3.54.66 \(\int \frac {-1+e^{x^2+2 x^4+x^6} (2 x+8 x^3+6 x^5)}{9+e^{2 x^2+4 x^4+2 x^6}+6 x+x^2+(6+2 x) \log (2) \log (5)+\log ^2(2) \log ^2(5)+e^{x^2+2 x^4+x^6} (-6-2 x-2 \log (2) \log (5))} \, dx\) [5366]

Optimal. Leaf size=27 \[ \frac {1}{3-e^{x^2 \left (-1-x^2\right )^2}+x+\log (2) \log (5)} \]

[Out]

1/(x+ln(2)*ln(5)+3-exp(x^2*(-x^2-1)^2))

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Rubi [A]
time = 0.32, antiderivative size = 21, normalized size of antiderivative = 0.78, number of steps used = 2, number of rules used = 2, integrand size = 105, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {6820, 6818} \begin {gather*} \frac {1}{-e^{\left (x^3+x\right )^2}+x+3+\log (2) \log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + E^(x^2 + 2*x^4 + x^6)*(2*x + 8*x^3 + 6*x^5))/(9 + E^(2*x^2 + 4*x^4 + 2*x^6) + 6*x + x^2 + (6 + 2*x)*
Log[2]*Log[5] + Log[2]^2*Log[5]^2 + E^(x^2 + 2*x^4 + x^6)*(-6 - 2*x - 2*Log[2]*Log[5])),x]

[Out]

(3 - E^(x + x^3)^2 + x + Log[2]*Log[5])^(-1)

Rule 6818

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*(y^(m + 1)/(m + 1)), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 21, normalized size = 0.78 \begin {gather*} \frac {1}{3-e^{\left (x+x^3\right )^2}+x+\log (2) \log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + E^(x^2 + 2*x^4 + x^6)*(2*x + 8*x^3 + 6*x^5))/(9 + E^(2*x^2 + 4*x^4 + 2*x^6) + 6*x + x^2 + (6 +
2*x)*Log[2]*Log[5] + Log[2]^2*Log[5]^2 + E^(x^2 + 2*x^4 + x^6)*(-6 - 2*x - 2*Log[2]*Log[5])),x]

[Out]

(3 - E^(x + x^3)^2 + x + Log[2]*Log[5])^(-1)

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Maple [A]
time = 6.29, size = 25, normalized size = 0.93

method	result	size
risch	\(\frac {1}{\ln \left (2\right ) \ln \left (5\right )+x -{\mathrm e}^{x^{2} \left (x^{2}+1\right )^{2}}+3}\)	\(25\)
norman	\(\frac {1}{\ln \left (2\right ) \ln \left (5\right )+x -{\mathrm e}^{x^{6}+2 x^{4}+x^{2}}+3}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((6*x^5+8*x^3+2*x)*exp(x^6+2*x^4+x^2)-1)/(exp(x^6+2*x^4+x^2)^2+(-2*ln(2)*ln(5)-2*x-6)*exp(x^6+2*x^4+x^2)+l
n(2)^2*ln(5)^2+(2*x+6)*ln(2)*ln(5)+x^2+6*x+9),x,method=_RETURNVERBOSE)

[Out]

1/(ln(2)*ln(5)+x-exp(x^2*(x^2+1)^2)+3)

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Maxima [A]
time = 0.55, size = 25, normalized size = 0.93 \begin {gather*} \frac {1}{\log \left (5\right ) \log \left (2\right ) + x - e^{\left (x^{6} + 2 \, x^{4} + x^{2}\right )} + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^5+8*x^3+2*x)*exp(x^6+2*x^4+x^2)-1)/(exp(x^6+2*x^4+x^2)^2+(-2*log(2)*log(5)-2*x-6)*exp(x^6+2*x^
4+x^2)+log(2)^2*log(5)^2+(2*x+6)*log(2)*log(5)+x^2+6*x+9),x, algorithm="maxima")

[Out]

1/(log(5)*log(2) + x - e^(x^6 + 2*x^4 + x^2) + 3)

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Fricas [A]
time = 0.36, size = 25, normalized size = 0.93 \begin {gather*} \frac {1}{\log \left (5\right ) \log \left (2\right ) + x - e^{\left (x^{6} + 2 \, x^{4} + x^{2}\right )} + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^5+8*x^3+2*x)*exp(x^6+2*x^4+x^2)-1)/(exp(x^6+2*x^4+x^2)^2+(-2*log(2)*log(5)-2*x-6)*exp(x^6+2*x^
4+x^2)+log(2)^2*log(5)^2+(2*x+6)*log(2)*log(5)+x^2+6*x+9),x, algorithm="fricas")

[Out]

1/(log(5)*log(2) + x - e^(x^6 + 2*x^4 + x^2) + 3)

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Sympy [A]
time = 0.07, size = 26, normalized size = 0.96 \begin {gather*} - \frac {1}{- x + e^{x^{6} + 2 x^{4} + x^{2}} - 3 - \log {\left (2 \right )} \log {\left (5 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x**5+8*x**3+2*x)*exp(x**6+2*x**4+x**2)-1)/(exp(x**6+2*x**4+x**2)**2+(-2*ln(2)*ln(5)-2*x-6)*exp(x
**6+2*x**4+x**2)+ln(2)**2*ln(5)**2+(2*x+6)*ln(2)*ln(5)+x**2+6*x+9),x)

[Out]

-1/(-x + exp(x**6 + 2*x**4 + x**2) - 3 - log(2)*log(5))

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Giac [A]
time = 0.42, size = 25, normalized size = 0.93 \begin {gather*} \frac {1}{\log \left (5\right ) \log \left (2\right ) + x - e^{\left (x^{6} + 2 \, x^{4} + x^{2}\right )} + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^5+8*x^3+2*x)*exp(x^6+2*x^4+x^2)-1)/(exp(x^6+2*x^4+x^2)^2+(-2*log(2)*log(5)-2*x-6)*exp(x^6+2*x^
4+x^2)+log(2)^2*log(5)^2+(2*x+6)*log(2)*log(5)+x^2+6*x+9),x, algorithm="giac")

[Out]

1/(log(5)*log(2) + x - e^(x^6 + 2*x^4 + x^2) + 3)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^{x^6+2\,x^4+x^2}\,\left (6\,x^5+8\,x^3+2\,x\right )-1}{6\,x+{\mathrm {e}}^{2\,x^6+4\,x^4+2\,x^2}-{\mathrm {e}}^{x^6+2\,x^4+x^2}\,\left (2\,x+2\,\ln \left (2\right )\,\ln \left (5\right )+6\right )+x^2+{\ln \left (2\right )}^2\,{\ln \left (5\right )}^2+\ln \left (2\right )\,\ln \left (5\right )\,\left (2\,x+6\right )+9} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x^2 + 2*x^4 + x^6)*(2*x + 8*x^3 + 6*x^5) - 1)/(6*x + exp(2*x^2 + 4*x^4 + 2*x^6) - exp(x^2 + 2*x^4 + x
^6)*(2*x + 2*log(2)*log(5) + 6) + x^2 + log(2)^2*log(5)^2 + log(2)*log(5)*(2*x + 6) + 9),x)

[Out]

int((exp(x^2 + 2*x^4 + x^6)*(2*x + 8*x^3 + 6*x^5) - 1)/(6*x + exp(2*x^2 + 4*x^4 + 2*x^6) - exp(x^2 + 2*x^4 + x
^6)*(2*x + 2*log(2)*log(5) + 6) + x^2 + log(2)^2*log(5)^2 + log(2)*log(5)*(2*x + 6) + 9), x)

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