3.35.98 \(\int \frac {8 e^{\frac {1+16 x^3}{8 x^2}} x^3-16 e^{\frac {1+16 x^3}{16 x^2}} x^4+8 x^5+e^{e^x} (8 x^3-8 e^x x^4+e^{\frac {1+16 x^3}{16 x^2}} (1-8 x^3+8 e^x x^3))}{8 e^{\frac {1+16 x^3}{8 x^2}} x^3-16 e^{\frac {1+16 x^3}{16 x^2}} x^4+8 x^5} \, dx\)

Optimal. Leaf size=29 \[ -\frac {7}{4}+x-\frac {e^{e^x}}{-e^{\frac {1}{16 x^2}+x}+x} \]

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Rubi [F]  time = 4.23, 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 {8 e^{\frac {1+16 x^3}{8 x^2}} x^3-16 e^{\frac {1+16 x^3}{16 x^2}} x^4+8 x^5+e^{e^x} \left (8 x^3-8 e^x x^4+e^{\frac {1+16 x^3}{16 x^2}} \left (1-8 x^3+8 e^x x^3\right )\right )}{8 e^{\frac {1+16 x^3}{8 x^2}} x^3-16 e^{\frac {1+16 x^3}{16 x^2}} x^4+8 x^5} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(8*E^((1 + 16*x^3)/(8*x^2))*x^3 - 16*E^((1 + 16*x^3)/(16*x^2))*x^4 + 8*x^5 + E^E^x*(8*x^3 - 8*E^x*x^4 + E^
((1 + 16*x^3)/(16*x^2))*(1 - 8*x^3 + 8*E^x*x^3)))/(8*E^((1 + 16*x^3)/(8*x^2))*x^3 - 16*E^((1 + 16*x^3)/(16*x^2
))*x^4 + 8*x^5),x]

[Out]

