3.35.51 \(\int \frac {-7+e^x-4 x-\log (4)+(7-e^x+x-4 x^2+(1-x) \log (4)) \log (x)}{(-7 x+e^x x-4 x^2-x \log (4)) \log (x)} \, dx\)

Optimal. Leaf size=28 \[ \log \left (\frac {e^x \log (x)}{x+x^2+\frac {1}{4} x \left (3-e^x+\log (4)\right )}\right ) \]

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Rubi [F]  time = 1.09, 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 {-7+e^x-4 x-\log (4)+\left (7-e^x+x-4 x^2+(1-x) \log (4)\right ) \log (x)}{\left (-7 x+e^x x-4 x^2-x \log (4)\right ) \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-7 + E^x - 4*x - Log[4] + (7 - E^x + x - 4*x^2 + (1 - x)*Log[4])*Log[x])/((-7*x + E^x*x - 4*x^2 - x*Log[4
])*Log[x]),x]

[Out]

-Log[x] + Log[Log[x]] + (3 + Log[4])*Defer[Int][(-E^x + 4*x + 7*(1 + (2*Log[2])/7))^(-1), x] + 4*Defer[Int][x/
(-E^x + 4*x + 7*(1 + (2*Log[2])/7)), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-7+e^x-4 x-\log (4)+\left (7-e^x+x-4 x^2+(1-x) \log (4)\right ) \log (x)}{\left (e^x x-4 x^2+x (-7-\log (4))\right ) \log (x)} \, dx\\ &=\int \frac {e^x-4 x-7 \left (1+\frac {2 \log (2)}{7}\right )+\left (7-e^x+x-4 x^2+(1-x) \log (4)\right ) \log (x)}{\left (e^x x-4 x^2+x (-7-\log (4))\right ) \log (x)} \, dx\\ &=\int \left (\frac {-3-4 x-\log (4)}{e^x-4 x-7 \left (1+\frac {2 \log (2)}{7}\right )}+\frac {1-\log (x)}{x \log (x)}\right ) \, dx\\ &=\int \frac {-3-4 x-\log (4)}{e^x-4 x-7 \left (1+\frac {2 \log (2)}{7}\right )} \, dx+\int \frac {1-\log (x)}{x \log (x)} \, dx\\ &=\int \left (\frac {4 x}{-e^x+4 x+7 \left (1+\frac {2 \log (2)}{7}\right )}+\frac {3 \left (1+\frac {2 \log (2)}{3}\right )}{-e^x+4 x+7 \left (1+\frac {2 \log (2)}{7}\right )}\right ) \, dx+\operatorname {Subst}\left (\int \frac {1-x}{x} \, dx,x,\log (x)\right )\\ &=4 \int \frac {x}{-e^x+4 x+7 \left (1+\frac {2 \log (2)}{7}\right )} \, dx+(3+\log (4)) \int \frac {1}{-e^x+4 x+7 \left (1+\frac {2 \log (2)}{7}\right )} \, dx+\operatorname {Subst}\left (\int \left (-1+\frac {1}{x}\right ) \, dx,x,\log (x)\right )\\ &=-\log (x)+\log (\log (x))+4 \int \frac {x}{-e^x+4 x+7 \left (1+\frac {2 \log (2)}{7}\right )} \, dx+(3+\log (4)) \int \frac {1}{-e^x+4 x+7 \left (1+\frac {2 \log (2)}{7}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.45, size = 24, normalized size = 0.86 \begin {gather*} x-\log (x)-\log \left (7-e^x+4 x+\log (4)\right )+\log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-7 + E^x - 4*x - Log[4] + (7 - E^x + x - 4*x^2 + (1 - x)*Log[4])*Log[x])/((-7*x + E^x*x - 4*x^2 - x
*Log[4])*Log[x]),x]

[Out]

x - Log[x] - Log[7 - E^x + 4*x + Log[4]] + Log[Log[x]]

