\(\int \frac {e^{-1+x} (-504+504 x-504 x^2+168 x^3)}{225 x^2+150 x^4+25 x^6} \, dx\) [8743]

Optimal result

Integrand size = 39, antiderivative size = 19 \[ \int \frac {e^{-1+x} \left (-504+504 x-504 x^2+168 x^3\right )}{225 x^2+150 x^4+25 x^6} \, dx=\frac {168 e^{-1+x}}{25 x \left (3+x^2\right )} \]

[Out]

168/25/(x^2+3)/x/exp(1-x)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1608, 28, 2326} \[ \int \frac {e^{-1+x} \left (-504+504 x-504 x^2+168 x^3\right )}{225 x^2+150 x^4+25 x^6} \, dx=\frac {168 e^{x-1} \left (x^3+3 x\right )}{25 x^2 \left (x^2+3\right )^2} \]

[In]

Int[(E^(-1 + x)*(-504 + 504*x - 504*x^2 + 168*x^3))/(225*x^2 + 150*x^4 + 25*x^6),x]

[Out]

(168*E^(-1 + x)*(3*x + x^3))/(25*x^2*(3 + x^2)^2)

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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]

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-1+x} \left (-504+504 x-504 x^2+168 x^3\right )}{x^2 \left (225+150 x^2+25 x^4\right )} \, dx \\ & = 25 \int \frac {e^{-1+x} \left (-504+504 x-504 x^2+168 x^3\right )}{x^2 \left (75+25 x^2\right )^2} \, dx \\ & = \frac {168 e^{-1+x} \left (3 x+x^3\right )}{25 x^2 \left (3+x^2\right )^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {e^{-1+x} \left (-504+504 x-504 x^2+168 x^3\right )}{225 x^2+150 x^4+25 x^6} \, dx=\frac {168 e^{-1+x}}{25 \left (3 x+x^3\right )} \]

[In]

Integrate[(E^(-1 + x)*(-504 + 504*x - 504*x^2 + 168*x^3))/(225*x^2 + 150*x^4 + 25*x^6),x]

[Out]

(168*E^(-1 + x))/(25*(3*x + x^3))

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89

method	result	size
risch	\(\frac {168 \,{\mathrm e}^{-1+x}}{25 \left (x^{2}+3\right ) x}\)	\(17\)
gosper	\(\frac {168 \,{\mathrm e}^{-1+x}}{25 \left (x^{2}+3\right ) x}\)	\(21\)
norman	\(\frac {168 \,{\mathrm e}^{-1+x}}{25 \left (x^{2}+3\right ) x}\)	\(21\)
parallelrisch	\(\frac {168 \,{\mathrm e}^{-1+x}}{25 \left (x^{2}+3\right ) x}\)	\(21\)
derivativedivides	\(-\frac {56 \,{\mathrm e}^{-1+x} \left (\left (1-x \right )^{2}+1+2 x \right )}{25 \left (\left (1-x \right )^{3}-3 \left (1-x \right )^{2}+2-6 x \right )}+\frac {56 \,{\mathrm e}^{-1+x} \left (1-x \right ) \left (-1-x \right )}{25 \left (\left (1-x \right )^{3}-3 \left (1-x \right )^{2}+2-6 x \right )}\)	\(82\)
default	\(-\frac {56 \,{\mathrm e}^{-1+x} \left (\left (1-x \right )^{2}+1+2 x \right )}{25 \left (\left (1-x \right )^{3}-3 \left (1-x \right )^{2}+2-6 x \right )}+\frac {56 \,{\mathrm e}^{-1+x} \left (1-x \right ) \left (-1-x \right )}{25 \left (\left (1-x \right )^{3}-3 \left (1-x \right )^{2}+2-6 x \right )}\)	\(82\)

[In]

int((168*x^3-504*x^2+504*x-504)/(25*x^6+150*x^4+225*x^2)/exp(1-x),x,method=_RETURNVERBOSE)

[Out]

168/25/(x^2+3)/x*exp(-1+x)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {e^{-1+x} \left (-504+504 x-504 x^2+168 x^3\right )}{225 x^2+150 x^4+25 x^6} \, dx=\frac {168 \, e^{\left (x - 1\right )}}{25 \, {\left (x^{3} + 3 \, x\right )}} \]

[In]

integrate((168*x^3-504*x^2+504*x-504)/(25*x^6+150*x^4+225*x^2)/exp(1-x),x, algorithm="fricas")

[Out]

168/25*e^(x - 1)/(x^3 + 3*x)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {e^{-1+x} \left (-504+504 x-504 x^2+168 x^3\right )}{225 x^2+150 x^4+25 x^6} \, dx=\frac {168 e^{x - 1}}{25 x^{3} + 75 x} \]

[In]

integrate((168*x**3-504*x**2+504*x-504)/(25*x**6+150*x**4+225*x**2)/exp(1-x),x)

[Out]

168*exp(x - 1)/(25*x**3 + 75*x)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {e^{-1+x} \left (-504+504 x-504 x^2+168 x^3\right )}{225 x^2+150 x^4+25 x^6} \, dx=\frac {168 \, e^{x}}{25 \, {\left (x^{3} e + 3 \, x e\right )}} \]

[In]

integrate((168*x^3-504*x^2+504*x-504)/(25*x^6+150*x^4+225*x^2)/exp(1-x),x, algorithm="maxima")

[Out]

168/25*e^x/(x^3*e + 3*x*e)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {e^{-1+x} \left (-504+504 x-504 x^2+168 x^3\right )}{225 x^2+150 x^4+25 x^6} \, dx=\frac {168 \, e^{x}}{25 \, {\left (x^{3} e + 3 \, x e\right )}} \]

[In]

integrate((168*x^3-504*x^2+504*x-504)/(25*x^6+150*x^4+225*x^2)/exp(1-x),x, algorithm="giac")

[Out]

168/25*e^x/(x^3*e + 3*x*e)

Mupad [B] (verification not implemented)

Time = 13.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {e^{-1+x} \left (-504+504 x-504 x^2+168 x^3\right )}{225 x^2+150 x^4+25 x^6} \, dx=\frac {168\,{\mathrm {e}}^{-1}\,{\mathrm {e}}^x}{25\,\left (x^3+3\,x\right )} \]

[In]

int((exp(x - 1)*(504*x - 504*x^2 + 168*x^3 - 504))/(225*x^2 + 150*x^4 + 25*x^6),x)

[Out]

(168*exp(-1)*exp(x))/(25*(3*x + x^3))