3.99.78 \(\int \frac {e^{5+e^{x^2}} (-3+6 e^{x^2} x^2)}{x^2} \, dx\)

Optimal. Leaf size=22 \[ -1+\frac {3 e^{5+e^{x^2}}}{x}-\log ^2(2) \]

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Rubi [A]  time = 0.05, antiderivative size = 14, normalized size of antiderivative = 0.64, number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {2288} \begin {gather*} \frac {3 e^{e^{x^2}+5}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(5 + E^x^2)*(-3 + 6*E^x^2*x^2))/x^2,x]

[Out]

(3*E^(5 + E^x^2))/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 {3 e^{5+e^{x^2}}}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 14, normalized size = 0.64 \begin {gather*} \frac {3 e^{5+e^{x^2}}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(5 + E^x^2)*(-3 + 6*E^x^2*x^2))/x^2,x]

[Out]

(3*E^(5 + E^x^2))/x

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fricas [A]  time = 0.56, size = 12, normalized size = 0.55 \begin {gather*} \frac {3 \, e^{\left (e^{\left (x^{2}\right )} + 5\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x^2*exp(x^2)-3)*exp(5+exp(x^2))/x^2,x, algorithm="fricas")

[Out]

3*e^(e^(x^2) + 5)/x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {3 \, {\left (2 \, x^{2} e^{\left (x^{2}\right )} - 1\right )} e^{\left (e^{\left (x^{2}\right )} + 5\right )}}{x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x^2*exp(x^2)-3)*exp(5+exp(x^2))/x^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.04, size = 13, normalized size = 0.59

method	result	size
norman	\(\frac {3 \,{\mathrm e}^{5+{\mathrm e}^{x^{2}}}}{x}\)	\(13\)
risch	\(\frac {3 \,{\mathrm e}^{5+{\mathrm e}^{x^{2}}}}{x}\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*x^2*exp(x^2)-3)*exp(5+exp(x^2))/x^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.40, size = 12, normalized size = 0.55 \begin {gather*} \frac {3 \, e^{\left (e^{\left (x^{2}\right )} + 5\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x^2*exp(x^2)-3)*exp(5+exp(x^2))/x^2,x, algorithm="maxima")

[Out]

3*e^(e^(x^2) + 5)/x

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mupad [B]  time = 0.10, size = 12, normalized size = 0.55 \begin {gather*} \frac {3\,{\mathrm {e}}^5\,{\mathrm {e}}^{{\mathrm {e}}^{x^2}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(x^2) + 5)*(6*x^2*exp(x^2) - 3))/x^2,x)

[Out]

(3*exp(5)*exp(exp(x^2)))/x

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sympy [A]  time = 0.28, size = 10, normalized size = 0.45 \begin {gather*} \frac {3 e^{e^{x^{2}} + 5}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x**2*exp(x**2)-3)*exp(5+exp(x**2))/x**2,x)

[Out]

3*exp(exp(x**2) + 5)/x

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