\(\int \frac {-800-200 x^2+e^{25+10 x+x^2} (-200-2000 x+1050 x^2+800 x^3+100 x^4)}{16-24 x+x^2+6 x^3+x^4} \, dx\) [7972]

Optimal result

Integrand size = 57, antiderivative size = 22 \[ \int \frac {-800-200 x^2+e^{25+10 x+x^2} \left (-200-2000 x+1050 x^2+800 x^3+100 x^4\right )}{16-24 x+x^2+6 x^3+x^4} \, dx=\frac {50 \left (4+e^{(5+x)^2}\right ) x}{(-1+x) (4+x)} \]

[Out]

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

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(53\) vs. \(2(22)=44\).

Time = 0.44 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.41, number of steps used = 14, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {6874, 908, 2259, 2235, 2274} \[ \int \frac {-800-200 x^2+e^{25+10 x+x^2} \left (-200-2000 x+1050 x^2+800 x^3+100 x^4\right )}{16-24 x+x^2+6 x^3+x^4} \, dx=\frac {40 e^{x^2+10 x+25}}{x+4}-\frac {10 e^{x^2+10 x+25}}{1-x}+\frac {160}{x+4}-\frac {40}{1-x} \]

[In]

Int[(-800 - 200*x^2 + E^(25 + 10*x + x^2)*(-200 - 2000*x + 1050*x^2 + 800*x^3 + 100*x^4))/(16 - 24*x + x^2 + 6
*x^3 + x^4),x]

[Out]

-40/(1 - x) - (10*E^(25 + 10*x + x^2))/(1 - x) + 160/(4 + x) + (40*E^(25 + 10*x + x^2))/(4 + x)

Rule 908

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

Rule 2235

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2
]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2259

Int[(u_.)*(F_)^((a_.) + (b_.)*(v_)), x_Symbol] :> Int[u*F^(a + b*NormalizePowerOfLinear[v, x]), x] /; FreeQ[{F
, a, b}, x] && PolynomialQ[u, x] && PowerOfLinearQ[v, x] &&  !PowerOfLinearMatchQ[v, x]

Rule 2274

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_))^(m_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(F
^(a + b*x + c*x^2)/(e*(m + 1))), x] + (-Dist[2*c*(Log[F]/(e^2*(m + 1))), Int[(d + e*x)^(m + 2)*F^(a + b*x + c*
x^2), x], x] - Dist[(b*e - 2*c*d)*(Log[F]/(e^2*(m + 1))), Int[(d + e*x)^(m + 1)*F^(a + b*x + c*x^2), x], x]) /
; FreeQ[{F, a, b, c, d, e}, x] && NeQ[b*e - 2*c*d, 0] && LtQ[m, -1]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {200 \left (4+x^2\right )}{(-1+x)^2 (4+x)^2}+\frac {50 e^{25+10 x+x^2} \left (-4-40 x+21 x^2+16 x^3+2 x^4\right )}{(1-x)^2 (4+x)^2}\right ) \, dx \\ & = 50 \int \frac {e^{25+10 x+x^2} \left (-4-40 x+21 x^2+16 x^3+2 x^4\right )}{(1-x)^2 (4+x)^2} \, dx-200 \int \frac {4+x^2}{(-1+x)^2 (4+x)^2} \, dx \\ & = 50 \int \left (2 e^{25+10 x+x^2}-\frac {e^{25+10 x+x^2}}{5 (-1+x)^2}+\frac {12 e^{25+10 x+x^2}}{5 (-1+x)}-\frac {4 e^{25+10 x+x^2}}{5 (4+x)^2}+\frac {8 e^{25+10 x+x^2}}{5 (4+x)}\right ) \, dx-200 \int \left (\frac {1}{5 (-1+x)^2}+\frac {4}{5 (4+x)^2}\right ) \, dx \\ & = -\frac {40}{1-x}+\frac {160}{4+x}-10 \int \frac {e^{25+10 x+x^2}}{(-1+x)^2} \, dx-40 \int \frac {e^{25+10 x+x^2}}{(4+x)^2} \, dx+80 \int \frac {e^{25+10 x+x^2}}{4+x} \, dx+100 \int e^{25+10 x+x^2} \, dx+120 \int \frac {e^{25+10 x+x^2}}{-1+x} \, dx \\ & = -\frac {40}{1-x}-\frac {10 e^{25+10 x+x^2}}{1-x}+\frac {160}{4+x}+\frac {40 e^{25+10 x+x^2}}{4+x}-20 \int e^{25+10 x+x^2} \, dx-80 \int e^{25+10 x+x^2} \, dx+100 \int e^{(5+x)^2} \, dx \\ & = -\frac {40}{1-x}-\frac {10 e^{25+10 x+x^2}}{1-x}+\frac {160}{4+x}+\frac {40 e^{25+10 x+x^2}}{4+x}+50 \sqrt {\pi } \text {erfi}(5+x)-20 \int e^{(5+x)^2} \, dx-80 \int e^{(5+x)^2} \, dx \\ & = -\frac {40}{1-x}-\frac {10 e^{25+10 x+x^2}}{1-x}+\frac {160}{4+x}+\frac {40 e^{25+10 x+x^2}}{4+x} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.39 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {-800-200 x^2+e^{25+10 x+x^2} \left (-200-2000 x+1050 x^2+800 x^3+100 x^4\right )}{16-24 x+x^2+6 x^3+x^4} \, dx=\frac {50 \left (4+e^{(5+x)^2}\right ) x}{-4+3 x+x^2} \]

[In]

