3.8.48 \(\int \frac {5 x^3+10 e^{2 x} x^3+e^{x^2} (8+2 x-8 x^2-2 x^3+e^{2 x} (2-2 x^2))}{e^{x^2} (-4 x-e^{2 x} x-x^2)+(20 x^3+5 e^{2 x} x^3+5 x^4) \log (4+e^{2 x}+x)} \, dx\)

Optimal. Leaf size=25 \[ \log \left (\frac {e^{x^2}}{5 x^2}-\log \left (4+e^{2 x}+x\right )\right ) \]

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Rubi [F]  time = 11.08, 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 {5 x^3+10 e^{2 x} x^3+e^{x^2} \left (8+2 x-8 x^2-2 x^3+e^{2 x} \left (2-2 x^2\right )\right )}{e^{x^2} \left (-4 x-e^{2 x} x-x^2\right )+\left (20 x^3+5 e^{2 x} x^3+5 x^4\right ) \log \left (4+e^{2 x}+x\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(5*x^3 + 10*E^(2*x)*x^3 + E^x^2*(8 + 2*x - 8*x^2 - 2*x^3 + E^(2*x)*(2 - 2*x^2)))/(E^x^2*(-4*x - E^(2*x)*x
- x^2) + (20*x^3 + 5*E^(2*x)*x^3 + 5*x^4)*Log[4 + E^(2*x) + x]),x]

[Out]

