\(\int \frac {12 x^2+e^{\frac {1-2 x^3+x^2 \log (5)}{x^2}} (-2+x^2-2 x^3)}{20 x^2} \, dx\) [8669]

Optimal result

Integrand size = 43, antiderivative size = 22 \[ \int \frac {12 x^2+e^{\frac {1-2 x^3+x^2 \log (5)}{x^2}} \left (-2+x^2-2 x^3\right )}{20 x^2} \, dx=x-\frac {1}{5} \left (2-\frac {5}{4} e^{\frac {1}{x^2}-2 x}\right ) x \]

[Out]

x-1/5*x*(2-1/4*exp(ln(5)-2*x+1/x^2))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.55, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {12, 14, 2326} \[ \int \frac {12 x^2+e^{\frac {1-2 x^3+x^2 \log (5)}{x^2}} \left (-2+x^2-2 x^3\right )}{20 x^2} \, dx=\frac {e^{\frac {1}{x^2}-2 x} \left (x^3+1\right )}{4 \left (\frac {1}{x^3}+1\right ) x^2}+\frac {3 x}{5} \]

[In]

Int[(12*x^2 + E^((1 - 2*x^3 + x^2*Log[5])/x^2)*(-2 + x^2 - 2*x^3))/(20*x^2),x]

[Out]

(3*x)/5 + (E^(x^(-2) - 2*x)*(1 + x^3))/(4*(1 + x^(-3))*x^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 2326

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{align*} \text {integral}& = \frac {1}{20} \int \frac {12 x^2+e^{\frac {1-2 x^3+x^2 \log (5)}{x^2}} \left (-2+x^2-2 x^3\right )}{x^2} \, dx \\ & = \frac {1}{20} \int \left (12-\frac {5 e^{\frac {1}{x^2}-2 x} \left (2-x^2+2 x^3\right )}{x^2}\right ) \, dx \\ & = \frac {3 x}{5}-\frac {1}{4} \int \frac {e^{\frac {1}{x^2}-2 x} \left (2-x^2+2 x^3\right )}{x^2} \, dx \\ & = \frac {3 x}{5}+\frac {e^{\frac {1}{x^2}-2 x} \left (1+x^3\right )}{4 \left (1+\frac {1}{x^3}\right ) x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {12 x^2+e^{\frac {1-2 x^3+x^2 \log (5)}{x^2}} \left (-2+x^2-2 x^3\right )}{20 x^2} \, dx=\left (\frac {3}{5}+\frac {1}{4} e^{\frac {1}{x^2}-2 x}\right ) x \]

[In]

Integrate[(12*x^2 + E^((1 - 2*x^3 + x^2*Log[5])/x^2)*(-2 + x^2 - 2*x^3))/(20*x^2),x]

[Out]

(3/5 + E^(x^(-2) - 2*x)/4)*x

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95

method	result	size
risch	\(\frac {x \,{\mathrm e}^{-\frac {2 x^{3}-1}{x^{2}}}}{4}+\frac {3 x}{5}\)	\(21\)
parallelrisch	\(\frac {x \,{\mathrm e}^{\frac {x^{2} \ln \left (5\right )-2 x^{3}+1}{x^{2}}}}{20}+\frac {3 x}{5}\)	\(26\)
parts	\(\frac {x \,{\mathrm e}^{\frac {x^{2} \ln \left (5\right )-2 x^{3}+1}{x^{2}}}}{20}+\frac {3 x}{5}\)	\(26\)
norman	\(\frac {\frac {3 x^{2}}{5}+\frac {{\mathrm e}^{\frac {x^{2} \ln \left (5\right )-2 x^{3}+1}{x^{2}}} x^{2}}{20}}{x}\)	\(34\)

[In]

int(1/20*((-2*x^3+x^2-2)*exp((x^2*ln(5)-2*x^3+1)/x^2)+12*x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {12 x^2+e^{\frac {1-2 x^3+x^2 \log (5)}{x^2}} \left (-2+x^2-2 x^3\right )}{20 x^2} \, dx=\frac {1}{20} \, x e^{\left (-\frac {2 \, x^{3} - x^{2} \log \left (5\right ) - 1}{x^{2}}\right )} + \frac {3}{5} \, x \]

[In]

integrate(1/20*((-2*x^3+x^2-2)*exp((x^2*log(5)-2*x^3+1)/x^2)+12*x^2)/x^2,x, algorithm="fricas")

[Out]

1/20*x*e^(-(2*x^3 - x^2*log(5) - 1)/x^2) + 3/5*x

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {12 x^2+e^{\frac {1-2 x^3+x^2 \log (5)}{x^2}} \left (-2+x^2-2 x^3\right )}{20 x^2} \, dx=\frac {x e^{\frac {- 2 x^{3} + x^{2} \log {\left (5 \right )} + 1}{x^{2}}}}{20} + \frac {3 x}{5} \]

[In]

integrate(1/20*((-2*x**3+x**2-2)*exp((x**2*ln(5)-2*x**3+1)/x**2)+12*x**2)/x**2,x)

[Out]

x*exp((-2*x**3 + x**2*log(5) + 1)/x**2)/20 + 3*x/5

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {12 x^2+e^{\frac {1-2 x^3+x^2 \log (5)}{x^2}} \left (-2+x^2-2 x^3\right )}{20 x^2} \, dx=\frac {1}{4} \, x e^{\left (-2 \, x + \frac {1}{x^{2}}\right )} + \frac {3}{5} \, x \]

[In]

integrate(1/20*((-2*x^3+x^2-2)*exp((x^2*log(5)-2*x^3+1)/x^2)+12*x^2)/x^2,x, algorithm="maxima")

[Out]

1/4*x*e^(-2*x + 1/x^2) + 3/5*x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {12 x^2+e^{\frac {1-2 x^3+x^2 \log (5)}{x^2}} \left (-2+x^2-2 x^3\right )}{20 x^2} \, dx=\frac {1}{4} \, x e^{\left (-\frac {2 \, x^{3} - 1}{x^{2}}\right )} + \frac {3}{5} \, x \]

[In]

integrate(1/20*((-2*x^3+x^2-2)*exp((x^2*log(5)-2*x^3+1)/x^2)+12*x^2)/x^2,x, algorithm="giac")

[Out]

1/4*x*e^(-(2*x^3 - 1)/x^2) + 3/5*x

Mupad [B] (verification not implemented)

Time = 12.84 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {12 x^2+e^{\frac {1-2 x^3+x^2 \log (5)}{x^2}} \left (-2+x^2-2 x^3\right )}{20 x^2} \, dx=\frac {x\,\left (5\,{\mathrm {e}}^{\frac {1}{x^2}-2\,x}+12\right )}{20} \]

[In]

int(-((exp((x^2*log(5) - 2*x^3 + 1)/x^2)*(2*x^3 - x^2 + 2))/20 - (3*x^2)/5)/x^2,x)

[Out]

(x*(5*exp(1/x^2 - 2*x) + 12))/20