3.22.25 \(\int \frac {\frac {5+18 x}{e^4 \log (\frac {1}{5} (5 x+9 x^2))}+(-5 x-9 x^2) \log (\frac {1}{5} (5 x+9 x^2))}{(5 x+9 x^2) \log (\frac {1}{5} (5 x+9 x^2))} \, dx\)

Optimal. Leaf size=22 \[ 2-x-\frac {1}{e^4 \log \left (x+\frac {9 x^2}{5}\right )} \]

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Rubi [F]  time = 0.73, 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 {\frac {5+18 x}{e^4 \log \left (\frac {1}{5} \left (5 x+9 x^2\right )\right )}+\left (-5 x-9 x^2\right ) \log \left (\frac {1}{5} \left (5 x+9 x^2\right )\right )}{\left (5 x+9 x^2\right ) \log \left (\frac {1}{5} \left (5 x+9 x^2\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((5 + 18*x)/(E^4*Log[(5*x + 9*x^2)/5]) + (-5*x - 9*x^2)*Log[(5*x + 9*x^2)/5])/((5*x + 9*x^2)*Log[(5*x + 9*
x^2)/5]),x]

[Out]

-x + Defer[Int][(5 + 18*x)/(x*(5 + 9*x)*Log[x*(1 + (9*x)/5)]^2), x]/E^4

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\frac {5+18 x}{e^4 \log \left (\frac {1}{5} \left (5 x+9 x^2\right )\right )}+\left (-5 x-9 x^2\right ) \log \left (\frac {1}{5} \left (5 x+9 x^2\right )\right )}{x (5+9 x) \log \left (\frac {1}{5} \left (5 x+9 x^2\right )\right )} \, dx\\ &=\int \frac {5+18 x-e^4 x (5+9 x) \log ^2\left (x+\frac {9 x^2}{5}\right )}{e^4 x (5+9 x) \log ^2\left (x \left (1+\frac {9 x}{5}\right )\right )} \, dx\\ &=\frac {\int \frac {5+18 x-e^4 x (5+9 x) \log ^2\left (x+\frac {9 x^2}{5}\right )}{x (5+9 x) \log ^2\left (x \left (1+\frac {9 x}{5}\right )\right )} \, dx}{e^4}\\ &=\frac {\int \left (-e^4+\frac {5+18 x}{x (5+9 x) \log ^2\left (x \left (1+\frac {9 x}{5}\right )\right )}\right ) \, dx}{e^4}\\ &=-x+\frac {\int \frac {5+18 x}{x (5+9 x) \log ^2\left (x \left (1+\frac {9 x}{5}\right )\right )} \, dx}{e^4}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 25, normalized size = 1.14 \begin {gather*} \frac {-e^4 x-\frac {1}{\log \left (x+\frac {9 x^2}{5}\right )}}{e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((5 + 18*x)/(E^4*Log[(5*x + 9*x^2)/5]) + (-5*x - 9*x^2)*Log[(5*x + 9*x^2)/5])/((5*x + 9*x^2)*Log[(5*
x + 9*x^2)/5]),x]

[Out]

(-(E^4*x) - Log[x + (9*x^2)/5]^(-1))/E^4

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fricas [A]  time = 0.83, size = 28, normalized size = 1.27 \begin {gather*} -\frac {{\left (x e^{4} \log \left (\frac {9}{5} \, x^{2} + x\right ) + 1\right )} e^{\left (-4\right )}}{\log \left (\frac {9}{5} \, x^{2} + x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((18*x+5)*exp(-log(log(9/5*x^2+x))-4)+(-9*x^2-5*x)*log(9/5*x^2+x))/(9*x^2+5*x)/log(9/5*x^2+x),x, alg
orithm="fricas")

[Out]

-(x*e^4*log(9/5*x^2 + x) + 1)*e^(-4)/log(9/5*x^2 + x)

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giac [B]  time = 0.26, size = 52, normalized size = 2.36 \begin {gather*} -\frac {x e^{4} \log \relax (5) - x e^{4} \log \left (9 \, x + 5\right ) - x e^{4} \log \relax (x) - 1}{e^{4} \log \relax (5) - e^{4} \log \left (9 \, x + 5\right ) - e^{4} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((18*x+5)*exp(-log(log(9/5*x^2+x))-4)+(-9*x^2-5*x)*log(9/5*x^2+x))/(9*x^2+5*x)/log(9/5*x^2+x),x, alg
orithm="giac")

[Out]

-(x*e^4*log(5) - x*e^4*log(9*x + 5) - x*e^4*log(x) - 1)/(e^4*log(5) - e^4*log(9*x + 5) - e^4*log(x))

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maple [A]  time = 0.36, size = 24, normalized size = 1.09

method	result	size
default	\(-x -\frac {{\mathrm e}^{-4}}{-\ln \relax (5)+\ln \left (x \left (9 x +5\right )\right )}\)	\(24\)
norman	\(\frac {-{\mathrm e}^{-4}-x \ln \left (\frac {9}{5} x^{2}+x \right )}{\ln \left (\frac {9}{5} x^{2}+x \right )}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((18*x+5)*exp(-ln(ln(9/5*x^2+x))-4)+(-9*x^2-5*x)*ln(9/5*x^2+x))/(9*x^2+5*x)/ln(9/5*x^2+x),x,method=_RETURN
VERBOSE)

[Out]

-x-exp(-4)/(-ln(5)+ln(x*(9*x+5)))

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maxima [A]  time = 0.88, size = 28, normalized size = 1.27 \begin {gather*} -x + \frac {1}{e^{4} \log \relax (5) - e^{4} \log \left (9 \, x + 5\right ) - e^{4} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((18*x+5)*exp(-log(log(9/5*x^2+x))-4)+(-9*x^2-5*x)*log(9/5*x^2+x))/(9*x^2+5*x)/log(9/5*x^2+x),x, alg
orithm="maxima")

[Out]

-x + 1/(e^4*log(5) - e^4*log(9*x + 5) - e^4*log(x))

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mupad [B]  time = 1.51, size = 18, normalized size = 0.82 \begin {gather*} -x-\frac {{\mathrm {e}}^{-4}}{\ln \left (\frac {9\,x^2}{5}+x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x + (9*x^2)/5)*(5*x + 9*x^2) - exp(- log(log(x + (9*x^2)/5)) - 4)*(18*x + 5))/(log(x + (9*x^2)/5)*(5
*x + 9*x^2)),x)

[Out]

- x - exp(-4)/log(x + (9*x^2)/5)

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sympy [A]  time = 0.14, size = 17, normalized size = 0.77 \begin {gather*} - x - \frac {1}{e^{4} \log {\left (\frac {9 x^{2}}{5} + x \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((18*x+5)*exp(-ln(ln(9/5*x**2+x))-4)+(-9*x**2-5*x)*ln(9/5*x**2+x))/(9*x**2+5*x)/ln(9/5*x**2+x),x)

[Out]

-x - exp(-4)/log(9*x**2/5 + x)

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