\(\int \frac {-2 e^{\frac {1}{x^2}}+x^2+3 x^3}{x^3} \, dx\) [5865]

Optimal result

Integrand size = 20, antiderivative size = 15 \[ \int \frac {-2 e^{\frac {1}{x^2}}+x^2+3 x^3}{x^3} \, dx=e^{\frac {1}{x^2}}+3 x-\log (5)+\log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {14, 2240, 45} \[ \int \frac {-2 e^{\frac {1}{x^2}}+x^2+3 x^3}{x^3} \, dx=e^{\frac {1}{x^2}}+3 x+\log (x) \]

[In]

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

[Out]

E^x^(-2) + 3*x + Log[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 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2240

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^n*(
F^(a + b*(c + d*x)^n)/(b*f*n*(c + d*x)^n*Log[F])), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {2 e^{\frac {1}{x^2}}}{x^3}+\frac {1+3 x}{x}\right ) \, dx \\ & = -\left (2 \int \frac {e^{\frac {1}{x^2}}}{x^3} \, dx\right )+\int \frac {1+3 x}{x} \, dx \\ & = e^{\frac {1}{x^2}}+\int \left (3+\frac {1}{x}\right ) \, dx \\ & = e^{\frac {1}{x^2}}+3 x+\log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \frac {-2 e^{\frac {1}{x^2}}+x^2+3 x^3}{x^3} \, dx=e^{\frac {1}{x^2}}+3 x+\log (x) \]

[In]

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

[Out]

E^x^(-2) + 3*x + Log[x]

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73

method	result	size
risch	\(3 x +\ln \left (x \right )+{\mathrm e}^{\frac {1}{x^{2}}}\)	\(11\)
parallelrisch	\(3 x +\ln \left (x \right )+{\mathrm e}^{\frac {1}{x^{2}}}\)	\(11\)
parts	\(3 x +\ln \left (x \right )+{\mathrm e}^{\frac {1}{x^{2}}}\)	\(11\)
derivativedivides	\(-\ln \left (\frac {1}{x}\right )+3 x +{\mathrm e}^{\frac {1}{x^{2}}}\)	\(15\)
default	\(-\ln \left (\frac {1}{x}\right )+3 x +{\mathrm e}^{\frac {1}{x^{2}}}\)	\(15\)
norman	\(\frac {x^{2} {\mathrm e}^{\frac {1}{x^{2}}}+3 x^{3}}{x^{2}}+\ln \left (x \right )\)	\(22\)

[In]

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

[Out]

3*x+ln(x)+exp(1/x^2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.67 \[ \int \frac {-2 e^{\frac {1}{x^2}}+x^2+3 x^3}{x^3} \, dx=3 \, x + e^{\left (\frac {1}{x^{2}}\right )} + \log \left (x\right ) \]

[In]

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

[Out]

3*x + e^(x^(-2)) + log(x)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \frac {-2 e^{\frac {1}{x^2}}+x^2+3 x^3}{x^3} \, dx=3 x + e^{\frac {1}{x^{2}}} + \log {\left (x \right )} \]

[In]

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

[Out]

3*x + exp(x**(-2)) + log(x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.67 \[ \int \frac {-2 e^{\frac {1}{x^2}}+x^2+3 x^3}{x^3} \, dx=3 \, x + e^{\left (\frac {1}{x^{2}}\right )} + \log \left (x\right ) \]

[In]

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

[Out]

3*x + e^(x^(-2)) + log(x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.67 \[ \int \frac {-2 e^{\frac {1}{x^2}}+x^2+3 x^3}{x^3} \, dx=3 \, x + e^{\left (\frac {1}{x^{2}}\right )} + \log \left (x\right ) \]

[In]

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

[Out]

3*x + e^(x^(-2)) + log(x)

Mupad [B] (verification not implemented)

Time = 12.74 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.67 \[ \int \frac {-2 e^{\frac {1}{x^2}}+x^2+3 x^3}{x^3} \, dx=3\,x+{\mathrm {e}}^{\frac {1}{x^2}}+\ln \left (x\right ) \]

[In]

int((x^2 - 2*exp(1/x^2) + 3*x^3)/x^3,x)

[Out]

3*x + exp(1/x^2) + log(x)