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

Optimal. Leaf size=14 \[ e^{2 x+\frac {\log \left (x^3\right )}{e^3}} \]

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Rubi [A]  time = 0.27, antiderivative size = 13, normalized size of antiderivative = 0.93, number of steps used = 3, number of rules used = 3, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.073, Rules used = {2274, 15, 2197} \begin {gather*} e^{2 x} \left (x^3\right )^{\frac {1}{e^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(2*x)*(x^3)^E^(-3)

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rule 2274

Int[(u_.)*(F_)^((a_.)*(Log[z_]*(b_.) + (v_.))), x_Symbol] :> Int[u*F^(a*v)*z^(a*b*Log[F]), x] /; FreeQ[{F, a,
b}, x]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 13, normalized size = 0.93 \begin {gather*} e^{2 x} \left (x^3\right )^{\frac {1}{e^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(2*x)*(x^3)^E^(-3)

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fricas [A]  time = 0.94, size = 14, normalized size = 1.00 \begin {gather*} e^{\left ({\left (2 \, x e^{3} + \log \left (x^{3}\right )\right )} e^{\left (-3\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^((2*x*e^3 + log(x^3))*e^(-3))

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giac [A]  time = 0.26, size = 14, normalized size = 1.00 \begin {gather*} e^{\left ({\left (2 \, x e^{3} + 3 \, \log \relax (x)\right )} e^{\left (-3\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^((2*x*e^3 + 3*log(x))*e^(-3))

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maple [A]  time = 0.12, size = 13, normalized size = 0.93

method	result	size
default	\({\mathrm e}^{\ln \left (x^{3}\right ) {\mathrm e}^{-3}+2 x}\)	\(13\)
norman	\({\mathrm e}^{\ln \left (x^{3}\right ) {\mathrm e}^{-3}+2 x}\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(ln(x^3)*exp(-3)+2*x)

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maxima [A]  time = 0.72, size = 11, normalized size = 0.79 \begin {gather*} e^{\left (3 \, e^{\left (-3\right )} \log \relax (x) + 2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(3*e^(-3)*log(x) + 2*x)

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mupad [B]  time = 1.22, size = 11, normalized size = 0.79 \begin {gather*} {\mathrm {e}}^{2\,x}\,{\left (x^3\right )}^{{\mathrm {e}}^{-3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(2*x)*(x^3)^exp(-3)

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sympy [A]  time = 4.45, size = 12, normalized size = 0.86 \begin {gather*} \left (x^{3}\right )^{e^{-3}} e^{2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x**3)**exp(-3)*exp(2*x)

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