3.60.78 \(\int \frac {5+e^{2-x} (2 x^2-2 x^3)-10 \log (\frac {221}{25})}{10 x^2} \, dx\)

Optimal. Leaf size=28 \[ \frac {1}{5} e^{2-x} x+\frac {-\frac {1}{2}-x+\log \left (\frac {221}{25}\right )}{x} \]

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Rubi [A]  time = 0.04, antiderivative size = 43, normalized size of antiderivative = 1.54, number of steps used = 5, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 14, 2176, 2194} \begin {gather*} -\frac {1}{5} e^{2-x} (1-x)+\frac {e^{2-x}}{5}-\frac {1-2 \log \left (\frac {221}{25}\right )}{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5 + E^(2 - x)*(2*x^2 - 2*x^3) - 10*Log[221/25])/(10*x^2),x]

[Out]

E^(2 - x)/5 - (E^(2 - x)*(1 - x))/5 - (1 - 2*Log[221/25])/(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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{10} \int \frac {5+e^{2-x} \left (2 x^2-2 x^3\right )-10 \log \left (\frac {221}{25}\right )}{x^2} \, dx\\ &=\frac {1}{10} \int \left (-2 e^{2-x} (-1+x)-\frac {5 \left (-1+2 \log \left (\frac {221}{25}\right )\right )}{x^2}\right ) \, dx\\ &=-\frac {1-2 \log \left (\frac {221}{25}\right )}{2 x}-\frac {1}{5} \int e^{2-x} (-1+x) \, dx\\ &=-\frac {1}{5} e^{2-x} (1-x)-\frac {1-2 \log \left (\frac {221}{25}\right )}{2 x}-\frac {1}{5} \int e^{2-x} \, dx\\ &=\frac {e^{2-x}}{5}-\frac {1}{5} e^{2-x} (1-x)-\frac {1-2 \log \left (\frac {221}{25}\right )}{2 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 27, normalized size = 0.96 \begin {gather*} \frac {-5+2 e^{2-x} x^2+10 \log \left (\frac {221}{25}\right )}{10 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 + E^(2 - x)*(2*x^2 - 2*x^3) - 10*Log[221/25])/(10*x^2),x]

[Out]

(-5 + 2*E^(2 - x)*x^2 + 10*Log[221/25])/(10*x)

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fricas [A]  time = 0.72, size = 22, normalized size = 0.79 \begin {gather*} \frac {2 \, x^{2} e^{\left (-x + 2\right )} + 10 \, \log \left (\frac {221}{25}\right ) - 5}{10 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((-2*x^3+2*x^2)*exp(2-x)-10*log(221/25)+5)/x^2,x, algorithm="fricas")

[Out]

1/10*(2*x^2*e^(-x + 2) + 10*log(221/25) - 5)/x

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giac [A]  time = 0.15, size = 22, normalized size = 0.79 \begin {gather*} \frac {2 \, x^{2} e^{\left (-x + 2\right )} + 10 \, \log \left (\frac {221}{25}\right ) - 5}{10 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((-2*x^3+2*x^2)*exp(2-x)-10*log(221/25)+5)/x^2,x, algorithm="giac")

[Out]

1/10*(2*x^2*e^(-x + 2) + 10*log(221/25) - 5)/x

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maple [A]  time = 0.10, size = 24, normalized size = 0.86

method	result	size
norman	\(\frac {\frac {x^{2} {\mathrm e}^{2-x}}{5}+\ln \left (221\right )-2 \ln \relax (5)-\frac {1}{2}}{x}\)	\(24\)
risch	\(-\frac {2 \ln \relax (5)}{x}+\frac {\ln \left (13\right )}{x}+\frac {\ln \left (17\right )}{x}-\frac {1}{2 x}+\frac {x \,{\mathrm e}^{2-x}}{5}\)	\(35\)
derivativedivides	\(-\frac {1}{2 x}-\frac {2 \ln \relax (5)}{x}+\frac {\ln \left (221\right )}{x}+\frac {2 \,{\mathrm e}^{2-x}}{5}-\frac {{\mathrm e}^{2-x} \left (2-x \right )}{5}\)	\(41\)
default	\(-\frac {1}{2 x}-\frac {2 \ln \relax (5)}{x}+\frac {\ln \left (221\right )}{x}+\frac {2 \,{\mathrm e}^{2-x}}{5}-\frac {{\mathrm e}^{2-x} \left (2-x \right )}{5}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/10*((-2*x^3+2*x^2)*exp(2-x)-10*ln(221/25)+5)/x^2,x,method=_RETURNVERBOSE)

[Out]

(1/5*x^2*exp(2-x)+ln(221)-2*ln(5)-1/2)/x

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maxima [A]  time = 0.35, size = 33, normalized size = 1.18 \begin {gather*} \frac {1}{5} \, {\left (x e^{2} + e^{2}\right )} e^{\left (-x\right )} + \frac {\log \left (\frac {221}{25}\right )}{x} - \frac {1}{2 \, x} - \frac {1}{5} \, e^{\left (-x + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((-2*x^3+2*x^2)*exp(2-x)-10*log(221/25)+5)/x^2,x, algorithm="maxima")

[Out]

1/5*(x*e^2 + e^2)*e^(-x) + log(221/25)/x - 1/2/x - 1/5*e^(-x + 2)

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mupad [B]  time = 0.08, size = 18, normalized size = 0.64 \begin {gather*} \frac {\ln \left (\frac {221}{25}\right )-\frac {1}{2}}{x}+\frac {x\,{\mathrm {e}}^{2-x}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(2 - x)*(2*x^2 - 2*x^3))/10 - log(221/25) + 1/2)/x^2,x)

[Out]

(log(221/25) - 1/2)/x + (x*exp(2 - x))/5

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sympy [A]  time = 0.13, size = 20, normalized size = 0.71 \begin {gather*} \frac {x e^{2 - x}}{5} - \frac {- \log {\left (221 \right )} + \frac {1}{2} + 2 \log {\relax (5 )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((-2*x**3+2*x**2)*exp(2-x)-10*ln(221/25)+5)/x**2,x)

[Out]

x*exp(2 - x)/5 - (-log(221) + 1/2 + 2*log(5))/x

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