3.95.9 \(\int \frac {e^{2 x+e^{x^2} x^2-4 x^3} (-5+10 x-60 x^3+e^{x^2} (10 x^2+10 x^4))}{x^2} \, dx\)

Optimal. Leaf size=30 \[ \frac {5 e^{-x \left (1+x \left (-e^{x^2}-\frac {3}{x}+4 x\right )\right )}}{x} \]

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Rubi [B]  time = 0.30, antiderivative size = 70, normalized size of antiderivative = 2.33, number of steps used = 1, number of rules used = 1, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.020, Rules used = {2288} \begin {gather*} \frac {5 e^{-4 x^3+e^{x^2} x^2+2 x} \left (-6 x^3+e^{x^2} \left (x^4+x^2\right )+x\right )}{x^2 \left (-6 x^2+e^{x^2} x+e^{x^2} x^3+1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(2*x + E^x^2*x^2 - 4*x^3)*(-5 + 10*x - 60*x^3 + E^x^2*(10*x^2 + 10*x^4)))/x^2,x]

[Out]

(5*E^(2*x + E^x^2*x^2 - 4*x^3)*(x - 6*x^3 + E^x^2*(x^2 + x^4)))/(x^2*(1 + E^x^2*x - 6*x^2 + E^x^2*x^3))

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

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

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Mathematica [A]  time = 0.46, size = 23, normalized size = 0.77 \begin {gather*} \frac {5 e^{x \left (2+e^{x^2} x-4 x^2\right )}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*x + E^x^2*x^2 - 4*x^3)*(-5 + 10*x - 60*x^3 + E^x^2*(10*x^2 + 10*x^4)))/x^2,x]

[Out]

(5*E^(x*(2 + E^x^2*x - 4*x^2)))/x

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fricas [A]  time = 0.69, size = 23, normalized size = 0.77 \begin {gather*} \frac {5 \, e^{\left (-4 \, x^{3} + x^{2} e^{\left (x^{2}\right )} + 2 \, x\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^4+10*x^2)*exp(x^2)-60*x^3+10*x-5)/x^2/exp(-x^2*exp(x^2)+4*x^3-2*x),x, algorithm="fricas")

[Out]

5*e^(-4*x^3 + x^2*e^(x^2) + 2*x)/x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {5 \, {\left (12 \, x^{3} - 2 \, {\left (x^{4} + x^{2}\right )} e^{\left (x^{2}\right )} - 2 \, x + 1\right )} e^{\left (-4 \, x^{3} + x^{2} e^{\left (x^{2}\right )} + 2 \, x\right )}}{x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^4+10*x^2)*exp(x^2)-60*x^3+10*x-5)/x^2/exp(-x^2*exp(x^2)+4*x^3-2*x),x, algorithm="giac")

[Out]

integrate(-5*(12*x^3 - 2*(x^4 + x^2)*e^(x^2) - 2*x + 1)*e^(-4*x^3 + x^2*e^(x^2) + 2*x)/x^2, x)

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maple [A]  time = 0.05, size = 24, normalized size = 0.80




method result size



risch \(\frac {5 \,{\mathrm e}^{-x \left (-{\mathrm e}^{x^{2}} x +4 x^{2}-2\right )}}{x}\) \(24\)
norman \(\frac {5 \,{\mathrm e}^{x^{2} {\mathrm e}^{x^{2}}-4 x^{3}+2 x}}{x}\) \(27\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((10*x^4+10*x^2)*exp(x^2)-60*x^3+10*x-5)/x^2/exp(-x^2*exp(x^2)+4*x^3-2*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.42, size = 23, normalized size = 0.77 \begin {gather*} \frac {5 \, e^{\left (-4 \, x^{3} + x^{2} e^{\left (x^{2}\right )} + 2 \, x\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^4+10*x^2)*exp(x^2)-60*x^3+10*x-5)/x^2/exp(-x^2*exp(x^2)+4*x^3-2*x),x, algorithm="maxima")

[Out]

5*e^(-4*x^3 + x^2*e^(x^2) + 2*x)/x

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mupad [B]  time = 0.34, size = 24, normalized size = 0.80 \begin {gather*} \frac {5\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-4\,x^3}\,{\mathrm {e}}^{x^2\,{\mathrm {e}}^{x^2}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x + x^2*exp(x^2) - 4*x^3)*(10*x + exp(x^2)*(10*x^2 + 10*x^4) - 60*x^3 - 5))/x^2,x)

[Out]

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

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sympy [A]  time = 0.33, size = 20, normalized size = 0.67 \begin {gather*} \frac {5 e^{- 4 x^{3} + x^{2} e^{x^{2}} + 2 x}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x**4+10*x**2)*exp(x**2)-60*x**3+10*x-5)/x**2/exp(-x**2*exp(x**2)+4*x**3-2*x),x)

[Out]

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

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