3.16.48 \(\int \frac {-3+5 e^x-5 x+(-5+5 x) \log (x \log (2))}{-3-5 x-5 e^x x+5 x \log (x \log (2))} \, dx\)

Optimal. Leaf size=26 \[ -2+x+\log \left (\frac {1}{3+5 \left (x-x \left (-e^x+\log (x \log (2))\right )\right )}\right ) \]

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Rubi [F]  time = 1.25, 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 {-3+5 e^x-5 x+(-5+5 x) \log (x \log (2))}{-3-5 x-5 e^x x+5 x \log (x \log (2))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-3 + 5*E^x - 5*x + (-5 + 5*x)*Log[x*Log[2]])/(-3 - 5*x - 5*E^x*x + 5*x*Log[x*Log[2]]),x]

[Out]

-Log[x] + 8*Defer[Int][(3 + 5*x + 5*E^x*x - 5*x*Log[x*Log[2]])^(-1), x] + 3*Defer[Int][1/(x*(3 + 5*x + 5*E^x*x
- 5*x*Log[x*Log[2]])), x] + 5*Defer[Int][x/(3 + 5*x + 5*E^x*x - 5*x*Log[x*Log[2]]), x] + 5*Defer[Int][(x*Log[
x*Log[2]])/(-3 - 5*x - 5*E^x*x + 5*x*Log[x*Log[2]]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {1}{x}-\frac {-3-8 x-5 x^2+5 x^2 \log (x \log (2))}{x \left (3+5 x+5 e^x x-5 x \log (x \log (2))\right )}\right ) \, dx\\ &=-\log (x)-\int \frac {-3-8 x-5 x^2+5 x^2 \log (x \log (2))}{x \left (3+5 x+5 e^x x-5 x \log (x \log (2))\right )} \, dx\\ &=-\log (x)-\int \left (-\frac {8}{3+5 x+5 e^x x-5 x \log (x \log (2))}-\frac {3}{x \left (3+5 x+5 e^x x-5 x \log (x \log (2))\right )}-\frac {5 x}{3+5 x+5 e^x x-5 x \log (x \log (2))}-\frac {5 x \log (x \log (2))}{-3-5 x-5 e^x x+5 x \log (x \log (2))}\right ) \, dx\\ &=-\log (x)+3 \int \frac {1}{x \left (3+5 x+5 e^x x-5 x \log (x \log (2))\right )} \, dx+5 \int \frac {x}{3+5 x+5 e^x x-5 x \log (x \log (2))} \, dx+5 \int \frac {x \log (x \log (2))}{-3-5 x-5 e^x x+5 x \log (x \log (2))} \, dx+8 \int \frac {1}{3+5 x+5 e^x x-5 x \log (x \log (2))} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.30, size = 24, normalized size = 0.92 \begin {gather*} x-\log \left (3+5 x+5 e^x x-5 x \log (x \log (2))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3 + 5*E^x - 5*x + (-5 + 5*x)*Log[x*Log[2]])/(-3 - 5*x - 5*E^x*x + 5*x*Log[x*Log[2]]),x]

[Out]

x - Log[3 + 5*x + 5*E^x*x - 5*x*Log[x*Log[2]]]

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fricas [A]  time = 0.60, size = 35, normalized size = 1.35 \begin {gather*} x - \log \left (x \log \relax (2)\right ) - \log \left (-\frac {5 \, x e^{x} - 5 \, x \log \left (x \log \relax (2)\right ) + 5 \, x + 3}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x-5)*log(x*log(2))+5*exp(x)-5*x-3)/(5*x*log(x*log(2))-5*exp(x)*x-5*x-3),x, algorithm="fricas")

[Out]

x - log(x*log(2)) - log(-(5*x*e^x - 5*x*log(x*log(2)) + 5*x + 3)/x)

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giac [A]  time = 0.15, size = 23, normalized size = 0.88 \begin {gather*} x - \log \left (-5 \, x e^{x} + 5 \, x \log \left (x \log \relax (2)\right ) - 5 \, x - 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x-5)*log(x*log(2))+5*exp(x)-5*x-3)/(5*x*log(x*log(2))-5*exp(x)*x-5*x-3),x, algorithm="giac")

[Out]

x - log(-5*x*e^x + 5*x*log(x*log(2)) - 5*x - 3)

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maple [A]  time = 0.04, size = 24, normalized size = 0.92

method	result	size
norman	\(x -\ln \left (-5 x \ln \left (x \ln \relax (2)\right )+5 \,{\mathrm e}^{x} x +5 x +3\right )\)	\(24\)
risch	\(x -\ln \relax (x )-\ln \left (\ln \left (x \ln \relax (2)\right )-\frac {5 \,{\mathrm e}^{x} x +5 x +3}{5 x}\right )\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*x-5)*ln(x*ln(2))+5*exp(x)-5*x-3)/(5*x*ln(x*ln(2))-5*exp(x)*x-5*x-3),x,method=_RETURNVERBOSE)

[Out]

x-ln(-5*x*ln(x*ln(2))+5*exp(x)*x+5*x+3)

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maxima [A]  time = 0.51, size = 34, normalized size = 1.31 \begin {gather*} x - \log \relax (x) - \log \left (-\frac {5 \, x {\left (\log \left (\log \relax (2)\right ) - 1\right )} - 5 \, x e^{x} + 5 \, x \log \relax (x) - 3}{5 \, x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x-5)*log(x*log(2))+5*exp(x)-5*x-3)/(5*x*log(x*log(2))-5*exp(x)*x-5*x-3),x, algorithm="maxima")

[Out]

x - log(x) - log(-1/5*(5*x*(log(log(2)) - 1) - 5*x*e^x + 5*x*log(x) - 3)/x)

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mupad [B]  time = 1.22, size = 24, normalized size = 0.92 \begin {gather*} x-\ln \left (5\,x+5\,x\,{\mathrm {e}}^x-5\,x\,\left (\ln \left (\ln \relax (2)\right )+\ln \relax (x)\right )+3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x - 5*exp(x) - log(x*log(2))*(5*x - 5) + 3)/(5*x - 5*x*log(x*log(2)) + 5*x*exp(x) + 3),x)

[Out]

x - log(5*x + 5*x*exp(x) - 5*x*(log(log(2)) + log(x)) + 3)

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sympy [A]  time = 0.38, size = 27, normalized size = 1.04 \begin {gather*} x - \log {\relax (x )} - \log {\left (e^{x} + \frac {- 5 x \log {\left (x \log {\relax (2 )} \right )} + 5 x + 3}{5 x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x-5)*ln(x*ln(2))+5*exp(x)-5*x-3)/(5*x*ln(x*ln(2))-5*exp(x)*x-5*x-3),x)

[Out]

x - log(x) - log(exp(x) + (-5*x*log(x*log(2)) + 5*x + 3)/(5*x))

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