3.54.93 \(\int \frac {-4 e-8 x+e^x (-64-56 x+4 x^2)+32 e^x \log (x)+e^x (16+16 x) \log ^2(x)}{-e x-x^2+e^x (-16 x+x^2)+4 e^x x \log ^2(x)} \, dx\)

Optimal. Leaf size=26 \[ 4 \log \left (x \left (-e-x+e^x \left (x+4 \left (-4+\log ^2(x)\right )\right )\right )\right ) \]

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Rubi [F]  time = 5.98, 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 {-4 e-8 x+e^x \left (-64-56 x+4 x^2\right )+32 e^x \log (x)+e^x (16+16 x) \log ^2(x)}{-e x-x^2+e^x \left (-16 x+x^2\right )+4 e^x x \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-4*E - 8*x + E^x*(-64 - 56*x + 4*x^2) + 32*E^x*Log[x] + E^x*(16 + 16*x)*Log[x]^2)/(-(E*x) - x^2 + E^x*(-1
6*x + x^2) + 4*E^x*x*Log[x]^2),x]

[Out]

4*x + 4*Log[x] + 4*Log[16 - x - 4*Log[x]^2] - 4*(16 - 15*E)*Defer[Int][1/((-16 + x + 4*Log[x]^2)*(E + 16*E^x +
x - E^x*x - 4*E^x*Log[x]^2)), x] + 4*(16 - E)*Defer[Int][x/((-16 + x + 4*Log[x]^2)*(E + 16*E^x + x - E^x*x -
4*E^x*Log[x]^2)), x] - 32*E*Defer[Int][Log[x]/(x*(-16 + x + 4*Log[x]^2)*(E + 16*E^x + x - E^x*x - 4*E^x*Log[x]
^2)), x] + 16*(1 - E)*Defer[Int][Log[x]^2/((-16 + x + 4*Log[x]^2)*(E + 16*E^x + x - E^x*x - 4*E^x*Log[x]^2)),
x] + 4*Defer[Int][x^2/((-16 + x + 4*Log[x]^2)*(-E - 16*E^x - x + E^x*x + 4*E^x*Log[x]^2)), x] + 32*Defer[Int][
Log[x]/((-16 + x + 4*Log[x]^2)*(-E - 16*E^x - x + E^x*x + 4*E^x*Log[x]^2)), x] + 16*Defer[Int][(x*Log[x]^2)/((
-16 + x + 4*Log[x]^2)*(-E - 16*E^x - x + E^x*x + 4*E^x*Log[x]^2)), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {4 \left (-16-14 x+x^2+8 \log (x)+4 \log ^2(x)+4 x \log ^2(x)\right )}{x \left (-16+x+4 \log ^2(x)\right )}+\frac {4 \left (16 \left (1-\frac {15 e}{16}\right ) x-16 \left (1-\frac {e}{16}\right ) x^2+x^3+8 e \log (x)+8 x \log (x)-4 (1-e) x \log ^2(x)+4 x^2 \log ^2(x)\right )}{x \left (16-x-4 \log ^2(x)\right ) \left (e+16 e^x+x-e^x x-4 e^x \log ^2(x)\right )}\right ) \, dx\\ &=4 \int \frac {-16-14 x+x^2+8 \log (x)+4 \log ^2(x)+4 x \log ^2(x)}{x \left (-16+x+4 \log ^2(x)\right )} \, dx+4 \int \frac {16 \left (1-\frac {15 e}{16}\right ) x-16 \left (1-\frac {e}{16}\right ) x^2+x^3+8 e \log (x)+8 x \log (x)-4 (1-e) x \log ^2(x)+4 x^2 \log ^2(x)}{x \left (16-x-4 \log ^2(x)\right ) \left (e+16 e^x+x-e^x x-4 e^x \log ^2(x)\right )} \, dx\\ &=4 \int \frac {x \left (16+e (-15+x)-16 x+x^2\right )+8 (e+x) \log (x)+4 x (-1+e+x) \log ^2(x)}{x \left (16-x-4 \log ^2(x)\right ) \left (e-e^x (-16+x)+x-4 e^x \log ^2(x)\right )} \, dx+4 \int \left (\frac {1+x}{x}+\frac {x+8 \log (x)}{x \left (-16+x+4 \log ^2(x)\right )}\right ) \, dx\\ &=4 \int \frac {1+x}{x} \, dx+4 \int \frac {x+8 \log (x)}{x \left (-16+x+4 \log ^2(x)\right )} \, dx+4 \int \left (\frac {15 \left (1-\frac {16}{15 e}\right ) e}{\left (-16+x+4 \log ^2(x)\right ) \left (e+16 e^x+x-e^x x-4 e^x \log ^2(x)\right )}-\frac {\left (1-\frac {16}{e}\right ) e x}{\left (-16+x+4 \log ^2(x)\right ) \left (e+16 e^x+x-e^x x-4 e^x \log ^2(x)\right )}-\frac {8 e \log (x)}{x \left (-16+x+4 \log ^2(x)\right ) \left (e+16 e^x+x-e^x x-4 e^x \log ^2(x)\right )}-\frac {4 \left (1-\frac {1}{e}\right ) e \log ^2(x)}{\left (-16+x+4 \log ^2(x)\right ) \left (e+16 e^x+x-e^x x-4 e^x \log ^2(x)\right )}+\frac {x^2}{\left (-16+x+4 \log ^2(x)\right ) \left (-e-16 e^x-x+e^x x+4 e^x \log ^2(x)\right )}+\frac {8 \log (x)}{\left (-16+x+4 \log ^2(x)\right ) \left (-e-16 e^x-x+e^x x+4 e^x \log ^2(x)\right )}+\frac {4 x \log ^2(x)}{\left (-16+x+4 \log ^2(x)\right ) \left (-e-16 e^x-x+e^x x+4 e^x \log ^2(x)\right )}\right ) \, dx\\ &=4 \log \left (16-x-4 \log ^2(x)\right )+4 \int \left (1+\frac {1}{x}\right ) \, dx+4 \int \frac {x^2}{\left (-16+x+4 \log ^2(x)\right ) \left (-e-16 e^x-x+e^x x+4 e^x \log ^2(x)\right )} \, dx+16 \int \frac {x \log ^2(x)}{\left (-16+x+4 \log ^2(x)\right ) \left (-e-16 e^x-x+e^x x+4 e^x \log ^2(x)\right )} \, dx+32 \int \frac {\log (x)}{\left (-16+x+4 \log ^2(x)\right ) \left (-e-16 e^x-x+e^x x+4 e^x \log ^2(x)\right )} \, dx-(4 (16-15 e)) \int \frac {1}{\left (-16+x+4 \log ^2(x)\right ) \left (e+16 e^x+x-e^x x-4 e^x \log ^2(x)\right )} \, dx+(16 (1-e)) \int \frac {\log ^2(x)}{\left (-16+x+4 \log ^2(x)\right ) \left (e+16 e^x+x-e^x x-4 e^x \log ^2(x)\right )} \, dx+(4 (16-e)) \int \frac {x}{\left (-16+x+4 \log ^2(x)\right ) \left (e+16 e^x+x-e^x x-4 e^x \log ^2(x)\right )} \, dx-(32 e) \int \frac {\log (x)}{x \left (-16+x+4 \log ^2(x)\right ) \left (e+16 e^x+x-e^x x-4 e^x \log ^2(x)\right )} \, dx\\ &=4 x+4 \log (x)+4 \log \left (16-x-4 \log ^2(x)\right )+4 \int \frac {x^2}{\left (-16+x+4 \log ^2(x)\right ) \left (-e-16 e^x-x+e^x x+4 e^x \log ^2(x)\right )} \, dx+16 \int \frac {x \log ^2(x)}{\left (-16+x+4 \log ^2(x)\right ) \left (-e-16 e^x-x+e^x x+4 e^x \log ^2(x)\right )} \, dx+32 \int \frac {\log (x)}{\left (-16+x+4 \log ^2(x)\right ) \left (-e-16 e^x-x+e^x x+4 e^x \log ^2(x)\right )} \, dx-(4 (16-15 e)) \int \frac {1}{\left (-16+x+4 \log ^2(x)\right ) \left (e+16 e^x+x-e^x x-4 e^x \log ^2(x)\right )} \, dx+(16 (1-e)) \int \frac {\log ^2(x)}{\left (-16+x+4 \log ^2(x)\right ) \left (e+16 e^x+x-e^x x-4 e^x \log ^2(x)\right )} \, dx+(4 (16-e)) \int \frac {x}{\left (-16+x+4 \log ^2(x)\right ) \left (e+16 e^x+x-e^x x-4 e^x \log ^2(x)\right )} \, dx-(32 e) \int \frac {\log (x)}{x \left (-16+x+4 \log ^2(x)\right ) \left (e+16 e^x+x-e^x x-4 e^x \log ^2(x)\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.21, size = 36, normalized size = 1.38 \begin {gather*} -4 \left (-\log (x)-\log \left (-e-16 e^x-x+e^x x+4 e^x \log ^2(x)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4*E - 8*x + E^x*(-64 - 56*x + 4*x^2) + 32*E^x*Log[x] + E^x*(16 + 16*x)*Log[x]^2)/(-(E*x) - x^2 + E
^x*(-16*x + x^2) + 4*E^x*x*Log[x]^2),x]

[Out]

-4*(-Log[x] - Log[-E - 16*E^x - x + E^x*x + 4*E^x*Log[x]^2])