x + Defer[Int][E^(E^x - 1/(16*x^2)), x] + Defer[Int][E^E^x/(E^(1/(16*x^2) + x) - x)^2, x] - Defer[Int][E^E^x/(
E^(1/(16*x^2) + x) - x), x] + Defer[Int][E^E^x/((E^(1/(16*x^2) + x) - x)*x^3), x]/8 + Defer[Int][E^E^x/((E^(1/
(16*x^2) + x) - x)^2*x^2), x]/8 - Defer[Int][(E^E^x*x)/(E^(1/(16*x^2) + x) - x)^2, x] + Defer[Int][(E^(E^x - 1
/(16*x^2))*x)/(E^(1/(16*x^2) + x) - x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {8 e^{\frac {1+16 x^3}{8 x^2}} x^3-16 e^{\frac {1+16 x^3}{16 x^2}} x^4+8 x^5+e^{e^x} \left (8 x^3-8 e^x x^4+e^{\frac {1+16 x^3}{16 x^2}} \left (1-8 x^3+8 e^x x^3\right )\right )}{8 \left (e^{\frac {1}{16 x^2}+x}-x\right )^2 x^3} \, dx\\ &=\frac {1}{8} \int \frac {8 e^{\frac {1+16 x^3}{8 x^2}} x^3-16 e^{\frac {1+16 x^3}{16 x^2}} x^4+8 x^5+e^{e^x} \left (8 x^3-8 e^x x^4+e^{\frac {1+16 x^3}{16 x^2}} \left (1-8 x^3+8 e^x x^3\right )\right )}{\left (e^{\frac {1}{16 x^2}+x}-x\right )^2 x^3} \, dx\\ &=\frac {1}{8} \int \left (8 e^{-\frac {1}{16 x^2}} \left (e^{e^x}+e^{\frac {1}{16 x^2}}\right )-\frac {e^{e^x} \left (-1-8 x^2+8 x^3\right )}{\left (e^{\frac {1}{16 x^2}+x}-x\right )^2 x^2}-\frac {e^{e^x-\frac {1}{16 x^2}} \left (-e^{\frac {1}{16 x^2}}+8 e^{\frac {1}{16 x^2}} x^3-8 x^4\right )}{\left (e^{\frac {1}{16 x^2}+x}-x\right ) x^3}\right ) \, dx\\ &=-\left (\frac {1}{8} \int \frac {e^{e^x} \left (-1-8 x^2+8 x^3\right )}{\left (e^{\frac {1}{16 x^2}+x}-x\right )^2 x^2} \, dx\right )-\frac {1}{8} \int \frac {e^{e^x-\frac {1}{16 x^2}} \left (-e^{\frac {1}{16 x^2}}+8 e^{\frac {1}{16 x^2}} x^3-8 x^4\right )}{\left (e^{\frac {1}{16 x^2}+x}-x\right ) x^3} \, dx+\int e^{-\frac {1}{16 x^2}} \left (e^{e^x}+e^{\frac {1}{16 x^2}}\right ) \, dx\\ &=-\left (\frac {1}{8} \int \left (-\frac {8 e^{e^x}}{\left (e^{\frac {1}{16 x^2}+x}-x\right )^2}-\frac {e^{e^x}}{\left (e^{\frac {1}{16 x^2}+x}-x\right )^2 x^2}+\frac {8 e^{e^x} x}{\left (e^{\frac {1}{16 x^2}+x}-x\right )^2}\right ) \, dx\right )-\frac {1}{8} \int \frac {e^{e^x} \left (-1+8 x^3-8 e^{-\frac {1}{16 x^2}} x^4\right )}{\left (e^{\frac {1}{16 x^2}+x}-x\right ) x^3} \, dx+\int \left (1+e^{e^x-\frac {1}{16 x^2}}\right ) \, dx\\ &=x+\frac {1}{8} \int \frac {e^{e^x}}{\left (e^{\frac {1}{16 x^2}+x}-x\right )^2 x^2} \, dx-\frac {1}{8} \int \left (\frac {8 e^{e^x}}{e^{\frac {1}{16 x^2}+x}-x}-\frac {e^{e^x}}{\left (e^{\frac {1}{16 x^2}+x}-x\right ) x^3}-\frac {8 e^{e^x-\frac {1}{16 x^2}} x}{e^{\frac {1}{16 x^2}+x}-x}\right ) \, dx+\int e^{e^x-\frac {1}{16 x^2}} \, dx+\int \frac {e^{e^x}}{\left (e^{\frac {1}{16 x^2}+x}-x\right )^2} \, dx-\int \frac {e^{e^x} x}{\left (e^{\frac {1}{16 x^2}+x}-x\right )^2} \, dx\\ &=x+\frac {1}{8} \int \frac {e^{e^x}}{\left (e^{\frac {1}{16 x^2}+x}-x\right ) x^3} \, dx+\frac {1}{8} \int \frac {e^{e^x}}{\left (e^{\frac {1}{16 x^2}+x}-x\right )^2 x^2} \, dx+\int e^{e^x-\frac {1}{16 x^2}} \, dx+\int \frac {e^{e^x}}{\left (e^{\frac {1}{16 x^2}+x}-x\right )^2} \, dx-\int \frac {e^{e^x}}{e^{\frac {1}{16 x^2}+x}-x} \, dx-\int \frac {e^{e^x} x}{\left (e^{\frac {1}{16 x^2}+x}-x\right )^2} \, dx+\int \frac {e^{e^x-\frac {1}{16 x^2}} x}{e^{\frac {1}{16 x^2}+x}-x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 32, normalized size = 1.10 \begin {gather*} \frac {1}{8} \left (\frac {8 e^{e^x}}{e^{\frac {1}{16 x^2}+x}-x}+8 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8*E^((1 + 16*x^3)/(8*x^2))*x^3 - 16*E^((1 + 16*x^3)/(16*x^2))*x^4 + 8*x^5 + E^E^x*(8*x^3 - 8*E^x*x^
4 + E^((1 + 16*x^3)/(16*x^2))*(1 - 8*x^3 + 8*E^x*x^3)))/(8*E^((1 + 16*x^3)/(8*x^2))*x^3 - 16*E^((1 + 16*x^3)/(
16*x^2))*x^4 + 8*x^5),x]

[Out]

((8*E^E^x)/(E^(1/(16*x^2) + x) - x) + 8*x)/8

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fricas [B]  time = 0.87, size = 45, normalized size = 1.55 \begin {gather*} \frac {x^{2} - x e^{\left (\frac {16 \, x^{3} + 1}{16 \, x^{2}}\right )} - e^{\left (e^{x}\right )}}{x - e^{\left (\frac {16 \, x^{3} + 1}{16 \, x^{2}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*exp(x)*x^3-8*x^3+1)*exp(1/16*(16*x^3+1)/x^2)-8*exp(x)*x^4+8*x^3)*exp(exp(x))+8*x^3*exp(1/16*(16
*x^3+1)/x^2)^2-16*x^4*exp(1/16*(16*x^3+1)/x^2)+8*x^5)/(8*x^3*exp(1/16*(16*x^3+1)/x^2)^2-16*x^4*exp(1/16*(16*x^
3+1)/x^2)+8*x^5),x, algorithm="fricas")

[Out]

(x^2 - x*e^(1/16*(16*x^3 + 1)/x^2) - e^(e^x))/(x - e^(1/16*(16*x^3 + 1)/x^2))

