3.102.72 \(\int \frac {-4 x^2-4 e^{x^2} x^3+(8 x^2+e^{x^2} (2 x+4 x^3)+2 x \log (8)) \log (e^{x^2}+2 x+\log (8))+(-4 e^{x^2} x-8 x^2-4 x \log (8)) \log ^2(e^{x^2}+2 x+\log (8))+(2 e^{x^2} x+4 x^2+2 x \log (8)) \log ^3(e^{x^2}+2 x+\log (8))}{(e^{x^2}+2 x+\log (8)) \log ^3(e^{x^2}+2 x+\log (8))} \, dx\)

Optimal. Leaf size=22 \[ \left (-x+\frac {x}{\log \left (e^{x^2}+2 x+\log (8)\right )}\right )^2 \]

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Rubi [F]  time = 6.69, 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 x^2-4 e^{x^2} x^3+\left (8 x^2+e^{x^2} \left (2 x+4 x^3\right )+2 x \log (8)\right ) \log \left (e^{x^2}+2 x+\log (8)\right )+\left (-4 e^{x^2} x-8 x^2-4 x \log (8)\right ) \log ^2\left (e^{x^2}+2 x+\log (8)\right )+\left (2 e^{x^2} x+4 x^2+2 x \log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

x^2 - 4*Defer[Int][x^3/Log[E^x^2 + 2*x + Log[8]]^3, x] - 4*Defer[Int][x^2/((E^x^2 + 2*x + Log[8])*Log[E^x^2 +
2*x + Log[8]]^3), x] + 4*Log[8]*Defer[Int][x^3/((E^x^2 + 2*x + Log[8])*Log[E^x^2 + 2*x + Log[8]]^3), x] + 8*De
fer[Int][x^4/((E^x^2 + 2*x + Log[8])*Log[E^x^2 + 2*x + Log[8]]^3), x] + 2*Defer[Int][x/Log[E^x^2 + 2*x + Log[8
]]^2, x] + 4*Defer[Int][x^3/Log[E^x^2 + 2*x + Log[8]]^2, x] + 4*Defer[Int][x^2/((E^x^2 + 2*x + Log[8])*Log[E^x
^2 + 2*x + Log[8]]^2), x] - 4*Log[8]*Defer[Int][x^3/((E^x^2 + 2*x + Log[8])*Log[E^x^2 + 2*x + Log[8]]^2), x] -
8*Defer[Int][x^4/((E^x^2 + 2*x + Log[8])*Log[E^x^2 + 2*x + Log[8]]^2), x] - 4*Defer[Int][x/Log[E^x^2 + 2*x +
Log[8]], x]

Rubi steps

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

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Mathematica [B]  time = 0.06, size = 50, normalized size = 2.27 \begin {gather*} 2 \left (\frac {x^2}{2}+\frac {x^2}{2 \log ^2\left (e^{x^2}+2 x+\log (8)\right )}-\frac {x^2}{\log \left (e^{x^2}+2 x+\log (8)\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4*x^2 - 4*E^x^2*x^3 + (8*x^2 + E^x^2*(2*x + 4*x^3) + 2*x*Log[8])*Log[E^x^2 + 2*x + Log[8]] + (-4*E
^x^2*x - 8*x^2 - 4*x*Log[8])*Log[E^x^2 + 2*x + Log[8]]^2 + (2*E^x^2*x + 4*x^2 + 2*x*Log[8])*Log[E^x^2 + 2*x +
Log[8]]^3)/((E^x^2 + 2*x + Log[8])*Log[E^x^2 + 2*x + Log[8]]^3),x]

[Out]

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

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fricas [B]  time = 0.52, size = 57, normalized size = 2.59 \begin {gather*} \frac {x^{2} \log \left (2 \, x + e^{\left (x^{2}\right )} + 3 \, \log \relax (2)\right )^{2} - 2 \, x^{2} \log \left (2 \, x + e^{\left (x^{2}\right )} + 3 \, \log \relax (2)\right ) + x^{2}}{\log \left (2 \, x + e^{\left (x^{2}\right )} + 3 \, \log \relax (2)\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x^2)*x+6*x*log(2)+4*x^2)*log(exp(x^2)+3*log(2)+2*x)^3+(-4*exp(x^2)*x-12*x*log(2)-8*x^2)*log(
exp(x^2)+3*log(2)+2*x)^2+((4*x^3+2*x)*exp(x^2)+6*x*log(2)+8*x^2)*log(exp(x^2)+3*log(2)+2*x)-4*x^3*exp(x^2)-4*x
^2)/(exp(x^2)+3*log(2)+2*x)/log(exp(x^2)+3*log(2)+2*x)^3,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.91, size = 57, normalized size = 2.59 \begin {gather*} \frac {x^{2} \log \left (2 \, x + e^{\left (x^{2}\right )} + 3 \, \log \relax (2)\right )^{2} - 2 \, x^{2} \log \left (2 \, x + e^{\left (x^{2}\right )} + 3 \, \log \relax (2)\right ) + x^{2}}{\log \left (2 \, x + e^{\left (x^{2}\right )} + 3 \, \log \relax (2)\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x^2)*x+6*x*log(2)+4*x^2)*log(exp(x^2)+3*log(2)+2*x)^3+(-4*exp(x^2)*x-12*x*log(2)-8*x^2)*log(
exp(x^2)+3*log(2)+2*x)^2+((4*x^3+2*x)*exp(x^2)+6*x*log(2)+8*x^2)*log(exp(x^2)+3*log(2)+2*x)-4*x^3*exp(x^2)-4*x
^2)/(exp(x^2)+3*log(2)+2*x)/log(exp(x^2)+3*log(2)+2*x)^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.04, size = 42, normalized size = 1.91

