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

Optimal. Leaf size=24 \[ \frac {1}{5} x \left (-1+e^4+\frac {5 \left (2+x+\log \left (\frac {3}{x}\right )\right )}{x}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 15, normalized size of antiderivative = 0.62, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {6, 12, 43} \begin {gather*} \frac {1}{5} \left (4+e^4\right ) x-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5 + 4*x + E^4*x)/(5*x),x]

[Out]

((4 + E^4)*x)/5 - 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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-5+\left (4+e^4\right ) x}{5 x} \, dx\\ &=\frac {1}{5} \int \frac {-5+\left (4+e^4\right ) x}{x} \, dx\\ &=\frac {1}{5} \int \left (4+e^4-\frac {5}{x}\right ) \, dx\\ &=\frac {1}{5} \left (4+e^4\right ) x-\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 17, normalized size = 0.71 \begin {gather*} \frac {1}{5} \left (4 x+e^4 x-5 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5 + 4*x + E^4*x)/(5*x),x]

[Out]

(4*x + E^4*x - 5*Log[x])/5

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fricas [A]  time = 0.55, size = 13, normalized size = 0.54 \begin {gather*} \frac {1}{5} \, x e^{4} + \frac {4}{5} \, x - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*x*e^4 + 4/5*x - log(x)

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giac [A]  time = 0.19, size = 14, normalized size = 0.58 \begin {gather*} \frac {1}{5} \, x e^{4} + \frac {4}{5} \, x - \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*x*e^4 + 4/5*x - log(abs(x))

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maple [A]  time = 0.02, size = 14, normalized size = 0.58

method	result	size
default	\(\frac {4 x}{5}+\frac {x \,{\mathrm e}^{4}}{5}-\ln \relax (x )\)	\(14\)
norman	\(\left (\frac {4}{5}+\frac {{\mathrm e}^{4}}{5}\right ) x -\ln \relax (x )\)	\(14\)
risch	\(\frac {4 x}{5}+\frac {x \,{\mathrm e}^{4}}{5}-\ln \relax (x )\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/5*x+1/5*x*exp(4)-ln(x)

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maxima [A]  time = 0.35, size = 12, normalized size = 0.50 \begin {gather*} \frac {1}{5} \, x {\left (e^{4} + 4\right )} - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*x*(e^4 + 4) - log(x)

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mupad [B]  time = 0.03, size = 13, normalized size = 0.54 \begin {gather*} x\,\left (\frac {{\mathrm {e}}^4}{5}+\frac {4}{5}\right )-\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x)/5 + (x*exp(4))/5 - 1)/x,x)

[Out]

x*(exp(4)/5 + 4/5) - log(x)

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sympy [A]  time = 0.08, size = 10, normalized size = 0.42 \begin {gather*} \frac {x \left (4 + e^{4}\right )}{5} - \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*(4 + exp(4))/5 - log(x)

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