Integrate[(-800 - 200*x^2 + E^(25 + 10*x + x^2)*(-200 - 2000*x + 1050*x^2 + 800*x^3 + 100*x^4))/(16 - 24*x + x
^2 + 6*x^3 + x^4),x]

[Out]

(50*(4 + E^(5 + x)^2)*x)/(-4 + 3*x + x^2)

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27

method	result	size
norman	\(\frac {200 x +50 \,{\mathrm e}^{x^{2}+10 x +25} x}{x^{2}+3 x -4}\)	\(28\)
parallelrisch	\(\frac {200 x +50 \,{\mathrm e}^{x^{2}+10 x +25} x}{x^{2}+3 x -4}\)	\(28\)
risch	\(\frac {200 x}{x^{2}+3 x -4}+\frac {50 x \,{\mathrm e}^{\left (5+x \right )^{2}}}{x^{2}+3 x -4}\)	\(34\)
parts	\(\frac {40}{-1+x}+\frac {160}{4+x}+\frac {50 x \,{\mathrm e}^{x^{2}+10 x +25}}{x^{2}+3 x -4}\)	\(38\)

[In]

int(((100*x^4+800*x^3+1050*x^2-2000*x-200)*exp(x^2+10*x+25)-200*x^2-800)/(x^4+6*x^3+x^2-24*x+16),x,method=_RET
URNVERBOSE)

[Out]

(200*x+50*exp(x^2+10*x+25)*x)/(x^2+3*x-4)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {-800-200 x^2+e^{25+10 x+x^2} \left (-200-2000 x+1050 x^2+800 x^3+100 x^4\right )}{16-24 x+x^2+6 x^3+x^4} \, dx=\frac {50 \, {\left (x e^{\left (x^{2} + 10 \, x + 25\right )} + 4 \, x\right )}}{x^{2} + 3 \, x - 4} \]

[In]

integrate(((100*x^4+800*x^3+1050*x^2-2000*x-200)*exp(x^2+10*x+25)-200*x^2-800)/(x^4+6*x^3+x^2-24*x+16),x, algo
rithm="fricas")

[Out]

50*(x*e^(x^2 + 10*x + 25) + 4*x)/(x^2 + 3*x - 4)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.45 \[ \int \frac {-800-200 x^2+e^{25+10 x+x^2} \left (-200-2000 x+1050 x^2+800 x^3+100 x^4\right )}{16-24 x+x^2+6 x^3+x^4} \, dx=\frac {50 x e^{x^{2} + 10 x + 25}}{x^{2} + 3 x - 4} + \frac {200 x}{x^{2} + 3 x - 4} \]

[In]

integrate(((100*x**4+800*x**3+1050*x**2-2000*x-200)*exp(x**2+10*x+25)-200*x**2-800)/(x**4+6*x**3+x**2-24*x+16)
,x)

[Out]

50*x*exp(x**2 + 10*x + 25)/(x**2 + 3*x - 4) + 200*x/(x**2 + 3*x - 4)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (21) = 42\).

Time = 0.25 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.59 \[ \int \frac {-800-200 x^2+e^{25+10 x+x^2} \left (-200-2000 x+1050 x^2+800 x^3+100 x^4\right )}{16-24 x+x^2+6 x^3+x^4} \, dx=\frac {50 \, x e^{\left (x^{2} + 10 \, x + 25\right )}}{x^{2} + 3 \, x - 4} + \frac {8 \, {\left (17 \, x - 12\right )}}{x^{2} + 3 \, x - 4} + \frac {32 \, {\left (2 \, x + 3\right )}}{x^{2} + 3 \, x - 4} \]

[In]

integrate(((100*x^4+800*x^3+1050*x^2-2000*x-200)*exp(x^2+10*x+25)-200*x^2-800)/(x^4+6*x^3+x^2-24*x+16),x, algo
rithm="maxima")

[Out]

50*x*e^(x^2 + 10*x + 25)/(x^2 + 3*x - 4) + 8*(17*x - 12)/(x^2 + 3*x - 4) + 32*(2*x + 3)/(x^2 + 3*x - 4)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {-800-200 x^2+e^{25+10 x+x^2} \left (-200-2000 x+1050 x^2+800 x^3+100 x^4\right )}{16-24 x+x^2+6 x^3+x^4} \, dx=\frac {50 \, {\left (x e^{\left (x^{2} + 10 \, x + 25\right )} + 4 \, x\right )}}{x^{2} + 3 \, x - 4} \]

[In]

integrate(((100*x^4+800*x^3+1050*x^2-2000*x-200)*exp(x^2+10*x+25)-200*x^2-800)/(x^4+6*x^3+x^2-24*x+16),x, algo
rithm="giac")

[Out]

50*(x*e^(x^2 + 10*x + 25) + 4*x)/(x^2 + 3*x - 4)

Mupad [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {-800-200 x^2+e^{25+10 x+x^2} \left (-200-2000 x+1050 x^2+800 x^3+100 x^4\right )}{16-24 x+x^2+6 x^3+x^4} \, dx=\frac {50\,x\,\left ({\mathrm {e}}^{x^2+10\,x+25}+4\right )}{x^2+3\,x-4} \]

[In]

int(-(200*x^2 - exp(10*x + x^2 + 25)*(1050*x^2 - 2000*x + 800*x^3 + 100*x^4 - 200) + 800)/(x^2 - 24*x + 6*x^3
+ x^4 + 16),x)

[Out]

(50*x*(exp(10*x + x^2 + 25) + 4))/(3*x + x^2 - 4)