3.29.33 \(\int \frac {24+12 x-4 e^x x^2 \log (2)-4 x^3 \log (2)+(-12 x+12 x^3 \log (2)+e^x (8 x^2+4 x^3) \log (2)) \log (x)}{3 x \log ^2(x)} \, dx\)

Optimal. Leaf size=25 \[ \frac {4 \left (-2-x+\frac {1}{3} x^2 \left (e^x+x\right ) \log (2)\right )}{\log (x)} \]

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Rubi [F]  time = 0.79, 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 {24+12 x-4 e^x x^2 \log (2)-4 x^3 \log (2)+\left (-12 x+12 x^3 \log (2)+e^x \left (8 x^2+4 x^3\right ) \log (2)\right ) \log (x)}{3 x \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(24 + 12*x - 4*E^x*x^2*Log[2] - 4*x^3*Log[2] + (-12*x + 12*x^3*Log[2] + E^x*(8*x^2 + 4*x^3)*Log[2])*Log[x]
)/(3*x*Log[x]^2),x]

[Out]

(4*ExpIntegralEi[3*Log[x]]*Log[8])/3 + (4*E^x*x^2*Log[2])/(3*Log[x]) - 4*LogIntegral[x] + (4*Defer[Int][(6 + 3
*x - x^3*Log[2])/(x*Log[x]^2), x])/3

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int \frac {24+12 x-4 e^x x^2 \log (2)-4 x^3 \log (2)+\left (-12 x+12 x^3 \log (2)+e^x \left (8 x^2+4 x^3\right ) \log (2)\right ) \log (x)}{x \log ^2(x)} \, dx\\ &=\frac {1}{3} \int \left (\frac {4 e^x x \log (2) (-1+2 \log (x)+x \log (x))}{\log ^2(x)}+\frac {4 \left (6+3 x-x^3 \log (2)-3 x \log (x)+x^3 \log (8) \log (x)\right )}{x \log ^2(x)}\right ) \, dx\\ &=\frac {4}{3} \int \frac {6+3 x-x^3 \log (2)-3 x \log (x)+x^3 \log (8) \log (x)}{x \log ^2(x)} \, dx+\frac {1}{3} (4 \log (2)) \int \frac {e^x x (-1+2 \log (x)+x \log (x))}{\log ^2(x)} \, dx\\ &=\frac {4 e^x x^2 \log (2)}{3 \log (x)}+\frac {4}{3} \int \left (\frac {6+3 x-x^3 \log (2)}{x \log ^2(x)}+\frac {-3+x^2 \log (8)}{\log (x)}\right ) \, dx\\ &=\frac {4 e^x x^2 \log (2)}{3 \log (x)}+\frac {4}{3} \int \frac {6+3 x-x^3 \log (2)}{x \log ^2(x)} \, dx+\frac {4}{3} \int \frac {-3+x^2 \log (8)}{\log (x)} \, dx\\ &=\frac {4 e^x x^2 \log (2)}{3 \log (x)}+\frac {4}{3} \int \left (-\frac {3}{\log (x)}+\frac {x^2 \log (8)}{\log (x)}\right ) \, dx+\frac {4}{3} \int \frac {6+3 x-x^3 \log (2)}{x \log ^2(x)} \, dx\\ &=\frac {4 e^x x^2 \log (2)}{3 \log (x)}+\frac {4}{3} \int \frac {6+3 x-x^3 \log (2)}{x \log ^2(x)} \, dx-4 \int \frac {1}{\log (x)} \, dx+\frac {1}{3} (4 \log (8)) \int \frac {x^2}{\log (x)} \, dx\\ &=\frac {4 e^x x^2 \log (2)}{3 \log (x)}-4 \text {li}(x)+\frac {4}{3} \int \frac {6+3 x-x^3 \log (2)}{x \log ^2(x)} \, dx+\frac {1}{3} (4 \log (8)) \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (x)\right )\\ &=\frac {4}{3} \text {Ei}(3 \log (x)) \log (8)+\frac {4 e^x x^2 \log (2)}{3 \log (x)}-4 \text {li}(x)+\frac {4}{3} \int \frac {6+3 x-x^3 \log (2)}{x \log ^2(x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.24, size = 28, normalized size = 1.12 \begin {gather*} \frac {4 \left (-6-3 x+e^x x^2 \log (2)+x^3 \log (2)\right )}{3 \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(24 + 12*x - 4*E^x*x^2*Log[2] - 4*x^3*Log[2] + (-12*x + 12*x^3*Log[2] + E^x*(8*x^2 + 4*x^3)*Log[2])*
Log[x])/(3*x*Log[x]^2),x]

[Out]

(4*(-6 - 3*x + E^x*x^2*Log[2] + x^3*Log[2]))/(3*Log[x])

