3.74.30 \(\int \frac {e^{3 x} (-1-x+3 x^2)+3 e^{3 x} x \log (2 x)}{x^3+2 x^2 \log (2 x)+x \log ^2(2 x)} \, dx\)

Optimal. Leaf size=23 \[ -\frac {1}{5}+\log \left (4 e^{\frac {e^{3 x}}{x+\log (2 x)}}\right ) \]

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Rubi [A]  time = 0.26, antiderivative size = 27, normalized size of antiderivative = 1.17, number of steps used = 2, number of rules used = 2, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {6688, 2288} \begin {gather*} \frac {e^{3 x} \left (x^2+x \log (2 x)\right )}{x (x+\log (2 x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(3*x)*(-1 - x + 3*x^2) + 3*E^(3*x)*x*Log[2*x])/(x^3 + 2*x^2*Log[2*x] + x*Log[2*x]^2),x]

[Out]

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

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 14, normalized size = 0.61 \begin {gather*} \frac {e^{3 x}}{x+\log (2 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(3*x)*(-1 - x + 3*x^2) + 3*E^(3*x)*x*Log[2*x])/(x^3 + 2*x^2*Log[2*x] + x*Log[2*x]^2),x]

[Out]

E^(3*x)/(x + Log[2*x])

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fricas [A]  time = 0.59, size = 13, normalized size = 0.57 \begin {gather*} \frac {e^{\left (3 \, x\right )}}{x + \log \left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x*exp(3*x)*log(2*x)+(3*x^2-x-1)*exp(3*x))/(x*log(2*x)^2+2*x^2*log(2*x)+x^3),x, algorithm="fricas"
)

[Out]

e^(3*x)/(x + log(2*x))

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giac [A]  time = 0.19, size = 13, normalized size = 0.57 \begin {gather*} \frac {e^{\left (3 \, x\right )}}{x + \log \left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x*exp(3*x)*log(2*x)+(3*x^2-x-1)*exp(3*x))/(x*log(2*x)^2+2*x^2*log(2*x)+x^3),x, algorithm="giac")

[Out]

e^(3*x)/(x + log(2*x))

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maple [A]  time = 0.02, size = 14, normalized size = 0.61

method	result	size
risch	\(\frac {{\mathrm e}^{3 x}}{\ln \left (2 x \right )+x}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x*exp(3*x)*ln(2*x)+(3*x^2-x-1)*exp(3*x))/(x*ln(2*x)^2+2*x^2*ln(2*x)+x^3),x,method=_RETURNVERBOSE)

[Out]

exp(3*x)/(ln(2*x)+x)

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maxima [A]  time = 0.49, size = 13, normalized size = 0.57 \begin {gather*} \frac {e^{\left (3 \, x\right )}}{x + \log \relax (2) + \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x*exp(3*x)*log(2*x)+(3*x^2-x-1)*exp(3*x))/(x*log(2*x)^2+2*x^2*log(2*x)+x^3),x, algorithm="maxima"
)

[Out]

e^(3*x)/(x + log(2) + log(x))

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mupad [B]  time = 4.58, size = 13, normalized size = 0.57 \begin {gather*} \frac {{\mathrm {e}}^{3\,x}}{x+\ln \left (2\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(3*x)*(x - 3*x^2 + 1) - 3*x*log(2*x)*exp(3*x))/(x*log(2*x)^2 + 2*x^2*log(2*x) + x^3),x)

[Out]

exp(3*x)/(x + log(2*x))

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sympy [A]  time = 0.29, size = 10, normalized size = 0.43 \begin {gather*} \frac {e^{3 x}}{x + \log {\left (2 x \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x*exp(3*x)*ln(2*x)+(3*x**2-x-1)*exp(3*x))/(x*ln(2*x)**2+2*x**2*ln(2*x)+x**3),x)

[Out]

exp(3*x)/(x + log(2*x))

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