3.64.90 \(\int \frac {e^{x+x^2}+e^x (-2-x^2)+(e^{x+x^2} (2-x+2 x^2)+e^x (-4+2 x-4 x^2+x^3)) \log (x)+(4 x+e^{2 x^2} x+4 x^3+x^5+e^{x^2} (-4 x-2 x^3)) \log ^2(x)}{(4 x^3+e^{2 x^2} x^3+4 x^5+x^7+e^{x^2} (-4 x^3-2 x^5)) \log ^2(x)} \, dx\)

Optimal. Leaf size=32 \[ 25-\frac {x-\frac {e^x}{\left (2-e^{x^2}+x^2\right ) \log (x)}}{x^2} \]

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Rubi [F]  time = 5.65, 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 {e^{x+x^2}+e^x \left (-2-x^2\right )+\left (e^{x+x^2} \left (2-x+2 x^2\right )+e^x \left (-4+2 x-4 x^2+x^3\right )\right ) \log (x)+\left (4 x+e^{2 x^2} x+4 x^3+x^5+e^{x^2} \left (-4 x-2 x^3\right )\right ) \log ^2(x)}{\left (4 x^3+e^{2 x^2} x^3+4 x^5+x^7+e^{x^2} \left (-4 x^3-2 x^5\right )\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(x + x^2) + E^x*(-2 - x^2) + (E^(x + x^2)*(2 - x + 2*x^2) + E^x*(-4 + 2*x - 4*x^2 + x^3))*Log[x] + (4*x
+ E^(2*x^2)*x + 4*x^3 + x^5 + E^x^2*(-4*x - 2*x^3))*Log[x]^2)/((4*x^3 + E^(2*x^2)*x^3 + 4*x^5 + x^7 + E^x^2*(
-4*x^3 - 2*x^5))*Log[x]^2),x]

[Out]

-x^(-1) - Defer[Int][E^x/(x^3*(2 - E^x^2 + x^2)*Log[x]^2), x] + 2*Defer[Int][E^x/(x*(2 - E^x^2 + x^2)^2*Log[x]
), x] + 2*Defer[Int][(E^x*x)/((2 - E^x^2 + x^2)^2*Log[x]), x] - 2*Defer[Int][E^x/(x^3*(2 - E^x^2 + x^2)*Log[x]
), x] + Defer[Int][E^x/(x^2*(2 - E^x^2 + x^2)*Log[x]), x] - 2*Defer[Int][E^x/(x*(2 - E^x^2 + x^2)*Log[x]), x]

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 31, normalized size = 0.97 \begin {gather*} -\frac {1}{x}+\frac {e^x}{x^2 \left (2-e^{x^2}+x^2\right ) \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(x + x^2) + E^x*(-2 - x^2) + (E^(x + x^2)*(2 - x + 2*x^2) + E^x*(-4 + 2*x - 4*x^2 + x^3))*Log[x]
+ (4*x + E^(2*x^2)*x + 4*x^3 + x^5 + E^x^2*(-4*x - 2*x^3))*Log[x]^2)/((4*x^3 + E^(2*x^2)*x^3 + 4*x^5 + x^7 + E
^x^2*(-4*x^3 - 2*x^5))*Log[x]^2),x]

[Out]

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

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fricas [B]  time = 0.68, size = 66, normalized size = 2.06 \begin {gather*} -\frac {{\left (x e^{\left (2 \, x^{2}\right )} - {\left (x^{3} + 2 \, x\right )} e^{\left (x^{2}\right )}\right )} \log \relax (x) + e^{\left (x^{2} + x\right )}}{{\left (x^{2} e^{\left (2 \, x^{2}\right )} - {\left (x^{4} + 2 \, x^{2}\right )} e^{\left (x^{2}\right )}\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(x^2)^2+(-2*x^3-4*x)*exp(x^2)+x^5+4*x^3+4*x)*log(x)^2+((2*x^2-x+2)*exp(x)*exp(x^2)+(x^3-4*x^2
+2*x-4)*exp(x))*log(x)+exp(x)*exp(x^2)+(-x^2-2)*exp(x))/(x^3*exp(x^2)^2+(-2*x^5-4*x^3)*exp(x^2)+x^7+4*x^5+4*x^
3)/log(x)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 54, normalized size = 1.69 \begin {gather*} -\frac {x^{3} \log \relax (x) - x e^{\left (x^{2}\right )} \log \relax (x) + 2 \, x \log \relax (x) - e^{x}}{x^{4} \log \relax (x) - x^{2} e^{\left (x^{2}\right )} \log \relax (x) + 2 \, x^{2} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(x^2)^2+(-2*x^3-4*x)*exp(x^2)+x^5+4*x^3+4*x)*log(x)^2+((2*x^2-x+2)*exp(x)*exp(x^2)+(x^3-4*x^2
+2*x-4)*exp(x))*log(x)+exp(x)*exp(x^2)+(-x^2-2)*exp(x))/(x^3*exp(x^2)^2+(-2*x^5-4*x^3)*exp(x^2)+x^7+4*x^5+4*x^
3)/log(x)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.18, size = 30, normalized size = 0.94

