3.39.97 \(\int \frac {e^{-2+x} (-32+32 x+108 x^2+20 x^3)}{5 x^2} \, dx\)

Optimal. Leaf size=20 \[ \frac {2 e^{-2+x} (4+x) \left (\frac {4}{5}+2 x\right )}{x} \]

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Rubi [A]  time = 0.10, antiderivative size = 30, normalized size of antiderivative = 1.50, number of steps used = 9, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {12, 2199, 2194, 2177, 2178, 2176} \begin {gather*} 4 e^{x-2} x+\frac {88 e^{x-2}}{5}+\frac {32 e^{x-2}}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-2 + x)*(-32 + 32*x + 108*x^2 + 20*x^3))/(5*x^2),x]

[Out]

(88*E^(-2 + x))/5 + (32*E^(-2 + x))/(5*x) + 4*E^(-2 + x)*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 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} &=\frac {1}{5} \int \frac {e^{-2+x} \left (-32+32 x+108 x^2+20 x^3\right )}{x^2} \, dx\\ &=\frac {1}{5} \int \left (108 e^{-2+x}-\frac {32 e^{-2+x}}{x^2}+\frac {32 e^{-2+x}}{x}+20 e^{-2+x} x\right ) \, dx\\ &=4 \int e^{-2+x} x \, dx-\frac {32}{5} \int \frac {e^{-2+x}}{x^2} \, dx+\frac {32}{5} \int \frac {e^{-2+x}}{x} \, dx+\frac {108}{5} \int e^{-2+x} \, dx\\ &=\frac {108 e^{-2+x}}{5}+\frac {32 e^{-2+x}}{5 x}+4 e^{-2+x} x+\frac {32 \text {Ei}(x)}{5 e^2}-4 \int e^{-2+x} \, dx-\frac {32}{5} \int \frac {e^{-2+x}}{x} \, dx\\ &=\frac {88 e^{-2+x}}{5}+\frac {32 e^{-2+x}}{5 x}+4 e^{-2+x} x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 22, normalized size = 1.10 \begin {gather*} \frac {4 e^{-2+x} \left (8+22 x+5 x^2\right )}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-2 + x)*(-32 + 32*x + 108*x^2 + 20*x^3))/(5*x^2),x]

[Out]

(4*E^(-2 + x)*(8 + 22*x + 5*x^2))/(5*x)

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fricas [A]  time = 0.53, size = 19, normalized size = 0.95 \begin {gather*} \frac {4 \, {\left (5 \, x^{2} + 22 \, x + 8\right )} e^{\left (x - 2\right )}}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(20*x^3+108*x^2+32*x-32)*exp(x-2)/x^2,x, algorithm="fricas")

[Out]

4/5*(5*x^2 + 22*x + 8)*e^(x - 2)/x

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giac [A]  time = 0.13, size = 24, normalized size = 1.20 \begin {gather*} \frac {4 \, {\left (5 \, x^{2} e^{x} + 22 \, x e^{x} + 8 \, e^{x}\right )} e^{\left (-2\right )}}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(20*x^3+108*x^2+32*x-32)*exp(x-2)/x^2,x, algorithm="giac")

[Out]

4/5*(5*x^2*e^x + 22*x*e^x + 8*e^x)*e^(-2)/x

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maple [A]  time = 0.06, size = 20, normalized size = 1.00

method	result	size
gosper	\(\frac {4 \,{\mathrm e}^{x -2} \left (5 x^{2}+22 x +8\right )}{5 x}\)	\(20\)
risch	\(\frac {4 \,{\mathrm e}^{x -2} \left (5 x^{2}+22 x +8\right )}{5 x}\)	\(20\)
derivativedivides	\(\frac {32 \,{\mathrm e}^{x -2}}{5 x}+\frac {128 \,{\mathrm e}^{x -2}}{5}+4 \,{\mathrm e}^{x -2} \left (x -2\right )\)	\(26\)
default	\(\frac {32 \,{\mathrm e}^{x -2}}{5 x}+\frac {128 \,{\mathrm e}^{x -2}}{5}+4 \,{\mathrm e}^{x -2} \left (x -2\right )\)	\(26\)
norman	\(\frac {\frac {88 x \,{\mathrm e}^{x -2}}{5}+4 x^{2} {\mathrm e}^{x -2}+\frac {32 \,{\mathrm e}^{x -2}}{5}}{x}\)	\(28\)
meijerg	\(\frac {32 \,{\mathrm e}^{-2} \left (\frac {1}{x}+1-\ln \relax (x )-i \pi -\frac {2 x +2}{2 x}+\frac {{\mathrm e}^{x}}{x}+\ln \left (-x \right )+\expIntegralEi \left (1, -x \right )\right )}{5}+4 \,{\mathrm e}^{-2} \left (1-\frac {\left (-2 x +2\right ) {\mathrm e}^{x}}{2}\right )-\frac {108 \,{\mathrm e}^{-2} \left (1-{\mathrm e}^{x}\right )}{5}+\frac {32 \,{\mathrm e}^{-2} \left (\ln \relax (x )+i \pi -\ln \left (-x \right )-\expIntegralEi \left (1, -x \right )\right )}{5}\)	\(93\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(20*x^3+108*x^2+32*x-32)*exp(x-2)/x^2,x,method=_RETURNVERBOSE)

[Out]

4/5*exp(x-2)*(5*x^2+22*x+8)/x

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maxima [C]  time = 0.38, size = 31, normalized size = 1.55 \begin {gather*} \frac {32}{5} \, {\rm Ei}\relax (x) e^{\left (-2\right )} + 4 \, {\left (x - 1\right )} e^{\left (x - 2\right )} - \frac {32}{5} \, e^{\left (-2\right )} \Gamma \left (-1, -x\right ) + \frac {108}{5} \, e^{\left (x - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(20*x^3+108*x^2+32*x-32)*exp(x-2)/x^2,x, algorithm="maxima")

[Out]

32/5*Ei(x)*e^(-2) + 4*(x - 1)*e^(x - 2) - 32/5*e^(-2)*gamma(-1, -x) + 108/5*e^(x - 2)

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mupad [B]  time = 0.06, size = 19, normalized size = 0.95 \begin {gather*} \frac {4\,{\mathrm {e}}^{x-2}\,\left (5\,x^2+22\,x+8\right )}{5\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x - 2)*(32*x + 108*x^2 + 20*x^3 - 32))/(5*x^2),x)

[Out]

(4*exp(x - 2)*(22*x + 5*x^2 + 8))/(5*x)

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sympy [A]  time = 0.10, size = 17, normalized size = 0.85 \begin {gather*} \frac {\left (20 x^{2} + 88 x + 32\right ) e^{x - 2}}{5 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(20*x**3+108*x**2+32*x-32)*exp(x-2)/x**2,x)

[Out]

(20*x**2 + 88*x + 32)*exp(x - 2)/(5*x)

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