3.49.97 \(\int \frac {e^{\frac {1}{5} (-4-x)} (-180 x^2+42 x^3-3 x^4)}{160-80 x+10 x^2} \, dx\)

Optimal. Leaf size=23 \[ \frac {3 e^{\frac {1}{5} (-4-x)} x^3}{2 (-4+x)} \]

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Rubi [B]  time = 0.33, antiderivative size = 66, normalized size of antiderivative = 2.87, number of steps used = 14, number of rules used = 8, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {27, 12, 1594, 2199, 2194, 2177, 2178, 2176} \begin {gather*} \frac {3}{2} e^{\frac {1}{5} (-x-4)} x^2+6 e^{\frac {1}{5} (-x-4)} x+24 e^{\frac {1}{5} (-x-4)}-\frac {96 e^{\frac {1}{5} (-x-4)}}{4-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((-4 - x)/5)*(-180*x^2 + 42*x^3 - 3*x^4))/(160 - 80*x + 10*x^2),x]

[Out]

24*E^((-4 - x)/5) - (96*E^((-4 - x)/5))/(4 - x) + 6*E^((-4 - x)/5)*x + (3*E^((-4 - x)/5)*x^2)/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 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 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), 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] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {1}{5} (-4-x)} \left (-180 x^2+42 x^3-3 x^4\right )}{10 (-4+x)^2} \, dx\\ &=\frac {1}{10} \int \frac {e^{\frac {1}{5} (-4-x)} \left (-180 x^2+42 x^3-3 x^4\right )}{(-4+x)^2} \, dx\\ &=\frac {1}{10} \int \frac {e^{\frac {1}{5} (-4-x)} x^2 \left (-180+42 x-3 x^2\right )}{(-4+x)^2} \, dx\\ &=\frac {1}{10} \int \left (12 e^{\frac {1}{5} (-4-x)}-\frac {960 e^{\frac {1}{5} (-4-x)}}{(-4+x)^2}-\frac {192 e^{\frac {1}{5} (-4-x)}}{-4+x}+18 e^{\frac {1}{5} (-4-x)} x-3 e^{\frac {1}{5} (-4-x)} x^2\right ) \, dx\\ &=-\left (\frac {3}{10} \int e^{\frac {1}{5} (-4-x)} x^2 \, dx\right )+\frac {6}{5} \int e^{\frac {1}{5} (-4-x)} \, dx+\frac {9}{5} \int e^{\frac {1}{5} (-4-x)} x \, dx-\frac {96}{5} \int \frac {e^{\frac {1}{5} (-4-x)}}{-4+x} \, dx-96 \int \frac {e^{\frac {1}{5} (-4-x)}}{(-4+x)^2} \, dx\\ &=-6 e^{\frac {1}{5} (-4-x)}-\frac {96 e^{\frac {1}{5} (-4-x)}}{4-x}-9 e^{\frac {1}{5} (-4-x)} x+\frac {3}{2} e^{\frac {1}{5} (-4-x)} x^2-\frac {96 \text {Ei}\left (\frac {4-x}{5}\right )}{5 e^{8/5}}-3 \int e^{\frac {1}{5} (-4-x)} x \, dx+9 \int e^{\frac {1}{5} (-4-x)} \, dx+\frac {96}{5} \int \frac {e^{\frac {1}{5} (-4-x)}}{-4+x} \, dx\\ &=-51 e^{\frac {1}{5} (-4-x)}-\frac {96 e^{\frac {1}{5} (-4-x)}}{4-x}+6 e^{\frac {1}{5} (-4-x)} x+\frac {3}{2} e^{\frac {1}{5} (-4-x)} x^2-15 \int e^{\frac {1}{5} (-4-x)} \, dx\\ &=24 e^{\frac {1}{5} (-4-x)}-\frac {96 e^{\frac {1}{5} (-4-x)}}{4-x}+6 e^{\frac {1}{5} (-4-x)} x+\frac {3}{2} e^{\frac {1}{5} (-4-x)} x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 23, normalized size = 1.00 \begin {gather*} \frac {3 e^{-\frac {4}{5}-\frac {x}{5}} x^3}{2 (-4+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-4 - x)/5)*(-180*x^2 + 42*x^3 - 3*x^4))/(160 - 80*x + 10*x^2),x]

[Out]

(3*E^(-4/5 - x/5)*x^3)/(2*(-4 + x))