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giac [B]  time = 0.29, size = 326, normalized size = 11.24 \begin {gather*} \frac {8 \, x^{5} e^{\left (\frac {16 \, x^{3} + 1}{16 \, x^{2}}\right )} - 8 \, x^{4} e^{\left (\frac {16 \, x^{3} + 1}{8 \, x^{2}}\right )} - 8 \, x^{4} e^{\left (\frac {16 \, x^{3} + 1}{16 \, x^{2}}\right )} + 8 \, x^{3} e^{\left (x + \frac {16 \, x^{3} + 16 \, x^{2} e^{x} + 1}{16 \, x^{2}}\right )} - 8 \, x^{3} e^{\left (x + \frac {16 \, x^{3} + 1}{16 \, x^{2}} + e^{x}\right )} - 8 \, x^{3} e^{\left (\frac {16 \, x^{3} + 16 \, x^{2} e^{x} + 1}{16 \, x^{2}}\right )} + 8 \, x^{3} e^{\left (\frac {16 \, x^{3} + 1}{8 \, x^{2}}\right )} + 8 \, x^{2} e^{\left (\frac {16 \, x^{3} + 1}{16 \, x^{2}} + e^{x}\right )} - x^{2} e^{\left (\frac {16 \, x^{3} + 1}{16 \, x^{2}}\right )} + x e^{\left (\frac {16 \, x^{3} + 1}{8 \, x^{2}}\right )} + e^{\left (\frac {16 \, x^{3} + 16 \, x^{2} e^{x} + 1}{16 \, x^{2}}\right )}}{8 \, x^{4} e^{\left (\frac {16 \, x^{3} + 1}{16 \, x^{2}}\right )} - 8 \, x^{3} e^{\left (\frac {16 \, x^{3} + 1}{8 \, x^{2}}\right )} - 8 \, x^{3} e^{\left (\frac {16 \, x^{3} + 1}{16 \, x^{2}}\right )} + 8 \, x^{2} e^{\left (\frac {16 \, x^{3} + 1}{8 \, x^{2}}\right )} - x e^{\left (\frac {16 \, x^{3} + 1}{16 \, x^{2}}\right )} + e^{\left (\frac {16 \, x^{3} + 1}{8 \, x^{2}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*exp(x)*x^3-8*x^3+1)*exp(1/16*(16*x^3+1)/x^2)-8*exp(x)*x^4+8*x^3)*exp(exp(x))+8*x^3*exp(1/16*(16
*x^3+1)/x^2)^2-16*x^4*exp(1/16*(16*x^3+1)/x^2)+8*x^5)/(8*x^3*exp(1/16*(16*x^3+1)/x^2)^2-16*x^4*exp(1/16*(16*x^
3+1)/x^2)+8*x^5),x, algorithm="giac")

[Out]

(8*x^5*e^(1/16*(16*x^3 + 1)/x^2) - 8*x^4*e^(1/8*(16*x^3 + 1)/x^2) - 8*x^4*e^(1/16*(16*x^3 + 1)/x^2) + 8*x^3*e^
(x + 1/16*(16*x^3 + 16*x^2*e^x + 1)/x^2) - 8*x^3*e^(x + 1/16*(16*x^3 + 1)/x^2 + e^x) - 8*x^3*e^(1/16*(16*x^3 +
16*x^2*e^x + 1)/x^2) + 8*x^3*e^(1/8*(16*x^3 + 1)/x^2) + 8*x^2*e^(1/16*(16*x^3 + 1)/x^2 + e^x) - x^2*e^(1/16*(
16*x^3 + 1)/x^2) + x*e^(1/8*(16*x^3 + 1)/x^2) + e^(1/16*(16*x^3 + 16*x^2*e^x + 1)/x^2))/(8*x^4*e^(1/16*(16*x^3
+ 1)/x^2) - 8*x^3*e^(1/8*(16*x^3 + 1)/x^2) - 8*x^3*e^(1/16*(16*x^3 + 1)/x^2) + 8*x^2*e^(1/8*(16*x^3 + 1)/x^2)
- x*e^(1/16*(16*x^3 + 1)/x^2) + e^(1/8*(16*x^3 + 1)/x^2))

