3.63.76 \(\int \frac {4-4 x+9 x^2-x^3-x^4+x^5+e^{4 x} (1+x)+e^{2 x} (6 x+2 x^3)+(-2 x+2 x^3+e^{2 x} (2+2 x)) \log (x)+(1+x) \log ^2(x)}{e^{4 x} x+x^3-2 x^4+x^5+e^{2 x} (-2 x^2+2 x^3)+(2 e^{2 x} x-2 x^2+2 x^3) \log (x)+x \log ^2(x)} \, dx\)

Optimal. Leaf size=25 \[ \frac {10}{3}+x+\log (x)-\frac {4}{e^{2 x}-x+x^2+\log (x)} \]

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Rubi [F]  time = 2.04, 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 {4-4 x+9 x^2-x^3-x^4+x^5+e^{4 x} (1+x)+e^{2 x} \left (6 x+2 x^3\right )+\left (-2 x+2 x^3+e^{2 x} (2+2 x)\right ) \log (x)+(1+x) \log ^2(x)}{e^{4 x} x+x^3-2 x^4+x^5+e^{2 x} \left (-2 x^2+2 x^3\right )+\left (2 e^{2 x} x-2 x^2+2 x^3\right ) \log (x)+x \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(4 - 4*x + 9*x^2 - x^3 - x^4 + x^5 + E^(4*x)*(1 + x) + E^(2*x)*(6*x + 2*x^3) + (-2*x + 2*x^3 + E^(2*x)*(2
+ 2*x))*Log[x] + (1 + x)*Log[x]^2)/(E^(4*x)*x + x^3 - 2*x^4 + x^5 + E^(2*x)*(-2*x^2 + 2*x^3) + (2*E^(2*x)*x -
2*x^2 + 2*x^3)*Log[x] + x*Log[x]^2),x]

[Out]

x + Log[x] - 4*Defer[Int][(E^(2*x) - x + x^2 + Log[x])^(-2), x] + 4*Defer[Int][1/(x*(E^(2*x) - x + x^2 + Log[x
])^2), x] + 16*Defer[Int][x/(E^(2*x) - x + x^2 + Log[x])^2, x] - 8*Defer[Int][x^2/(E^(2*x) - x + x^2 + Log[x])
^2, x] - 8*Defer[Int][Log[x]/(E^(2*x) - x + x^2 + Log[x])^2, x] + 8*Defer[Int][(E^(2*x) - x + x^2 + Log[x])^(-
1), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4-4 x+9 x^2-x^3-x^4+x^5+e^{4 x} (1+x)+2 e^{2 x} x \left (3+x^2\right )+2 (1+x) \left (e^{2 x}+(-1+x) x\right ) \log (x)+(1+x) \log ^2(x)}{x \left (e^{2 x}+(-1+x) x+\log (x)\right )^2} \, dx\\ &=\int \left (\frac {1+x}{x}+\frac {8}{e^{2 x}-x+x^2+\log (x)}-\frac {4 \left (-1+x-4 x^2+2 x^3+2 x \log (x)\right )}{x \left (e^{2 x}-x+x^2+\log (x)\right )^2}\right ) \, dx\\ &=-\left (4 \int \frac {-1+x-4 x^2+2 x^3+2 x \log (x)}{x \left (e^{2 x}-x+x^2+\log (x)\right )^2} \, dx\right )+8 \int \frac {1}{e^{2 x}-x+x^2+\log (x)} \, dx+\int \frac {1+x}{x} \, dx\\ &=-\left (4 \int \left (\frac {1}{\left (e^{2 x}-x+x^2+\log (x)\right )^2}-\frac {1}{x \left (e^{2 x}-x+x^2+\log (x)\right )^2}-\frac {4 x}{\left (e^{2 x}-x+x^2+\log (x)\right )^2}+\frac {2 x^2}{\left (e^{2 x}-x+x^2+\log (x)\right )^2}+\frac {2 \log (x)}{\left (e^{2 x}-x+x^2+\log (x)\right )^2}\right ) \, dx\right )+8 \int \frac {1}{e^{2 x}-x+x^2+\log (x)} \, dx+\int \left (1+\frac {1}{x}\right ) \, dx\\ &=x+\log (x)-4 \int \frac {1}{\left (e^{2 x}-x+x^2+\log (x)\right )^2} \, dx+4 \int \frac {1}{x \left (e^{2 x}-x+x^2+\log (x)\right )^2} \, dx-8 \int \frac {x^2}{\left (e^{2 x}-x+x^2+\log (x)\right )^2} \, dx-8 \int \frac {\log (x)}{\left (e^{2 x}-x+x^2+\log (x)\right )^2} \, dx+8 \int \frac {1}{e^{2 x}-x+x^2+\log (x)} \, dx+16 \int \frac {x}{\left (e^{2 x}-x+x^2+\log (x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 22, normalized size = 0.88 \begin {gather*} x+\log (x)-\frac {4}{e^{2 x}-x+x^2+\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 - 4*x + 9*x^2 - x^3 - x^4 + x^5 + E^(4*x)*(1 + x) + E^(2*x)*(6*x + 2*x^3) + (-2*x + 2*x^3 + E^(2*
x)*(2 + 2*x))*Log[x] + (1 + x)*Log[x]^2)/(E^(4*x)*x + x^3 - 2*x^4 + x^5 + E^(2*x)*(-2*x^2 + 2*x^3) + (2*E^(2*x
)*x - 2*x^2 + 2*x^3)*Log[x] + x*Log[x]^2),x]

[Out]

