3.92.42 \(\int \frac {-240 e^4-120 e^2 x-15 x^2+e^{\frac {4+3 x+2 x^2}{4 e^2+x}} (-4 x^2+2 x^4+e^2 (12 x^2+16 x^3))}{16 e^4 x^2+8 e^2 x^3+x^4} \, dx\)

Optimal. Leaf size=35 \[ e^{\frac {-x+x^2+(2+x)^2}{4 e^2+x}}+\frac {3 (5-x)}{x} \]

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Rubi [F]  time = 1.37, 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 {-240 e^4-120 e^2 x-15 x^2+e^{\frac {4+3 x+2 x^2}{4 e^2+x}} \left (-4 x^2+2 x^4+e^2 \left (12 x^2+16 x^3\right )\right )}{16 e^4 x^2+8 e^2 x^3+x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-240*E^4 - 120*E^2*x - 15*x^2 + E^((4 + 3*x + 2*x^2)/(4*E^2 + x))*(-4*x^2 + 2*x^4 + E^2*(12*x^2 + 16*x^3)
))/(16*E^4*x^2 + 8*E^2*x^3 + x^4),x]

[Out]

15/x + 2*Defer[Int][E^((4 + 3*x + 2*x^2)/(4*E^2 + x)), x] - 4*(1 - 3*E^2 + 8*E^4)*Defer[Int][E^((4 + 3*x + 2*x
^2)/(4*E^2 + x))/(4*E^2 + x)^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-240 e^4-120 e^2 x-15 x^2+e^{\frac {4+3 x+2 x^2}{4 e^2+x}} \left (-4 x^2+2 x^4+e^2 \left (12 x^2+16 x^3\right )\right )}{x^2 \left (16 e^4+8 e^2 x+x^2\right )} \, dx\\ &=\int \frac {-240 e^4-120 e^2 x-15 x^2+e^{\frac {4+3 x+2 x^2}{4 e^2+x}} \left (-4 x^2+2 x^4+e^2 \left (12 x^2+16 x^3\right )\right )}{x^2 \left (4 e^2+x\right )^2} \, dx\\ &=\int \left (-\frac {15}{x^2}+\frac {2 e^{\frac {4+3 x+2 x^2}{4 e^2+x}} \left (-2 \left (1-3 e^2\right )+8 e^2 x+x^2\right )}{\left (4 e^2+x\right )^2}\right ) \, dx\\ &=\frac {15}{x}+2 \int \frac {e^{\frac {4+3 x+2 x^2}{4 e^2+x}} \left (-2 \left (1-3 e^2\right )+8 e^2 x+x^2\right )}{\left (4 e^2+x\right )^2} \, dx\\ &=\frac {15}{x}+2 \int \left (e^{\frac {4+3 x+2 x^2}{4 e^2+x}}-\frac {2 e^{\frac {4+3 x+2 x^2}{4 e^2+x}} \left (1-3 e^2+8 e^4\right )}{\left (4 e^2+x\right )^2}\right ) \, dx\\ &=\frac {15}{x}+2 \int e^{\frac {4+3 x+2 x^2}{4 e^2+x}} \, dx-\left (4 \left (1-3 e^2+8 e^4\right )\right ) \int \frac {e^{\frac {4+3 x+2 x^2}{4 e^2+x}}}{\left (4 e^2+x\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.29, size = 28, normalized size = 0.80 \begin {gather*} e^{\frac {4+3 x+2 x^2}{4 e^2+x}}+\frac {15}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-240*E^4 - 120*E^2*x - 15*x^2 + E^((4 + 3*x + 2*x^2)/(4*E^2 + x))*(-4*x^2 + 2*x^4 + E^2*(12*x^2 + 1
6*x^3)))/(16*E^4*x^2 + 8*E^2*x^3 + x^4),x]

[Out]

E^((4 + 3*x + 2*x^2)/(4*E^2 + x)) + 15/x

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fricas [A]  time = 0.58, size = 28, normalized size = 0.80 \begin {gather*} \frac {x e^{\left (\frac {2 \, x^{2} + 3 \, x + 4}{x + 4 \, e^{2}}\right )} + 15}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x^3+12*x^2)*exp(1)^2+2*x^4-4*x^2)*exp((2*x^2+3*x+4)/(4*exp(1)^2+x))-240*exp(1)^4-120*x*exp(1)^
2-15*x^2)/(16*x^2*exp(1)^4+8*x^3*exp(1)^2+x^4),x, algorithm="fricas")

[Out]

(x*e^((2*x^2 + 3*x + 4)/(x + 4*e^2)) + 15)/x

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giac [A]  time = 0.33, size = 52, normalized size = 1.49 \begin {gather*} \frac {{\left (x e^{\left ({\left (2 \, e^{2} + 1\right )} e^{\left (-2\right )} + \frac {2 \, x^{2} e^{2} + 3 \, x e^{2} - x}{x e^{2} + 4 \, e^{4}}\right )} + 15 \, e^{2}\right )} e^{\left (-2\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x^3+12*x^2)*exp(1)^2+2*x^4-4*x^2)*exp((2*x^2+3*x+4)/(4*exp(1)^2+x))-240*exp(1)^4-120*x*exp(1)^
2-15*x^2)/(16*x^2*exp(1)^4+8*x^3*exp(1)^2+x^4),x, algorithm="giac")

