3.48.33 \(\int \frac {8 x+e^{x^2} (8+16 x-16 x^2-16 x^3)+(8+16 x) \log (x)}{x+x^2+e^{x^2} (x+x^2)+(x+x^2) \log (x)} \, dx\)

Optimal. Leaf size=23 \[ 4 \log \left (\frac {625 \left (x+x^2\right )^2}{\left (1+e^{x^2}+\log (x)\right )^2}\right ) \]

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Rubi [F]  time = 1.39, 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 {8 x+e^{x^2} \left (8+16 x-16 x^2-16 x^3\right )+(8+16 x) \log (x)}{x+x^2+e^{x^2} \left (x+x^2\right )+\left (x+x^2\right ) \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(8*x + E^x^2*(8 + 16*x - 16*x^2 - 16*x^3) + (8 + 16*x)*Log[x])/(x + x^2 + E^x^2*(x + x^2) + (x + x^2)*Log[
x]),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.23, size = 21, normalized size = 0.91 \begin {gather*} 8 \left (\log (x)+\log (1+x)-\log \left (1+e^{x^2}+\log (x)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8*x + E^x^2*(8 + 16*x - 16*x^2 - 16*x^3) + (8 + 16*x)*Log[x])/(x + x^2 + E^x^2*(x + x^2) + (x + x^2
)*Log[x]),x]

[Out]

8*(Log[x] + Log[1 + x] - Log[1 + E^x^2 + Log[x]])

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fricas [A]  time = 0.61, size = 20, normalized size = 0.87 \begin {gather*} 8 \, \log \left (x^{2} + x\right ) - 8 \, \log \left (e^{\left (x^{2}\right )} + \log \relax (x) + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x+8)*log(x)+(-16*x^3-16*x^2+16*x+8)*exp(x^2)+8*x)/((x^2+x)*log(x)+(x^2+x)*exp(x^2)+x^2+x),x, al
gorithm="fricas")

[Out]

8*log(x^2 + x) - 8*log(e^(x^2) + log(x) + 1)

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giac [A]  time = 0.20, size = 22, normalized size = 0.96 \begin {gather*} 8 \, \log \left (x + 1\right ) + 8 \, \log \relax (x) - 8 \, \log \left (e^{\left (x^{2}\right )} + \log \relax (x) + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x+8)*log(x)+(-16*x^3-16*x^2+16*x+8)*exp(x^2)+8*x)/((x^2+x)*log(x)+(x^2+x)*exp(x^2)+x^2+x),x, al
gorithm="giac")

[Out]

8*log(x + 1) + 8*log(x) - 8*log(e^(x^2) + log(x) + 1)

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maple [A]  time = 0.06, size = 21, normalized size = 0.91

method	result	size
risch	\(8 \ln \left (x^{2}+x \right )-8 \ln \left (\ln \relax (x )+{\mathrm e}^{x^{2}}+1\right )\)	\(21\)
norman	\(8 \ln \relax (x )+8 \ln \left (x +1\right )-8 \ln \left (\ln \relax (x )+{\mathrm e}^{x^{2}}+1\right )\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((16*x+8)*ln(x)+(-16*x^3-16*x^2+16*x+8)*exp(x^2)+8*x)/((x^2+x)*ln(x)+(x^2+x)*exp(x^2)+x^2+x),x,method=_RET
URNVERBOSE)

[Out]

8*ln(x^2+x)-8*ln(ln(x)+exp(x^2)+1)

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maxima [A]  time = 0.40, size = 22, normalized size = 0.96 \begin {gather*} 8 \, \log \left (x + 1\right ) + 8 \, \log \relax (x) - 8 \, \log \left (e^{\left (x^{2}\right )} + \log \relax (x) + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x+8)*log(x)+(-16*x^3-16*x^2+16*x+8)*exp(x^2)+8*x)/((x^2+x)*log(x)+(x^2+x)*exp(x^2)+x^2+x),x, al
gorithm="maxima")

[Out]

8*log(x + 1) + 8*log(x) - 8*log(e^(x^2) + log(x) + 1)

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mupad [B]  time = 3.40, size = 20, normalized size = 0.87 \begin {gather*} 8\,\ln \left (x\,\left (x+1\right )\right )-8\,\ln \left ({\mathrm {e}}^{x^2}+\ln \relax (x)+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x + exp(x^2)*(16*x - 16*x^2 - 16*x^3 + 8) + log(x)*(16*x + 8))/(x + exp(x^2)*(x + x^2) + x^2 + log(x)*(
x + x^2)),x)

[Out]

8*log(x*(x + 1)) - 8*log(exp(x^2) + log(x) + 1)

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sympy [A]  time = 0.34, size = 20, normalized size = 0.87 \begin {gather*} 8 \log {\left (x^{2} + x \right )} - 8 \log {\left (e^{x^{2}} + \log {\relax (x )} + 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x+8)*ln(x)+(-16*x**3-16*x**2+16*x+8)*exp(x**2)+8*x)/((x**2+x)*ln(x)+(x**2+x)*exp(x**2)+x**2+x),
x)

[Out]

8*log(x**2 + x) - 8*log(exp(x**2) + log(x) + 1)

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