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fricas [A]  time = 0.73, size = 23, normalized size = 0.82 \begin {gather*} x - \log \relax (x) - \log \left (-4 \, x + e^{x} - 2 \, \log \relax (2) - 7\right ) + \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)+2*(-x+1)*log(2)-4*x^2+x+7)*log(x)+exp(x)-2*log(2)-4*x-7)/(exp(x)*x-2*x*log(2)-4*x^2-7*x)/l
og(x),x, algorithm="fricas")

[Out]

x - log(x) - log(-4*x + e^x - 2*log(2) - 7) + log(log(x))

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giac [A]  time = 0.23, size = 25, normalized size = 0.89 \begin {gather*} x - \log \left (4 \, x - e^{x} + 2 \, \log \relax (2) + 7\right ) - \log \relax (x) + \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)+2*(-x+1)*log(2)-4*x^2+x+7)*log(x)+exp(x)-2*log(2)-4*x-7)/(exp(x)*x-2*x*log(2)-4*x^2-7*x)/l
og(x),x, algorithm="giac")

[Out]

x - log(4*x - e^x + 2*log(2) + 7) - log(x) + log(log(x))

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maple [A]  time = 0.09, size = 24, normalized size = 0.86

method	result	size
risch	\(-\ln \relax (x )+x -\ln \left ({\mathrm e}^{x}-2 \ln \relax (2)-4 x -7\right )+\ln \left (\ln \relax (x )\right )\)	\(24\)
norman	\(x -\ln \relax (x )-\ln \left (2 \ln \relax (2)+4 x -{\mathrm e}^{x}+7\right )+\ln \left (\ln \relax (x )\right )\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-exp(x)+2*(1-x)*ln(2)-4*x^2+x+7)*ln(x)+exp(x)-2*ln(2)-4*x-7)/(exp(x)*x-2*x*ln(2)-4*x^2-7*x)/ln(x),x,meth
od=_RETURNVERBOSE)

[Out]

-ln(x)+x-ln(exp(x)-2*ln(2)-4*x-7)+ln(ln(x))

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maxima [A]  time = 0.57, size = 23, normalized size = 0.82 \begin {gather*} x - \log \relax (x) - \log \left (-4 \, x + e^{x} - 2 \, \log \relax (2) - 7\right ) + \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)+2*(-x+1)*log(2)-4*x^2+x+7)*log(x)+exp(x)-2*log(2)-4*x-7)/(exp(x)*x-2*x*log(2)-4*x^2-7*x)/l
og(x),x, algorithm="maxima")

[Out]

x - log(x) - log(-4*x + e^x - 2*log(2) - 7) + log(log(x))

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mupad [B]  time = 2.35, size = 23, normalized size = 0.82 \begin {gather*} x-\ln \left ({\mathrm {e}}^x-\ln \relax (4)-4\,x-7\right )+\ln \left (\ln \relax (x)\right )-\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x + 2*log(2) - exp(x) + log(x)*(exp(x) - x + 2*log(2)*(x - 1) + 4*x^2 - 7) + 7)/(log(x)*(7*x + 2*x*log(
2) - x*exp(x) + 4*x^2)),x)

[Out]

x - log(exp(x) - log(4) - 4*x - 7) + log(log(x)) - log(x)

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sympy [A]  time = 0.33, size = 24, normalized size = 0.86 \begin {gather*} x - \log {\relax (x )} - \log {\left (- 4 x + e^{x} - 7 - 2 \log {\relax (2 )} \right )} + \log {\left (\log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)+2*(-x+1)*ln(2)-4*x**2+x+7)*ln(x)+exp(x)-2*ln(2)-4*x-7)/(exp(x)*x-2*x*ln(2)-4*x**2-7*x)/ln(
x),x)

[Out]

x - log(x) - log(-4*x + exp(x) - 7 - 2*log(2)) + log(log(x))

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