\(\int \frac {40+16 x-56 x^2}{e^3 (25 x^2+10 x^3+x^4)} \, dx\) [3085]

Optimal result

Integrand size = 30, antiderivative size = 20 \[ \int \frac {40+16 x-56 x^2}{e^3 \left (25 x^2+10 x^3+x^4\right )} \, dx=-\frac {8 (1-x)^2}{e^3 x (5+x)} \]

[Out]

-8/exp(3)/x/(5+x)*(1-x)^2

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {12, 1608, 27, 907} \[ \int \frac {40+16 x-56 x^2}{e^3 \left (25 x^2+10 x^3+x^4\right )} \, dx=\frac {288}{5 e^3 (x+5)}-\frac {8}{5 e^3 x} \]

[In]

Int[(40 + 16*x - 56*x^2)/(E^3*(25*x^2 + 10*x^3 + x^4)),x]

[Out]

-8/(5*E^3*x) + 288/(5*E^3*(5 + 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 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 907

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 1608

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {40+16 x-56 x^2}{25 x^2+10 x^3+x^4} \, dx}{e^3} \\ & = \frac {\int \frac {40+16 x-56 x^2}{x^2 \left (25+10 x+x^2\right )} \, dx}{e^3} \\ & = \frac {\int \frac {40+16 x-56 x^2}{x^2 (5+x)^2} \, dx}{e^3} \\ & = \frac {\int \left (\frac {8}{5 x^2}-\frac {288}{5 (5+x)^2}\right ) \, dx}{e^3} \\ & = -\frac {8}{5 e^3 x}+\frac {288}{5 e^3 (5+x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {40+16 x-56 x^2}{e^3 \left (25 x^2+10 x^3+x^4\right )} \, dx=\frac {8 (-1+7 x)}{e^3 x (5+x)} \]

[In]

Integrate[(40 + 16*x - 56*x^2)/(E^3*(25*x^2 + 10*x^3 + x^4)),x]

[Out]

(8*(-1 + 7*x))/(E^3*x*(5 + x))

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85

method	result	size
risch	\(\frac {{\mathrm e}^{-3} \left (56 x -8\right )}{\left (5+x \right ) x}\)	\(17\)
parallelrisch	\(\frac {{\mathrm e}^{-3} \left (56 x -8\right )}{\left (5+x \right ) x}\)	\(19\)
gosper	\(\frac {8 \left (-1+7 x \right ) {\mathrm e}^{-3}}{x \left (5+x \right )}\)	\(20\)
default	\(8 \,{\mathrm e}^{-3} \left (-\frac {1}{5 x}+\frac {36}{5 \left (5+x \right )}\right )\)	\(20\)
norman	\(\frac {56 \,{\mathrm e}^{-3} x -8 \,{\mathrm e}^{-3}}{\left (5+x \right ) x}\)	\(24\)

[In]

int((-56*x^2+16*x+40)/(x^4+10*x^3+25*x^2)/exp(3),x,method=_RETURNVERBOSE)

[Out]

exp(-3)*(56*x-8)/(5+x)/x

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {40+16 x-56 x^2}{e^3 \left (25 x^2+10 x^3+x^4\right )} \, dx=\frac {8 \, {\left (7 \, x - 1\right )} e^{\left (-3\right )}}{x^{2} + 5 \, x} \]

[In]

integrate((-56*x^2+16*x+40)/(x^4+10*x^3+25*x^2)/exp(3),x, algorithm="fricas")

[Out]

8*(7*x - 1)*e^(-3)/(x^2 + 5*x)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {40+16 x-56 x^2}{e^3 \left (25 x^2+10 x^3+x^4\right )} \, dx=- \frac {8 - 56 x}{x^{2} e^{3} + 5 x e^{3}} \]

[In]

integrate((-56*x**2+16*x+40)/(x**4+10*x**3+25*x**2)/exp(3),x)

[Out]

-(8 - 56*x)/(x**2*exp(3) + 5*x*exp(3))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {40+16 x-56 x^2}{e^3 \left (25 x^2+10 x^3+x^4\right )} \, dx=\frac {8 \, {\left (7 \, x - 1\right )} e^{\left (-3\right )}}{x^{2} + 5 \, x} \]

[In]

integrate((-56*x^2+16*x+40)/(x^4+10*x^3+25*x^2)/exp(3),x, algorithm="maxima")

[Out]

8*(7*x - 1)*e^(-3)/(x^2 + 5*x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {40+16 x-56 x^2}{e^3 \left (25 x^2+10 x^3+x^4\right )} \, dx=\frac {8 \, {\left (7 \, x - 1\right )} e^{\left (-3\right )}}{x^{2} + 5 \, x} \]

[In]

integrate((-56*x^2+16*x+40)/(x^4+10*x^3+25*x^2)/exp(3),x, algorithm="giac")

[Out]

8*(7*x - 1)*e^(-3)/(x^2 + 5*x)

Mupad [B] (verification not implemented)

Time = 9.52 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {40+16 x-56 x^2}{e^3 \left (25 x^2+10 x^3+x^4\right )} \, dx=\frac {8\,{\mathrm {e}}^{-3}\,\left (7\,x-1\right )}{x\,\left (x+5\right )} \]

[In]

int((exp(-3)*(16*x - 56*x^2 + 40))/(25*x^2 + 10*x^3 + x^4),x)

[Out]

(8*exp(-3)*(7*x - 1))/(x*(x + 5))