3.56.10 \(\int \frac {-2 e^{4 x}-8 e^{3 x} x+8 x^2-16 x^3-2 x^4+e^{2 x} (4 x-32 x^2+4 e^5 x^2)+e^5 (-2 x^2+4 x^3)+e^x (-12 x^2-28 x^3+e^5 (4 x^2+4 x^3))}{e^{2+4 x} x^2+4 e^{2+3 x} x^3+e^{2+2 x} (-2 x^3+6 x^4)+e^{2+x} (-4 x^4+4 x^5)+e^2 (x^4-2 x^5+x^6)} \, dx\)

Optimal. Leaf size=32 \[ \frac {\frac {2}{x}-\frac {2 \left (5-e^5\right )}{x-\left (e^x+x\right )^2}}{e^2} \]

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Rubi [F]  time = 1.85, 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 {-2 e^{4 x}-8 e^{3 x} x+8 x^2-16 x^3-2 x^4+e^{2 x} \left (4 x-32 x^2+4 e^5 x^2\right )+e^5 \left (-2 x^2+4 x^3\right )+e^x \left (-12 x^2-28 x^3+e^5 \left (4 x^2+4 x^3\right )\right )}{e^{2+4 x} x^2+4 e^{2+3 x} x^3+e^{2+2 x} \left (-2 x^3+6 x^4\right )+e^{2+x} \left (-4 x^4+4 x^5\right )+e^2 \left (x^4-2 x^5+x^6\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-2*E^(4*x) - 8*E^(3*x)*x + 8*x^2 - 16*x^3 - 2*x^4 + E^(2*x)*(4*x - 32*x^2 + 4*E^5*x^2) + E^5*(-2*x^2 + 4*
x^3) + E^x*(-12*x^2 - 28*x^3 + E^5*(4*x^2 + 4*x^3)))/(E^(2 + 4*x)*x^2 + 4*E^(2 + 3*x)*x^3 + E^(2 + 2*x)*(-2*x^
3 + 6*x^4) + E^(2 + x)*(-4*x^4 + 4*x^5) + E^2*(x^4 - 2*x^5 + x^6)),x]

[Out]

