3.14.9 \(\int \frac {-x^3+2 e^{e^{2 x}+2 x} x^3-4 \log (x^2)+2 \log ^2(x^2)}{x^3} \, dx\)

Optimal. Leaf size=22 \[ e^{e^{2 x}}-x-\frac {\log ^2\left (x^2\right )}{x^2} \]

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Rubi [A]  time = 0.07, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {14, 2282, 2194, 2304, 2305} \begin {gather*} -\frac {\log ^2\left (x^2\right )}{x^2}+e^{e^{2 x}}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-x^3 + 2*E^(E^(2*x) + 2*x)*x^3 - 4*Log[x^2] + 2*Log[x^2]^2)/x^3,x]

[Out]

E^E^(2*x) - x - Log[x^2]^2/x^2

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 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2305

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Lo
g[c*x^n])^p)/(d*(m + 1)), x] - Dist[(b*n*p)/(m + 1), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (2 e^{e^{2 x}+2 x}+\frac {-x^3-4 \log \left (x^2\right )+2 \log ^2\left (x^2\right )}{x^3}\right ) \, dx\\ &=2 \int e^{e^{2 x}+2 x} \, dx+\int \frac {-x^3-4 \log \left (x^2\right )+2 \log ^2\left (x^2\right )}{x^3} \, dx\\ &=\int \left (-1-\frac {4 \log \left (x^2\right )}{x^3}+\frac {2 \log ^2\left (x^2\right )}{x^3}\right ) \, dx+\operatorname {Subst}\left (\int e^x \, dx,x,e^{2 x}\right )\\ &=e^{e^{2 x}}-x+2 \int \frac {\log ^2\left (x^2\right )}{x^3} \, dx-4 \int \frac {\log \left (x^2\right )}{x^3} \, dx\\ &=e^{e^{2 x}}+\frac {2}{x^2}-x+\frac {2 \log \left (x^2\right )}{x^2}-\frac {\log ^2\left (x^2\right )}{x^2}+4 \int \frac {\log \left (x^2\right )}{x^3} \, dx\\ &=e^{e^{2 x}}-x-\frac {\log ^2\left (x^2\right )}{x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 22, normalized size = 1.00 \begin {gather*} e^{e^{2 x}}-x-\frac {\log ^2\left (x^2\right )}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-x^3 + 2*E^(E^(2*x) + 2*x)*x^3 - 4*Log[x^2] + 2*Log[x^2]^2)/x^3,x]

[Out]

E^E^(2*x) - x - Log[x^2]^2/x^2

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fricas [B]  time = 0.56, size = 43, normalized size = 1.95 \begin {gather*} -\frac {{\left (x^{3} e^{\left (2 \, x\right )} - x^{2} e^{\left (2 \, x + e^{\left (2 \, x\right )}\right )} + e^{\left (2 \, x\right )} \log \left (x^{2}\right )^{2}\right )} e^{\left (-2 \, x\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^3*exp(2*x)*exp(exp(2*x))+2*log(x^2)^2-4*log(x^2)-x^3)/x^3,x, algorithm="fricas")

[Out]

-(x^3*e^(2*x) - x^2*e^(2*x + e^(2*x)) + e^(2*x)*log(x^2)^2)*e^(-2*x)/x^2

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giac [B]  time = 0.46, size = 43, normalized size = 1.95 \begin {gather*} -\frac {{\left (x^{3} e^{\left (2 \, x\right )} - x^{2} e^{\left (2 \, x + e^{\left (2 \, x\right )}\right )} + e^{\left (2 \, x\right )} \log \left (x^{2}\right )^{2}\right )} e^{\left (-2 \, x\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^3*exp(2*x)*exp(exp(2*x))+2*log(x^2)^2-4*log(x^2)-x^3)/x^3,x, algorithm="giac")

[Out]

-(x^3*e^(2*x) - x^2*e^(2*x + e^(2*x)) + e^(2*x)*log(x^2)^2)*e^(-2*x)/x^2

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maple [B]  time = 0.06, size = 44, normalized size = 2.00

method	result	size
default	\(-x +\frac {2 \ln \left (x^{2}\right )}{x^{2}}+\frac {2}{x^{2}}+\frac {-2-\ln \left (x^{2}\right )^{2}-2 \ln \left (x^{2}\right )}{x^{2}}+{\mathrm e}^{{\mathrm e}^{2 x}}\)	\(44\)
risch	\(-\frac {4 \ln \relax (x )^{2}}{x^{2}}+\frac {2 i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (\mathrm {csgn}\left (i x \right )^{2}-2 \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )+\mathrm {csgn}\left (i x^{2}\right )^{2}\right ) \ln \relax (x )}{x^{2}}+\frac {\pi ^{2} \mathrm {csgn}\left (i x \right )^{4} \mathrm {csgn}\left (i x^{2}\right )^{2}-4 \pi ^{2} \mathrm {csgn}\left (i x \right )^{3} \mathrm {csgn}\left (i x^{2}\right )^{3}+6 \pi ^{2} \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )^{4}-4 \pi ^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{5}+\pi ^{2} \mathrm {csgn}\left (i x^{2}\right )^{6}-4 x^{3}}{4 x^{2}}+{\mathrm e}^{{\mathrm e}^{2 x}}\)	\(168\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^3*exp(2*x)*exp(exp(2*x))+2*ln(x^2)^2-4*ln(x^2)-x^3)/x^3,x,method=_RETURNVERBOSE)

[Out]

-x+2*ln(x^2)/x^2+2/x^2+(-2-ln(x^2)^2-2*ln(x^2))/x^2+exp(exp(2*x))

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maxima [A]  time = 0.34, size = 20, normalized size = 0.91 \begin {gather*} -x - \frac {\log \left (x^{2}\right )^{2}}{x^{2}} + e^{\left (e^{\left (2 \, x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^3*exp(2*x)*exp(exp(2*x))+2*log(x^2)^2-4*log(x^2)-x^3)/x^3,x, algorithm="maxima")

[Out]

-x - log(x^2)^2/x^2 + e^(e^(2*x))

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mupad [B]  time = 0.97, size = 20, normalized size = 0.91 \begin {gather*} {\mathrm {e}}^{{\mathrm {e}}^{2\,x}}-x-\frac {{\ln \left (x^2\right )}^2}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*log(x^2) - 2*log(x^2)^2 + x^3 - 2*x^3*exp(2*x)*exp(exp(2*x)))/x^3,x)

[Out]

exp(exp(2*x)) - x - log(x^2)^2/x^2

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sympy [A]  time = 0.33, size = 17, normalized size = 0.77 \begin {gather*} - x + e^{e^{2 x}} - \frac {\log {\left (x^{2} \right )}^{2}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**3*exp(2*x)*exp(exp(2*x))+2*ln(x**2)**2-4*ln(x**2)-x**3)/x**3,x)

[Out]

-x + exp(exp(2*x)) - log(x**2)**2/x**2

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