3.29.54 \(\int \frac {e^{\frac {2 (5-5 x)}{x}+2 x^2} (50-15 x+2 x^2-20 x^3+4 x^4)}{25 x} \, dx\)

Optimal. Leaf size=34 \[ 3-\frac {1}{25} e^{\frac {2 (5-5 x)}{x}+2 x^2} (5-x) x-\log (4) \]

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Rubi [F]  time = 0.55, 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^{\frac {2 (5-5 x)}{x}+2 x^2} \left (50-15 x+2 x^2-20 x^3+4 x^4\right )}{25 x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((2*(5 - 5*x))/x + 2*x^2)*(50 - 15*x + 2*x^2 - 20*x^3 + 4*x^4))/(25*x),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{25} \int \frac {e^{\frac {2 (5-5 x)}{x}+2 x^2} \left (50-15 x+2 x^2-20 x^3+4 x^4\right )}{x} \, dx\\ &=\frac {1}{25} \int \frac {e^{\frac {2 \left (5-5 x+x^3\right )}{x}} \left (50-15 x+2 x^2-20 x^3+4 x^4\right )}{x} \, dx\\ &=\frac {1}{25} \int \left (-15 e^{\frac {2 \left (5-5 x+x^3\right )}{x}}+\frac {50 e^{\frac {2 \left (5-5 x+x^3\right )}{x}}}{x}+2 e^{\frac {2 \left (5-5 x+x^3\right )}{x}} x-20 e^{\frac {2 \left (5-5 x+x^3\right )}{x}} x^2+4 e^{\frac {2 \left (5-5 x+x^3\right )}{x}} x^3\right ) \, dx\\ &=\frac {2}{25} \int e^{\frac {2 \left (5-5 x+x^3\right )}{x}} x \, dx+\frac {4}{25} \int e^{\frac {2 \left (5-5 x+x^3\right )}{x}} x^3 \, dx-\frac {3}{5} \int e^{\frac {2 \left (5-5 x+x^3\right )}{x}} \, dx-\frac {4}{5} \int e^{\frac {2 \left (5-5 x+x^3\right )}{x}} x^2 \, dx+2 \int \frac {e^{\frac {2 \left (5-5 x+x^3\right )}{x}}}{x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 23, normalized size = 0.68 \begin {gather*} \frac {1}{25} e^{\frac {2 \left (5-5 x+x^3\right )}{x}} (-5+x) x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((2*(5 - 5*x))/x + 2*x^2)*(50 - 15*x + 2*x^2 - 20*x^3 + 4*x^4))/(25*x),x]

[Out]

(E^((2*(5 - 5*x + x^3))/x)*(-5 + x)*x)/25

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fricas [A]  time = 0.53, size = 23, normalized size = 0.68 \begin {gather*} \frac {1}{25} \, {\left (x^{2} - 5 \, x\right )} e^{\left (\frac {2 \, {\left (x^{3} - 5 \, x + 5\right )}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(4*x^4-20*x^3+2*x^2-15*x+50)*exp((-5*x+5)/x)^2*exp(x^2)^2/x,x, algorithm="fricas")

[Out]

1/25*(x^2 - 5*x)*e^(2*(x^3 - 5*x + 5)/x)

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giac [A]  time = 0.19, size = 37, normalized size = 1.09 \begin {gather*} \frac {1}{25} \, x^{2} e^{\left (\frac {2 \, {\left (x^{3} - 5 \, x + 5\right )}}{x}\right )} - \frac {1}{5} \, x e^{\left (\frac {2 \, {\left (x^{3} - 5 \, x + 5\right )}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(4*x^4-20*x^3+2*x^2-15*x+50)*exp((-5*x+5)/x)^2*exp(x^2)^2/x,x, algorithm="giac")

[Out]

1/25*x^2*e^(2*(x^3 - 5*x + 5)/x) - 1/5*x*e^(2*(x^3 - 5*x + 5)/x)

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maple [A]  time = 0.10, size = 21, normalized size = 0.62

method	result	size
risch	\(\frac {\left (x -5\right ) x \,{\mathrm e}^{\frac {2 x^{3}-10 x +10}{x}}}{25}\)	\(21\)
gosper	\(\frac {x \left (x -5\right ) {\mathrm e}^{-\frac {10 \left (x -1\right )}{x}} {\mathrm e}^{2 x^{2}}}{25}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/25*(4*x^4-20*x^3+2*x^2-15*x+50)*exp((-5*x+5)/x)^2*exp(x^2)^2/x,x,method=_RETURNVERBOSE)

[Out]

1/25*(x-5)*x*exp(2*(x^3-5*x+5)/x)

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maxima [A]  time = 0.82, size = 22, normalized size = 0.65 \begin {gather*} \frac {1}{25} \, {\left (x^{2} - 5 \, x\right )} e^{\left (2 \, x^{2} + \frac {10}{x} - 10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(4*x^4-20*x^3+2*x^2-15*x+50)*exp((-5*x+5)/x)^2*exp(x^2)^2/x,x, algorithm="maxima")

[Out]

1/25*(x^2 - 5*x)*e^(2*x^2 + 10/x - 10)

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mupad [B]  time = 1.70, size = 20, normalized size = 0.59 \begin {gather*} \frac {x\,{\mathrm {e}}^{-10}\,{\mathrm {e}}^{2\,x^2}\,{\mathrm {e}}^{10/x}\,\left (x-5\right )}{25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x^2)*exp(-(2*(5*x - 5))/x)*(2*x^2 - 15*x - 20*x^3 + 4*x^4 + 50))/(25*x),x)

[Out]

(x*exp(-10)*exp(2*x^2)*exp(10/x)*(x - 5))/25

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sympy [A]  time = 24.56, size = 29, normalized size = 0.85 \begin {gather*} \frac {\left (x^{2} e^{2 x^{2}} - 5 x e^{2 x^{2}}\right ) e^{\frac {2 \left (5 - 5 x\right )}{x}}}{25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(4*x**4-20*x**3+2*x**2-15*x+50)*exp((-5*x+5)/x)**2*exp(x**2)**2/x,x)

[Out]

(x**2*exp(2*x**2) - 5*x*exp(2*x**2))*exp(2*(5 - 5*x)/x)/25

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