\(\int \frac {-50-2 e^{10}+e^5 (20-60 x)+300 x}{81 x^3} \, dx\) [9646]

Optimal result

Integrand size = 26, antiderivative size = 27 \[ \int \frac {-50-2 e^{10}+e^5 (20-60 x)+300 x}{81 x^3} \, dx=\left (4+\frac {-x+\frac {1}{3} \left (\frac {1}{3} \left (-5+e^5\right )+x\right )}{x}\right )^2 \]

[Out]

((1/9*exp(5)-5/9-2/3*x)/x+4)^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {12, 192, 37} \[ \int \frac {-50-2 e^{10}+e^5 (20-60 x)+300 x}{81 x^3} \, dx=\frac {\left (\left (e^5-5\right )^2-30 \left (5-e^5\right ) x\right )^2}{81 \left (5-e^5\right )^2 x^2} \]

[In]

Int[(-50 - 2*E^10 + E^5*(20 - 60*x) + 300*x)/(81*x^3),x]

[Out]

((-5 + E^5)^2 - 30*(5 - E^5)*x)^2/(81*(5 - E^5)^2*x^2)

Rule 12

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

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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}& = \frac {1}{81} \int \frac {-50-2 e^{10}+e^5 (20-60 x)+300 x}{x^3} \, dx \\ & = \frac {1}{81} \int \frac {-2 \left (5-e^5\right )^2+60 \left (5-e^5\right ) x}{x^3} \, dx \\ & = \frac {\left (\left (-5+e^5\right )^2-30 \left (5-e^5\right ) x\right )^2}{81 \left (5-e^5\right )^2 x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {-50-2 e^{10}+e^5 (20-60 x)+300 x}{81 x^3} \, dx=\frac {\left (-5+e^5\right ) \left (-5+e^5+60 x\right )}{81 x^2} \]

[In]

Integrate[(-50 - 2*E^10 + E^5*(20 - 60*x) + 300*x)/(81*x^3),x]

[Out]

((-5 + E^5)*(-5 + E^5 + 60*x))/(81*x^2)

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.63

method	result	size
gosper	\(\frac {\left ({\mathrm e}^{5}-5\right ) \left ({\mathrm e}^{5}+60 x -5\right )}{81 x^{2}}\)	\(17\)
risch	\(\frac {\left (60 \,{\mathrm e}^{5}-300\right ) x +{\mathrm e}^{10}-10 \,{\mathrm e}^{5}+25}{81 x^{2}}\)	\(22\)
parallelrisch	\(\frac {60 x \,{\mathrm e}^{5}+{\mathrm e}^{10}-10 \,{\mathrm e}^{5}-300 x +25}{81 x^{2}}\)	\(24\)
norman	\(\frac {\left (\frac {20 \,{\mathrm e}^{5}}{27}-\frac {100}{27}\right ) x +\frac {{\mathrm e}^{10}}{81}-\frac {10 \,{\mathrm e}^{5}}{81}+\frac {25}{81}}{x^{2}}\)	\(25\)
default	\(\frac {\left (2 \,{\mathrm e}^{5}-10\right ) \left (\frac {30}{x}-\frac {5-{\mathrm e}^{5}}{2 x^{2}}\right )}{81}\)	\(26\)

[In]

int(1/81*(-2*exp(5)^2+(-60*x+20)*exp(5)+300*x-50)/x^3,x,method=_RETURNVERBOSE)

[Out]

1/81*(exp(5)-5)*(exp(5)+60*x-5)/x^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {-50-2 e^{10}+e^5 (20-60 x)+300 x}{81 x^3} \, dx=\frac {10 \, {\left (6 \, x - 1\right )} e^{5} - 300 \, x + e^{10} + 25}{81 \, x^{2}} \]

[In]

integrate(1/81*(-2*exp(5)^2+(-60*x+20)*exp(5)+300*x-50)/x^3,x, algorithm="fricas")

[Out]

1/81*(10*(6*x - 1)*e^5 - 300*x + e^10 + 25)/x^2

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {-50-2 e^{10}+e^5 (20-60 x)+300 x}{81 x^3} \, dx=\frac {x \left (-300 + 60 e^{5}\right ) - 10 e^{5} + 25 + e^{10}}{81 x^{2}} \]

[In]

integrate(1/81*(-2*exp(5)**2+(-60*x+20)*exp(5)+300*x-50)/x**3,x)

[Out]

(x*(-300 + 60*exp(5)) - 10*exp(5) + 25 + exp(10))/(81*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {-50-2 e^{10}+e^5 (20-60 x)+300 x}{81 x^3} \, dx=\frac {60 \, x {\left (e^{5} - 5\right )} + e^{10} - 10 \, e^{5} + 25}{81 \, x^{2}} \]

[In]

integrate(1/81*(-2*exp(5)^2+(-60*x+20)*exp(5)+300*x-50)/x^3,x, algorithm="maxima")

[Out]

1/81*(60*x*(e^5 - 5) + e^10 - 10*e^5 + 25)/x^2

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {-50-2 e^{10}+e^5 (20-60 x)+300 x}{81 x^3} \, dx=\frac {60 \, x e^{5} - 300 \, x + e^{10} - 10 \, e^{5} + 25}{81 \, x^{2}} \]

[In]

integrate(1/81*(-2*exp(5)^2+(-60*x+20)*exp(5)+300*x-50)/x^3,x, algorithm="giac")

[Out]

1/81*(60*x*e^5 - 300*x + e^10 - 10*e^5 + 25)/x^2

Mupad [B] (verification not implemented)

Time = 13.47 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {-50-2 e^{10}+e^5 (20-60 x)+300 x}{81 x^3} \, dx=\frac {{\mathrm {e}}^{10}-10\,{\mathrm {e}}^5+x\,\left (60\,{\mathrm {e}}^5-300\right )+25}{81\,x^2} \]

[In]

int(-((2*exp(10))/81 - (100*x)/27 + (exp(5)*(60*x - 20))/81 + 50/81)/x^3,x)

[Out]

(exp(10) - 10*exp(5) + x*(60*exp(5) - 300) + 25)/(81*x^2)