method	result	size
risch	\(x^{2}-\frac {\left (2 \ln \left ({\mathrm e}^{x^{2}}+3 \ln \relax (2)+2 x \right )-1\right ) x^{2}}{\ln \left ({\mathrm e}^{x^{2}}+3 \ln \relax (2)+2 x \right )^{2}}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*exp(x^2)*x+6*x*ln(2)+4*x^2)*ln(exp(x^2)+3*ln(2)+2*x)^3+(-4*exp(x^2)*x-12*x*ln(2)-8*x^2)*ln(exp(x^2)+3*
ln(2)+2*x)^2+((4*x^3+2*x)*exp(x^2)+6*x*ln(2)+8*x^2)*ln(exp(x^2)+3*ln(2)+2*x)-4*x^3*exp(x^2)-4*x^2)/(exp(x^2)+3
*ln(2)+2*x)/ln(exp(x^2)+3*ln(2)+2*x)^3,x,method=_RETURNVERBOSE)

[Out]

x^2-(2*ln(exp(x^2)+3*ln(2)+2*x)-1)*x^2/ln(exp(x^2)+3*ln(2)+2*x)^2

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maxima [B]  time = 0.48, size = 57, normalized size = 2.59 \begin {gather*} \frac {x^{2} \log \left (2 \, x + e^{\left (x^{2}\right )} + 3 \, \log \relax (2)\right )^{2} - 2 \, x^{2} \log \left (2 \, x + e^{\left (x^{2}\right )} + 3 \, \log \relax (2)\right ) + x^{2}}{\log \left (2 \, x + e^{\left (x^{2}\right )} + 3 \, \log \relax (2)\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x^2)*x+6*x*log(2)+4*x^2)*log(exp(x^2)+3*log(2)+2*x)^3+(-4*exp(x^2)*x-12*x*log(2)-8*x^2)*log(
exp(x^2)+3*log(2)+2*x)^2+((4*x^3+2*x)*exp(x^2)+6*x*log(2)+8*x^2)*log(exp(x^2)+3*log(2)+2*x)-4*x^3*exp(x^2)-4*x
^2)/(exp(x^2)+3*log(2)+2*x)/log(exp(x^2)+3*log(2)+2*x)^3,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} -\int \frac {{\ln \left (2\,x+{\mathrm {e}}^{x^2}+3\,\ln \relax (2)\right )}^2\,\left (4\,x\,{\mathrm {e}}^{x^2}+12\,x\,\ln \relax (2)+8\,x^2\right )-{\ln \left (2\,x+{\mathrm {e}}^{x^2}+3\,\ln \relax (2)\right )}^3\,\left (2\,x\,{\mathrm {e}}^{x^2}+6\,x\,\ln \relax (2)+4\,x^2\right )+4\,x^3\,{\mathrm {e}}^{x^2}+4\,x^2-\ln \left (2\,x+{\mathrm {e}}^{x^2}+3\,\ln \relax (2)\right )\,\left ({\mathrm {e}}^{x^2}\,\left (4\,x^3+2\,x\right )+6\,x\,\ln \relax (2)+8\,x^2\right )}{{\ln \left (2\,x+{\mathrm {e}}^{x^2}+3\,\ln \relax (2)\right )}^3\,\left (2\,x+{\mathrm {e}}^{x^2}+3\,\ln \relax (2)\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(2*x + exp(x^2) + 3*log(2))^2*(4*x*exp(x^2) + 12*x*log(2) + 8*x^2) - log(2*x + exp(x^2) + 3*log(2))^3
*(2*x*exp(x^2) + 6*x*log(2) + 4*x^2) + 4*x^3*exp(x^2) + 4*x^2 - log(2*x + exp(x^2) + 3*log(2))*(exp(x^2)*(2*x
+ 4*x^3) + 6*x*log(2) + 8*x^2))/(log(2*x + exp(x^2) + 3*log(2))^3*(2*x + exp(x^2) + 3*log(2))),x)

[Out]

-int((log(2*x + exp(x^2) + 3*log(2))^2*(4*x*exp(x^2) + 12*x*log(2) + 8*x^2) - log(2*x + exp(x^2) + 3*log(2))^3
*(2*x*exp(x^2) + 6*x*log(2) + 4*x^2) + 4*x^3*exp(x^2) + 4*x^2 - log(2*x + exp(x^2) + 3*log(2))*(exp(x^2)*(2*x
+ 4*x^3) + 6*x*log(2) + 8*x^2))/(log(2*x + exp(x^2) + 3*log(2))^3*(2*x + exp(x^2) + 3*log(2))), x)

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sympy [B]  time = 0.37, size = 42, normalized size = 1.91 \begin {gather*} x^{2} + \frac {- 2 x^{2} \log {\left (2 x + e^{x^{2}} + 3 \log {\relax (2 )} \right )} + x^{2}}{\log {\left (2 x + e^{x^{2}} + 3 \log {\relax (2 )} \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x**2)*x+6*x*ln(2)+4*x**2)*ln(exp(x**2)+3*ln(2)+2*x)**3+(-4*exp(x**2)*x-12*x*ln(2)-8*x**2)*ln
(exp(x**2)+3*ln(2)+2*x)**2+((4*x**3+2*x)*exp(x**2)+6*x*ln(2)+8*x**2)*ln(exp(x**2)+3*ln(2)+2*x)-4*x**3*exp(x**2
)-4*x**2)/(exp(x**2)+3*ln(2)+2*x)/ln(exp(x**2)+3*ln(2)+2*x)**3,x)

[Out]

x**2 + (-2*x**2*log(2*x + exp(x**2) + 3*log(2)) + x**2)/log(2*x + exp(x**2) + 3*log(2))**2

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