3.49.15 \(\int \frac {e^x (-32+56 x-32 x^2+4 x^3)}{-64 x^2+48 x^3-12 x^4+x^5} \, dx\)

Optimal. Leaf size=18 \[ 25+\frac {4 e^x (-2+x)}{(-4+x)^2 x} \]

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Rubi [B]  time = 0.41, antiderivative size = 37, normalized size of antiderivative = 2.06, number of steps used = 13, number of rules used = 5, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6688, 12, 6742, 2177, 2178} \begin {gather*} -\frac {e^x}{2 x}-\frac {e^x}{2 (4-x)}+\frac {2 e^x}{(4-x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(-32 + 56*x - 32*x^2 + 4*x^3))/(-64*x^2 + 48*x^3 - 12*x^4 + x^5),x]

[Out]

(2*E^x)/(4 - x)^2 - E^x/(2*(4 - x)) - E^x/(2*x)

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 {4 e^x \left (8-14 x+8 x^2-x^3\right )}{(4-x)^3 x^2} \, dx\\ &=4 \int \frac {e^x \left (8-14 x+8 x^2-x^3\right )}{(4-x)^3 x^2} \, dx\\ &=4 \int \left (-\frac {e^x}{(-4+x)^3}+\frac {3 e^x}{8 (-4+x)^2}+\frac {e^x}{8 (-4+x)}+\frac {e^x}{8 x^2}-\frac {e^x}{8 x}\right ) \, dx\\ &=\frac {1}{2} \int \frac {e^x}{-4+x} \, dx+\frac {1}{2} \int \frac {e^x}{x^2} \, dx-\frac {1}{2} \int \frac {e^x}{x} \, dx+\frac {3}{2} \int \frac {e^x}{(-4+x)^2} \, dx-4 \int \frac {e^x}{(-4+x)^3} \, dx\\ &=\frac {2 e^x}{(4-x)^2}+\frac {3 e^x}{2 (4-x)}-\frac {e^x}{2 x}+\frac {1}{2} e^4 \text {Ei}(-4+x)-\frac {\text {Ei}(x)}{2}+\frac {1}{2} \int \frac {e^x}{x} \, dx+\frac {3}{2} \int \frac {e^x}{-4+x} \, dx-2 \int \frac {e^x}{(-4+x)^2} \, dx\\ &=\frac {2 e^x}{(4-x)^2}-\frac {e^x}{2 (4-x)}-\frac {e^x}{2 x}+2 e^4 \text {Ei}(-4+x)-2 \int \frac {e^x}{-4+x} \, dx\\ &=\frac {2 e^x}{(4-x)^2}-\frac {e^x}{2 (4-x)}-\frac {e^x}{2 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 16, normalized size = 0.89 \begin {gather*} \frac {4 e^x (-2+x)}{(-4+x)^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-32 + 56*x - 32*x^2 + 4*x^3))/(-64*x^2 + 48*x^3 - 12*x^4 + x^5),x]

[Out]

(4*E^x*(-2 + x))/((-4 + x)^2*x)

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fricas [A]  time = 0.63, size = 21, normalized size = 1.17 \begin {gather*} \frac {4 \, {\left (x - 2\right )} e^{x}}{x^{3} - 8 \, x^{2} + 16 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^3-32*x^2+56*x-32)*exp(x)/(x^5-12*x^4+48*x^3-64*x^2),x, algorithm="fricas")

[Out]

4*(x - 2)*e^x/(x^3 - 8*x^2 + 16*x)

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giac [A]  time = 0.15, size = 25, normalized size = 1.39 \begin {gather*} \frac {4 \, {\left (x e^{x} - 2 \, e^{x}\right )}}{x^{3} - 8 \, x^{2} + 16 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^3-32*x^2+56*x-32)*exp(x)/(x^5-12*x^4+48*x^3-64*x^2),x, algorithm="giac")

[Out]

4*(x*e^x - 2*e^x)/(x^3 - 8*x^2 + 16*x)

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maple [A]  time = 0.05, size = 16, normalized size = 0.89

method	result	size
risch	\(\frac {4 \,{\mathrm e}^{x} \left (x -2\right )}{\left (x -4\right )^{2} x}\)	\(16\)
norman	\(\frac {4 \,{\mathrm e}^{x} x -8 \,{\mathrm e}^{x}}{x \left (x -4\right )^{2}}\)	\(20\)
gosper	\(\frac {4 \left (x -2\right ) {\mathrm e}^{x}}{x \left (x^{2}-8 x +16\right )}\)	\(21\)
default	\(\frac {{\mathrm e}^{x}}{2 x -8}+\frac {2 \,{\mathrm e}^{x}}{\left (x -4\right )^{2}}-\frac {{\mathrm e}^{x}}{2 x}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^3-32*x^2+56*x-32)*exp(x)/(x^5-12*x^4+48*x^3-64*x^2),x,method=_RETURNVERBOSE)

[Out]

4/(x-4)^2*exp(x)*(x-2)/x

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maxima [A]  time = 0.39, size = 21, normalized size = 1.17 \begin {gather*} \frac {4 \, {\left (x - 2\right )} e^{x}}{x^{3} - 8 \, x^{2} + 16 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^3-32*x^2+56*x-32)*exp(x)/(x^5-12*x^4+48*x^3-64*x^2),x, algorithm="maxima")

[Out]

4*(x - 2)*e^x/(x^3 - 8*x^2 + 16*x)

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mupad [B]  time = 3.37, size = 15, normalized size = 0.83 \begin {gather*} \frac {4\,{\mathrm {e}}^x\,\left (x-2\right )}{x\,{\left (x-4\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*(56*x - 32*x^2 + 4*x^3 - 32))/(64*x^2 - 48*x^3 + 12*x^4 - x^5),x)

[Out]

(4*exp(x)*(x - 2))/(x*(x - 4)^2)

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sympy [A]  time = 0.11, size = 19, normalized size = 1.06 \begin {gather*} \frac {\left (4 x - 8\right ) e^{x}}{x^{3} - 8 x^{2} + 16 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x**3-32*x**2+56*x-32)*exp(x)/(x**5-12*x**4+48*x**3-64*x**2),x)

[Out]

(4*x - 8)*exp(x)/(x**3 - 8*x**2 + 16*x)

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