3.3.28 \(\int \frac {e^{2 e^x-2 i \pi } (-4+8 e^x x)}{x^2 \log ^2(5)} \, dx\)

Optimal. Leaf size=24 \[ 6+\frac {4 e^{2 e^x-2 i \pi }}{x \log ^2(5)} \]

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Rubi [A]  time = 0.05, antiderivative size = 22, normalized size of antiderivative = 0.92, number of steps used = 2, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {12, 2288} \begin {gather*} \frac {4 e^{2 e^x-2 i \pi }}{x \log ^2(5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(2*E^x - (2*I)*Pi)*(-4 + 8*E^x*x))/(x^2*Log[5]^2),x]

[Out]

(4*E^(2*E^x - (2*I)*Pi))/(x*Log[5]^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {e^{2 e^x-2 i \pi } \left (-4+8 e^x x\right )}{x^2} \, dx}{\log ^2(5)}\\ &=\frac {4 e^{2 e^x-2 i \pi }}{x \log ^2(5)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 16, normalized size = 0.67 \begin {gather*} \frac {4 e^{2 e^x}}{x \log ^2(5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*E^x - (2*I)*Pi)*(-4 + 8*E^x*x))/(x^2*Log[5]^2),x]

[Out]

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

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fricas [A]  time = 1.28, size = 18, normalized size = 0.75 \begin {gather*} \frac {4 \, e^{\left (2 \, e^{x} - 2 \, \log \left (-\log \relax (5)\right )\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*exp(x)*x-4)*exp(-log(-log(5))+exp(x))^2/x^2,x, algorithm="fricas")

[Out]

4*e^(2*e^x - 2*log(-log(5)))/x

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giac [A]  time = 0.31, size = 14, normalized size = 0.58 \begin {gather*} \frac {4 \, e^{\left (2 \, e^{x}\right )}}{x \log \relax (5)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*exp(x)*x-4)*exp(-log(-log(5))+exp(x))^2/x^2,x, algorithm="giac")

[Out]

4*e^(2*e^x)/(x*log(5)^2)

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maple [A]  time = 0.06, size = 15, normalized size = 0.62

method	result	size
risch	\(\frac {4 \,{\mathrm e}^{2 \,{\mathrm e}^{x}}}{x \ln \relax (5)^{2}}\)	\(15\)
norman	\(\frac {4 \,{\mathrm e}^{2 \,{\mathrm e}^{x}}}{x \ln \relax (5)^{2}}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*exp(x)*x-4)*exp(-ln(-ln(5))+exp(x))^2/x^2,x,method=_RETURNVERBOSE)

[Out]

4/x/ln(5)^2*exp(2*exp(x))

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maxima [A]  time = 0.50, size = 14, normalized size = 0.58 \begin {gather*} \frac {4 \, e^{\left (2 \, e^{x}\right )}}{x \log \relax (5)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*exp(x)*x-4)*exp(-log(-log(5))+exp(x))^2/x^2,x, algorithm="maxima")

[Out]

4*e^(2*e^x)/(x*log(5)^2)

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mupad [B]  time = 0.32, size = 14, normalized size = 0.58 \begin {gather*} \frac {4\,{\mathrm {e}}^{2\,{\mathrm {e}}^x}}{x\,{\ln \relax (5)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*exp(x) - 2*log(-log(5)))*(8*x*exp(x) - 4))/x^2,x)

[Out]

(4*exp(2*exp(x)))/(x*log(5)^2)

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sympy [A]  time = 0.12, size = 14, normalized size = 0.58 \begin {gather*} \frac {4 e^{2 e^{x}}}{x \log {\relax (5 )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*exp(x)*x-4)*exp(-ln(-ln(5))+exp(x))**2/x**2,x)

[Out]

4*exp(2*exp(x))/(x*log(5)**2)

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