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

Optimal. Leaf size=22 \[ \frac {50 \left (4+e^{(5+x)^2}\right ) x}{(-1+x) (4+x)} \]

________________________________________________________________________________________

Rubi [B]  time = 0.70, antiderivative size = 53, normalized size of antiderivative = 2.41, number of steps used = 14, number of rules used = 5, integrand size = 57, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {6742, 894, 2227, 2204, 2242} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[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 894

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 2204

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 2227

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 2242

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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.46, size = 22, normalized size = 1.00 \begin {gather*} \frac {50 \left (4+e^{(5+x)^2}\right ) x}{-4+3 x+x^2} \end {gather*}

Antiderivative was successfully verified.

[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)

________________________________________________________________________________________

fricas [A]  time = 0.48, size = 27, normalized size = 1.23 \begin {gather*} \frac {50 \, {\left (x e^{\left (x^{2} + 10 \, x + 25\right )} + 4 \, x\right )}}{x^{2} + 3 \, x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

giac [A]  time = 0.21, size = 27, normalized size = 1.23 \begin {gather*} \frac {50 \, {\left (x e^{\left (x^{2} + 10 \, x + 25\right )} + 4 \, x\right )}}{x^{2} + 3 \, x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [A]  time = 0.13, size = 28, normalized size = 1.27

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maxima [B]  time = 0.44, size = 57, normalized size = 2.59 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [B]  time = 0.23, size = 24, normalized size = 1.09 \begin {gather*} \frac {50\,x\,\left ({\mathrm {e}}^{x^2+10\,x+25}+4\right )}{x^2+3\,x-4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [A]  time = 0.15, size = 32, normalized size = 1.45 \begin {gather*} \frac {50 x e^{x^{2} + 10 x + 25}}{x^{2} + 3 x - 4} + \frac {200 x}{x^{2} + 3 x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________