3.31.86 \(\int \frac {e^{\frac {3 x+(12+x) \log (x)}{3 \log (x)}} (3 x-3 x \log (x)+(3-x) \log ^2(x))}{3 x^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=22 \[ -4-\frac {e^{4+\frac {x}{3}+\frac {x}{\log (x)}}}{x} \]

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Rubi [F]  time = 0.88, 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 {e^{\frac {3 x+(12+x) \log (x)}{3 \log (x)}} \left (3 x-3 x \log (x)+(3-x) \log ^2(x)\right )}{3 x^2 \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((3*x + (12 + x)*Log[x])/(3*Log[x]))*(3*x - 3*x*Log[x] + (3 - x)*Log[x]^2))/(3*x^2*Log[x]^2),x]

[Out]

Defer[Int][E^((3*x + (12 + x)*Log[x])/(3*Log[x]))/x^2, x] - Defer[Int][E^((3*x + (12 + x)*Log[x])/(3*Log[x]))/
x, x]/3 + Defer[Int][E^((3*x + (12 + x)*Log[x])/(3*Log[x]))/(x*Log[x]^2), x] - Defer[Int][E^((3*x + (12 + x)*L
og[x])/(3*Log[x]))/(x*Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int \frac {e^{\frac {3 x+(12+x) \log (x)}{3 \log (x)}} \left (3 x-3 x \log (x)+(3-x) \log ^2(x)\right )}{x^2 \log ^2(x)} \, dx\\ &=\frac {1}{3} \int \left (\frac {e^{\frac {3 x+(12+x) \log (x)}{3 \log (x)}} (3-x)}{x^2}+\frac {3 e^{\frac {3 x+(12+x) \log (x)}{3 \log (x)}}}{x \log ^2(x)}-\frac {3 e^{\frac {3 x+(12+x) \log (x)}{3 \log (x)}}}{x \log (x)}\right ) \, dx\\ &=\frac {1}{3} \int \frac {e^{\frac {3 x+(12+x) \log (x)}{3 \log (x)}} (3-x)}{x^2} \, dx+\int \frac {e^{\frac {3 x+(12+x) \log (x)}{3 \log (x)}}}{x \log ^2(x)} \, dx-\int \frac {e^{\frac {3 x+(12+x) \log (x)}{3 \log (x)}}}{x \log (x)} \, dx\\ &=\frac {1}{3} \int \left (\frac {3 e^{\frac {3 x+(12+x) \log (x)}{3 \log (x)}}}{x^2}-\frac {e^{\frac {3 x+(12+x) \log (x)}{3 \log (x)}}}{x}\right ) \, dx+\int \frac {e^{\frac {3 x+(12+x) \log (x)}{3 \log (x)}}}{x \log ^2(x)} \, dx-\int \frac {e^{\frac {3 x+(12+x) \log (x)}{3 \log (x)}}}{x \log (x)} \, dx\\ &=-\left (\frac {1}{3} \int \frac {e^{\frac {3 x+(12+x) \log (x)}{3 \log (x)}}}{x} \, dx\right )+\int \frac {e^{\frac {3 x+(12+x) \log (x)}{3 \log (x)}}}{x^2} \, dx+\int \frac {e^{\frac {3 x+(12+x) \log (x)}{3 \log (x)}}}{x \log ^2(x)} \, dx-\int \frac {e^{\frac {3 x+(12+x) \log (x)}{3 \log (x)}}}{x \log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 20, normalized size = 0.91 \begin {gather*} -\frac {e^{4+\frac {x}{3}+\frac {x}{\log (x)}}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((3*x + (12 + x)*Log[x])/(3*Log[x]))*(3*x - 3*x*Log[x] + (3 - x)*Log[x]^2))/(3*x^2*Log[x]^2),x]

[Out]

-(E^(4 + x/3 + x/Log[x])/x)

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fricas [A]  time = 0.93, size = 22, normalized size = 1.00 \begin {gather*} -\frac {e^{\left (\frac {{\left (x + 12\right )} \log \relax (x) + 3 \, x}{3 \, \log \relax (x)}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((3-x)*log(x)^2-3*x*log(x)+3*x)*exp(1/3*((x+12)*log(x)+3*x)/log(x))/x^2/log(x)^2,x, algorithm="f
ricas")

[Out]

-e^(1/3*((x + 12)*log(x) + 3*x)/log(x))/x

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giac [A]  time = 0.22, size = 24, normalized size = 1.09 \begin {gather*} -\frac {e^{\left (\frac {x \log \relax (x) + 3 \, x + 12 \, \log \relax (x)}{3 \, \log \relax (x)}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((3-x)*log(x)^2-3*x*log(x)+3*x)*exp(1/3*((x+12)*log(x)+3*x)/log(x))/x^2/log(x)^2,x, algorithm="g
iac")

[Out]

-e^(1/3*(x*log(x) + 3*x + 12*log(x))/log(x))/x

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maple [A]  time = 0.04, size = 23, normalized size = 1.05

method	result	size
norman	\(-\frac {{\mathrm e}^{\frac {\left (x +12\right ) \ln \relax (x )+3 x}{3 \ln \relax (x )}}}{x}\)	\(23\)
risch	\(-\frac {{\mathrm e}^{\frac {x \ln \relax (x )+12 \ln \relax (x )+3 x}{3 \ln \relax (x )}}}{x}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*((3-x)*ln(x)^2-3*x*ln(x)+3*x)*exp(1/3*((x+12)*ln(x)+3*x)/ln(x))/x^2/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

-exp(1/3*((x+12)*ln(x)+3*x)/ln(x))/x

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maxima [A]  time = 0.89, size = 17, normalized size = 0.77 \begin {gather*} -\frac {e^{\left (\frac {1}{3} \, x + \frac {x}{\log \relax (x)} + 4\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((3-x)*log(x)^2-3*x*log(x)+3*x)*exp(1/3*((x+12)*log(x)+3*x)/log(x))/x^2/log(x)^2,x, algorithm="m
axima")

[Out]

-e^(1/3*x + x/log(x) + 4)/x

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mupad [B]  time = 1.89, size = 18, normalized size = 0.82 \begin {gather*} -\frac {{\mathrm {e}}^{x/3}\,{\mathrm {e}}^4\,{\mathrm {e}}^{\frac {x}{\ln \relax (x)}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((x + (log(x)*(x + 12))/3)/log(x))*(log(x)^2*(x - 3) - 3*x + 3*x*log(x)))/(3*x^2*log(x)^2),x)

[Out]

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

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sympy [A]  time = 0.30, size = 17, normalized size = 0.77 \begin {gather*} - \frac {e^{\frac {x + \frac {\left (x + 12\right ) \log {\relax (x )}}{3}}{\log {\relax (x )}}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((3-x)*ln(x)**2-3*x*ln(x)+3*x)*exp(1/3*((x+12)*ln(x)+3*x)/ln(x))/x**2/ln(x)**2,x)

[Out]

-exp((x + (x + 12)*log(x)/3)/log(x))/x

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