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (8 \,{\mathrm e}^{x} x^{3}-8 x^{3}+1\right ) {\mathrm e}^{\frac {16 x^{3}+1}{16 x^{2}}}-8 \,{\mathrm e}^{x} x^{4}+8 x^{3}\right ) {\mathrm e}^{{\mathrm e}^{x}}+8 x^{3} {\mathrm e}^{\frac {16 x^{3}+1}{8 x^{2}}}-16 x^{4} {\mathrm e}^{\frac {16 x^{3}+1}{16 x^{2}}}+8 x^{5}}{8 x^{3} {\mathrm e}^{\frac {16 x^{3}+1}{8 x^{2}}}-16 x^{4} {\mathrm e}^{\frac {16 x^{3}+1}{16 x^{2}}}+8 x^{5}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((8*exp(x)*x^3-8*x^3+1)*exp(1/16*(16*x^3+1)/x^2)-8*exp(x)*x^4+8*x^3)*exp(exp(x))+8*x^3*exp(1/16*(16*x^3+1
)/x^2)^2-16*x^4*exp(1/16*(16*x^3+1)/x^2)+8*x^5)/(8*x^3*exp(1/16*(16*x^3+1)/x^2)^2-16*x^4*exp(1/16*(16*x^3+1)/x
^2)+8*x^5),x)

[Out]

int((((8*exp(x)*x^3-8*x^3+1)*exp(1/16*(16*x^3+1)/x^2)-8*exp(x)*x^4+8*x^3)*exp(exp(x))+8*x^3*exp(1/16*(16*x^3+1
)/x^2)^2-16*x^4*exp(1/16*(16*x^3+1)/x^2)+8*x^5)/(8*x^3*exp(1/16*(16*x^3+1)/x^2)^2-16*x^4*exp(1/16*(16*x^3+1)/x
^2)+8*x^5),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} x + \frac {1}{8} \, \int \frac {{\left (8 \, x^{3} e^{\left (2 \, x\right )} - {\left (8 \, x^{3} - 1\right )} e^{x}\right )} e^{\left (\frac {1}{16 \, x^{2}} + e^{x}\right )} - 8 \, {\left (x^{4} e^{x} - x^{3}\right )} e^{\left (e^{x}\right )}}{x^{5} - 2 \, x^{4} e^{\left (x + \frac {1}{16 \, x^{2}}\right )} + x^{3} e^{\left (2 \, x + \frac {1}{8 \, x^{2}}\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*exp(x)*x^3-8*x^3+1)*exp(1/16*(16*x^3+1)/x^2)-8*exp(x)*x^4+8*x^3)*exp(exp(x))+8*x^3*exp(1/16*(16
*x^3+1)/x^2)^2-16*x^4*exp(1/16*(16*x^3+1)/x^2)+8*x^5)/(8*x^3*exp(1/16*(16*x^3+1)/x^2)^2-16*x^4*exp(1/16*(16*x^
3+1)/x^2)+8*x^5),x, algorithm="maxima")

[Out]

x + 1/8*integrate(((8*x^3*e^(2*x) - (8*x^3 - 1)*e^x)*e^(1/16/x^2 + e^x) - 8*(x^4*e^x - x^3)*e^(e^x))/(x^5 - 2*
x^4*e^(x + 1/16/x^2) + x^3*e^(2*x + 1/8/x^2)), x)

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mupad [B]  time = 2.09, size = 21, normalized size = 0.72 \begin {gather*} x-\frac {{\mathrm {e}}^{{\mathrm {e}}^x}}{x-{\mathrm {e}}^{x+\frac {1}{16\,x^2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x^3*exp((2*(x^3 + 1/16))/x^2) - 16*x^4*exp((x^3 + 1/16)/x^2) + exp(exp(x))*(8*x^3 - 8*x^4*exp(x) + exp(
(x^3 + 1/16)/x^2)*(8*x^3*exp(x) - 8*x^3 + 1)) + 8*x^5)/(8*x^3*exp((2*(x^3 + 1/16))/x^2) - 16*x^4*exp((x^3 + 1/
16)/x^2) + 8*x^5),x)

[Out]

x - exp(exp(x))/(x - exp(x + 1/(16*x^2)))

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sympy [A]  time = 0.29, size = 19, normalized size = 0.66 \begin {gather*} x - \frac {e^{e^{x}}}{x - e^{\frac {x^{3} + \frac {1}{16}}{x^{2}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*exp(x)*x**3-8*x**3+1)*exp(1/16*(16*x**3+1)/x**2)-8*exp(x)*x**4+8*x**3)*exp(exp(x))+8*x**3*exp(1
/16*(16*x**3+1)/x**2)**2-16*x**4*exp(1/16*(16*x**3+1)/x**2)+8*x**5)/(8*x**3*exp(1/16*(16*x**3+1)/x**2)**2-16*x
**4*exp(1/16*(16*x**3+1)/x**2)+8*x**5),x)

[Out]

x - exp(exp(x))/(x - exp((x**3 + 1/16)/x**2))

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