3.97.28 \(\int \frac {3208 x^2-11212 x^3+12006 x^4-5201 x^5+800 x^6+e^x (-2406+6015 x-6003 x^2+1200 x^3)}{-3200 x^2+4800 x^3-2400 x^4+400 x^5} \, dx\)

Optimal. Leaf size=23 \[ \left (-\frac {401}{400}+x\right ) \left (\frac {3 e^x}{(2-x)^2 x}+x\right ) \]

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Rubi [B]  time = 0.46, antiderivative size = 47, normalized size of antiderivative = 2.04, number of steps used = 14, number of rules used = 5, integrand size = 69, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.072, Rules used = {6688, 12, 6742, 2177, 2178} \begin {gather*} x^2-\frac {401 x}{400}-\frac {1203 e^x}{1600 (2-x)}+\frac {1197 e^x}{800 (2-x)^2}-\frac {1203 e^x}{1600 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3208*x^2 - 11212*x^3 + 12006*x^4 - 5201*x^5 + 800*x^6 + E^x*(-2406 + 6015*x - 6003*x^2 + 1200*x^3))/(-320
0*x^2 + 4800*x^3 - 2400*x^4 + 400*x^5),x]

[Out]

(1197*E^x)/(800*(2 - x)^2) - (1203*E^x)/(1600*(2 - x)) - (1203*E^x)/(1600*x) - (401*x)/400 + 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 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 6688

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

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 \frac {1}{400} \left (-401+800 x+\frac {3 e^x \left (-802+2005 x-2001 x^2+400 x^3\right )}{(-2+x)^3 x^2}\right ) \, dx\\ &=\frac {1}{400} \int \left (-401+800 x+\frac {3 e^x \left (-802+2005 x-2001 x^2+400 x^3\right )}{(-2+x)^3 x^2}\right ) \, dx\\ &=-\frac {401 x}{400}+x^2+\frac {3}{400} \int \frac {e^x \left (-802+2005 x-2001 x^2+400 x^3\right )}{(-2+x)^3 x^2} \, dx\\ &=-\frac {401 x}{400}+x^2+\frac {3}{400} \int \left (-\frac {399 e^x}{(-2+x)^3}+\frac {397 e^x}{4 (-2+x)^2}+\frac {401 e^x}{4 (-2+x)}+\frac {401 e^x}{4 x^2}-\frac {401 e^x}{4 x}\right ) \, dx\\ &=-\frac {401 x}{400}+x^2+\frac {1191 \int \frac {e^x}{(-2+x)^2} \, dx}{1600}+\frac {1203 \int \frac {e^x}{-2+x} \, dx}{1600}+\frac {1203 \int \frac {e^x}{x^2} \, dx}{1600}-\frac {1203 \int \frac {e^x}{x} \, dx}{1600}-\frac {1197}{400} \int \frac {e^x}{(-2+x)^3} \, dx\\ &=\frac {1197 e^x}{800 (2-x)^2}+\frac {1191 e^x}{1600 (2-x)}-\frac {1203 e^x}{1600 x}-\frac {401 x}{400}+x^2+\frac {1203 e^2 \text {Ei}(-2+x)}{1600}-\frac {1203 \text {Ei}(x)}{1600}+\frac {1191 \int \frac {e^x}{-2+x} \, dx}{1600}+\frac {1203 \int \frac {e^x}{x} \, dx}{1600}-\frac {1197}{800} \int \frac {e^x}{(-2+x)^2} \, dx\\ &=\frac {1197 e^x}{800 (2-x)^2}-\frac {1203 e^x}{1600 (2-x)}-\frac {1203 e^x}{1600 x}-\frac {401 x}{400}+x^2+\frac {1197}{800} e^2 \text {Ei}(-2+x)-\frac {1197}{800} \int \frac {e^x}{-2+x} \, dx\\ &=\frac {1197 e^x}{800 (2-x)^2}-\frac {1203 e^x}{1600 (2-x)}-\frac {1203 e^x}{1600 x}-\frac {401 x}{400}+x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 32, normalized size = 1.39 \begin {gather*} \frac {(-401+400 x) \left (3 e^x+(-2+x)^2 x^2\right )}{400 (-2+x)^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3208*x^2 - 11212*x^3 + 12006*x^4 - 5201*x^5 + 800*x^6 + E^x*(-2406 + 6015*x - 6003*x^2 + 1200*x^3))
/(-3200*x^2 + 4800*x^3 - 2400*x^4 + 400*x^5),x]

[Out]

((-401 + 400*x)*(3*E^x + (-2 + x)^2*x^2))/(400*(-2 + x)^2*x)

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fricas [B]  time = 0.52, size = 46, normalized size = 2.00 \begin {gather*} \frac {400 \, x^{5} - 2001 \, x^{4} + 3204 \, x^{3} - 1604 \, x^{2} + 3 \, {\left (400 \, x - 401\right )} e^{x}}{400 \, {\left (x^{3} - 4 \, x^{2} + 4 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1200*x^3-6003*x^2+6015*x-2406)*exp(x)+800*x^6-5201*x^5+12006*x^4-11212*x^3+3208*x^2)/(400*x^5-2400
*x^4+4800*x^3-3200*x^2),x, algorithm="fricas")

[Out]

