3.28.10 \(\int \frac {-2 e^2 (\frac {1}{x})^{2 e^2}-2 e^x x+2 e^{2 x} x}{x} \, dx\)

Optimal. Leaf size=20 \[ -4+\left (1-e^x\right )^2+\left (\frac {1}{x}\right )^{2 e^2} \]

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Rubi [A]  time = 0.03, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {14, 2246, 15, 30} \begin {gather*} \left (\frac {1}{x}\right )^{2 e^2}-2 e^x+e^{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*E^2*(x^(-1))^(2*E^2) - 2*E^x*x + 2*E^(2*x)*x)/x,x]

[Out]

-2*E^x + E^(2*x) + (x^(-1))^(2*E^2)

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 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2246

Int[((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)*((a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.))^(p_.),
x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int[(a + b*x)^p, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b,
c, d, e, n, p}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (2 e^x \left (-1+e^x\right )-2 e^2 \left (\frac {1}{x}\right )^{1+2 e^2}\right ) \, dx\\ &=2 \int e^x \left (-1+e^x\right ) \, dx-\left (2 e^2\right ) \int \left (\frac {1}{x}\right )^{1+2 e^2} \, dx\\ &=2 \operatorname {Subst}\left (\int (-1+x) \, dx,x,e^x\right )-\left (2 e^2 \left (\frac {1}{x}\right )^{2 e^2} x^{2 e^2}\right ) \int x^{-1-2 e^2} \, dx\\ &=-2 e^x+e^{2 x}+\left (\frac {1}{x}\right )^{2 e^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 20, normalized size = 1.00 \begin {gather*} -2 e^x+e^{2 x}+\left (\frac {1}{x}\right )^{2 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*E^2*(x^(-1))^(2*E^2) - 2*E^x*x + 2*E^(2*x)*x)/x,x]

[Out]

-2*E^x + E^(2*x) + (x^(-1))^(2*E^2)

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fricas [A]  time = 0.48, size = 17, normalized size = 0.85 \begin {gather*} \frac {1}{x}^{2 \, e^{2}} + e^{\left (2 \, x\right )} - 2 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1/x)^(2*e^2) + e^(2*x) - 2*e^x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {2 \, {\left (\frac {1}{x}^{2 \, e^{2}} e^{2} - x e^{\left (2 \, x\right )} + x e^{x}\right )}}{x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(-2*((1/x)^(2*e^2)*e^2 - x*e^(2*x) + x*e^x)/x, x)

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maple [A]  time = 0.07, size = 16, normalized size = 0.80

method	result	size
risch	\({\mathrm e}^{2 x}-2 \,{\mathrm e}^{x}+x^{-2 \,{\mathrm e}^{2}}\)	\(16\)
derivativedivides	\({\mathrm e}^{2 x}+{\mathrm e}^{2 \,{\mathrm e}^{2} \ln \left (\frac {1}{x}\right )}-2 \,{\mathrm e}^{x}\)	\(19\)
default	\({\mathrm e}^{2 x}+{\mathrm e}^{2 \,{\mathrm e}^{2} \ln \left (\frac {1}{x}\right )}-2 \,{\mathrm e}^{x}\)	\(19\)
norman	\({\mathrm e}^{2 x}+{\mathrm e}^{2 \,{\mathrm e}^{2} \ln \left (\frac {1}{x}\right )}-2 \,{\mathrm e}^{x}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(2*x)-2*exp(x)+x^(-2*exp(2))

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maxima [A]  time = 0.36, size = 17, normalized size = 0.85 \begin {gather*} \frac {1}{x^{2 \, e^{2}}} + e^{\left (2 \, x\right )} - 2 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/x^(2*e^2) + e^(2*x) - 2*e^x

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mupad [B]  time = 1.52, size = 17, normalized size = 0.85 \begin {gather*} {\mathrm {e}}^{2\,x}-2\,{\mathrm {e}}^x+{\left (\frac {1}{x}\right )}^{2\,{\mathrm {e}}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x*exp(x) - 2*x*exp(2*x) + 2*exp(2)*(1/x)^(2*exp(2)))/x,x)

[Out]

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

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sympy [A]  time = 0.80, size = 17, normalized size = 0.85 \begin {gather*} \left (\frac {1}{x}\right )^{2 e^{2}} + e^{2 x} - 2 e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(2)*exp(2*exp(2)*ln(1/x))+2*x*exp(x)**2-2*exp(x)*x)/x,x)

[Out]

(1/x)**(2*exp(2)) + exp(2*x) - 2*exp(x)

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