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fricas [A]  time = 1.02, size = 25, normalized size = 1.00 \begin {gather*} \frac {4 \, {\left (x^{3} \log \relax (2) + x^{2} e^{x} \log \relax (2) - 3 \, x - 6\right )}}{3 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(((4*x^3+8*x^2)*log(2)*exp(x)+12*x^3*log(2)-12*x)*log(x)-4*x^2*log(2)*exp(x)-4*x^3*log(2)+12*x+2
4)/x/log(x)^2,x, algorithm="fricas")

[Out]

4/3*(x^3*log(2) + x^2*e^x*log(2) - 3*x - 6)/log(x)

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giac [A]  time = 0.27, size = 25, normalized size = 1.00 \begin {gather*} \frac {4 \, {\left (x^{3} \log \relax (2) + x^{2} e^{x} \log \relax (2) - 3 \, x - 6\right )}}{3 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(((4*x^3+8*x^2)*log(2)*exp(x)+12*x^3*log(2)-12*x)*log(x)-4*x^2*log(2)*exp(x)-4*x^3*log(2)+12*x+2
4)/x/log(x)^2,x, algorithm="giac")

[Out]

4/3*(x^3*log(2) + x^2*e^x*log(2) - 3*x - 6)/log(x)

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maple [A]  time = 0.06, size = 26, normalized size = 1.04

method	result	size
risch	\(\frac {\frac {4 x^{3} \ln \relax (2)}{3}+\frac {4 x^{2} \ln \relax (2) {\mathrm e}^{x}}{3}-4 x -8}{\ln \relax (x )}\)	\(26\)
default	\(\frac {4 x^{2} \ln \relax (2) {\mathrm e}^{x}}{3 \ln \relax (x )}+\frac {4 x^{3} \ln \relax (2)}{3 \ln \relax (x )}-\frac {4 x}{\ln \relax (x )}-\frac {8}{\ln \relax (x )}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*(((4*x^3+8*x^2)*ln(2)*exp(x)+12*x^3*ln(2)-12*x)*ln(x)-4*x^2*ln(2)*exp(x)-4*x^3*ln(2)+12*x+24)/x/ln(x)^
2,x,method=_RETURNVERBOSE)

[Out]

4/3*(x^3*ln(2)+x^2*ln(2)*exp(x)-3*x-6)/ln(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {4 \, x^{2} e^{x} \log \relax (2)}{3 \, \log \relax (x)} - 4 \, \Gamma \left (-1, -3 \, \log \relax (x)\right ) \log \relax (2) - \frac {8}{\log \relax (x)} + 4 \, \Gamma \left (-1, -\log \relax (x)\right ) + \frac {4}{3} \, \int \frac {3 \, {\left (x^{2} \log \relax (2) - 1\right )}}{\log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(((4*x^3+8*x^2)*log(2)*exp(x)+12*x^3*log(2)-12*x)*log(x)-4*x^2*log(2)*exp(x)-4*x^3*log(2)+12*x+2
4)/x/log(x)^2,x, algorithm="maxima")

[Out]

4/3*x^2*e^x*log(2)/log(x) - 4*gamma(-1, -3*log(x))*log(2) - 8/log(x) + 4*gamma(-1, -log(x)) + 4/3*integrate(3*
(x^2*log(2) - 1)/log(x), x)

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mupad [B]  time = 1.86, size = 38, normalized size = 1.52 \begin {gather*} \frac {4\,x^3\,\ln \relax (2)}{3\,\ln \relax (x)}-\frac {8}{\ln \relax (x)}-\frac {4\,x}{\ln \relax (x)}+\frac {4\,x^2\,{\mathrm {e}}^x\,\ln \relax (2)}{3\,\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x + (log(x)*(12*x^3*log(2) - 12*x + exp(x)*log(2)*(8*x^2 + 4*x^3)))/3 - (4*x^3*log(2))/3 - (4*x^2*exp(x
)*log(2))/3 + 8)/(x*log(x)^2),x)

[Out]

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

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sympy [A]  time = 0.30, size = 34, normalized size = 1.36 \begin {gather*} \frac {4 x^{2} e^{x} \log {\relax (2 )}}{3 \log {\relax (x )}} + \frac {4 x^{3} \log {\relax (2 )} - 12 x - 24}{3 \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(((4*x**3+8*x**2)*ln(2)*exp(x)+12*x**3*ln(2)-12*x)*ln(x)-4*x**2*ln(2)*exp(x)-4*x**3*ln(2)+12*x+2
4)/x/ln(x)**2,x)

[Out]

4*x**2*exp(x)*log(2)/(3*log(x)) + (4*x**3*log(2) - 12*x - 24)/(3*log(x))

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