x^2 - 2*Log[x] + 5*Defer[Int][x^2/((4 + E^(2*x) + x)*(-E^x^2 + 5*x^2*Log[4 + E^(2*x) + x])), x] + 10*Defer[Int
][(E^(2*x)*x^2)/((4 + E^(2*x) + x)*(-E^x^2 + 5*x^2*Log[4 + E^(2*x) + x])), x] + 40*Defer[Int][(x*Log[4 + E^(2*
x) + x])/((4 + E^(2*x) + x)*(-E^x^2 + 5*x^2*Log[4 + E^(2*x) + x])), x] + 10*Defer[Int][(E^(2*x)*x*Log[4 + E^(2
*x) + x])/((4 + E^(2*x) + x)*(-E^x^2 + 5*x^2*Log[4 + E^(2*x) + x])), x] + 10*Defer[Int][(x^2*Log[4 + E^(2*x) +
x])/((4 + E^(2*x) + x)*(-E^x^2 + 5*x^2*Log[4 + E^(2*x) + x])), x] - 40*Defer[Int][(x^3*Log[4 + E^(2*x) + x])/
((4 + E^(2*x) + x)*(-E^x^2 + 5*x^2*Log[4 + E^(2*x) + x])), x] - 10*Defer[Int][(E^(2*x)*x^3*Log[4 + E^(2*x) + x
])/((4 + E^(2*x) + x)*(-E^x^2 + 5*x^2*Log[4 + E^(2*x) + x])), x] - 10*Defer[Int][(x^4*Log[4 + E^(2*x) + x])/((
4 + E^(2*x) + x)*(-E^x^2 + 5*x^2*Log[4 + E^(2*x) + x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-5 x^3-10 e^{2 x} x^3-e^{x^2} \left (8+2 x-8 x^2-2 x^3+e^{2 x} \left (2-2 x^2\right )\right )}{x \left (4+e^{2 x}+x\right ) \left (e^{x^2}-5 x^2 \log \left (4+e^{2 x}+x\right )\right )} \, dx\\ &=\int \frac {-5 x^3-10 e^{2 x} x^3+2 e^{x^2} \left (4+e^{2 x}+x\right ) \left (-1+x^2\right )}{x \left (4+e^{2 x}+x\right ) \left (e^{x^2}-5 x^2 \log \left (4+e^{2 x}+x\right )\right )} \, dx\\ &=\int \left (\frac {2 \left (-1+x^2\right )}{x}-\frac {5 x \left (-x-2 e^{2 x} x-8 \log \left (4+e^{2 x}+x\right )-2 e^{2 x} \log \left (4+e^{2 x}+x\right )-2 x \log \left (4+e^{2 x}+x\right )+8 x^2 \log \left (4+e^{2 x}+x\right )+2 e^{2 x} x^2 \log \left (4+e^{2 x}+x\right )+2 x^3 \log \left (4+e^{2 x}+x\right )\right )}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )}\right ) \, dx\\ &=2 \int \frac {-1+x^2}{x} \, dx-5 \int \frac {x \left (-x-2 e^{2 x} x-8 \log \left (4+e^{2 x}+x\right )-2 e^{2 x} \log \left (4+e^{2 x}+x\right )-2 x \log \left (4+e^{2 x}+x\right )+8 x^2 \log \left (4+e^{2 x}+x\right )+2 e^{2 x} x^2 \log \left (4+e^{2 x}+x\right )+2 x^3 \log \left (4+e^{2 x}+x\right )\right )}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )} \, dx\\ &=2 \int \left (-\frac {1}{x}+x\right ) \, dx-5 \int \frac {x \left (\left (1+2 e^{2 x}\right ) x-2 \left (4+e^{2 x}+x\right ) \left (-1+x^2\right ) \log \left (4+e^{2 x}+x\right )\right )}{\left (4+e^{2 x}+x\right ) \left (e^{x^2}-5 x^2 \log \left (4+e^{2 x}+x\right )\right )} \, dx\\ &=x^2-2 \log (x)-5 \int \left (-\frac {x^2}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )}-\frac {2 e^{2 x} x^2}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )}-\frac {8 x \log \left (4+e^{2 x}+x\right )}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )}-\frac {2 e^{2 x} x \log \left (4+e^{2 x}+x\right )}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )}-\frac {2 x^2 \log \left (4+e^{2 x}+x\right )}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )}+\frac {8 x^3 \log \left (4+e^{2 x}+x\right )}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )}+\frac {2 e^{2 x} x^3 \log \left (4+e^{2 x}+x\right )}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )}+\frac {2 x^4 \log \left (4+e^{2 x}+x\right )}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )}\right ) \, dx\\ &=x^2-2 \log (x)+5 \int \frac {x^2}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )} \, dx+10 \int \frac {e^{2 x} x^2}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )} \, dx+10 \int \frac {e^{2 x} x \log \left (4+e^{2 x}+x\right )}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )} \, dx+10 \int \frac {x^2 \log \left (4+e^{2 x}+x\right )}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )} \, dx-10 \int \frac {e^{2 x} x^3 \log \left (4+e^{2 x}+x\right )}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )} \, dx-10 \int \frac {x^4 \log \left (4+e^{2 x}+x\right )}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )} \, dx+40 \int \frac {x \log \left (4+e^{2 x}+x\right )}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )} \, dx-40 \int \frac {x^3 \log \left (4+e^{2 x}+x\right )}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 38, normalized size = 1.52 \begin {gather*} -2 x-2 \log (x)+\log \left (e^{2 x+x^2}-5 e^{2 x} x^2 \log \left (4+e^{2 x}+x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5*x^3 + 10*E^(2*x)*x^3 + E^x^2*(8 + 2*x - 8*x^2 - 2*x^3 + E^(2*x)*(2 - 2*x^2)))/(E^x^2*(-4*x - E^(2
*x)*x - x^2) + (20*x^3 + 5*E^(2*x)*x^3 + 5*x^4)*Log[4 + E^(2*x) + x]),x]

[Out]

-2*x - 2*Log[x] + Log[E^(2*x + x^2) - 5*E^(2*x)*x^2*Log[4 + E^(2*x) + x]]

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fricas [A]  time = 0.56, size = 25, normalized size = 1.00 \begin {gather*} \log \left (\frac {5 \, x^{2} \log \left (x + e^{\left (2 \, x\right )} + 4\right ) - e^{\left (x^{2}\right )}}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2+2)*exp(x)^2-2*x^3-8*x^2+2*x+8)*exp(x^2)+10*exp(x)^2*x^3+5*x^3)/((5*exp(x)^2*x^3+5*x^4+20*x
^3)*log(exp(x)^2+4+x)+(-x*exp(x)^2-x^2-4*x)*exp(x^2)),x, algorithm="fricas")