[Out]

(x*e^((2*e^2 + 1)*e^(-2) + (2*x^2*e^2 + 3*x*e^2 - x)/(x*e^2 + 4*e^4)) + 15*e^2)*e^(-2)/x

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maple [A]  time = 0.35, size = 27, normalized size = 0.77

method	result	size
risch	\(\frac {15}{x}+{\mathrm e}^{\frac {2 x^{2}+3 x +4}{4 \,{\mathrm e}^{2}+x}}\)	\(27\)
norman	\(\frac {x^{2} {\mathrm e}^{\frac {2 x^{2}+3 x +4}{4 \,{\mathrm e}^{2}+x}}+15 x +60 \,{\mathrm e}^{2}+4 x \,{\mathrm e}^{2} {\mathrm e}^{\frac {2 x^{2}+3 x +4}{4 \,{\mathrm e}^{2}+x}}}{x \left (4 \,{\mathrm e}^{2}+x \right )}\)	\(80\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((16*x^3+12*x^2)*exp(1)^2+2*x^4-4*x^2)*exp((2*x^2+3*x+4)/(4*exp(1)^2+x))-240*exp(1)^4-120*x*exp(1)^2-15*x
^2)/(16*x^2*exp(1)^4+8*x^3*exp(1)^2+x^4),x,method=_RETURNVERBOSE)

[Out]

15/x+exp((2*x^2+3*x+4)/(4*exp(2)+x))

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maxima [B]  time = 0.47, size = 132, normalized size = 3.77 \begin {gather*} -\frac {15}{2} \, {\left (e^{\left (-6\right )} \log \left (x + 4 \, e^{2}\right ) - e^{\left (-6\right )} \log \relax (x) - \frac {4 \, {\left (x + 2 \, e^{2}\right )}}{x^{2} e^{4} + 4 \, x e^{6}}\right )} e^{4} + \frac {15}{2} \, {\left (e^{\left (-4\right )} \log \left (x + 4 \, e^{2}\right ) - e^{\left (-4\right )} \log \relax (x) - \frac {4}{x e^{2} + 4 \, e^{4}}\right )} e^{2} + \frac {15}{x + 4 \, e^{2}} + e^{\left (2 \, x + \frac {32 \, e^{4}}{x + 4 \, e^{2}} - \frac {12 \, e^{2}}{x + 4 \, e^{2}} + \frac {4}{x + 4 \, e^{2}} - 8 \, e^{2} + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x^3+12*x^2)*exp(1)^2+2*x^4-4*x^2)*exp((2*x^2+3*x+4)/(4*exp(1)^2+x))-240*exp(1)^4-120*x*exp(1)^
2-15*x^2)/(16*x^2*exp(1)^4+8*x^3*exp(1)^2+x^4),x, algorithm="maxima")

[Out]

-15/2*(e^(-6)*log(x + 4*e^2) - e^(-6)*log(x) - 4*(x + 2*e^2)/(x^2*e^4 + 4*x*e^6))*e^4 + 15/2*(e^(-4)*log(x + 4
*e^2) - e^(-4)*log(x) - 4/(x*e^2 + 4*e^4))*e^2 + 15/(x + 4*e^2) + e^(2*x + 32*e^4/(x + 4*e^2) - 12*e^2/(x + 4*
e^2) + 4/(x + 4*e^2) - 8*e^2 + 3)

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mupad [B]  time = 7.63, size = 44, normalized size = 1.26 \begin {gather*} \frac {15}{x}+{\mathrm {e}}^{\frac {3\,x}{x+4\,{\mathrm {e}}^2}}\,{\mathrm {e}}^{\frac {2\,x^2}{x+4\,{\mathrm {e}}^2}}\,{\mathrm {e}}^{\frac {4}{x+4\,{\mathrm {e}}^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(240*exp(4) + 120*x*exp(2) + 15*x^2 - exp((3*x + 2*x^2 + 4)/(x + 4*exp(2)))*(exp(2)*(12*x^2 + 16*x^3) - 4
*x^2 + 2*x^4))/(8*x^3*exp(2) + 16*x^2*exp(4) + x^4),x)

[Out]

15/x + exp((3*x)/(x + 4*exp(2)))*exp((2*x^2)/(x + 4*exp(2)))*exp(4/(x + 4*exp(2)))

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sympy [A]  time = 0.31, size = 20, normalized size = 0.57 \begin {gather*} e^{\frac {2 x^{2} + 3 x + 4}{x + 4 e^{2}}} + \frac {15}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x**3+12*x**2)*exp(1)**2+2*x**4-4*x**2)*exp((2*x**2+3*x+4)/(4*exp(1)**2+x))-240*exp(1)**4-120*x
*exp(1)**2-15*x**2)/(16*x**2*exp(1)**4+8*x**3*exp(1)**2+x**4),x)

[Out]

exp((2*x**2 + 3*x + 4)/(x + 4*exp(2))) + 15/x

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