3.63.18 \(\int \frac {-2+e^x (5-5 x)+e^{3 e^{x^2}-3 x} (-5-15 x+30 e^{x^2} x^2)}{5 x^2} \, dx\)

Optimal. Leaf size=29 \[ \frac {\frac {2}{5}+e^{3 \left (e^{x^2}-x\right )}-e^x-x}{x} \]

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Rubi [F]  time = 1.07, 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^x (5-5 x)+e^{3 e^{x^2}-3 x} \left (-5-15 x+30 e^{x^2} x^2\right )}{5 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-2 + E^x*(5 - 5*x) + E^(3*E^x^2 - 3*x)*(-5 - 15*x + 30*E^x^2*x^2))/(5*x^2),x]

[Out]

2/(5*x) - E^x/x + 6*Defer[Int][E^(3*E^x^2 + (-3 + x)*x), x] - Defer[Int][E^(3*(E^x^2 - x))/x^2, x] - 3*Defer[I
nt][E^(3*(E^x^2 - x))/x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {-2+e^x (5-5 x)+e^{3 e^{x^2}-3 x} \left (-5-15 x+30 e^{x^2} x^2\right )}{x^2} \, dx\\ &=\frac {1}{5} \int \left (30 e^{3 e^{x^2}+(-3+x) x}-\frac {e^{-3 x} \left (5 e^{3 e^{x^2}}+2 e^{3 x}-5 e^{4 x}+15 e^{3 e^{x^2}} x+5 e^{4 x} x\right )}{x^2}\right ) \, dx\\ &=-\left (\frac {1}{5} \int \frac {e^{-3 x} \left (5 e^{3 e^{x^2}}+2 e^{3 x}-5 e^{4 x}+15 e^{3 e^{x^2}} x+5 e^{4 x} x\right )}{x^2} \, dx\right )+6 \int e^{3 e^{x^2}+(-3+x) x} \, dx\\ &=-\left (\frac {1}{5} \int \frac {2+5 e^x (-1+x)+5 e^{3 e^{x^2}-3 x} (1+3 x)}{x^2} \, dx\right )+6 \int e^{3 e^{x^2}+(-3+x) x} \, dx\\ &=-\left (\frac {1}{5} \int \left (\frac {2}{x^2}+\frac {5 e^x (-1+x)}{x^2}+\frac {5 e^{3 \left (e^{x^2}-x\right )} (1+3 x)}{x^2}\right ) \, dx\right )+6 \int e^{3 e^{x^2}+(-3+x) x} \, dx\\ &=\frac {2}{5 x}+6 \int e^{3 e^{x^2}+(-3+x) x} \, dx-\int \frac {e^x (-1+x)}{x^2} \, dx-\int \frac {e^{3 \left (e^{x^2}-x\right )} (1+3 x)}{x^2} \, dx\\ &=\frac {2}{5 x}-\frac {e^x}{x}+6 \int e^{3 e^{x^2}+(-3+x) x} \, dx-\int \left (\frac {e^{3 \left (e^{x^2}-x\right )}}{x^2}+\frac {3 e^{3 \left (e^{x^2}-x\right )}}{x}\right ) \, dx\\ &=\frac {2}{5 x}-\frac {e^x}{x}-3 \int \frac {e^{3 \left (e^{x^2}-x\right )}}{x} \, dx+6 \int e^{3 e^{x^2}+(-3+x) x} \, dx-\int \frac {e^{3 \left (e^{x^2}-x\right )}}{x^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 29, normalized size = 1.00 \begin {gather*} \frac {2+5 e^{3 e^{x^2}-3 x}-5 e^x}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 + E^x*(5 - 5*x) + E^(3*E^x^2 - 3*x)*(-5 - 15*x + 30*E^x^2*x^2))/(5*x^2),x]

[Out]

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

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fricas [A]  time = 1.11, size = 24, normalized size = 0.83 \begin {gather*} -\frac {5 \, e^{x} - 5 \, e^{\left (-3 \, x + 3 \, e^{\left (x^{2}\right )}\right )} - 2}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((30*x^2*exp(x^2)-15*x-5)*exp(3*exp(x^2)-3*x)+(-5*x+5)*exp(x)-2)/x^2,x, algorithm="fricas")

[Out]

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

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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^{x} - 5 \, {\left (6 \, x^{2} e^{\left (x^{2}\right )} - 3 \, x - 1\right )} e^{\left (-3 \, x + 3 \, e^{\left (x^{2}\right )}\right )} + 2}{5 \, x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((30*x^2*exp(x^2)-15*x-5)*exp(3*exp(x^2)-3*x)+(-5*x+5)*exp(x)-2)/x^2,x, algorithm="giac")

[Out]

integrate(-1/5*(5*(x - 1)*e^x - 5*(6*x^2*e^(x^2) - 3*x - 1)*e^(-3*x + 3*e^(x^2)) + 2)/x^2, x)

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maple [A]  time = 0.06, size = 22, normalized size = 0.76

method	result	size
norman	\(\frac {\frac {2}{5}-{\mathrm e}^{x}+{\mathrm e}^{3 \,{\mathrm e}^{x^{2}}-3 x}}{x}\)	\(22\)
risch	\(\frac {2}{5 x}-\frac {{\mathrm e}^{x}}{x}+\frac {{\mathrm e}^{3 \,{\mathrm e}^{x^{2}}-3 x}}{x}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((30*x^2*exp(x^2)-15*x-5)*exp(3*exp(x^2)-3*x)+(-5*x+5)*exp(x)-2)/x^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [C]  time = 0.42, size = 30, normalized size = 1.03 \begin {gather*} \frac {e^{\left (-3 \, x + 3 \, e^{\left (x^{2}\right )}\right )}}{x} + \frac {2}{5 \, x} - {\rm Ei}\relax (x) + \Gamma \left (-1, -x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((30*x^2*exp(x^2)-15*x-5)*exp(3*exp(x^2)-3*x)+(-5*x+5)*exp(x)-2)/x^2,x, algorithm="maxima")

[Out]

e^(-3*x + 3*e^(x^2))/x + 2/5/x - Ei(x) + gamma(-1, -x)

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mupad [B]  time = 4.32, size = 21, normalized size = 0.72 \begin {gather*} \frac {{\mathrm {e}}^{3\,{\mathrm {e}}^{x^2}-3\,x}-{\mathrm {e}}^x+\frac {2}{5}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((exp(3*exp(x^2) - 3*x)*(15*x - 30*x^2*exp(x^2) + 5))/5 + (exp(x)*(5*x - 5))/5 + 2/5)/x^2,x)

[Out]

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

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sympy [A]  time = 0.21, size = 22, normalized size = 0.76 \begin {gather*} - \frac {e^{x}}{x} + \frac {e^{- 3 x + 3 e^{x^{2}}}}{x} + \frac {2}{5 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((30*x**2*exp(x**2)-15*x-5)*exp(3*exp(x**2)-3*x)+(-5*x+5)*exp(x)-2)/x**2,x)

[Out]

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

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