\(\int \frac {3 e^5 x-4 x^3+2 x^4+e^{2 x^2} (1-2 x^2)}{3 x^3} \, dx\) [8060]

Optimal result

Integrand size = 39, antiderivative size = 35 \[ \int \frac {3 e^5 x-4 x^3+2 x^4+e^{2 x^2} \left (1-2 x^2\right )}{3 x^3} \, dx=9-\frac {e^5+\frac {e^{2 x^2}}{6 x}}{x}-\frac {1}{3} (4-x) x \]

[Out]

9-1/3*(-x+4)*x-(1/6*exp(2*x^2)/x+exp(5))/x

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {12, 14, 2326} \[ \int \frac {3 e^5 x-4 x^3+2 x^4+e^{2 x^2} \left (1-2 x^2\right )}{3 x^3} \, dx=\frac {x^2}{3}-\frac {e^{2 x^2}}{6 x^2}-\frac {4 x}{3}-\frac {e^5}{x} \]

[In]

Int[(3*E^5*x - 4*x^3 + 2*x^4 + E^(2*x^2)*(1 - 2*x^2))/(3*x^3),x]

[Out]

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

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}{3} \int \frac {3 e^5 x-4 x^3+2 x^4+e^{2 x^2} \left (1-2 x^2\right )}{x^3} \, dx \\ & = \frac {1}{3} \int \left (-\frac {e^{2 x^2} \left (-1+2 x^2\right )}{x^3}+\frac {3 e^5-4 x^2+2 x^3}{x^2}\right ) \, dx \\ & = -\left (\frac {1}{3} \int \frac {e^{2 x^2} \left (-1+2 x^2\right )}{x^3} \, dx\right )+\frac {1}{3} \int \frac {3 e^5-4 x^2+2 x^3}{x^2} \, dx \\ & = -\frac {e^{2 x^2}}{6 x^2}+\frac {1}{3} \int \left (-4+\frac {3 e^5}{x^2}+2 x\right ) \, dx \\ & = -\frac {e^{2 x^2}}{6 x^2}-\frac {e^5}{x}-\frac {4 x}{3}+\frac {x^2}{3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.94 \[ \int \frac {3 e^5 x-4 x^3+2 x^4+e^{2 x^2} \left (1-2 x^2\right )}{3 x^3} \, dx=\frac {1}{3} \left (-\frac {e^{2 x^2}}{2 x^2}-\frac {3 e^5}{x}-4 x+x^2\right ) \]

[In]

Integrate[(3*E^5*x - 4*x^3 + 2*x^4 + E^(2*x^2)*(1 - 2*x^2))/(3*x^3),x]

[Out]

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

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.80

method	result	size
default	\(\frac {x^{2}}{3}-\frac {4 x}{3}-\frac {{\mathrm e}^{5}}{x}-\frac {{\mathrm e}^{2 x^{2}}}{6 x^{2}}\)	\(28\)
risch	\(\frac {x^{2}}{3}-\frac {4 x}{3}-\frac {{\mathrm e}^{5}}{x}-\frac {{\mathrm e}^{2 x^{2}}}{6 x^{2}}\)	\(28\)
parallelrisch	\(-\frac {-2 x^{4}+8 x^{3}+6 x \,{\mathrm e}^{5}+{\mathrm e}^{2 x^{2}}}{6 x^{2}}\)	\(28\)
parts	\(\frac {x^{2}}{3}-\frac {4 x}{3}-\frac {{\mathrm e}^{5}}{x}-\frac {{\mathrm e}^{2 x^{2}}}{6 x^{2}}\)	\(28\)
norman	\(\frac {-\frac {4 x^{3}}{3}+\frac {x^{4}}{3}-x \,{\mathrm e}^{5}-\frac {{\mathrm e}^{2 x^{2}}}{6}}{x^{2}}\)	\(29\)

[In]

int(1/3*((-2*x^2+1)*exp(2*x^2)+3*x*exp(5)+2*x^4-4*x^3)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int \frac {3 e^5 x-4 x^3+2 x^4+e^{2 x^2} \left (1-2 x^2\right )}{3 x^3} \, dx=\frac {2 \, x^{4} - 8 \, x^{3} - 6 \, x e^{5} - e^{\left (2 \, x^{2}\right )}}{6 \, x^{2}} \]

[In]

integrate(1/3*((-2*x^2+1)*exp(2*x^2)+3*x*exp(5)+2*x^4-4*x^3)/x^3,x, algorithm="fricas")

[Out]

1/6*(2*x^4 - 8*x^3 - 6*x*e^5 - e^(2*x^2))/x^2

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.74 \[ \int \frac {3 e^5 x-4 x^3+2 x^4+e^{2 x^2} \left (1-2 x^2\right )}{3 x^3} \, dx=\frac {x^{2}}{3} - \frac {4 x}{3} - \frac {e^{5}}{x} - \frac {e^{2 x^{2}}}{6 x^{2}} \]

[In]

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

[Out]

x**2/3 - 4*x/3 - exp(5)/x - exp(2*x**2)/(6*x**2)

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.22 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.94 \[ \int \frac {3 e^5 x-4 x^3+2 x^4+e^{2 x^2} \left (1-2 x^2\right )}{3 x^3} \, dx=\frac {1}{3} \, x^{2} - \frac {4}{3} \, x - \frac {e^{5}}{x} - \frac {1}{3} \, {\rm Ei}\left (2 \, x^{2}\right ) + \frac {1}{3} \, \Gamma \left (-1, -2 \, x^{2}\right ) \]

[In]

integrate(1/3*((-2*x^2+1)*exp(2*x^2)+3*x*exp(5)+2*x^4-4*x^3)/x^3,x, algorithm="maxima")

[Out]

1/3*x^2 - 4/3*x - e^5/x - 1/3*Ei(2*x^2) + 1/3*gamma(-1, -2*x^2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int \frac {3 e^5 x-4 x^3+2 x^4+e^{2 x^2} \left (1-2 x^2\right )}{3 x^3} \, dx=\frac {2 \, x^{4} - 8 \, x^{3} - 6 \, x e^{5} - e^{\left (2 \, x^{2}\right )}}{6 \, x^{2}} \]

[In]

integrate(1/3*((-2*x^2+1)*exp(2*x^2)+3*x*exp(5)+2*x^4-4*x^3)/x^3,x, algorithm="giac")

[Out]

1/6*(2*x^4 - 8*x^3 - 6*x*e^5 - e^(2*x^2))/x^2

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.77 \[ \int \frac {3 e^5 x-4 x^3+2 x^4+e^{2 x^2} \left (1-2 x^2\right )}{3 x^3} \, dx=-\frac {{\mathrm {e}}^{2\,x^2}+6\,x\,{\mathrm {e}}^5+8\,x^3-2\,x^4}{6\,x^2} \]

[In]

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

[Out]

-(exp(2*x^2) + 6*x*exp(5) + 8*x^3 - 2*x^4)/(6*x^2)