2/(E^2*x) + (2*(5 - E^5)*Defer[Int][(E^(2*x) - x + 2*E^x*x + x^2)^(-2), x])/E^2 - (4*(5 - E^5)*Defer[Int][E^x/
(E^(2*x) - x + 2*E^x*x + x^2)^2, x])/E^2 - (8*(5 - E^5)*Defer[Int][x/(E^(2*x) - x + 2*E^x*x + x^2)^2, x])/E^2
+ (4*(5 - E^5)*Defer[Int][(E^x*x)/(E^(2*x) - x + 2*E^x*x + x^2)^2, x])/E^2 + (4*(5 - E^5)*Defer[Int][x^2/(E^(2
*x) - x + 2*E^x*x + x^2)^2, x])/E^2 - (4*(5 - E^5)*Defer[Int][(E^(2*x) - x + 2*E^x*x + x^2)^(-1), x])/E^2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \left (-e^{4 x}-4 e^{3 x} x+2 e^{5+2 x} x^2+2 e^{5+x} x^2 (1+x)-2 e^x x^2 (3+7 x)-2 e^{2 x} x (-1+8 x)-x^2 \left (-4+8 x+x^2\right )-e^5 \left (x^2-2 x^3\right )\right )}{e^2 x^2 \left (e^{2 x}+2 e^x x+(-1+x) x\right )^2} \, dx\\ &=\frac {2 \int \frac {-e^{4 x}-4 e^{3 x} x+2 e^{5+2 x} x^2+2 e^{5+x} x^2 (1+x)-2 e^x x^2 (3+7 x)-2 e^{2 x} x (-1+8 x)-x^2 \left (-4+8 x+x^2\right )-e^5 \left (x^2-2 x^3\right )}{x^2 \left (e^{2 x}+2 e^x x+(-1+x) x\right )^2} \, dx}{e^2}\\ &=\frac {2 \int \left (-\frac {1}{x^2}+\frac {2 \left (-5+e^5\right )}{e^{2 x}-x+2 e^x x+x^2}-\frac {\left (-5+e^5\right ) \left (1-2 e^x-4 x+2 e^x x+2 x^2\right )}{\left (e^{2 x}-x+2 e^x x+x^2\right )^2}\right ) \, dx}{e^2}\\ &=\frac {2}{e^2 x}+\frac {\left (2 \left (5-e^5\right )\right ) \int \frac {1-2 e^x-4 x+2 e^x x+2 x^2}{\left (e^{2 x}-x+2 e^x x+x^2\right )^2} \, dx}{e^2}-\frac {\left (4 \left (5-e^5\right )\right ) \int \frac {1}{e^{2 x}-x+2 e^x x+x^2} \, dx}{e^2}\\ &=\frac {2}{e^2 x}+\frac {\left (2 \left (5-e^5\right )\right ) \int \left (\frac {1}{\left (e^{2 x}-x+2 e^x x+x^2\right )^2}-\frac {2 e^x}{\left (e^{2 x}-x+2 e^x x+x^2\right )^2}-\frac {4 x}{\left (e^{2 x}-x+2 e^x x+x^2\right )^2}+\frac {2 e^x x}{\left (e^{2 x}-x+2 e^x x+x^2\right )^2}+\frac {2 x^2}{\left (e^{2 x}-x+2 e^x x+x^2\right )^2}\right ) \, dx}{e^2}-\frac {\left (4 \left (5-e^5\right )\right ) \int \frac {1}{e^{2 x}-x+2 e^x x+x^2} \, dx}{e^2}\\ &=\frac {2}{e^2 x}+\frac {\left (2 \left (5-e^5\right )\right ) \int \frac {1}{\left (e^{2 x}-x+2 e^x x+x^2\right )^2} \, dx}{e^2}-\frac {\left (4 \left (5-e^5\right )\right ) \int \frac {e^x}{\left (e^{2 x}-x+2 e^x x+x^2\right )^2} \, dx}{e^2}+\frac {\left (4 \left (5-e^5\right )\right ) \int \frac {e^x x}{\left (e^{2 x}-x+2 e^x x+x^2\right )^2} \, dx}{e^2}+\frac {\left (4 \left (5-e^5\right )\right ) \int \frac {x^2}{\left (e^{2 x}-x+2 e^x x+x^2\right )^2} \, dx}{e^2}-\frac {\left (4 \left (5-e^5\right )\right ) \int \frac {1}{e^{2 x}-x+2 e^x x+x^2} \, dx}{e^2}-\frac {\left (8 \left (5-e^5\right )\right ) \int \frac {x}{\left (e^{2 x}-x+2 e^x x+x^2\right )^2} \, dx}{e^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 37, normalized size = 1.16 \begin {gather*} -\frac {2 \left (-\frac {1}{x}+\frac {-5+e^5}{e^{2 x}-x+2 e^x x+x^2}\right )}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*E^(4*x) - 8*E^(3*x)*x + 8*x^2 - 16*x^3 - 2*x^4 + E^(2*x)*(4*x - 32*x^2 + 4*E^5*x^2) + E^5*(-2*x^
2 + 4*x^3) + E^x*(-12*x^2 - 28*x^3 + E^5*(4*x^2 + 4*x^3)))/(E^(2 + 4*x)*x^2 + 4*E^(2 + 3*x)*x^3 + E^(2 + 2*x)*
(-2*x^3 + 6*x^4) + E^(2 + x)*(-4*x^4 + 4*x^5) + E^2*(x^4 - 2*x^5 + x^6)),x]

[Out]

