3.37.57 \(\int \frac {e^{e^{\frac {e^{x^2}}{x}}} (10 x^2+2 x^3+e^{\frac {e^{x^2}}{x}+x^2} (-25-10 x+49 x^2+20 x^3+2 x^4))}{x^2} \, dx\)

Optimal. Leaf size=21 \[ -2+e^{e^{\frac {e^{x^2}}{x}}} (5+x)^2 \]

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Rubi [B]  time = 0.16, antiderivative size = 62, normalized size of antiderivative = 2.95, number of steps used = 1, number of rules used = 1, integrand size = 64, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.016, Rules used = {2288} \begin {gather*} -\frac {e^{x^2+e^{\frac {e^{x^2}}{x}}} \left (-2 x^4-20 x^3-49 x^2+10 x+25\right )}{\left (2 e^{x^2}-\frac {e^{x^2}}{x^2}\right ) x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^E^(E^x^2/x)*(10*x^2 + 2*x^3 + E^(E^x^2/x + x^2)*(-25 - 10*x + 49*x^2 + 20*x^3 + 2*x^4)))/x^2,x]

[Out]

-((E^(E^(E^x^2/x) + x^2)*(25 + 10*x - 49*x^2 - 20*x^3 - 2*x^4))/((2*E^x^2 - E^x^2/x^2)*x^2))

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^{e^{\frac {e^{x^2}}{x}}+x^2} \left (25+10 x-49 x^2-20 x^3-2 x^4\right )}{\left (2 e^{x^2}-\frac {e^{x^2}}{x^2}\right ) x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 19, normalized size = 0.90 \begin {gather*} e^{e^{\frac {e^{x^2}}{x}}} (5+x)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^E^(E^x^2/x)*(10*x^2 + 2*x^3 + E^(E^x^2/x + x^2)*(-25 - 10*x + 49*x^2 + 20*x^3 + 2*x^4)))/x^2,x]

[Out]

E^E^(E^x^2/x)*(5 + x)^2

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fricas [A]  time = 0.70, size = 29, normalized size = 1.38 \begin {gather*} {\left (x^{2} + 10 \, x + 25\right )} e^{\left (e^{\left (-x^{2} + \frac {x^{3} + e^{\left (x^{2}\right )}}{x}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^4+20*x^3+49*x^2-10*x-25)*exp(x^2)*exp(exp(x^2)/x)+2*x^3+10*x^2)*exp(exp(exp(x^2)/x))/x^2,x, al
gorithm="fricas")

[Out]

(x^2 + 10*x + 25)*e^(e^(-x^2 + (x^3 + e^(x^2))/x))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (2 \, x^{3} + 10 \, x^{2} + {\left (2 \, x^{4} + 20 \, x^{3} + 49 \, x^{2} - 10 \, x - 25\right )} e^{\left (x^{2} + \frac {e^{\left (x^{2}\right )}}{x}\right )}\right )} e^{\left (e^{\left (\frac {e^{\left (x^{2}\right )}}{x}\right )}\right )}}{x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^4+20*x^3+49*x^2-10*x-25)*exp(x^2)*exp(exp(x^2)/x)+2*x^3+10*x^2)*exp(exp(exp(x^2)/x))/x^2,x, al
gorithm="giac")

[Out]

integrate((2*x^3 + 10*x^2 + (2*x^4 + 20*x^3 + 49*x^2 - 10*x - 25)*e^(x^2 + e^(x^2)/x))*e^(e^(e^(x^2)/x))/x^2,
x)

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maple [A]  time = 0.05, size = 20, normalized size = 0.95

method	result	size
risch	\(\left (x^{2}+10 x +25\right ) {\mathrm e}^{{\mathrm e}^{\frac {{\mathrm e}^{x^{2}}}{x}}}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^4+20*x^3+49*x^2-10*x-25)*exp(x^2)*exp(exp(x^2)/x)+2*x^3+10*x^2)*exp(exp(exp(x^2)/x))/x^2,x,method=_R
ETURNVERBOSE)

[Out]

(x^2+10*x+25)*exp(exp(exp(x^2)/x))

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maxima [A]  time = 0.62, size = 19, normalized size = 0.90 \begin {gather*} {\left (x^{2} + 10 \, x + 25\right )} e^{\left (e^{\left (\frac {e^{\left (x^{2}\right )}}{x}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^4+20*x^3+49*x^2-10*x-25)*exp(x^2)*exp(exp(x^2)/x)+2*x^3+10*x^2)*exp(exp(exp(x^2)/x))/x^2,x, al
gorithm="maxima")

[Out]

(x^2 + 10*x + 25)*e^(e^(e^(x^2)/x))

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mupad [B]  time = 2.19, size = 19, normalized size = 0.90 \begin {gather*} {\mathrm {e}}^{{\mathrm {e}}^{\frac {{\mathrm {e}}^{x^2}}{x}}}\,\left (x^2+10\,x+25\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(exp(x^2)/x))*(10*x^2 + 2*x^3 + exp(x^2)*exp(exp(x^2)/x)*(49*x^2 - 10*x + 20*x^3 + 2*x^4 - 25)))/x
^2,x)

[Out]

exp(exp(exp(x^2)/x))*(10*x + x^2 + 25)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**4+20*x**3+49*x**2-10*x-25)*exp(x**2)*exp(exp(x**2)/x)+2*x**3+10*x**2)*exp(exp(exp(x**2)/x))/x
**2,x)

[Out]

Timed out

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