3.81.46 \(\int \frac {e^{25} (1+4 x)}{4 (-4 x+4 x^2+x \log (6 x))} \, dx\) [8046]

Optimal. Leaf size=17 \[ \frac {1}{4} e^{25} \log (-4+4 x+\log (6 x)) \]

[Out]

exp(ln(ln(ln(6*x)+4*x-4))-2*ln(2)+25)

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Rubi [F]
time = 0.09, 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^{25} (1+4 x)}{4 \left (-4 x+4 x^2+x \log (6 x)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^25*(1 + 4*x))/(4*(-4*x + 4*x^2 + x*Log[6*x])),x]

[Out]

(E^25*Defer[Int][1/(x*(-4 + 4*x + Log[6*x])), x])/4 + (E^25*Defer[Subst][Defer[Int][(-4 + 2*x + Log[3*x])^(-1)
, x], x, 2*x])/2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} e^{25} \int \frac {1+4 x}{-4 x+4 x^2+x \log (6 x)} \, dx\\ &=\frac {1}{4} e^{25} \int \left (\frac {4}{-4+4 x+\log (6 x)}+\frac {1}{x (-4+4 x+\log (6 x))}\right ) \, dx\\ &=\frac {1}{4} e^{25} \int \frac {1}{x (-4+4 x+\log (6 x))} \, dx+e^{25} \int \frac {1}{-4+4 x+\log (6 x)} \, dx\\ &=\frac {1}{4} e^{25} \int \frac {1}{x (-4+4 x+\log (6 x))} \, dx+\frac {1}{2} e^{25} \text {Subst}\left (\int \frac {1}{-4+2 x+\log (3 x)} \, dx,x,2 x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.05, size = 19, normalized size = 1.12 \begin {gather*} \frac {1}{4} e^{25} \log (12-12 x-3 \log (6 x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^25*(1 + 4*x))/(4*(-4*x + 4*x^2 + x*Log[6*x])),x]

[Out]

(E^25*Log[12 - 12*x - 3*Log[6*x]])/4

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Maple [A]
time = 2.18, size = 17, normalized size = 1.00

method	result	size
norman	\(\frac {\ln \left (\ln \left (6 x \right )+4 x -4\right ) {\mathrm e}^{25}}{4}\)	\(15\)
risch	\(\frac {\ln \left (\ln \left (6 x \right )+4 x -4\right ) {\mathrm e}^{25}}{4}\)	\(15\)
default	\(\frac {{\mathrm e}^{25} \ln \left (\ln \left (2\right )+\ln \left (3\right )+\ln \left (x \right )+4 x -4\right )}{4}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x+1)*exp(ln(ln(ln(6*x)+4*x-4))-2*ln(2)+25)/(x*ln(6*x)+4*x^2-4*x)/ln(ln(6*x)+4*x-4),x,method=_RETURNVERB
OSE)

[Out]

1/4*exp(25)*ln(ln(2)+ln(3)+ln(x)+4*x-4)

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Maxima [A]
time = 0.53, size = 16, normalized size = 0.94 \begin {gather*} \frac {1}{4} \, e^{25} \log \left (4 \, x + \log \left (3\right ) + \log \left (2\right ) + \log \left (x\right ) - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+4*x)*exp(log(log(log(6*x)+4*x-4))-2*log(2)+25)/(x*log(6*x)+4*x^2-4*x)/log(log(6*x)+4*x-4),x, algo
rithm="maxima")

[Out]

1/4*e^25*log(4*x + log(3) + log(2) + log(x) - 4)

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Fricas [A]
time = 0.37, size = 18, normalized size = 1.06 \begin {gather*} e^{\left (-2 \, \log \left (2\right ) + 25\right )} \log \left (4 \, x + \log \left (6 \, x\right ) - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+4*x)*exp(log(log(log(6*x)+4*x-4))-2*log(2)+25)/(x*log(6*x)+4*x^2-4*x)/log(log(6*x)+4*x-4),x, algo
rithm="fricas")

[Out]

e^(-2*log(2) + 25)*log(4*x + log(6*x) - 4)

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Sympy [A]
time = 0.07, size = 15, normalized size = 0.88 \begin {gather*} \frac {e^{25} \log {\left (4 x + \log {\left (6 x \right )} - 4 \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+4*x)*exp(ln(ln(ln(6*x)+4*x-4))-2*ln(2)+25)/(x*ln(6*x)+4*x**2-4*x)/ln(ln(6*x)+4*x-4),x)

[Out]

exp(25)*log(4*x + log(6*x) - 4)/4

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Giac [A]
time = 0.41, size = 14, normalized size = 0.82 \begin {gather*} \frac {1}{4} \, e^{25} \log \left (4 \, x + \log \left (6\right ) + \log \left (x\right ) - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+4*x)*exp(log(log(log(6*x)+4*x-4))-2*log(2)+25)/(x*log(6*x)+4*x^2-4*x)/log(log(6*x)+4*x-4),x, algo
rithm="giac")

[Out]

1/4*e^25*log(4*x + log(6) + log(x) - 4)

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Mupad [B]
time = 6.57, size = 14, normalized size = 0.82 \begin {gather*} \frac {{\mathrm {e}}^{25}\,\ln \left (4\,x+\ln \left (6\,x\right )-4\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(log(log(4*x + log(6*x) - 4)) - 2*log(2) + 25)*(4*x + 1))/(log(4*x + log(6*x) - 4)*(x*log(6*x) - 4*x +
4*x^2)),x)

[Out]

(exp(25)*log(4*x + log(6*x) - 4))/4

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