ({\mathrm e}^{2}-1\right ) x}}{{\mathrm e}^{2}-1}-5 x \ln \relax (x )+x \,{\mathrm e}^{6}\)	\(144\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-5*ln(x)+(-5*exp(2)*x+5*x-5)*exp(exp(2)*x-x-5)+exp(6)-5,x,method=_RETURNVERBOSE)

[Out]

x*exp(6)-5*x*ln(x)-5*exp(exp(2)*x-x-5)*x

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maxima [A]  time = 0.38, size = 23, normalized size = 0.72 \begin {gather*} x e^{6} - 5 \, x e^{\left (x e^{2} - x - 5\right )} - 5 \, x \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5*log(x)+(-5*exp(2)*x+5*x-5)*exp(exp(2)*x-x-5)+exp(6)-5,x, algorithm="maxima")

[Out]

x*e^6 - 5*x*e^(x*e^2 - x - 5) - 5*x*log(x)

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mupad [B]  time = 3.52, size = 24, normalized size = 0.75 \begin {gather*} x\,{\mathrm {e}}^6-5\,x\,\ln \relax (x)-5\,x\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-5}\,{\mathrm {e}}^{x\,{\mathrm {e}}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(6) - 5*log(x) - exp(x*exp(2) - x - 5)*(5*x*exp(2) - 5*x + 5) - 5,x)

[Out]

x*exp(6) - 5*x*log(x) - 5*x*exp(-x)*exp(-5)*exp(x*exp(2))

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sympy [A]  time = 0.28, size = 24, normalized size = 0.75 \begin {gather*} - 5 x e^{- x + x e^{2} - 5} - 5 x \log {\relax (x )} + x e^{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5*ln(x)+(-5*exp(2)*x+5*x-5)*exp(exp(2)*x-x-5)+exp(6)-5,x)

[Out]

-5*x*exp(-x + x*exp(2) - 5) - 5*x*log(x) + x*exp(6)

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