3.63.70 \(\int \frac {e^{-2 x} (e^5 (3 x^3-2 x^4)+e^{2 x} (2+e^5 (-x-2 x^2)))}{2 x} \, dx\)

Optimal. Leaf size=30 \[ -\frac {1}{2} e^5 \left (x+e^{-2 x} \left (e^{2 x}-x\right ) x^2\right )+\log (x) \]

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Rubi [A]  time = 0.25, antiderivative size = 31, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 5, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.102, Rules used = {12, 6688, 2196, 2176, 2194} \begin {gather*} \frac {1}{2} e^{5-2 x} x^3-\frac {1}{8} e^5 (2 x+1)^2+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(5 - 2*x)*x^3)/2 - (E^5*(1 + 2*x)^2)/8 + Log[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 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 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} &=\frac {1}{2} \int \frac {e^{-2 x} \left (e^5 \left (3 x^3-2 x^4\right )+e^{2 x} \left (2+e^5 \left (-x-2 x^2\right )\right )\right )}{x} \, dx\\ &=\frac {1}{2} \int \left (\frac {2}{x}+e^{5-2 x} (3-2 x) x^2-e^5 (1+2 x)\right ) \, dx\\ &=-\frac {1}{8} e^5 (1+2 x)^2+\log (x)+\frac {1}{2} \int e^{5-2 x} (3-2 x) x^2 \, dx\\ &=-\frac {1}{8} e^5 (1+2 x)^2+\log (x)+\frac {1}{2} \int \left (3 e^{5-2 x} x^2-2 e^{5-2 x} x^3\right ) \, dx\\ &=-\frac {1}{8} e^5 (1+2 x)^2+\log (x)+\frac {3}{2} \int e^{5-2 x} x^2 \, dx-\int e^{5-2 x} x^3 \, dx\\ &=-\frac {3}{4} e^{5-2 x} x^2+\frac {1}{2} e^{5-2 x} x^3-\frac {1}{8} e^5 (1+2 x)^2+\log (x)+\frac {3}{2} \int e^{5-2 x} x \, dx-\frac {3}{2} \int e^{5-2 x} x^2 \, dx\\ &=-\frac {3}{4} e^{5-2 x} x+\frac {1}{2} e^{5-2 x} x^3-\frac {1}{8} e^5 (1+2 x)^2+\log (x)+\frac {3}{4} \int e^{5-2 x} \, dx-\frac {3}{2} \int e^{5-2 x} x \, dx\\ &=-\frac {3}{8} e^{5-2 x}+\frac {1}{2} e^{5-2 x} x^3-\frac {1}{8} e^5 (1+2 x)^2+\log (x)-\frac {3}{4} \int e^{5-2 x} \, dx\\ &=\frac {1}{2} e^{5-2 x} x^3-\frac {1}{8} e^5 (1+2 x)^2+\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 35, normalized size = 1.17 \begin {gather*} -\frac {e^5 x}{2}-\frac {e^5 x^2}{2}+\frac {1}{2} e^{5-2 x} x^3+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(E^5*x) - (E^5*x^2)/2 + (E^(5 - 2*x)*x^3)/2 + Log[x]

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fricas [A]  time = 0.75, size = 34, normalized size = 1.13 \begin {gather*} \frac {1}{2} \, {\left (x^{3} e^{5} - {\left (x^{2} + x\right )} e^{\left (2 \, x + 5\right )} + 2 \, e^{\left (2 \, x\right )} \log \relax (x)\right )} e^{\left (-2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.13, size = 26, normalized size = 0.87 \begin {gather*} \frac {1}{2} \, x^{3} e^{\left (-2 \, x + 5\right )} - \frac {1}{2} \, x^{2} e^{5} - \frac {1}{2} \, x e^{5} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 27, normalized size = 0.90

method	result	size
risch	\(-\frac {x^{2} {\mathrm e}^{5}}{2}-\frac {x \,{\mathrm e}^{5}}{2}+\ln \relax (x )+\frac {x^{3} {\mathrm e}^{-2 x +5}}{2}\)	\(27\)
norman	\(\left (\frac {x^{3} {\mathrm e}^{5}}{2}-\frac {x \,{\mathrm e}^{5} {\mathrm e}^{2 x}}{2}-\frac {{\mathrm e}^{5} {\mathrm e}^{2 x} x^{2}}{2}\right ) {\mathrm e}^{-2 x}+\ln \relax (x )\)	\(37\)
default	\(\ln \relax (x )-\frac {x^{2} {\mathrm e}^{5}}{2}+\frac {3 \,{\mathrm e}^{5} \left (-\frac {x^{2} {\mathrm e}^{-2 x}}{2}-\frac {x \,{\mathrm e}^{-2 x}}{2}-\frac {{\mathrm e}^{-2 x}}{4}\right )}{2}-{\mathrm e}^{5} \left (-\frac {x^{3} {\mathrm e}^{-2 x}}{2}-\frac {3 x^{2} {\mathrm e}^{-2 x}}{4}-\frac {3 x \,{\mathrm e}^{-2 x}}{4}-\frac {3 \,{\mathrm e}^{-2 x}}{8}\right )-\frac {x \,{\mathrm e}^{5}}{2}\)	\(79\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(((-2*x^2-x)*exp(5)+2)*exp(x)^2+(-2*x^4+3*x^3)*exp(5))/x/exp(x)^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.37, size = 66, normalized size = 2.20 \begin {gather*} -\frac {1}{2} \, x^{2} e^{5} - \frac {1}{2} \, x e^{5} + \frac {1}{8} \, {\left (4 \, x^{3} e^{5} + 6 \, x^{2} e^{5} + 6 \, x e^{5} + 3 \, e^{5}\right )} e^{\left (-2 \, x\right )} - \frac {3}{8} \, {\left (2 \, x^{2} e^{5} + 2 \, x e^{5} + e^{5}\right )} e^{\left (-2 \, x\right )} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x^2*e^5 - 1/2*x*e^5 + 1/8*(4*x^3*e^5 + 6*x^2*e^5 + 6*x*e^5 + 3*e^5)*e^(-2*x) - 3/8*(2*x^2*e^5 + 2*x*e^5 +
e^5)*e^(-2*x) + log(x)

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mupad [B]  time = 0.13, size = 26, normalized size = 0.87 \begin {gather*} \ln \relax (x)-\frac {x\,{\mathrm {e}}^5}{2}-\frac {x^2\,{\mathrm {e}}^5}{2}+\frac {x^3\,{\mathrm {e}}^{5-2\,x}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.16, size = 31, normalized size = 1.03 \begin {gather*} \frac {x^{3} e^{5} e^{- 2 x}}{2} - \frac {x^{2} e^{5}}{2} - \frac {x e^{5}}{2} + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(((-2*x**2-x)*exp(5)+2)*exp(x)**2+(-2*x**4+3*x**3)*exp(5))/x/exp(x)**2,x)

[Out]

x**3*exp(5)*exp(-2*x)/2 - x**2*exp(5)/2 - x*exp(5)/2 + log(x)

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