[Out]

log((5*x^2*log(x + e^(2*x) + 4) - e^(x^2))/x^2)

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giac [A]  time = 0.57, size = 24, normalized size = 0.96 \begin {gather*} \log \left (-5 \, x^{2} \log \left (x + e^{\left (2 \, x\right )} + 4\right ) + e^{\left (x^{2}\right )}\right ) - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2+2)*exp(x)^2-2*x^3-8*x^2+2*x+8)*exp(x^2)+10*exp(x)^2*x^3+5*x^3)/((5*exp(x)^2*x^3+5*x^4+20*x
^3)*log(exp(x)^2+4+x)+(-x*exp(x)^2-x^2-4*x)*exp(x^2)),x, algorithm="giac")

[Out]

log(-5*x^2*log(x + e^(2*x) + 4) + e^(x^2)) - 2*log(x)

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maple [A]  time = 0.04, size = 20, normalized size = 0.80

method	result	size
risch	\(\ln \left (\ln \left ({\mathrm e}^{2 x}+4+x \right )-\frac {{\mathrm e}^{x^{2}}}{5 x^{2}}\right )\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-2*x^2+2)*exp(x)^2-2*x^3-8*x^2+2*x+8)*exp(x^2)+10*exp(x)^2*x^3+5*x^3)/((5*exp(x)^2*x^3+5*x^4+20*x^3)*ln
(exp(x)^2+4+x)+(-x*exp(x)^2-x^2-4*x)*exp(x^2)),x,method=_RETURNVERBOSE)

[Out]

ln(ln(exp(2*x)+4+x)-1/5*exp(x^2)/x^2)

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maxima [A]  time = 0.83, size = 26, normalized size = 1.04 \begin {gather*} \log \left (\frac {5 \, x^{2} \log \left (x + e^{\left (2 \, x\right )} + 4\right ) - e^{\left (x^{2}\right )}}{5 \, x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2+2)*exp(x)^2-2*x^3-8*x^2+2*x+8)*exp(x^2)+10*exp(x)^2*x^3+5*x^3)/((5*exp(x)^2*x^3+5*x^4+20*x
^3)*log(exp(x)^2+4+x)+(-x*exp(x)^2-x^2-4*x)*exp(x^2)),x, algorithm="maxima")

[Out]

log(1/5*(5*x^2*log(x + e^(2*x) + 4) - e^(x^2))/x^2)

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mupad [B]  time = 0.76, size = 26, normalized size = 1.04 \begin {gather*} \ln \left (5\,x^2\,\ln \left (x+{\mathrm {e}}^{2\,x}+4\right )-{\mathrm {e}}^{x^2}\right )+\ln \left (\frac {1}{x^2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*x^3*exp(2*x) - exp(x^2)*(exp(2*x)*(2*x^2 - 2) - 2*x + 8*x^2 + 2*x^3 - 8) + 5*x^3)/(log(x + exp(2*x) +
4)*(5*x^3*exp(2*x) + 20*x^3 + 5*x^4) - exp(x^2)*(4*x + x*exp(2*x) + x^2)),x)

[Out]

log(5*x^2*log(x + exp(2*x) + 4) - exp(x^2)) + log(1/x^2)

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sympy [A]  time = 0.92, size = 20, normalized size = 0.80 \begin {gather*} \log {\left (\log {\left (x + e^{2 x} + 4 \right )} - \frac {e^{x^{2}}}{5 x^{2}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x**2+2)*exp(x)**2-2*x**3-8*x**2+2*x+8)*exp(x**2)+10*exp(x)**2*x**3+5*x**3)/((5*exp(x)**2*x**3+
5*x**4+20*x**3)*ln(exp(x)**2+4+x)+(-x*exp(x)**2-x**2-4*x)*exp(x**2)),x)

[Out]

log(log(x + exp(2*x) + 4) - exp(x**2)/(5*x**2))

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