(-2*(-x^(-1) + (-5 + E^5)/(E^(2*x) - x + 2*E^x*x + x^2)))/E^2

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fricas [B]  time = 0.82, size = 65, normalized size = 2.03 \begin {gather*} -\frac {2 \, {\left (x e^{9} - {\left (x^{2} + 4 \, x\right )} e^{4} - 2 \, x e^{\left (x + 4\right )} - e^{\left (2 \, x + 4\right )}\right )}}{2 \, x^{2} e^{\left (x + 6\right )} + {\left (x^{3} - x^{2}\right )} e^{6} + x e^{\left (2 \, x + 6\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(x)^4-8*x*exp(x)^3+(4*x^2*exp(5)-32*x^2+4*x)*exp(x)^2+((4*x^3+4*x^2)*exp(5)-28*x^3-12*x^2)*ex
p(x)+(4*x^3-2*x^2)*exp(5)-2*x^4-16*x^3+8*x^2)/(x^2*exp(2)*exp(x)^4+4*x^3*exp(2)*exp(x)^3+(6*x^4-2*x^3)*exp(2)*
exp(x)^2+(4*x^5-4*x^4)*exp(2)*exp(x)+(x^6-2*x^5+x^4)*exp(2)),x, algorithm="fricas")

[Out]

-2*(x*e^9 - (x^2 + 4*x)*e^4 - 2*x*e^(x + 4) - e^(2*x + 4))/(2*x^2*e^(x + 6) + (x^3 - x^2)*e^6 + x*e^(2*x + 6))

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giac [B]  time = 0.48, size = 65, normalized size = 2.03 \begin {gather*} \frac {2 \, {\left (x^{2} e^{2} - 2 \, x e^{7} + 9 \, x e^{2} + 2 \, x e^{\left (x + 2\right )} + e^{\left (2 \, x + 2\right )}\right )}}{x^{3} e^{4} - x^{2} e^{4} + 2 \, x^{2} e^{\left (x + 4\right )} + x e^{\left (2 \, x + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(x)^4-8*x*exp(x)^3+(4*x^2*exp(5)-32*x^2+4*x)*exp(x)^2+((4*x^3+4*x^2)*exp(5)-28*x^3-12*x^2)*ex
p(x)+(4*x^3-2*x^2)*exp(5)-2*x^4-16*x^3+8*x^2)/(x^2*exp(2)*exp(x)^4+4*x^3*exp(2)*exp(x)^3+(6*x^4-2*x^3)*exp(2)*
exp(x)^2+(4*x^5-4*x^4)*exp(2)*exp(x)+(x^6-2*x^5+x^4)*exp(2)),x, algorithm="giac")

[Out]

2*(x^2*e^2 - 2*x*e^7 + 9*x*e^2 + 2*x*e^(x + 2) + e^(2*x + 2))/(x^3*e^4 - x^2*e^4 + 2*x^2*e^(x + 4) + x*e^(2*x
+ 4))

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maple [A]  time = 0.08, size = 55, normalized size = 1.72

method	result	size
risch	\(\frac {2 \,{\mathrm e}^{-2}}{x}-\frac {2 \,{\mathrm e}^{-2} {\mathrm e}^{5}}{{\mathrm e}^{2 x}+2 \,{\mathrm e}^{x} x +x^{2}-x}+\frac {10 \,{\mathrm e}^{-2}}{{\mathrm e}^{2 x}+2 \,{\mathrm e}^{x} x +x^{2}-x}\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*exp(x)^4-8*x*exp(x)^3+(4*x^2*exp(5)-32*x^2+4*x)*exp(x)^2+((4*x^3+4*x^2)*exp(5)-28*x^3-12*x^2)*exp(x)+(
4*x^3-2*x^2)*exp(5)-2*x^4-16*x^3+8*x^2)/(x^2*exp(2)*exp(x)^4+4*x^3*exp(2)*exp(x)^3+(6*x^4-2*x^3)*exp(2)*exp(x)
^2+(4*x^5-4*x^4)*exp(2)*exp(x)+(x^6-2*x^5+x^4)*exp(2)),x,method=_RETURNVERBOSE)

