3.12.8 \(\int \frac {400+480 x-48 x^2+e^x (-80-80 x+16 x^2)}{125-50 x+5 x^2} \, dx\)

Optimal. Leaf size=27 \[ \frac {16 \left (3+\frac {5-e^x}{x}\right ) x^2}{5 (5-x)} \]

________________________________________________________________________________________

Rubi [A]  time = 0.23, antiderivative size = 34, normalized size of antiderivative = 1.26, number of steps used = 12, number of rules used = 8, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.216, Rules used = {27, 12, 6742, 2199, 2194, 2177, 2178, 683} \begin {gather*} -\frac {48 x}{5}+\frac {16 e^x}{5}-\frac {16 e^x}{5-x}+\frac {320}{5-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(400 + 480*x - 48*x^2 + E^x*(-80 - 80*x + 16*x^2))/(125 - 50*x + 5*x^2),x]

[Out]

(16*E^x)/5 + 320/(5 - x) - (16*E^x)/(5 - x) - (48*x)/5

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 683

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && IGtQ[p, 0] &&  !(EqQ[m, 3] && NeQ[p, 1])

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 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

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 {400+480 x-48 x^2+e^x \left (-80-80 x+16 x^2\right )}{5 (-5+x)^2} \, dx\\ &=\frac {1}{5} \int \frac {400+480 x-48 x^2+e^x \left (-80-80 x+16 x^2\right )}{(-5+x)^2} \, dx\\ &=\frac {1}{5} \int \left (\frac {16 e^x \left (-5-5 x+x^2\right )}{(-5+x)^2}-\frac {16 \left (-25-30 x+3 x^2\right )}{(-5+x)^2}\right ) \, dx\\ &=\frac {16}{5} \int \frac {e^x \left (-5-5 x+x^2\right )}{(-5+x)^2} \, dx-\frac {16}{5} \int \frac {-25-30 x+3 x^2}{(-5+x)^2} \, dx\\ &=-\left (\frac {16}{5} \int \left (3-\frac {100}{(-5+x)^2}\right ) \, dx\right )+\frac {16}{5} \int \left (e^x-\frac {5 e^x}{(-5+x)^2}+\frac {5 e^x}{-5+x}\right ) \, dx\\ &=\frac {320}{5-x}-\frac {48 x}{5}+\frac {16 \int e^x \, dx}{5}-16 \int \frac {e^x}{(-5+x)^2} \, dx+16 \int \frac {e^x}{-5+x} \, dx\\ &=\frac {16 e^x}{5}+\frac {320}{5-x}-\frac {16 e^x}{5-x}-\frac {48 x}{5}+16 e^5 \text {Ei}(-5+x)-16 \int \frac {e^x}{-5+x} \, dx\\ &=\frac {16 e^x}{5}+\frac {320}{5-x}-\frac {16 e^x}{5-x}-\frac {48 x}{5}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.05, size = 23, normalized size = 0.85 \begin {gather*} \frac {16 \left (-100+\left (15+e^x\right ) x-3 x^2\right )}{5 (-5+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(400 + 480*x - 48*x^2 + E^x*(-80 - 80*x + 16*x^2))/(125 - 50*x + 5*x^2),x]

[Out]

(16*(-100 + (15 + E^x)*x - 3*x^2))/(5*(-5 + x))

________________________________________________________________________________________

fricas [A]  time = 0.72, size = 22, normalized size = 0.81 \begin {gather*} -\frac {16 \, {\left (3 \, x^{2} - x e^{x} - 15 \, x + 100\right )}}{5 \, {\left (x - 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x^2-80*x-80)*exp(x)-48*x^2+480*x+400)/(5*x^2-50*x+125),x, algorithm="fricas")

[Out]

-16/5*(3*x^2 - x*e^x - 15*x + 100)/(x - 5)

________________________________________________________________________________________

giac [A]  time = 0.30, size = 22, normalized size = 0.81 \begin {gather*} -\frac {16 \, {\left (3 \, x^{2} - x e^{x} - 15 \, x + 100\right )}}{5 \, {\left (x - 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x^2-80*x-80)*exp(x)-48*x^2+480*x+400)/(5*x^2-50*x+125),x, algorithm="giac")

[Out]

-16/5*(3*x^2 - x*e^x - 15*x + 100)/(x - 5)

________________________________________________________________________________________

maple [A]  time = 0.23, size = 19, normalized size = 0.70

method	result	size
norman	\(\frac {-\frac {48 x^{2}}{5}+\frac {16 \,{\mathrm e}^{x} x}{5}-80}{x -5}\)	\(19\)
risch	\(-\frac {48 x}{5}-\frac {320}{x -5}+\frac {16 x \,{\mathrm e}^{x}}{5 \left (x -5\right )}\)	\(22\)
default	\(-\frac {320}{x -5}-\frac {48 x}{5}+\frac {16 \,{\mathrm e}^{x}}{x -5}+\frac {16 \,{\mathrm e}^{x}}{5}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((16*x^2-80*x-80)*exp(x)-48*x^2+480*x+400)/(5*x^2-50*x+125),x,method=_RETURNVERBOSE)

[Out]

(-48/5*x^2+16/5*exp(x)*x-80)/(x-5)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {48}{5} \, x + \frac {16 \, x e^{x}}{5 \, {\left (x - 5\right )}} + \frac {16 \, e^{5} E_{2}\left (-x + 5\right )}{x - 5} - \frac {320}{x - 5} + 16 \, \int \frac {e^{x}}{x^{2} - 10 \, x + 25}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x^2-80*x-80)*exp(x)-48*x^2+480*x+400)/(5*x^2-50*x+125),x, algorithm="maxima")

[Out]

-48/5*x + 16/5*x*e^x/(x - 5) + 16*e^5*exp_integral_e(2, -x + 5)/(x - 5) - 320/(x - 5) + 16*integrate(e^x/(x^2
- 10*x + 25), x)

________________________________________________________________________________________

mupad [B]  time = 0.10, size = 19, normalized size = 0.70 \begin {gather*} -\frac {16\,x\,\left (3\,x-{\mathrm {e}}^x+5\right )}{5\,\left (x-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((480*x - exp(x)*(80*x - 16*x^2 + 80) - 48*x^2 + 400)/(5*x^2 - 50*x + 125),x)

[Out]

-(16*x*(3*x - exp(x) + 5))/(5*(x - 5))

________________________________________________________________________________________

sympy [A]  time = 0.13, size = 20, normalized size = 0.74 \begin {gather*} - \frac {48 x}{5} + \frac {16 x e^{x}}{5 x - 25} - \frac {320}{x - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x**2-80*x-80)*exp(x)-48*x**2+480*x+400)/(5*x**2-50*x+125),x)

[Out]

-48*x/5 + 16*x*exp(x)/(5*x - 25) - 320/(x - 5)

________________________________________________________________________________________