3.29.23 \(\int \frac {-101+e^3-4 e^x+8 x+(-8 x+4 e^x x) \log (x)}{10201 x+e^6 x+16 e^{2 x} x-1616 x^2+64 x^3+e^x (808 x-8 e^3 x-64 x^2)+e^3 (-202 x+16 x^2)} \, dx\)

Optimal. Leaf size=20 \[ \frac {\log (x)}{-1+e^3-4 \left (25+e^x-2 x\right )} \]

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Rubi [F]  time = 1.36, 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 {-101+e^3-4 e^x+8 x+\left (-8 x+4 e^x x\right ) \log (x)}{10201 x+e^6 x+16 e^{2 x} x-1616 x^2+64 x^3+e^x \left (808 x-8 e^3 x-64 x^2\right )+e^3 \left (-202 x+16 x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-101 + E^3 - 4*E^x + 8*x + (-8*x + 4*E^x*x)*Log[x])/(10201*x + E^6*x + 16*E^(2*x)*x - 1616*x^2 + 64*x^3 +
E^x*(808*x - 8*E^3*x - 64*x^2) + E^3*(-202*x + 16*x^2)),x]

[Out]

-((109 - E^3)*Log[x]*Defer[Int][(4*E^x + 101*(1 - E^3/101) - 8*x)^(-2), x]) + Log[x]*Defer[Int][(4*E^x + 101*(
1 - E^3/101) - 8*x)^(-1), x] + 8*Log[x]*Defer[Int][x/(4*E^x + 101*(1 - E^3/101) - 8*x)^2, x] + Defer[Int][1/(x
*(-4*E^x - 101*(1 - E^3/101) + 8*x)), x] - Defer[Int][Defer[Int][(4*E^x + 101*(1 - E^3/101) - 8*x)^(-1), x]/x,
x] + (109 - E^3)*Defer[Int][Defer[Int][(-4*E^x - 101*(1 - E^3/101) + 8*x)^(-2), x]/x, x] - 8*Defer[Int][Defer
[Int][x/(-4*E^x - 101*(1 - E^3/101) + 8*x)^2, x]/x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-101+e^3-4 e^x+8 x+\left (-8 x+4 e^x x\right ) \log (x)}{16 e^{2 x} x+\left (10201+e^6\right ) x-1616 x^2+64 x^3+e^x \left (808 x-8 e^3 x-64 x^2\right )+e^3 \left (-202 x+16 x^2\right )} \, dx\\ &=\int \frac {-4 e^x-101 \left (1-\frac {e^3}{101}\right )+8 x+4 \left (-2+e^x\right ) x \log (x)}{\left (4 e^x+101 \left (1-\frac {e^3}{101}\right )-8 x\right )^2 x} \, dx\\ &=\int \left (\frac {\left (-109+e^3+8 x\right ) \log (x)}{\left (4 e^x+101 \left (1-\frac {e^3}{101}\right )-8 x\right )^2}+\frac {-1+x \log (x)}{\left (4 e^x+101 \left (1-\frac {e^3}{101}\right )-8 x\right ) x}\right ) \, dx\\ &=\int \frac {\left (-109+e^3+8 x\right ) \log (x)}{\left (4 e^x+101 \left (1-\frac {e^3}{101}\right )-8 x\right )^2} \, dx+\int \frac {-1+x \log (x)}{\left (4 e^x+101 \left (1-\frac {e^3}{101}\right )-8 x\right ) x} \, dx\\ &=(8 \log (x)) \int \frac {x}{\left (4 e^x+101 \left (1-\frac {e^3}{101}\right )-8 x\right )^2} \, dx-\left (\left (109-e^3\right ) \log (x)\right ) \int \frac {1}{\left (4 e^x+101 \left (1-\frac {e^3}{101}\right )-8 x\right )^2} \, dx+\int \left (\frac {1}{x \left (-4 e^x-101 \left (1-\frac {e^3}{101}\right )+8 x\right )}+\frac {\log (x)}{4 e^x+101 \left (1-\frac {e^3}{101}\right )-8 x}\right ) \, dx-\int \frac {\left (-109+e^3\right ) \int \frac {1}{\left (-101+e^3-4 e^x+8 x\right )^2} \, dx+8 \int \frac {x}{\left (-101+e^3-4 e^x+8 x\right )^2} \, dx}{x} \, dx\\ &=(8 \log (x)) \int \frac {x}{\left (4 e^x+101 \left (1-\frac {e^3}{101}\right )-8 x\right )^2} \, dx-\left (\left (109-e^3\right ) \log (x)\right ) \int \frac {1}{\left (4 e^x+101 \left (1-\frac {e^3}{101}\right )-8 x\right )^2} \, dx+\int \frac {1}{x \left (-4 e^x-101 \left (1-\frac {e^3}{101}\right )+8 x\right )} \, dx+\int \frac {\log (x)}{4 e^x+101 \left (1-\frac {e^3}{101}\right )-8 x} \, dx-\int \left (\frac {\left (-109+e^3\right ) \int \frac {1}{\left (-4 e^x-101 \left (1-\frac {e^3}{101}\right )+8 x\right )^2} \, dx}{x}+\frac {8 \int \frac {x}{\left (-4 e^x-101 \left (1-\frac {e^3}{101}\right )+8 x\right )^2} \, dx}{x}\right ) \, dx\\ &=-\left (8 \int \frac {\int \frac {x}{\left (-4 e^x-101 \left (1-\frac {e^3}{101}\right )+8 x\right )^2} \, dx}{x} \, dx\right )-\left (-109+e^3\right ) \int \frac {\int \frac {1}{\left (-4 e^x-101 \left (1-\frac {e^3}{101}\right )+8 x\right )^2} \, dx}{x} \, dx+\log (x) \int \frac {1}{4 e^x+101 \left (1-\frac {e^3}{101}\right )-8 x} \, dx+(8 \log (x)) \int \frac {x}{\left (4 e^x+101 \left (1-\frac {e^3}{101}\right )-8 x\right )^2} \, dx-\left (\left (109-e^3\right ) \log (x)\right ) \int \frac {1}{\left (4 e^x+101 \left (1-\frac {e^3}{101}\right )-8 x\right )^2} \, dx+\int \frac {1}{x \left (-4 e^x-101 \left (1-\frac {e^3}{101}\right )+8 x\right )} \, dx-\int \frac {\int \frac {1}{4 e^x+101 \left (1-\frac {e^3}{101}\right )-8 x} \, dx}{x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.35, size = 21, normalized size = 1.05 \begin {gather*} -\frac {\log (x)}{101-e^3+4 e^x-8 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-101 + E^3 - 4*E^x + 8*x + (-8*x + 4*E^x*x)*Log[x])/(10201*x + E^6*x + 16*E^(2*x)*x - 1616*x^2 + 64
*x^3 + E^x*(808*x - 8*E^3*x - 64*x^2) + E^3*(-202*x + 16*x^2)),x]

