3.53.30 \(\int \frac {-200+40 x+3 x^3+2 x^4+e^{-2+x^2} (-x+2 x^3)}{2 x^3} \, dx\)

Optimal. Leaf size=31 \[ \frac {1}{2} \left (\left (-2+\frac {10}{x}-x\right )^2+\frac {e^{-2+x^2}}{x}-x\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 37, normalized size of antiderivative = 1.19, number of steps used = 6, number of rules used = 3, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {12, 14, 2288} \begin {gather*} \frac {x^2}{2}+\frac {e^{x^2-2}}{2 x}+\frac {50}{x^2}+\frac {3 x}{2}-\frac {20}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-200 + 40*x + 3*x^3 + 2*x^4 + E^(-2 + x^2)*(-x + 2*x^3))/(2*x^3),x]

[Out]

50/x^2 - 20/x + E^(-2 + x^2)/(2*x) + (3*x)/2 + x^2/2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 {1}{2} \int \frac {-200+40 x+3 x^3+2 x^4+e^{-2+x^2} \left (-x+2 x^3\right )}{x^3} \, dx\\ &=\frac {1}{2} \int \left (\frac {e^{-2+x^2} \left (-1+2 x^2\right )}{x^2}+\frac {-200+40 x+3 x^3+2 x^4}{x^3}\right ) \, dx\\ &=\frac {1}{2} \int \frac {e^{-2+x^2} \left (-1+2 x^2\right )}{x^2} \, dx+\frac {1}{2} \int \frac {-200+40 x+3 x^3+2 x^4}{x^3} \, dx\\ &=\frac {e^{-2+x^2}}{2 x}+\frac {1}{2} \int \left (3-\frac {200}{x^3}+\frac {40}{x^2}+2 x\right ) \, dx\\ &=\frac {50}{x^2}-\frac {20}{x}+\frac {e^{-2+x^2}}{2 x}+\frac {3 x}{2}+\frac {x^2}{2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 28, normalized size = 0.90 \begin {gather*} \frac {100+\left (-40+e^{-2+x^2}\right ) x+3 x^3+x^4}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-200 + 40*x + 3*x^3 + 2*x^4 + E^(-2 + x^2)*(-x + 2*x^3))/(2*x^3),x]

[Out]

(100 + (-40 + E^(-2 + x^2))*x + 3*x^3 + x^4)/(2*x^2)

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fricas [A]  time = 0.55, size = 26, normalized size = 0.84 \begin {gather*} \frac {x^{4} + 3 \, x^{3} + x e^{\left (x^{2} - 2\right )} - 40 \, x + 100}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((2*x^3-x)*exp(2)*exp(x^2-4)+2*x^4+3*x^3+40*x-200)/x^3,x, algorithm="fricas")

[Out]

1/2*(x^4 + 3*x^3 + x*e^(x^2 - 2) - 40*x + 100)/x^2

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giac [A]  time = 0.16, size = 36, normalized size = 1.16 \begin {gather*} \frac {{\left (x^{4} e^{2} + 3 \, x^{3} e^{2} - 40 \, x e^{2} + x e^{\left (x^{2}\right )} + 100 \, e^{2}\right )} e^{\left (-2\right )}}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((2*x^3-x)*exp(2)*exp(x^2-4)+2*x^4+3*x^3+40*x-200)/x^3,x, algorithm="giac")

[Out]

1/2*(x^4*e^2 + 3*x^3*e^2 - 40*x*e^2 + x*e^(x^2) + 100*e^2)*e^(-2)/x^2

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maple [A]  time = 0.06, size = 31, normalized size = 1.00

method	result	size
norman	\(\frac {50-20 x +\frac {3 x^{3}}{2}+\frac {x^{4}}{2}+\frac {{\mathrm e}^{2} x \,{\mathrm e}^{x^{2}-4}}{2}}{x^{2}}\)	\(31\)
risch	\(\frac {x^{2}}{2}+\frac {3 x}{2}+\frac {-40 x +100}{2 x^{2}}+\frac {{\mathrm e}^{x^{2}-2}}{2 x}\)	\(31\)
default	\(\frac {x^{2}}{2}+\frac {3 x}{2}+\frac {50}{x^{2}}-\frac {20}{x}+\frac {{\mathrm e}^{2} {\mathrm e}^{-4} \sqrt {\pi }\, \erfi \relax (x )}{2}-\frac {{\mathrm e}^{2} {\mathrm e}^{-4} \left (-\frac {{\mathrm e}^{x^{2}}}{x}+\sqrt {\pi }\, \erfi \relax (x )\right )}{2}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((2*x^3-x)*exp(2)*exp(x^2-4)+2*x^4+3*x^3+40*x-200)/x^3,x,method=_RETURNVERBOSE)

[Out]

(50-20*x+3/2*x^3+1/2*x^4+1/2*exp(2)*x*exp(x^2-4))/x^2

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maxima [C]  time = 0.38, size = 51, normalized size = 1.65 \begin {gather*} -\frac {1}{2} i \, \sqrt {\pi } \operatorname {erf}\left (i \, x\right ) e^{\left (-2\right )} + \frac {1}{2} \, x^{2} + \frac {\sqrt {-x^{2}} e^{\left (-2\right )} \Gamma \left (-\frac {1}{2}, -x^{2}\right )}{4 \, x} + \frac {3}{2} \, x - \frac {20}{x} + \frac {50}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((2*x^3-x)*exp(2)*exp(x^2-4)+2*x^4+3*x^3+40*x-200)/x^3,x, algorithm="maxima")

[Out]

-1/2*I*sqrt(pi)*erf(I*x)*e^(-2) + 1/2*x^2 + 1/4*sqrt(-x^2)*e^(-2)*gamma(-1/2, -x^2)/x + 3/2*x - 20/x + 50/x^2

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mupad [B]  time = 0.09, size = 27, normalized size = 0.87 \begin {gather*} \frac {3\,x}{2}+\frac {x\,\left (\frac {{\mathrm {e}}^{x^2-2}}{2}-20\right )+50}{x^2}+\frac {x^2}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((20*x + (3*x^3)/2 + x^4 - (exp(2)*exp(x^2 - 4)*(x - 2*x^3))/2 - 100)/x^3,x)

[Out]

(3*x)/2 + (x*(exp(x^2 - 2)/2 - 20) + 50)/x^2 + x^2/2

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sympy [A]  time = 0.13, size = 32, normalized size = 1.03 \begin {gather*} \frac {x^{2}}{2} + \frac {3 x}{2} + \frac {e^{2} e^{x^{2} - 4}}{2 x} + \frac {100 - 40 x}{2 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((2*x**3-x)*exp(2)*exp(x**2-4)+2*x**4+3*x**3+40*x-200)/x**3,x)

[Out]

x**2/2 + 3*x/2 + exp(2)*exp(x**2 - 4)/(2*x) + (100 - 40*x)/(2*x**2)

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