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fricas [A]  time = 0.63, size = 16, normalized size = 0.70 \begin {gather*} \frac {3 \, x^{3} e^{\left (-\frac {1}{5} \, x - \frac {4}{5}\right )}}{2 \, {\left (x - 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x^4+42*x^3-180*x^2)/(10*x^2-80*x+160)/exp(4/5+1/5*x),x, algorithm="fricas")

[Out]

3/2*x^3*e^(-1/5*x - 4/5)/(x - 4)

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giac [A]  time = 0.17, size = 16, normalized size = 0.70 \begin {gather*} \frac {3 \, x^{3} e^{\left (-\frac {1}{5} \, x - \frac {4}{5}\right )}}{2 \, {\left (x - 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x^4+42*x^3-180*x^2)/(10*x^2-80*x+160)/exp(4/5+1/5*x),x, algorithm="giac")

[Out]

3/2*x^3*e^(-1/5*x - 4/5)/(x - 4)

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maple [A]  time = 0.20, size = 17, normalized size = 0.74

method	result	size
risch	\(\frac {3 x^{3} {\mathrm e}^{-\frac {4}{5}-\frac {x}{5}}}{2 \left (x -4\right )}\)	\(17\)
gosper	\(\frac {3 x^{3} {\mathrm e}^{-\frac {4}{5}-\frac {x}{5}}}{2 \left (x -4\right )}\)	\(19\)
norman	\(\frac {3 x^{3} {\mathrm e}^{-\frac {4}{5}-\frac {x}{5}}}{2 \left (x -4\right )}\)	\(19\)
derivativedivides	\(\frac {96 \,{\mathrm e}^{-\frac {4}{5}-\frac {x}{5}}}{x -4}+486 \,{\mathrm e}^{-\frac {4}{5}-\frac {x}{5}}-45 \left (x +25\right ) {\mathrm e}^{-\frac {4}{5}-\frac {x}{5}}+\frac {3 \left (25 \left (\frac {4}{5}+\frac {x}{5}\right )^{2}+426+26 x \right ) {\mathrm e}^{-\frac {4}{5}-\frac {x}{5}}}{2}\)	\(56\)
default	\(\frac {96 \,{\mathrm e}^{-\frac {4}{5}-\frac {x}{5}}}{x -4}+486 \,{\mathrm e}^{-\frac {4}{5}-\frac {x}{5}}-45 \left (x +25\right ) {\mathrm e}^{-\frac {4}{5}-\frac {x}{5}}+\frac {3 \left (25 \left (\frac {4}{5}+\frac {x}{5}\right )^{2}+426+26 x \right ) {\mathrm e}^{-\frac {4}{5}-\frac {x}{5}}}{2}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-3*x^4+42*x^3-180*x^2)/(10*x^2-80*x+160)/exp(4/5+1/5*x),x,method=_RETURNVERBOSE)

[Out]

3/2*x^3*exp(-4/5-1/5*x)/(x-4)

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maxima [A]  time = 0.42, size = 22, normalized size = 0.96 \begin {gather*} \frac {3 \, x^{3} e^{\left (-\frac {1}{5} \, x + \frac {1}{5}\right )}}{2 \, {\left (x e - 4 \, e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x^4+42*x^3-180*x^2)/(10*x^2-80*x+160)/exp(4/5+1/5*x),x, algorithm="maxima")

[Out]

3/2*x^3*e^(-1/5*x + 1/5)/(x*e - 4*e)

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mupad [B]  time = 3.21, size = 18, normalized size = 0.78 \begin {gather*} \frac {3\,x^3\,{\mathrm {e}}^{-\frac {x}{5}-\frac {4}{5}}}{2\,\left (x-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(- x/5 - 4/5)*(180*x^2 - 42*x^3 + 3*x^4))/(10*x^2 - 80*x + 160),x)

[Out]

(3*x^3*exp(- x/5 - 4/5))/(2*(x - 4))

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sympy [A]  time = 0.13, size = 19, normalized size = 0.83 \begin {gather*} \frac {3 x^{3} e^{- \frac {x}{5} - \frac {4}{5}}}{2 x - 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x**4+42*x**3-180*x**2)/(10*x**2-80*x+160)/exp(4/5+1/5*x),x)

[Out]

3*x**3*exp(-x/5 - 4/5)/(2*x - 8)

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