3.57.51 \(\int \frac {-4+4 x+5 e^{4+x} x}{5 x} \, dx\)

Optimal. Leaf size=19 \[ e^{4+x}+\frac {4}{5} (2+x-\log (4 x)) \]

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Rubi [A]  time = 0.01, antiderivative size = 17, normalized size of antiderivative = 0.89, number of steps used = 6, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {12, 14, 2194, 43} \begin {gather*} \frac {4 x}{5}+e^{x+4}-\frac {4 \log (x)}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4 + 4*x + 5*E^(4 + x)*x)/(5*x),x]

[Out]

E^(4 + x) + (4*x)/5 - (4*Log[x])/5

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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} &=\frac {1}{5} \int \frac {-4+4 x+5 e^{4+x} x}{x} \, dx\\ &=\frac {1}{5} \int \left (5 e^{4+x}+\frac {4 (-1+x)}{x}\right ) \, dx\\ &=\frac {4}{5} \int \frac {-1+x}{x} \, dx+\int e^{4+x} \, dx\\ &=e^{4+x}+\frac {4}{5} \int \left (1-\frac {1}{x}\right ) \, dx\\ &=e^{4+x}+\frac {4 x}{5}-\frac {4 \log (x)}{5}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 17, normalized size = 0.89 \begin {gather*} e^{4+x}+\frac {4 x}{5}-\frac {4 \log (x)}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 + 4*x + 5*E^(4 + x)*x)/(5*x),x]

[Out]

E^(4 + x) + (4*x)/5 - (4*Log[x])/5

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fricas [A]  time = 0.70, size = 12, normalized size = 0.63 \begin {gather*} \frac {4}{5} \, x + e^{\left (x + 4\right )} - \frac {4}{5} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(5*x*exp(4+x)+4*x-4)/x,x, algorithm="fricas")

[Out]

4/5*x + e^(x + 4) - 4/5*log(x)

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giac [A]  time = 0.14, size = 12, normalized size = 0.63 \begin {gather*} \frac {4}{5} \, x + e^{\left (x + 4\right )} - \frac {4}{5} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(5*x*exp(4+x)+4*x-4)/x,x, algorithm="giac")

[Out]

4/5*x + e^(x + 4) - 4/5*log(x)

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maple [A]  time = 0.07, size = 13, normalized size = 0.68




method result size



norman \(\frac {4 x}{5}+{\mathrm e}^{4+x}-\frac {4 \ln \relax (x )}{5}\) \(13\)
risch \(\frac {4 x}{5}+{\mathrm e}^{4+x}-\frac {4 \ln \relax (x )}{5}\) \(13\)
derivativedivides \(-\frac {4 \ln \relax (x )}{5}+\frac {16}{5}+\frac {4 x}{5}+{\mathrm e}^{4+x}\) \(14\)
default \(-\frac {4 \ln \relax (x )}{5}+\frac {16}{5}+\frac {4 x}{5}+{\mathrm e}^{4+x}\) \(14\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(5*x*exp(4+x)+4*x-4)/x,x,method=_RETURNVERBOSE)

[Out]

4/5*x+exp(4+x)-4/5*ln(x)

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maxima [A]  time = 0.36, size = 12, normalized size = 0.63 \begin {gather*} \frac {4}{5} \, x + e^{\left (x + 4\right )} - \frac {4}{5} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(5*x*exp(4+x)+4*x-4)/x,x, algorithm="maxima")

[Out]

4/5*x + e^(x + 4) - 4/5*log(x)

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mupad [B]  time = 3.50, size = 12, normalized size = 0.63 \begin {gather*} \frac {4\,x}{5}+{\mathrm {e}}^{x+4}-\frac {4\,\ln \relax (x)}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x)/5 + x*exp(x + 4) - 4/5)/x,x)

[Out]

(4*x)/5 + exp(x + 4) - (4*log(x))/5

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sympy [A]  time = 0.09, size = 15, normalized size = 0.79 \begin {gather*} \frac {4 x}{5} + e^{x + 4} - \frac {4 \log {\relax (x )}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(5*x*exp(4+x)+4*x-4)/x,x)

[Out]

4*x/5 + exp(x + 4) - 4*log(x)/5

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