3.52.58 \(\int \frac {e^5 (-1-2 x)+(-6-24 x-24 x^2) \log (5 x)-e^5 \log (5 x) \log (\log (5 x))}{(6+24 x+24 x^2) \log (5 x)} \, dx\)

Optimal. Leaf size=25 \[ -x-\frac {e^5 \log (\log (5 x))}{3 \left (4+\frac {2}{x}\right )} \]

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Rubi [F]  time = 0.17, 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^5 (-1-2 x)+\left (-6-24 x-24 x^2\right ) \log (5 x)-e^5 \log (5 x) \log (\log (5 x))}{\left (6+24 x+24 x^2\right ) \log (5 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^5*(-1 - 2*x) + (-6 - 24*x - 24*x^2)*Log[5*x] - E^5*Log[5*x]*Log[Log[5*x]])/((6 + 24*x + 24*x^2)*Log[5*x
]),x]

[Out]

-x - (E^5*Defer[Int][1/((1 + 2*x)*Log[5*x]), x])/6 - (E^5*Defer[Int][Log[Log[5*x]]/(1 + 2*x)^2, x])/6

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^5 (-1-2 x)+\left (-6-24 x-24 x^2\right ) \log (5 x)-e^5 \log (5 x) \log (\log (5 x))}{6 (1+2 x)^2 \log (5 x)} \, dx\\ &=\frac {1}{6} \int \frac {e^5 (-1-2 x)+\left (-6-24 x-24 x^2\right ) \log (5 x)-e^5 \log (5 x) \log (\log (5 x))}{(1+2 x)^2 \log (5 x)} \, dx\\ &=\frac {1}{6} \int \left (-6-\frac {e^5}{(1+2 x) \log (5 x)}-\frac {e^5 \log (\log (5 x))}{(1+2 x)^2}\right ) \, dx\\ &=-x-\frac {1}{6} e^5 \int \frac {1}{(1+2 x) \log (5 x)} \, dx-\frac {1}{6} e^5 \int \frac {\log (\log (5 x))}{(1+2 x)^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 24, normalized size = 0.96 \begin {gather*} -\frac {x \left (6+12 x+e^5 \log (\log (5 x))\right )}{6+12 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^5*(-1 - 2*x) + (-6 - 24*x - 24*x^2)*Log[5*x] - E^5*Log[5*x]*Log[Log[5*x]])/((6 + 24*x + 24*x^2)*L
og[5*x]),x]

[Out]

-((x*(6 + 12*x + E^5*Log[Log[5*x]]))/(6 + 12*x))

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fricas [A]  time = 1.19, size = 27, normalized size = 1.08 \begin {gather*} -\frac {x e^{5} \log \left (\log \left (5 \, x\right )\right ) + 12 \, x^{2} + 6 \, x}{6 \, {\left (2 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(5)*log(5*x)*log(log(5*x))+(-24*x^2-24*x-6)*log(5*x)+(-2*x-1)*exp(5))/(24*x^2+24*x+6)/log(5*x),
x, algorithm="fricas")

[Out]

-1/6*(x*e^5*log(log(5*x)) + 12*x^2 + 6*x)/(2*x + 1)

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giac [A]  time = 0.16, size = 27, normalized size = 1.08 \begin {gather*} -\frac {x e^{5} \log \left (\log \left (5 \, x\right )\right ) + 12 \, x^{2} + 6 \, x}{6 \, {\left (2 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(5)*log(5*x)*log(log(5*x))+(-24*x^2-24*x-6)*log(5*x)+(-2*x-1)*exp(5))/(24*x^2+24*x+6)/log(5*x),
x, algorithm="giac")

[Out]

-1/6*(x*e^5*log(log(5*x)) + 12*x^2 + 6*x)/(2*x + 1)

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maple [A]  time = 0.24, size = 28, normalized size = 1.12

method	result	size
norman	\(\frac {-x -\frac {{\mathrm e}^{5} x \ln \left (\ln \left (5 x \right )\right )}{6}-2 x^{2}}{2 x +1}\)	\(28\)
risch	\(\frac {{\mathrm e}^{5} \ln \left (\ln \left (5 x \right )\right )}{24 x +12}-x -\frac {{\mathrm e}^{5} \ln \left (\ln \left (5 x \right )\right )}{12}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-exp(5)*ln(5*x)*ln(ln(5*x))+(-24*x^2-24*x-6)*ln(5*x)+(-2*x-1)*exp(5))/(24*x^2+24*x+6)/ln(5*x),x,method=_R
ETURNVERBOSE)

[Out]

(-x-1/6*exp(5)*x*ln(ln(5*x))-2*x^2)/(2*x+1)

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maxima [A]  time = 0.50, size = 28, normalized size = 1.12 \begin {gather*} -\frac {x e^{5} \log \left (\log \relax (5) + \log \relax (x)\right ) + 12 \, x^{2} + 6 \, x}{6 \, {\left (2 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(5)*log(5*x)*log(log(5*x))+(-24*x^2-24*x-6)*log(5*x)+(-2*x-1)*exp(5))/(24*x^2+24*x+6)/log(5*x),
x, algorithm="maxima")

[Out]

-1/6*(x*e^5*log(log(5) + log(x)) + 12*x^2 + 6*x)/(2*x + 1)

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mupad [B]  time = 3.55, size = 23, normalized size = 0.92 \begin {gather*} -\frac {x\,\left (12\,x+{\mathrm {e}}^5\,\ln \left (\ln \left (5\,x\right )\right )+6\right )}{6\,\left (2\,x+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(5*x)*(24*x + 24*x^2 + 6) + exp(5)*(2*x + 1) + log(5*x)*exp(5)*log(log(5*x)))/(log(5*x)*(24*x + 24*x^
2 + 6)),x)

[Out]

-(x*(12*x + exp(5)*log(log(5*x)) + 6))/(6*(2*x + 1))

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sympy [A]  time = 0.39, size = 27, normalized size = 1.08 \begin {gather*} - x - \frac {e^{5} \log {\left (\log {\left (5 x \right )} \right )}}{12} + \frac {e^{5} \log {\left (\log {\left (5 x \right )} \right )}}{24 x + 12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(5)*ln(5*x)*ln(ln(5*x))+(-24*x**2-24*x-6)*ln(5*x)+(-2*x-1)*exp(5))/(24*x**2+24*x+6)/ln(5*x),x)

[Out]

-x - exp(5)*log(log(5*x))/12 + exp(5)*log(log(5*x))/(24*x + 12)

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