\(\int \frac {5+e^2 (-1-4 x)+20 x}{x} \, dx\) [6542]

Optimal result

Integrand size = 18, antiderivative size = 27 \[ \int \frac {5+e^2 (-1-4 x)+20 x}{x} \, dx=\left (5-e^2\right ) \left (-4+\frac {-4+(2+2 x)^2}{x}+\log (3 x)\right ) \]

[Out]

(5-exp(2))*(ln(3*x)-4+((2+2*x)^2-4)/x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {192, 45} \[ \int \frac {5+e^2 (-1-4 x)+20 x}{x} \, dx=4 \left (5-e^2\right ) x+\left (5-e^2\right ) \log (x) \]

[In]

Int[(5 + E^2*(-1 - 4*x) + 20*x)/x,x]

[Out]

4*(5 - E^2)*x + (5 - E^2)*Log[x]

Rule 45

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 192

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.48 \[ \int \frac {5+e^2 (-1-4 x)+20 x}{x} \, dx=-\left (\left (-5+e^2\right ) (4 x+\log (x))\right ) \]

[In]

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

[Out]

-((-5 + E^2)*(4*x + Log[x]))

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.52

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.63 \[ \int \frac {5+e^2 (-1-4 x)+20 x}{x} \, dx=-4 \, x e^{2} - {\left (e^{2} - 5\right )} \log \left (x\right ) + 20 \, x \]

[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

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.56 \[ \int \frac {5+e^2 (-1-4 x)+20 x}{x} \, dx=x \left (20 - 4 e^{2}\right ) + \left (5 - e^{2}\right ) \log {\left (x \right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.59 \[ \int \frac {5+e^2 (-1-4 x)+20 x}{x} \, dx=-4 \, x {\left (e^{2} - 5\right )} - {\left (e^{2} - 5\right )} \log \left (x\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67 \[ \int \frac {5+e^2 (-1-4 x)+20 x}{x} \, dx=-4 \, x e^{2} - {\left (e^{2} - 5\right )} \log \left ({\left | x \right |}\right ) + 20 \, x \]

[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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.44 \[ \int \frac {5+e^2 (-1-4 x)+20 x}{x} \, dx=-\left (4\,x+\ln \left (x\right )\right )\,\left ({\mathrm {e}}^2-5\right ) \]

[In]

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

[Out]

-(4*x + log(x))*(exp(2) - 5)