3.98.54 \(\int \frac {-e^{-6+x} x^2+2 e^x x^2+5 \log (2)}{x^2} \, dx\)

Optimal. Leaf size=20 \[ -e^{-6+x}+2 e^x-\frac {5 \log (2)}{x} \]

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Rubi [A]  time = 0.02, antiderivative size = 22, normalized size of antiderivative = 1.10, number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {14, 2194} \begin {gather*} \left (1-2 e^6\right ) \left (-e^{x-6}\right )-\frac {\log (32)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-(E^(-6 + x)*x^2) + 2*E^x*x^2 + 5*Log[2])/x^2,x]

[Out]

-(E^(-6 + x)*(1 - 2*E^6)) - Log[32]/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 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} &=\int \left (e^{-6+x} \left (-1+2 e^6\right )+\frac {\log (32)}{x^2}\right ) \, dx\\ &=-\frac {\log (32)}{x}+\left (-1+2 e^6\right ) \int e^{-6+x} \, dx\\ &=-e^{-6+x} \left (1-2 e^6\right )-\frac {\log (32)}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 20, normalized size = 1.00 \begin {gather*} -e^{-6+x}+2 e^x-\frac {\log (32)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-(E^(-6 + x)*x^2) + 2*E^x*x^2 + 5*Log[2])/x^2,x]

[Out]

-E^(-6 + x) + 2*E^x - Log[32]/x

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fricas [A]  time = 0.95, size = 25, normalized size = 1.25 \begin {gather*} \frac {{\left ({\left (2 \, x e^{6} - x\right )} e^{x} - 5 \, e^{6} \log \relax (2)\right )} e^{\left (-6\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(x)*x^2-x^2*exp(x-6)+5*log(2))/x^2,x, algorithm="fricas")

[Out]

((2*x*e^6 - x)*e^x - 5*e^6*log(2))*e^(-6)/x

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giac [A]  time = 0.21, size = 25, normalized size = 1.25 \begin {gather*} \frac {{\left (2 \, x e^{\left (x + 6\right )} - x e^{x} - 5 \, e^{6} \log \relax (2)\right )} e^{\left (-6\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(x)*x^2-x^2*exp(x-6)+5*log(2))/x^2,x, algorithm="giac")

[Out]

(2*x*e^(x + 6) - x*e^x - 5*e^6*log(2))*e^(-6)/x

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maple [A]  time = 0.04, size = 19, normalized size = 0.95

method	result	size
default	\(-{\mathrm e}^{x} {\mathrm e}^{-6}-\frac {5 \ln \relax (2)}{x}+2 \,{\mathrm e}^{x}\)	\(19\)
risch	\(-{\mathrm e}^{x} {\mathrm e}^{-6}-\frac {5 \ln \relax (2)}{x}+2 \,{\mathrm e}^{x}\)	\(19\)
norman	\(\frac {\left (2 \,{\mathrm e}^{6}-1\right ) {\mathrm e}^{-6} x \,{\mathrm e}^{x}-5 \ln \relax (2)}{x}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*exp(x)*x^2-x^2*exp(x-6)+5*ln(2))/x^2,x,method=_RETURNVERBOSE)

[Out]

-exp(x)*exp(-6)-5*ln(2)/x+2*exp(x)

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maxima [A]  time = 0.39, size = 18, normalized size = 0.90 \begin {gather*} -\frac {5 \, \log \relax (2)}{x} - e^{\left (x - 6\right )} + 2 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(x)*x^2-x^2*exp(x-6)+5*log(2))/x^2,x, algorithm="maxima")

[Out]

-5*log(2)/x - e^(x - 6) + 2*e^x

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mupad [B]  time = 5.92, size = 16, normalized size = 0.80 \begin {gather*} -{\mathrm {e}}^x\,\left ({\mathrm {e}}^{-6}-2\right )-\frac {\ln \left (32\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*log(2) + 2*x^2*exp(x) - x^2*exp(x - 6))/x^2,x)

[Out]

- exp(x)*(exp(-6) - 2) - log(32)/x

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sympy [A]  time = 0.13, size = 19, normalized size = 0.95 \begin {gather*} \frac {\left (-1 + 2 e^{6}\right ) e^{x}}{e^{6}} - \frac {5 \log {\relax (2 )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(x)*x**2-x**2*exp(x-6)+5*ln(2))/x**2,x)

[Out]

(-1 + 2*exp(6))*exp(-6)*exp(x) - 5*log(2)/x

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