3.95.85 \(\int \frac {-2 e^{4 x}+e^{4 x} (-4+8 x) \log (x)}{3 x^3 \log ^2(x)} \, dx\)

Optimal. Leaf size=16 \[ \frac {2 e^{4 x}}{3 x^2 \log (x)} \]

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Rubi [A]  time = 0.27, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {12, 6741, 2288} \begin {gather*} \frac {2 e^{4 x}}{3 x^2 \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*E^(4*x) + E^(4*x)*(-4 + 8*x)*Log[x])/(3*x^3*Log[x]^2),x]

[Out]

(2*E^(4*x))/(3*x^2*Log[x])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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]

Rule 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int \frac {-2 e^{4 x}+e^{4 x} (-4+8 x) \log (x)}{x^3 \log ^2(x)} \, dx\\ &=\frac {1}{3} \int \frac {2 e^{4 x} (-1-2 \log (x)+4 x \log (x))}{x^3 \log ^2(x)} \, dx\\ &=\frac {2}{3} \int \frac {e^{4 x} (-1-2 \log (x)+4 x \log (x))}{x^3 \log ^2(x)} \, dx\\ &=\frac {2 e^{4 x}}{3 x^2 \log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 16, normalized size = 1.00 \begin {gather*} \frac {2 e^{4 x}}{3 x^2 \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*E^(4*x) + E^(4*x)*(-4 + 8*x)*Log[x])/(3*x^3*Log[x]^2),x]

[Out]

(2*E^(4*x))/(3*x^2*Log[x])

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fricas [A]  time = 0.57, size = 13, normalized size = 0.81 \begin {gather*} \frac {2 \, e^{\left (4 \, x\right )}}{3 \, x^{2} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((8*x-4)*exp(4*x)*log(x)-2*exp(4*x))/x^3/log(x)^2,x, algorithm="fricas")

[Out]

2/3*e^(4*x)/(x^2*log(x))

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giac [A]  time = 0.18, size = 13, normalized size = 0.81 \begin {gather*} \frac {2 \, e^{\left (4 \, x\right )}}{3 \, x^{2} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((8*x-4)*exp(4*x)*log(x)-2*exp(4*x))/x^3/log(x)^2,x, algorithm="giac")

[Out]

2/3*e^(4*x)/(x^2*log(x))

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maple [A]  time = 0.02, size = 14, normalized size = 0.88

method	result	size
risch	\(\frac {2 \,{\mathrm e}^{4 x}}{3 \ln \relax (x ) x^{2}}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*((8*x-4)*exp(4*x)*ln(x)-2*exp(4*x))/x^3/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

2/3*exp(4*x)/ln(x)/x^2

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maxima [A]  time = 0.39, size = 13, normalized size = 0.81 \begin {gather*} \frac {2 \, e^{\left (4 \, x\right )}}{3 \, x^{2} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((8*x-4)*exp(4*x)*log(x)-2*exp(4*x))/x^3/log(x)^2,x, algorithm="maxima")

[Out]

2/3*e^(4*x)/(x^2*log(x))

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mupad [B]  time = 9.15, size = 13, normalized size = 0.81 \begin {gather*} \frac {2\,{\mathrm {e}}^{4\,x}}{3\,x^2\,\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((2*exp(4*x))/3 - (exp(4*x)*log(x)*(8*x - 4))/3)/(x^3*log(x)^2),x)

[Out]

(2*exp(4*x))/(3*x^2*log(x))

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sympy [A]  time = 0.24, size = 14, normalized size = 0.88 \begin {gather*} \frac {2 e^{4 x}}{3 x^{2} \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((8*x-4)*exp(4*x)*ln(x)-2*exp(4*x))/x**3/ln(x)**2,x)

[Out]

2*exp(4*x)/(3*x**2*log(x))

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