3.81.53 \(\int \frac {-1-256 e^{2+64 e^2 x^2} x^2}{x} \, dx\)

Optimal. Leaf size=24 \[ 11-\log \left (\frac {1}{5} e^{2 e^{64 e^2 x^2}} x\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 17, normalized size of antiderivative = 0.71, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {14, 2209} \begin {gather*} -2 e^{64 e^2 x^2}-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 - 256*E^(2 + 64*E^2*x^2)*x^2)/x,x]

[Out]

-2*E^(64*E^2*x^2) - Log[x]

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 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {1}{x}-256 e^{2+64 e^2 x^2} x\right ) \, dx\\ &=-\log (x)-256 \int e^{2+64 e^2 x^2} x \, dx\\ &=-2 e^{64 e^2 x^2}-\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 17, normalized size = 0.71 \begin {gather*} -2 e^{64 e^2 x^2}-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 - 256*E^(2 + 64*E^2*x^2)*x^2)/x,x]

[Out]

-2*E^(64*E^2*x^2) - Log[x]

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fricas [A]  time = 0.68, size = 22, normalized size = 0.92 \begin {gather*} -{\left (e^{2} \log \relax (x) + 2 \, e^{\left (64 \, x^{2} e^{2} + 2\right )}\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-256*x^2*exp(1)^2*exp(64*x^2*exp(1)^2)-1)/x,x, algorithm="fricas")

[Out]

-(e^2*log(x) + 2*e^(64*x^2*e^2 + 2))*e^(-2)

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giac [A]  time = 0.15, size = 28, normalized size = 1.17 \begin {gather*} -\frac {1}{2} \, {\left (e^{2} \log \left (64 \, x^{2} e^{2}\right ) + 4 \, e^{\left (64 \, x^{2} e^{2} + 2\right )}\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-256*x^2*exp(1)^2*exp(64*x^2*exp(1)^2)-1)/x,x, algorithm="giac")

[Out]

-1/2*(e^2*log(64*x^2*e^2) + 4*e^(64*x^2*e^2 + 2))*e^(-2)

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maple [A]  time = 0.04, size = 16, normalized size = 0.67

method	result	size
risch	\(-2 \,{\mathrm e}^{64 x^{2} {\mathrm e}^{2}}-\ln \relax (x )\)	\(16\)
norman	\(-2 \,{\mathrm e}^{64 x^{2} {\mathrm e}^{2}}-\ln \relax (x )\)	\(18\)
default	\(-2 \,{\mathrm e}^{64 x^{2} {\mathrm e}^{2}}-\ln \relax (x )\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-256*x^2*exp(1)^2*exp(64*x^2*exp(1)^2)-1)/x,x,method=_RETURNVERBOSE)

[Out]

-2*exp(64*x^2*exp(2))-ln(x)

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maxima [A]  time = 0.36, size = 15, normalized size = 0.62 \begin {gather*} -2 \, e^{\left (64 \, x^{2} e^{2}\right )} - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-256*x^2*exp(1)^2*exp(64*x^2*exp(1)^2)-1)/x,x, algorithm="maxima")

[Out]

-2*e^(64*x^2*e^2) - log(x)

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mupad [B]  time = 6.79, size = 15, normalized size = 0.62 \begin {gather*} -2\,{\mathrm {e}}^{64\,x^2\,{\mathrm {e}}^2}-\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(256*x^2*exp(64*x^2*exp(2))*exp(2) + 1)/x,x)

[Out]

- 2*exp(64*x^2*exp(2)) - log(x)

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sympy [A]  time = 0.11, size = 15, normalized size = 0.62 \begin {gather*} - 2 e^{64 x^{2} e^{2}} - \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-256*x**2*exp(1)**2*exp(64*x**2*exp(1)**2)-1)/x,x)

[Out]

-2*exp(64*x**2*exp(2)) - log(x)

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