1/400*(400*x^5 - 2001*x^4 + 3204*x^3 - 1604*x^2 + 3*(400*x - 401)*e^x)/(x^3 - 4*x^2 + 4*x)

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giac [B]  time = 0.18, size = 46, normalized size = 2.00 \begin {gather*} \frac {400 \, x^{5} - 2001 \, x^{4} + 3204 \, x^{3} - 1604 \, x^{2} + 1200 \, x e^{x} - 1203 \, e^{x}}{400 \, {\left (x^{3} - 4 \, x^{2} + 4 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1200*x^3-6003*x^2+6015*x-2406)*exp(x)+800*x^6-5201*x^5+12006*x^4-11212*x^3+3208*x^2)/(400*x^5-2400
*x^4+4800*x^3-3200*x^2),x, algorithm="giac")

[Out]

1/400*(400*x^5 - 2001*x^4 + 3204*x^3 - 1604*x^2 + 1200*x*e^x - 1203*e^x)/(x^3 - 4*x^2 + 4*x)

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maple [A]  time = 0.07, size = 25, normalized size = 1.09

method	result	size
risch	\(x^{2}-\frac {401 x}{400}+\frac {3 \left (400 x -401\right ) {\mathrm e}^{x}}{400 x \left (x -2\right )^{2}}\)	\(25\)
default	\(x^{2}-\frac {401 x}{400}+\frac {1197 \,{\mathrm e}^{x}}{800 \left (x -2\right )^{2}}+\frac {1203 \,{\mathrm e}^{x}}{1600 \left (x -2\right )}-\frac {1203 \,{\mathrm e}^{x}}{1600 x}\)	\(33\)
norman	\(\frac {x^{5}-\frac {801 x}{25}+\frac {2803 x^{2}}{100}-\frac {2001 x^{4}}{400}+3 \,{\mathrm e}^{x} x -\frac {1203 \,{\mathrm e}^{x}}{400}}{x \left (x -2\right )^{2}}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1200*x^3-6003*x^2+6015*x-2406)*exp(x)+800*x^6-5201*x^5+12006*x^4-11212*x^3+3208*x^2)/(400*x^5-2400*x^4+4
800*x^3-3200*x^2),x,method=_RETURNVERBOSE)

[Out]

x^2-401/400*x+3/400*(400*x-401)/x/(x-2)^2*exp(x)

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maxima [B]  time = 0.40, size = 108, normalized size = 4.70 \begin {gather*} x^{2} - \frac {401}{400} \, x + \frac {3 \, {\left (400 \, x - 401\right )} e^{x}}{400 \, {\left (x^{3} - 4 \, x^{2} + 4 \, x\right )}} - \frac {16 \, {\left (4 \, x - 7\right )}}{x^{2} - 4 \, x + 4} + \frac {5201 \, {\left (3 \, x - 5\right )}}{100 \, {\left (x^{2} - 4 \, x + 4\right )}} - \frac {6003 \, {\left (2 \, x - 3\right )}}{100 \, {\left (x^{2} - 4 \, x + 4\right )}} + \frac {2803 \, {\left (x - 1\right )}}{100 \, {\left (x^{2} - 4 \, x + 4\right )}} - \frac {401}{100 \, {\left (x^{2} - 4 \, x + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1200*x^3-6003*x^2+6015*x-2406)*exp(x)+800*x^6-5201*x^5+12006*x^4-11212*x^3+3208*x^2)/(400*x^5-2400
*x^4+4800*x^3-3200*x^2),x, algorithm="maxima")

[Out]

x^2 - 401/400*x + 3/400*(400*x - 401)*e^x/(x^3 - 4*x^2 + 4*x) - 16*(4*x - 7)/(x^2 - 4*x + 4) + 5201/100*(3*x -
5)/(x^2 - 4*x + 4) - 6003/100*(2*x - 3)/(x^2 - 4*x + 4) + 2803/100*(x - 1)/(x^2 - 4*x + 4) - 401/100/(x^2 - 4
*x + 4)

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mupad [B]  time = 5.56, size = 33, normalized size = 1.43 \begin {gather*} \frac {\left (400\,x-401\right )\,\left (3\,{\mathrm {e}}^x+4\,x^2-4\,x^3+x^4\right )}{400\,x\,{\left (x-2\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3208*x^2 - 11212*x^3 + 12006*x^4 - 5201*x^5 + 800*x^6 + exp(x)*(6015*x - 6003*x^2 + 1200*x^3 - 2406))/(3
200*x^2 - 4800*x^3 + 2400*x^4 - 400*x^5),x)

[Out]

((400*x - 401)*(3*exp(x) + 4*x^2 - 4*x^3 + x^4))/(400*x*(x - 2)^2)

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sympy [A]  time = 0.13, size = 29, normalized size = 1.26 \begin {gather*} x^{2} - \frac {401 x}{400} + \frac {\left (1200 x - 1203\right ) e^{x}}{400 x^{3} - 1600 x^{2} + 1600 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1200*x**3-6003*x**2+6015*x-2406)*exp(x)+800*x**6-5201*x**5+12006*x**4-11212*x**3+3208*x**2)/(400*x
**5-2400*x**4+4800*x**3-3200*x**2),x)

[Out]

x**2 - 401*x/400 + (1200*x - 1203)*exp(x)/(400*x**3 - 1600*x**2 + 1600*x)

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