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fricas [A]  time = 0.88, size = 38, normalized size = 1.46 \begin {gather*} 4 \, x + 4 \, \log \left ({\left (4 \, e^{x} \log \relax (x)^{2} + {\left (x - 16\right )} e^{x} - x - e\right )} e^{\left (-x\right )}\right ) + 4 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x+16)*exp(x)*log(x)^2+32*exp(x)*log(x)+(4*x^2-56*x-64)*exp(x)-4*exp(1)-8*x)/(4*x*exp(x)*log(x)^
2+(x^2-16*x)*exp(x)-x*exp(1)-x^2),x, algorithm="fricas")

[Out]

4*x + 4*log((4*e^x*log(x)^2 + (x - 16)*e^x - x - e)*e^(-x)) + 4*log(x)

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giac [A]  time = 0.27, size = 29, normalized size = 1.12 \begin {gather*} 4 \, \log \left (-4 \, e^{x} \log \relax (x)^{2} - x e^{x} + x + e + 16 \, e^{x}\right ) + 4 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x+16)*exp(x)*log(x)^2+32*exp(x)*log(x)+(4*x^2-56*x-64)*exp(x)-4*exp(1)-8*x)/(4*x*exp(x)*log(x)^
2+(x^2-16*x)*exp(x)-x*exp(1)-x^2),x, algorithm="giac")

[Out]

4*log(-4*e^x*log(x)^2 - x*e^x + x + e + 16*e^x) + 4*log(x)

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maple [A]  time = 0.11, size = 30, normalized size = 1.15

method	result	size
norman	\(4 \ln \relax (x )+4 \ln \left (-4 \,{\mathrm e}^{x} \ln \relax (x )^{2}-{\mathrm e}^{x} x +{\mathrm e}+x +16 \,{\mathrm e}^{x}\right )\)	\(30\)
risch	\(4 x +4 \ln \relax (x )+4 \ln \left (\ln \relax (x )^{2}-\frac {\left (-{\mathrm e}^{x} x +{\mathrm e}+x +16 \,{\mathrm e}^{x}\right ) {\mathrm e}^{-x}}{4}\right )\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((16*x+16)*exp(x)*ln(x)^2+32*exp(x)*ln(x)+(4*x^2-56*x-64)*exp(x)-4*exp(1)-8*x)/(4*x*exp(x)*ln(x)^2+(x^2-16
*x)*exp(x)-x*exp(1)-x^2),x,method=_RETURNVERBOSE)

[Out]

4*ln(x)+4*ln(-4*exp(x)*ln(x)^2-exp(x)*x+exp(1)+x+16*exp(x))

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maxima [B]  time = 0.41, size = 52, normalized size = 2.00 \begin {gather*} 4 \, \log \left (\log \relax (x)^{2} + \frac {1}{4} \, x - 4\right ) + 4 \, \log \relax (x) + 4 \, \log \left (\frac {{\left (4 \, \log \relax (x)^{2} + x - 16\right )} e^{x} - x - e}{4 \, \log \relax (x)^{2} + x - 16}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x+16)*exp(x)*log(x)^2+32*exp(x)*log(x)+(4*x^2-56*x-64)*exp(x)-4*exp(1)-8*x)/(4*x*exp(x)*log(x)^
2+(x^2-16*x)*exp(x)-x*exp(1)-x^2),x, algorithm="maxima")

[Out]

4*log(log(x)^2 + 1/4*x - 4) + 4*log(x) + 4*log(((4*log(x)^2 + x - 16)*e^x - x - e)/(4*log(x)^2 + x - 16))

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mupad [B]  time = 3.98, size = 32, normalized size = 1.23 \begin {gather*} 4\,\ln \left (x\,\mathrm {e}+{\mathrm {e}}^x\,\left (16\,x-x^2\right )+x^2-4\,x\,{\mathrm {e}}^x\,{\ln \relax (x)}^2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x + 4*exp(1) - 32*exp(x)*log(x) + exp(x)*(56*x - 4*x^2 + 64) - exp(x)*log(x)^2*(16*x + 16))/(x*exp(1) +
exp(x)*(16*x - x^2) + x^2 - 4*x*exp(x)*log(x)^2),x)

[Out]

4*log(x*exp(1) + exp(x)*(16*x - x^2) + x^2 - 4*x*exp(x)*log(x)^2)

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sympy [A]  time = 1.88, size = 41, normalized size = 1.58 \begin {gather*} 4 \log {\relax (x )} + 4 \log {\left (\frac {- x - e}{x + 4 \log {\relax (x )}^{2} - 16} + e^{x} \right )} + 4 \log {\left (\frac {x}{4} + \log {\relax (x )}^{2} - 4 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x+16)*exp(x)*ln(x)**2+32*exp(x)*ln(x)+(4*x**2-56*x-64)*exp(x)-4*exp(1)-8*x)/(4*x*exp(x)*ln(x)**
2+(x**2-16*x)*exp(x)-x*exp(1)-x**2),x)

[Out]

4*log(x) + 4*log((-x - E)/(x + 4*log(x)**2 - 16) + exp(x)) + 4*log(x/4 + log(x)**2 - 4)

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