3.66.42 \(\int \frac {5+e^2 (-1-4 x)+20 x}{x} \, dx\)

Optimal. Leaf size=27 \[ \left (5-e^2\right ) \left (-4+\frac {-4+(2+2 x)^2}{x}+\log (3 x)\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 21, normalized size of antiderivative = 0.78, number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {186, 43} \begin {gather*} 4 \left (5-e^2\right ) x+\left (5-e^2\right ) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5 + E^2*(-1 - 4*x) + 20*x)/x,x]

[Out]

4*(5 - E^2)*x + (5 - E^2)*Log[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])

Rule 186

Int[(u_)^(m_.)*(v_)^(n_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*ExpandToSum[v, x]^n, x] /; FreeQ[{m, n}, x] &&
LinearQ[{u, v}, x] &&  !LinearMatchQ[{u, v}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5-e^2+4 \left (5-e^2\right ) x}{x} \, dx\\ &=\int \left (4 \left (5-e^2\right )+\frac {5-e^2}{x}\right ) \, dx\\ &=4 \left (5-e^2\right ) x+\left (5-e^2\right ) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 13, normalized size = 0.48 \begin {gather*} -\left (\left (-5+e^2\right ) (4 x+\log (x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 + E^2*(-1 - 4*x) + 20*x)/x,x]

[Out]

-((-5 + E^2)*(4*x + Log[x]))

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fricas [A]  time = 1.02, size = 17, normalized size = 0.63 \begin {gather*} -4 \, x e^{2} - {\left (e^{2} - 5\right )} \log \relax (x) + 20 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x-1)*exp(2)+20*x+5)/x,x, algorithm="fricas")

[Out]

-4*x*e^2 - (e^2 - 5)*log(x) + 20*x

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giac [A]  time = 0.13, size = 18, normalized size = 0.67 \begin {gather*} -4 \, x e^{2} - {\left (e^{2} - 5\right )} \log \left ({\left | x \right |}\right ) + 20 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x-1)*exp(2)+20*x+5)/x,x, algorithm="giac")

[Out]

-4*x*e^2 - (e^2 - 5)*log(abs(x)) + 20*x

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maple [A]  time = 0.05, size = 14, normalized size = 0.52

method	result	size
default	\(\left ({\mathrm e}^{2}-5\right ) \left (-\ln \relax (x )-4 x \right )\)	\(14\)
norman	\(\left (-4 \,{\mathrm e}^{2}+20\right ) x +\left (5-{\mathrm e}^{2}\right ) \ln \relax (x )\)	\(19\)
risch	\(-4 \,{\mathrm e}^{2} x +20 x -{\mathrm e}^{2} \ln \relax (x )+5 \ln \relax (x )\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x-1)*exp(2)+20*x+5)/x,x,method=_RETURNVERBOSE)

[Out]

(exp(2)-5)*(-ln(x)-4*x)

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maxima [A]  time = 0.41, size = 16, normalized size = 0.59 \begin {gather*} -4 \, x {\left (e^{2} - 5\right )} - {\left (e^{2} - 5\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x-1)*exp(2)+20*x+5)/x,x, algorithm="maxima")

[Out]

-4*x*(e^2 - 5) - (e^2 - 5)*log(x)

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mupad [B]  time = 0.06, size = 12, normalized size = 0.44 \begin {gather*} -\left (4\,x+\ln \relax (x)\right )\,\left ({\mathrm {e}}^2-5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((20*x - exp(2)*(4*x + 1) + 5)/x,x)

[Out]

-(4*x + log(x))*(exp(2) - 5)

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sympy [A]  time = 0.10, size = 15, normalized size = 0.56 \begin {gather*} x \left (20 - 4 e^{2}\right ) + \left (5 - e^{2}\right ) \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x-1)*exp(2)+20*x+5)/x,x)

[Out]

x*(20 - 4*exp(2)) + (5 - exp(2))*log(x)

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