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

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fricas [B]  time = 1.12, size = 47, normalized size = 1.88 \begin {gather*} \frac {x^{3} - x^{2} + x e^{\left (2 \, x\right )} + {\left (x^{2} + e^{\left (2 \, x\right )}\right )} \log \relax (x) + \log \relax (x)^{2} - 4}{x^{2} - x + e^{\left (2 \, x\right )} + \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+1)*log(x)^2+((2*x+2)*exp(x)^2+2*x^3-2*x)*log(x)+(x+1)*exp(x)^4+(2*x^3+6*x)*exp(x)^2+x^5-x^4-x^3+
9*x^2-4*x+4)/(x*log(x)^2+(2*x*exp(x)^2+2*x^3-2*x^2)*log(x)+x*exp(x)^4+(2*x^3-2*x^2)*exp(x)^2+x^5-2*x^4+x^3),x,
algorithm="fricas")

[Out]

(x^3 - x^2 + x*e^(2*x) + (x^2 + e^(2*x))*log(x) + log(x)^2 - 4)/(x^2 - x + e^(2*x) + log(x))

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giac [B]  time = 0.26, size = 49, normalized size = 1.96 \begin {gather*} \frac {x^{3} + x^{2} \log \relax (x) - x^{2} + x e^{\left (2 \, x\right )} + e^{\left (2 \, x\right )} \log \relax (x) + \log \relax (x)^{2} - 4}{x^{2} - x + e^{\left (2 \, x\right )} + \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+1)*log(x)^2+((2*x+2)*exp(x)^2+2*x^3-2*x)*log(x)+(x+1)*exp(x)^4+(2*x^3+6*x)*exp(x)^2+x^5-x^4-x^3+
9*x^2-4*x+4)/(x*log(x)^2+(2*x*exp(x)^2+2*x^3-2*x^2)*log(x)+x*exp(x)^4+(2*x^3-2*x^2)*exp(x)^2+x^5-2*x^4+x^3),x,
algorithm="giac")

[Out]

(x^3 + x^2*log(x) - x^2 + x*e^(2*x) + e^(2*x)*log(x) + log(x)^2 - 4)/(x^2 - x + e^(2*x) + log(x))

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maple [A]  time = 0.05, size = 22, normalized size = 0.88

method	result	size
risch	\(x +\ln \relax (x )-\frac {4}{x^{2}+{\mathrm e}^{2 x}+\ln \relax (x )-x}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x+1)*ln(x)^2+((2*x+2)*exp(x)^2+2*x^3-2*x)*ln(x)+(x+1)*exp(x)^4+(2*x^3+6*x)*exp(x)^2+x^5-x^4-x^3+9*x^2-4*
x+4)/(x*ln(x)^2+(2*x*exp(x)^2+2*x^3-2*x^2)*ln(x)+x*exp(x)^4+(2*x^3-2*x^2)*exp(x)^2+x^5-2*x^4+x^3),x,method=_RE
TURNVERBOSE)

[Out]

x+ln(x)-4/(x^2+exp(2*x)+ln(x)-x)

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maxima [A]  time = 0.44, size = 39, normalized size = 1.56 \begin {gather*} \frac {x^{3} - x^{2} + x e^{\left (2 \, x\right )} + x \log \relax (x) - 4}{x^{2} - x + e^{\left (2 \, x\right )} + \log \relax (x)} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+1)*log(x)^2+((2*x+2)*exp(x)^2+2*x^3-2*x)*log(x)+(x+1)*exp(x)^4+(2*x^3+6*x)*exp(x)^2+x^5-x^4-x^3+
9*x^2-4*x+4)/(x*log(x)^2+(2*x*exp(x)^2+2*x^3-2*x^2)*log(x)+x*exp(x)^4+(2*x^3-2*x^2)*exp(x)^2+x^5-2*x^4+x^3),x,
algorithm="maxima")

[Out]

(x^3 - x^2 + x*e^(2*x) + x*log(x) - 4)/(x^2 - x + e^(2*x) + log(x)) + log(x)

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mupad [B]  time = 4.11, size = 21, normalized size = 0.84 \begin {gather*} x+\ln \relax (x)-\frac {4}{{\mathrm {e}}^{2\,x}-x+\ln \relax (x)+x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x)*(6*x + 2*x^3) - 4*x + exp(4*x)*(x + 1) + log(x)^2*(x + 1) + log(x)*(exp(2*x)*(2*x + 2) - 2*x + 2
*x^3) + 9*x^2 - x^3 - x^4 + x^5 + 4)/(x*exp(4*x) + x*log(x)^2 + log(x)*(2*x*exp(2*x) - 2*x^2 + 2*x^3) - exp(2*
x)*(2*x^2 - 2*x^3) + x^3 - 2*x^4 + x^5),x)

[Out]

x + log(x) - 4/(exp(2*x) - x + log(x) + x^2)

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sympy [A]  time = 0.32, size = 19, normalized size = 0.76 \begin {gather*} x + \log {\relax (x )} - \frac {4}{x^{2} - x + e^{2 x} + \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+1)*ln(x)**2+((2*x+2)*exp(x)**2+2*x**3-2*x)*ln(x)+(x+1)*exp(x)**4+(2*x**3+6*x)*exp(x)**2+x**5-x**
4-x**3+9*x**2-4*x+4)/(x*ln(x)**2+(2*x*exp(x)**2+2*x**3-2*x**2)*ln(x)+x*exp(x)**4+(2*x**3-2*x**2)*exp(x)**2+x**
5-2*x**4+x**3),x)

[Out]

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

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