3.15.65 \(\int \frac {-\frac {3}{5} e^{2 x} \log (x)+(\frac {e^{2 x}}{5}+\frac {2}{5} e^{2 x} x \log (x)) \log (x^3)}{5 x \log ^2(x^3)} \, dx\)

Optimal. Leaf size=17 \[ \frac {e^{2 x} \log (x)}{25 \log \left (x^3\right )} \]

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Rubi [F]  time = 0.83, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-\frac {3}{5} e^{2 x} \log (x)+\left (\frac {e^{2 x}}{5}+\frac {2}{5} e^{2 x} x \log (x)\right ) \log \left (x^3\right )}{5 x \log ^2\left (x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-3*E^(2*x)*Log[x])/5 + (E^(2*x)/5 + (2*E^(2*x)*x*Log[x])/5)*Log[x^3])/(5*x*Log[x^3]^2),x]

[Out]

(-3*Defer[Int][(E^(2*x)*Log[x])/(x*Log[x^3]^2), x])/25 + Defer[Int][E^(2*x)/(x*Log[x^3]), x]/25 + (2*Defer[Int
][(E^(2*x)*Log[x])/Log[x^3], x])/25

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {-\frac {3}{5} e^{2 x} \log (x)+\left (\frac {e^{2 x}}{5}+\frac {2}{5} e^{2 x} x \log (x)\right ) \log \left (x^3\right )}{x \log ^2\left (x^3\right )} \, dx\\ &=\frac {1}{5} \int \frac {e^{2 x} \left (\log \left (x^3\right )+\log (x) \left (-3+2 x \log \left (x^3\right )\right )\right )}{5 x \log ^2\left (x^3\right )} \, dx\\ &=\frac {1}{25} \int \frac {e^{2 x} \left (\log \left (x^3\right )+\log (x) \left (-3+2 x \log \left (x^3\right )\right )\right )}{x \log ^2\left (x^3\right )} \, dx\\ &=\frac {1}{25} \int \left (-\frac {3 e^{2 x} \log (x)}{x \log ^2\left (x^3\right )}+\frac {e^{2 x} (1+2 x \log (x))}{x \log \left (x^3\right )}\right ) \, dx\\ &=\frac {1}{25} \int \frac {e^{2 x} (1+2 x \log (x))}{x \log \left (x^3\right )} \, dx-\frac {3}{25} \int \frac {e^{2 x} \log (x)}{x \log ^2\left (x^3\right )} \, dx\\ &=\frac {1}{25} \int \left (\frac {e^{2 x}}{x \log \left (x^3\right )}+\frac {2 e^{2 x} \log (x)}{\log \left (x^3\right )}\right ) \, dx-\frac {3}{25} \int \frac {e^{2 x} \log (x)}{x \log ^2\left (x^3\right )} \, dx\\ &=\frac {1}{25} \int \frac {e^{2 x}}{x \log \left (x^3\right )} \, dx+\frac {2}{25} \int \frac {e^{2 x} \log (x)}{\log \left (x^3\right )} \, dx-\frac {3}{25} \int \frac {e^{2 x} \log (x)}{x \log ^2\left (x^3\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 17, normalized size = 1.00 \begin {gather*} \frac {e^{2 x} \log (x)}{25 \log \left (x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-3*E^(2*x)*Log[x])/5 + (E^(2*x)/5 + (2*E^(2*x)*x*Log[x])/5)*Log[x^3])/(5*x*Log[x^3]^2),x]

[Out]

(E^(2*x)*Log[x])/(25*Log[x^3])

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fricas [A]  time = 0.61, size = 11, normalized size = 0.65 \begin {gather*} \frac {1}{15} \, e^{\left (2 \, x - \log \relax (5)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((2*x*exp(-log(5)+2*x)*log(x)+exp(-log(5)+2*x))*log(x^3)-3*exp(-log(5)+2*x)*log(x))/x/log(x^3)^2
,x, algorithm="fricas")

[Out]

1/15*e^(2*x - log(5))

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giac [A]  time = 0.31, size = 6, normalized size = 0.35 \begin {gather*} \frac {1}{75} \, e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((2*x*exp(-log(5)+2*x)*log(x)+exp(-log(5)+2*x))*log(x^3)-3*exp(-log(5)+2*x)*log(x))/x/log(x^3)^2
,x, algorithm="giac")

[Out]

1/75*e^(2*x)

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maple [A]  time = 0.11, size = 20, normalized size = 1.18




method result size



default \(\frac {\ln \relax (x ) {\mathrm e}^{-\ln \relax (5)+2 x}}{5 \ln \left (x^{3}\right )}\) \(20\)
risch \(\frac {{\mathrm e}^{2 x}}{75}-\frac {\pi \,{\mathrm e}^{2 x} \left (\mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )-\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{3}\right )^{2}+\mathrm {csgn}\left (i x^{2}\right )^{3}-\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )^{2}+\mathrm {csgn}\left (i x^{3}\right )^{3}\right )}{75 \left (\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{3}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}-\pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )^{2}+\pi \mathrm {csgn}\left (i x^{3}\right )^{3}+6 i \ln \relax (x )\right )}\) \(239\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((2*x*exp(-ln(5)+2*x)*ln(x)+exp(-ln(5)+2*x))*ln(x^3)-3*exp(-ln(5)+2*x)*ln(x))/x/ln(x^3)^2,x,method=_RE
TURNVERBOSE)

[Out]

1/5*ln(x)/ln(x^3)*exp(-ln(5)+2*x)

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maxima [A]  time = 0.45, size = 6, normalized size = 0.35 \begin {gather*} \frac {1}{75} \, e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((2*x*exp(-log(5)+2*x)*log(x)+exp(-log(5)+2*x))*log(x^3)-3*exp(-log(5)+2*x)*log(x))/x/log(x^3)^2
,x, algorithm="maxima")

[Out]

1/75*e^(2*x)

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mupad [B]  time = 1.19, size = 14, normalized size = 0.82 \begin {gather*} \frac {{\mathrm {e}}^{2\,x}\,\ln \relax (x)}{25\,\ln \left (x^3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((log(x^3)*(exp(2*x - log(5)) + 2*x*exp(2*x - log(5))*log(x)))/5 - (3*exp(2*x - log(5))*log(x))/5)/(x*log(
x^3)^2),x)

[Out]

(exp(2*x)*log(x))/(25*log(x^3))

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sympy [A]  time = 0.25, size = 5, normalized size = 0.29 \begin {gather*} \frac {e^{2 x}}{75} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((2*x*exp(-ln(5)+2*x)*ln(x)+exp(-ln(5)+2*x))*ln(x**3)-3*exp(-ln(5)+2*x)*ln(x))/x/ln(x**3)**2,x)

[Out]

exp(2*x)/75

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