[Out]

2/x*exp(-2)-2*exp(-2)/(exp(2*x)+2*exp(x)*x+x^2-x)*exp(5)+10*exp(-2)/(exp(2*x)+2*exp(x)*x+x^2-x)

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maxima [B]  time = 0.46, size = 55, normalized size = 1.72 \begin {gather*} \frac {2 \, {\left (x^{2} - x {\left (e^{5} - 4\right )} + 2 \, x e^{x} + e^{\left (2 \, x\right )}\right )}}{x^{3} e^{2} - x^{2} e^{2} + 2 \, x^{2} e^{\left (x + 2\right )} + x e^{\left (2 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(x)^4-8*x*exp(x)^3+(4*x^2*exp(5)-32*x^2+4*x)*exp(x)^2+((4*x^3+4*x^2)*exp(5)-28*x^3-12*x^2)*ex
p(x)+(4*x^3-2*x^2)*exp(5)-2*x^4-16*x^3+8*x^2)/(x^2*exp(2)*exp(x)^4+4*x^3*exp(2)*exp(x)^3+(6*x^4-2*x^3)*exp(2)*
exp(x)^2+(4*x^5-4*x^4)*exp(2)*exp(x)+(x^6-2*x^5+x^4)*exp(2)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.76, size = 58, normalized size = 1.81 \begin {gather*} \frac {2\,{\mathrm {e}}^{2\,x}+x\,\left (4\,{\mathrm {e}}^x-2\,{\mathrm {e}}^5+8\right )+2\,x^2}{x\,{\mathrm {e}}^{2\,x+2}+2\,x^2\,{\mathrm {e}}^{x+2}-x^2\,{\mathrm {e}}^2+x^3\,{\mathrm {e}}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*exp(4*x) + 8*x*exp(3*x) + exp(5)*(2*x^2 - 4*x^3) + exp(x)*(12*x^2 - exp(5)*(4*x^2 + 4*x^3) + 28*x^3) -
exp(2*x)*(4*x + 4*x^2*exp(5) - 32*x^2) - 8*x^2 + 16*x^3 + 2*x^4)/(exp(2)*(x^4 - 2*x^5 + x^6) - exp(2)*exp(x)*
(4*x^4 - 4*x^5) - exp(2*x)*exp(2)*(2*x^3 - 6*x^4) + x^2*exp(4*x)*exp(2) + 4*x^3*exp(3*x)*exp(2)),x)

[Out]

(2*exp(2*x) + x*(4*exp(x) - 2*exp(5) + 8) + 2*x^2)/(x*exp(2*x + 2) + 2*x^2*exp(x + 2) - x^2*exp(2) + x^3*exp(2
))

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sympy [A]  time = 0.21, size = 42, normalized size = 1.31 \begin {gather*} \frac {10 - 2 e^{5}}{x^{2} e^{2} + 2 x e^{2} e^{x} - x e^{2} + e^{2} e^{2 x}} + \frac {2}{x e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(x)**4-8*x*exp(x)**3+(4*x**2*exp(5)-32*x**2+4*x)*exp(x)**2+((4*x**3+4*x**2)*exp(5)-28*x**3-12
*x**2)*exp(x)+(4*x**3-2*x**2)*exp(5)-2*x**4-16*x**3+8*x**2)/(x**2*exp(2)*exp(x)**4+4*x**3*exp(2)*exp(x)**3+(6*
x**4-2*x**3)*exp(2)*exp(x)**2+(4*x**5-4*x**4)*exp(2)*exp(x)+(x**6-2*x**5+x**4)*exp(2)),x)

[Out]

(10 - 2*exp(5))/(x**2*exp(2) + 2*x*exp(2)*exp(x) - x*exp(2) + exp(2)*exp(2*x)) + 2*exp(-2)/x

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