3.44.66 \(\int \frac {e^{\frac {1}{4} (-99+20 e^{e^x})} (-1+5 e^{e^x+x} x)}{x^2} \, dx\)

Optimal. Leaf size=21 \[ 25+\frac {e^{\frac {1}{4}+5 \left (-5+e^{e^x}\right )}}{x} \]

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Rubi [A]  time = 0.08, antiderivative size = 19, normalized size of antiderivative = 0.90, number of steps used = 1, number of rules used = 1, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {2288} \begin {gather*} \frac {e^{\frac {1}{4} \left (20 e^{e^x}-99\right )}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((-99 + 20*E^E^x)/4)*(-1 + 5*E^(E^x + x)*x))/x^2,x]

[Out]

E^((-99 + 20*E^E^x)/4)/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 {e^{\frac {1}{4} \left (-99+20 e^{e^x}\right )}}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 17, normalized size = 0.81 \begin {gather*} \frac {e^{-\frac {99}{4}+5 e^{e^x}}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-99 + 20*E^E^x)/4)*(-1 + 5*E^(E^x + x)*x))/x^2,x]

[Out]

E^(-99/4 + 5*E^E^x)/x

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fricas [A]  time = 0.60, size = 23, normalized size = 1.10 \begin {gather*} \frac {e^{\left (\frac {1}{4} \, {\left (20 \, e^{\left (x + e^{x}\right )} - 99 \, e^{x}\right )} e^{\left (-x\right )}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x*exp(x)*exp(exp(x))-1)*exp(5*exp(exp(x))-99/4)/x^2,x, algorithm="fricas")

[Out]

e^(1/4*(20*e^(x + e^x) - 99*e^x)*e^(-x))/x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (5 \, x e^{\left (x + e^{x}\right )} - 1\right )} e^{\left (5 \, e^{\left (e^{x}\right )} - \frac {99}{4}\right )}}{x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x*exp(x)*exp(exp(x))-1)*exp(5*exp(exp(x))-99/4)/x^2,x, algorithm="giac")

[Out]

integrate((5*x*e^(x + e^x) - 1)*e^(5*e^(e^x) - 99/4)/x^2, x)

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maple [A]  time = 0.04, size = 13, normalized size = 0.62

method	result	size
norman	\(\frac {{\mathrm e}^{5 \,{\mathrm e}^{{\mathrm e}^{x}}-\frac {99}{4}}}{x}\)	\(13\)
risch	\(\frac {{\mathrm e}^{5 \,{\mathrm e}^{{\mathrm e}^{x}}-\frac {99}{4}}}{x}\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x*exp(x)*exp(exp(x))-1)*exp(5*exp(exp(x))-99/4)/x^2,x,method=_RETURNVERBOSE)

[Out]

exp(5*exp(exp(x))-99/4)/x

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maxima [A]  time = 0.44, size = 12, normalized size = 0.57 \begin {gather*} \frac {e^{\left (5 \, e^{\left (e^{x}\right )} - \frac {99}{4}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x*exp(x)*exp(exp(x))-1)*exp(5*exp(exp(x))-99/4)/x^2,x, algorithm="maxima")

[Out]

e^(5*e^(e^x) - 99/4)/x

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mupad [B]  time = 3.26, size = 12, normalized size = 0.57 \begin {gather*} \frac {{\mathrm {e}}^{5\,{\mathrm {e}}^{{\mathrm {e}}^x}}\,{\mathrm {e}}^{-\frac {99}{4}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(5*exp(exp(x)) - 99/4)*(5*x*exp(exp(x))*exp(x) - 1))/x^2,x)

[Out]

(exp(5*exp(exp(x)))*exp(-99/4))/x

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sympy [A]  time = 0.15, size = 12, normalized size = 0.57 \begin {gather*} \frac {e^{5 e^{e^{x}} - \frac {99}{4}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x*exp(x)*exp(exp(x))-1)*exp(5*exp(exp(x))-99/4)/x**2,x)

[Out]

exp(5*exp(exp(x)) - 99/4)/x

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