3.50.43 \(\int \frac {e^{-2 x} (-7 e^{5+2 x}+e^5 (2 x^2-6 x^3+2 x^4)+e^{5+2 x} \log (x^2))}{x^2} \, dx\)

Optimal. Leaf size=31 \[ \frac {e^5 \left (5+2 x-e^{-2 x} (-2+x) x^2-\log \left (x^2\right )\right )}{x} \]

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Rubi [A]  time = 0.31, antiderivative size = 46, normalized size of antiderivative = 1.48, number of steps used = 13, number of rules used = 7, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.137, Rules used = {6688, 12, 14, 2196, 2194, 2176, 2304} \begin {gather*} -e^{5-2 x} x^2+\frac {e^5 \left (7-\log \left (x^2\right )\right )}{x}+2 e^{5-2 x} x-\frac {2 e^5}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-7*E^(5 + 2*x) + E^5*(2*x^2 - 6*x^3 + 2*x^4) + E^(5 + 2*x)*Log[x^2])/(E^(2*x)*x^2),x]

[Out]

(-2*E^5)/x + 2*E^(5 - 2*x)*x - E^(5 - 2*x)*x^2 + (E^5*(7 - Log[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 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]

Rule 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^5 \left (-7+2 e^{-2 x} x^2 \left (1-3 x+x^2\right )+\log \left (x^2\right )\right )}{x^2} \, dx\\ &=e^5 \int \frac {-7+2 e^{-2 x} x^2 \left (1-3 x+x^2\right )+\log \left (x^2\right )}{x^2} \, dx\\ &=e^5 \int \left (2 e^{-2 x} \left (1-3 x+x^2\right )+\frac {-7+\log \left (x^2\right )}{x^2}\right ) \, dx\\ &=e^5 \int \frac {-7+\log \left (x^2\right )}{x^2} \, dx+\left (2 e^5\right ) \int e^{-2 x} \left (1-3 x+x^2\right ) \, dx\\ &=-\frac {2 e^5}{x}+\frac {e^5 \left (7-\log \left (x^2\right )\right )}{x}+\left (2 e^5\right ) \int \left (e^{-2 x}-3 e^{-2 x} x+e^{-2 x} x^2\right ) \, dx\\ &=-\frac {2 e^5}{x}+\frac {e^5 \left (7-\log \left (x^2\right )\right )}{x}+\left (2 e^5\right ) \int e^{-2 x} \, dx+\left (2 e^5\right ) \int e^{-2 x} x^2 \, dx-\left (6 e^5\right ) \int e^{-2 x} x \, dx\\ &=-e^{5-2 x}-\frac {2 e^5}{x}+3 e^{5-2 x} x-e^{5-2 x} x^2+\frac {e^5 \left (7-\log \left (x^2\right )\right )}{x}+\left (2 e^5\right ) \int e^{-2 x} x \, dx-\left (3 e^5\right ) \int e^{-2 x} \, dx\\ &=\frac {1}{2} e^{5-2 x}-\frac {2 e^5}{x}+2 e^{5-2 x} x-e^{5-2 x} x^2+\frac {e^5 \left (7-\log \left (x^2\right )\right )}{x}+e^5 \int e^{-2 x} \, dx\\ &=-\frac {2 e^5}{x}+2 e^{5-2 x} x-e^{5-2 x} x^2+\frac {e^5 \left (7-\log \left (x^2\right )\right )}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.28, size = 26, normalized size = 0.84 \begin {gather*} -\frac {e^5 \left (-5+e^{-2 x} (-2+x) x^2+\log \left (x^2\right )\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-7*E^(5 + 2*x) + E^5*(2*x^2 - 6*x^3 + 2*x^4) + E^(5 + 2*x)*Log[x^2])/(E^(2*x)*x^2),x]

[Out]

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

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fricas [A]  time = 0.89, size = 43, normalized size = 1.39 \begin {gather*} -\frac {{\left ({\left (x^{3} - 2 \, x^{2}\right )} e^{10} + e^{\left (2 \, x + 10\right )} \log \left (x^{2}\right ) - 5 \, e^{\left (2 \, x + 10\right )}\right )} e^{\left (-2 \, x - 5\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(5)*exp(2*x)*log(x^2)-7*exp(5)*exp(2*x)+(2*x^4-6*x^3+2*x^2)*exp(5))/exp(2*x)/x^2,x, algorithm="f
ricas")

[Out]

