3.98.64 \(\int \frac {1+x+e^{1+e^2-x} x}{x} \, dx\)

Optimal. Leaf size=22 \[ -3+e^{e^4}-e^{1+e^2-x}+x+\log (x) \]

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Rubi [A]  time = 0.01, antiderivative size = 16, normalized size of antiderivative = 0.73, number of steps used = 5, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {14, 2194, 43} \begin {gather*} x-e^{-x+e^2+1}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + x + E^(1 + E^2 - x)*x)/x,x]

[Out]

-E^(1 + E^2 - x) + x + Log[x]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 16, normalized size = 0.73 \begin {gather*} -e^{1+e^2-x}+x+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + x + E^(1 + E^2 - x)*x)/x,x]

[Out]

-E^(1 + E^2 - x) + x + Log[x]

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fricas [A]  time = 0.93, size = 14, normalized size = 0.64 \begin {gather*} x - e^{\left (-x + e^{2} + 1\right )} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*exp(exp(2)-x+1)+x+1)/x,x, algorithm="fricas")

[Out]

x - e^(-x + e^2 + 1) + log(x)

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giac [A]  time = 0.16, size = 14, normalized size = 0.64 \begin {gather*} x - e^{\left (-x + e^{2} + 1\right )} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*exp(exp(2)-x+1)+x+1)/x,x, algorithm="giac")

[Out]

x - e^(-x + e^2 + 1) + log(x)

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maple [A]  time = 0.06, size = 15, normalized size = 0.68

method	result	size
norman	\(x -{\mathrm e}^{{\mathrm e}^{2}-x +1}+\ln \relax (x )\)	\(15\)
risch	\(x -{\mathrm e}^{{\mathrm e}^{2}-x +1}+\ln \relax (x )\)	\(15\)
derivativedivides	\({\mathrm e}^{2} \ln \relax (x )-{\mathrm e}^{{\mathrm e}^{2}+1} \expIntegralEi \left (1, x\right )-{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{2}+1} \expIntegralEi \left (1, x\right )+2 \ln \relax (x )-{\mathrm e}^{2}+x -1-\ln \left (-x \right ) {\mathrm e}^{2}-\ln \left (-x \right )-{\mathrm e}^{{\mathrm e}^{2}-x +1}-\left (-1-{\mathrm e}^{2}\right ) {\mathrm e}^{{\mathrm e}^{2}+1} \expIntegralEi \left (1, x\right )\)	\(79\)
default	\({\mathrm e}^{2} \ln \relax (x )-{\mathrm e}^{{\mathrm e}^{2}+1} \expIntegralEi \left (1, x\right )-{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{2}+1} \expIntegralEi \left (1, x\right )+2 \ln \relax (x )-{\mathrm e}^{2}+x -1-\ln \left (-x \right ) {\mathrm e}^{2}-\ln \left (-x \right )-{\mathrm e}^{{\mathrm e}^{2}-x +1}-\left (-1-{\mathrm e}^{2}\right ) {\mathrm e}^{{\mathrm e}^{2}+1} \expIntegralEi \left (1, x\right )\)	\(79\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*exp(exp(2)-x+1)+x+1)/x,x,method=_RETURNVERBOSE)

[Out]

x-exp(exp(2)-x+1)+ln(x)

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maxima [A]  time = 0.39, size = 14, normalized size = 0.64 \begin {gather*} x - e^{\left (-x + e^{2} + 1\right )} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*exp(exp(2)-x+1)+x+1)/x,x, algorithm="maxima")

[Out]

x - e^(-x + e^2 + 1) + log(x)

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mupad [B]  time = 0.06, size = 15, normalized size = 0.68 \begin {gather*} x+\ln \relax (x)-{\mathrm {e}}^{-x}\,\mathrm {e}\,{\mathrm {e}}^{{\mathrm {e}}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + x*exp(exp(2) - x + 1) + 1)/x,x)

[Out]

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

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sympy [A]  time = 0.11, size = 12, normalized size = 0.55 \begin {gather*} x - e^{- x + 1 + e^{2}} + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*exp(exp(2)-x+1)+x+1)/x,x)

[Out]

x - exp(-x + 1 + exp(2)) + log(x)

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