method	result	size
risch	\(-\frac {1}{x}+\frac {{\mathrm e}^{x}}{x^{2} \left (2+x^{2}-{\mathrm e}^{x^{2}}\right ) \ln \relax (x )}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x*exp(x^2)^2+(-2*x^3-4*x)*exp(x^2)+x^5+4*x^3+4*x)*ln(x)^2+((2*x^2-x+2)*exp(x)*exp(x^2)+(x^3-4*x^2+2*x-4)
*exp(x))*ln(x)+exp(x)*exp(x^2)+(-x^2-2)*exp(x))/(x^3*exp(x^2)^2+(-2*x^5-4*x^3)*exp(x^2)+x^7+4*x^5+4*x^3)/ln(x)
^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.50, size = 50, normalized size = 1.56 \begin {gather*} -\frac {x e^{\left (x^{2}\right )} \log \relax (x) - {\left (x^{3} + 2 \, x\right )} \log \relax (x) + e^{x}}{x^{2} e^{\left (x^{2}\right )} \log \relax (x) - {\left (x^{4} + 2 \, x^{2}\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(x^2)^2+(-2*x^3-4*x)*exp(x^2)+x^5+4*x^3+4*x)*log(x)^2+((2*x^2-x+2)*exp(x)*exp(x^2)+(x^3-4*x^2
+2*x-4)*exp(x))*log(x)+exp(x)*exp(x^2)+(-x^2-2)*exp(x))/(x^3*exp(x^2)^2+(-2*x^5-4*x^3)*exp(x^2)+x^7+4*x^5+4*x^
3)/log(x)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.32, size = 48, normalized size = 1.50 \begin {gather*} -\frac {x^3\,\ln \relax (x)-{\mathrm {e}}^x+x\,\left (2\,\ln \relax (x)-{\mathrm {e}}^{x^2}\,\ln \relax (x)\right )}{x^2\,\ln \relax (x)\,\left (x^2-{\mathrm {e}}^{x^2}+2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)*(exp(x)*(2*x - 4*x^2 + x^3 - 4) + exp(x^2)*exp(x)*(2*x^2 - x + 2)) + log(x)^2*(4*x - exp(x^2)*(4*x
+ 2*x^3) + x*exp(2*x^2) + 4*x^3 + x^5) - exp(x)*(x^2 + 2) + exp(x^2)*exp(x))/(log(x)^2*(x^3*exp(2*x^2) - exp(
x^2)*(4*x^3 + 2*x^5) + 4*x^3 + 4*x^5 + x^7)),x)

[Out]

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

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sympy [A]  time = 0.34, size = 34, normalized size = 1.06 \begin {gather*} - \frac {e^{x}}{- x^{4} \log {\relax (x )} + x^{2} e^{x^{2}} \log {\relax (x )} - 2 x^{2} \log {\relax (x )}} - \frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(x**2)**2+(-2*x**3-4*x)*exp(x**2)+x**5+4*x**3+4*x)*ln(x)**2+((2*x**2-x+2)*exp(x)*exp(x**2)+(x
**3-4*x**2+2*x-4)*exp(x))*ln(x)+exp(x)*exp(x**2)+(-x**2-2)*exp(x))/(x**3*exp(x**2)**2+(-2*x**5-4*x**3)*exp(x**
2)+x**7+4*x**5+4*x**3)/ln(x)**2,x)

[Out]

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

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