[Out]

-(Log[x]/(101 - E^3 + 4*E^x - 8*x))

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fricas [A]  time = 0.43, size = 16, normalized size = 0.80 \begin {gather*} \frac {\log \relax (x)}{8 \, x + e^{3} - 4 \, e^{x} - 101} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(x)*x-8*x)*log(x)-4*exp(x)+exp(3)+8*x-101)/(16*x*exp(x)^2+(-8*x*exp(3)-64*x^2+808*x)*exp(x)+x
*exp(3)^2+(16*x^2-202*x)*exp(3)+64*x^3-1616*x^2+10201*x),x, algorithm="fricas")

[Out]

log(x)/(8*x + e^3 - 4*e^x - 101)

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giac [A]  time = 0.33, size = 16, normalized size = 0.80 \begin {gather*} \frac {\log \relax (x)}{8 \, x + e^{3} - 4 \, e^{x} - 101} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(x)*x-8*x)*log(x)-4*exp(x)+exp(3)+8*x-101)/(16*x*exp(x)^2+(-8*x*exp(3)-64*x^2+808*x)*exp(x)+x
*exp(3)^2+(16*x^2-202*x)*exp(3)+64*x^3-1616*x^2+10201*x),x, algorithm="giac")

[Out]

log(x)/(8*x + e^3 - 4*e^x - 101)

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maple [A]  time = 0.03, size = 17, normalized size = 0.85

method	result	size
risch	\(\frac {\ln \relax (x )}{-4 \,{\mathrm e}^{x}+{\mathrm e}^{3}+8 x -101}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*exp(x)*x-8*x)*ln(x)-4*exp(x)+exp(3)+8*x-101)/(16*x*exp(x)^2+(-8*x*exp(3)-64*x^2+808*x)*exp(x)+x*exp(3)
^2+(16*x^2-202*x)*exp(3)+64*x^3-1616*x^2+10201*x),x,method=_RETURNVERBOSE)

[Out]

ln(x)/(-4*exp(x)+exp(3)+8*x-101)

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maxima [A]  time = 0.74, size = 16, normalized size = 0.80 \begin {gather*} \frac {\log \relax (x)}{8 \, x + e^{3} - 4 \, e^{x} - 101} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(x)*x-8*x)*log(x)-4*exp(x)+exp(3)+8*x-101)/(16*x*exp(x)^2+(-8*x*exp(3)-64*x^2+808*x)*exp(x)+x
*exp(3)^2+(16*x^2-202*x)*exp(3)+64*x^3-1616*x^2+10201*x),x, algorithm="maxima")

[Out]

log(x)/(8*x + e^3 - 4*e^x - 101)

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mupad [B]  time = 2.21, size = 16, normalized size = 0.80 \begin {gather*} \frac {\ln \relax (x)}{8\,x+{\mathrm {e}}^3-4\,{\mathrm {e}}^x-101} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*exp(x) - exp(3) - 8*x + log(x)*(8*x - 4*x*exp(x)) + 101)/(10201*x + 16*x*exp(2*x) - exp(3)*(202*x - 16
*x^2) + x*exp(6) - 1616*x^2 + 64*x^3 - exp(x)*(8*x*exp(3) - 808*x + 64*x^2)),x)

[Out]

log(x)/(8*x + exp(3) - 4*exp(x) - 101)

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sympy [A]  time = 0.27, size = 17, normalized size = 0.85 \begin {gather*} - \frac {\log {\relax (x )}}{- 8 x + 4 e^{x} - e^{3} + 101} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(x)*x-8*x)*ln(x)-4*exp(x)+exp(3)+8*x-101)/(16*x*exp(x)**2+(-8*x*exp(3)-64*x**2+808*x)*exp(x)+
x*exp(3)**2+(16*x**2-202*x)*exp(3)+64*x**3-1616*x**2+10201*x),x)

[Out]

-log(x)/(-8*x + 4*exp(x) - exp(3) + 101)

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