-((x^3 - 2*x^2)*e^10 + e^(2*x + 10)*log(x^2) - 5*e^(2*x + 10))*e^(-2*x - 5)/x

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giac [A]  time = 0.18, size = 38, normalized size = 1.23 \begin {gather*} -\frac {x^{3} e^{\left (-2 \, x + 5\right )} - 2 \, x^{2} e^{\left (-2 \, x + 5\right )} + e^{5} \log \left (x^{2}\right ) - 5 \, e^{5}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(5)*exp(2*x)*log(x^2)-7*exp(5)*exp(2*x)+(2*x^4-6*x^3+2*x^2)*exp(5))/exp(2*x)/x^2,x, algorithm="g
iac")

[Out]

-(x^3*e^(-2*x + 5) - 2*x^2*e^(-2*x + 5) + e^5*log(x^2) - 5*e^5)/x

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maple [A]  time = 0.11, size = 44, normalized size = 1.42

method	result	size
default	\(-\frac {{\mathrm e}^{5} \ln \left (x^{2}\right )}{x}+\frac {5 \,{\mathrm e}^{5}}{x}+2 \,{\mathrm e}^{5} \left (-\frac {x^{2} {\mathrm e}^{-2 x}}{2}+x \,{\mathrm e}^{-2 x}\right )\)	\(44\)
norman	\(\frac {\left (2 x^{2} {\mathrm e}^{5}-x^{3} {\mathrm e}^{5}+5 \,{\mathrm e}^{5} {\mathrm e}^{2 x}-{\mathrm e}^{5} {\mathrm e}^{2 x} \ln \left (x^{2}\right )\right ) {\mathrm e}^{-2 x}}{x}\)	\(46\)
risch	\(-\frac {2 \,{\mathrm e}^{5} \ln \relax (x )}{x}-\frac {\left (-i {\mathrm e}^{2 x} \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+2 i {\mathrm e}^{2 x} \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-i {\mathrm e}^{2 x} \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+2 x^{3}-4 x^{2}-10 \,{\mathrm e}^{2 x}\right ) {\mathrm e}^{-2 x +5}}{2 x}\)	\(100\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(5)*exp(2*x)*ln(x^2)-7*exp(5)*exp(2*x)+(2*x^4-6*x^3+2*x^2)*exp(5))/exp(2*x)/x^2,x,method=_RETURNVERBOS
E)

[Out]

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

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maxima [B]  time = 0.35, size = 69, normalized size = 2.23 \begin {gather*} -{\left (\frac {\log \left (x^{2}\right )}{x} + \frac {2}{x}\right )} e^{5} - \frac {1}{2} \, {\left (2 \, x^{2} e^{5} + 2 \, x e^{5} + e^{5}\right )} e^{\left (-2 \, x\right )} + \frac {3}{2} \, {\left (2 \, x e^{5} + e^{5}\right )} e^{\left (-2 \, x\right )} + \frac {7 \, e^{5}}{x} - e^{\left (-2 \, x + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(5)*exp(2*x)*log(x^2)-7*exp(5)*exp(2*x)+(2*x^4-6*x^3+2*x^2)*exp(5))/exp(2*x)/x^2,x, algorithm="m
axima")

[Out]

-(log(x^2)/x + 2/x)*e^5 - 1/2*(2*x^2*e^5 + 2*x*e^5 + e^5)*e^(-2*x) + 3/2*(2*x*e^5 + e^5)*e^(-2*x) + 7*e^5/x -
e^(-2*x + 5)

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mupad [B]  time = 3.41, size = 39, normalized size = 1.26 \begin {gather*} 2\,x\,{\mathrm {e}}^{5-2\,x}+\frac {5\,{\mathrm {e}}^5}{x}-x^2\,{\mathrm {e}}^{5-2\,x}-\frac {\ln \left (x^2\right )\,{\mathrm {e}}^5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-2*x)*(exp(5)*(2*x^2 - 6*x^3 + 2*x^4) - 7*exp(2*x)*exp(5) + log(x^2)*exp(2*x)*exp(5)))/x^2,x)

[Out]

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

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sympy [A]  time = 0.37, size = 34, normalized size = 1.10 \begin {gather*} \left (- x^{2} e^{5} + 2 x e^{5}\right ) e^{- 2 x} - \frac {e^{5} \log {\left (x^{2} \right )}}{x} + \frac {5 e^{5}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(5)*exp(2*x)*ln(x**2)-7*exp(5)*exp(2*x)+(2*x**4-6*x**3+2*x**2)*exp(5))/exp(2*x)/x**2,x)

[Out]

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

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