3.45.81 \(\int \frac {e^{x+2 e^8 x^2} (-5+5 x)+e^{x+e^8 x^2} (2 x-x^2-2 e^8 x^3)}{25 e^{3+2 e^8 x^2} x^2-10 e^{3+e^8 x^2} x^3+e^3 x^4} \, dx\)

Optimal. Leaf size=26 \[ \frac {e^{-3+x}}{x \left (5-e^{-e^8 x^2} x\right )} \]

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Rubi [F]  time = 2.86, 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+2 e^8 x^2} (-5+5 x)+e^{x+e^8 x^2} \left (2 x-x^2-2 e^8 x^3\right )}{25 e^{3+2 e^8 x^2} x^2-10 e^{3+e^8 x^2} x^3+e^3 x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(x + 2*E^8*x^2)*(-5 + 5*x) + E^(x + E^8*x^2)*(2*x - x^2 - 2*E^8*x^3))/(25*E^(3 + 2*E^8*x^2)*x^2 - 10*E^
(3 + E^8*x^2)*x^3 + E^3*x^4),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.59, size = 33, normalized size = 1.27 \begin {gather*} \frac {e^{-3+x+e^8 x^2}}{\left (5 e^{e^8 x^2}-x\right ) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(x + 2*E^8*x^2)*(-5 + 5*x) + E^(x + E^8*x^2)*(2*x - x^2 - 2*E^8*x^3))/(25*E^(3 + 2*E^8*x^2)*x^2 -
10*E^(3 + E^8*x^2)*x^3 + E^3*x^4),x]

[Out]

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

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fricas [A]  time = 0.51, size = 45, normalized size = 1.73 \begin {gather*} -\frac {e^{\left (2 \, x^{2} e^{8} + 2 \, x\right )}}{x^{2} e^{\left (x^{2} e^{8} + x + 3\right )} - 5 \, x e^{\left (2 \, x^{2} e^{8} + x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x-5)*exp(x)*exp(x^2*exp(4)^2)^2+(-2*x^3*exp(4)^2-x^2+2*x)*exp(x)*exp(x^2*exp(4)^2))/(25*x^2*exp(
3)*exp(x^2*exp(4)^2)^2-10*x^3*exp(3)*exp(x^2*exp(4)^2)+x^4*exp(3)),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {5 \, {\left (x - 1\right )} e^{\left (2 \, x^{2} e^{8} + x\right )} - {\left (2 \, x^{3} e^{8} + x^{2} - 2 \, x\right )} e^{\left (x^{2} e^{8} + x\right )}}{x^{4} e^{3} - 10 \, x^{3} e^{\left (x^{2} e^{8} + 3\right )} + 25 \, x^{2} e^{\left (2 \, x^{2} e^{8} + 3\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x-5)*exp(x)*exp(x^2*exp(4)^2)^2+(-2*x^3*exp(4)^2-x^2+2*x)*exp(x)*exp(x^2*exp(4)^2))/(25*x^2*exp(
3)*exp(x^2*exp(4)^2)^2-10*x^3*exp(3)*exp(x^2*exp(4)^2)+x^4*exp(3)),x, algorithm="giac")

[Out]

integrate((5*(x - 1)*e^(2*x^2*e^8 + x) - (2*x^3*e^8 + x^2 - 2*x)*e^(x^2*e^8 + x))/(x^4*e^3 - 10*x^3*e^(x^2*e^8
+ 3) + 25*x^2*e^(2*x^2*e^8 + 3)), x)

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maple [A]  time = 0.27, size = 30, normalized size = 1.15

method	result	size
risch	\(\frac {{\mathrm e}^{x -3}}{5 x}-\frac {{\mathrm e}^{x -3}}{5 \left (x -5 \,{\mathrm e}^{x^{2} {\mathrm e}^{8}}\right )}\)	\(30\)
norman	\(-\frac {{\mathrm e}^{x} {\mathrm e}^{-3} {\mathrm e}^{x^{2} {\mathrm e}^{8}}}{x \left (x -5 \,{\mathrm e}^{x^{2} {\mathrm e}^{8}}\right )}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*x-5)*exp(x)*exp(x^2*exp(4)^2)^2+(-2*x^3*exp(4)^2-x^2+2*x)*exp(x)*exp(x^2*exp(4)^2))/(25*x^2*exp(3)*exp
(x^2*exp(4)^2)^2-10*x^3*exp(3)*exp(x^2*exp(4)^2)+x^4*exp(3)),x,method=_RETURNVERBOSE)

[Out]

1/5/x*exp(x-3)-1/5/(x-5*exp(x^2*exp(8)))*exp(x-3)

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maxima [A]  time = 0.43, size = 32, normalized size = 1.23 \begin {gather*} -\frac {e^{\left (x^{2} e^{8} + x\right )}}{x^{2} e^{3} - 5 \, x e^{\left (x^{2} e^{8} + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x-5)*exp(x)*exp(x^2*exp(4)^2)^2+(-2*x^3*exp(4)^2-x^2+2*x)*exp(x)*exp(x^2*exp(4)^2))/(25*x^2*exp(
3)*exp(x^2*exp(4)^2)^2-10*x^3*exp(3)*exp(x^2*exp(4)^2)+x^4*exp(3)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.57, size = 50, normalized size = 1.92 \begin {gather*} \frac {{\mathrm {e}}^{x-3}}{5\,x}+\frac {{\mathrm {e}}^{-3}\,\left ({\mathrm {e}}^x-2\,x^2\,{\mathrm {e}}^{x+8}\right )}{5\,\left (x-5\,{\mathrm {e}}^{x^2\,{\mathrm {e}}^8}\right )\,\left (2\,x^2\,{\mathrm {e}}^8-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x^2*exp(8))*exp(x)*(5*x - 5) - exp(x^2*exp(8))*exp(x)*(2*x^3*exp(8) - 2*x + x^2))/(x^4*exp(3) - 10*
x^3*exp(x^2*exp(8))*exp(3) + 25*x^2*exp(2*x^2*exp(8))*exp(3)),x)

[Out]

exp(x - 3)/(5*x) + (exp(-3)*(exp(x) - 2*x^2*exp(x + 8)))/(5*(x - 5*exp(x^2*exp(8)))*(2*x^2*exp(8) - 1))

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sympy [A]  time = 0.19, size = 32, normalized size = 1.23 \begin {gather*} \frac {e^{x}}{- 5 x e^{3} + 25 e^{3} e^{x^{2} e^{8}}} + \frac {e^{x}}{5 x e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x-5)*exp(x)*exp(x**2*exp(4)**2)**2+(-2*x**3*exp(4)**2-x**2+2*x)*exp(x)*exp(x**2*exp(4)**2))/(25*
x**2*exp(3)*exp(x**2*exp(4)**2)**2-10*x**3*exp(3)*exp(x**2*exp(4)**2)+x**4*exp(3)),x)

[Out]

exp(x)/(-5*x*exp(3) + 25*exp(3)*exp(x**2*exp(8))) + exp(-3)*exp(x)/(5*x)

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