\(\int \frac {18 x^3+e^{\frac {20+9 x^2}{9 x^2}} (-40+9 x^2)}{9 x^2} \, dx\) [10138]

Optimal result

Integrand size = 37, antiderivative size = 15 \[ \int \frac {18 x^3+e^{\frac {20+9 x^2}{9 x^2}} \left (-40+9 x^2\right )}{9 x^2} \, dx=x \left (e^{1+\frac {20}{9 x^2}}+x\right ) \]

[Out]

(x+exp(20/9/x^2+1))*x

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.13, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {12, 14, 2326} \[ \int \frac {18 x^3+e^{\frac {20+9 x^2}{9 x^2}} \left (-40+9 x^2\right )}{9 x^2} \, dx=x^2+e^{\frac {20}{9 x^2}+1} x \]

[In]

Int[(18*x^3 + E^((20 + 9*x^2)/(9*x^2))*(-40 + 9*x^2))/(9*x^2),x]

[Out]

E^(1 + 20/(9*x^2))*x + 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}{9} \int \frac {18 x^3+e^{\frac {20+9 x^2}{9 x^2}} \left (-40+9 x^2\right )}{x^2} \, dx \\ & = \frac {1}{9} \int \left (18 x+\frac {e^{1+\frac {20}{9 x^2}} \left (-40+9 x^2\right )}{x^2}\right ) \, dx \\ & = x^2+\frac {1}{9} \int \frac {e^{1+\frac {20}{9 x^2}} \left (-40+9 x^2\right )}{x^2} \, dx \\ & = e^{1+\frac {20}{9 x^2}} x+x^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.13 \[ \int \frac {18 x^3+e^{\frac {20+9 x^2}{9 x^2}} \left (-40+9 x^2\right )}{9 x^2} \, dx=e^{1+\frac {20}{9 x^2}} x+x^2 \]

[In]

Integrate[(18*x^3 + E^((20 + 9*x^2)/(9*x^2))*(-40 + 9*x^2))/(9*x^2),x]

[Out]

E^(1 + 20/(9*x^2))*x + x^2

Maple [A] (verified)

Time = 0.66 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.33

method	result	size
risch	\(x^{2}+x \,{\mathrm e}^{\frac {9 x^{2}+20}{9 x^{2}}}\)	\(20\)
parallelrisch	\(x^{2}+x \,{\mathrm e}^{\frac {9 x^{2}+20}{9 x^{2}}}\)	\(20\)
norman	\(\frac {x^{3}+{\mathrm e}^{\frac {9 x^{2}+20}{9 x^{2}}} x^{2}}{x}\)	\(26\)
derivativedivides	\(x^{2}-\frac {2 i {\mathrm e} \sqrt {\pi }\, \sqrt {5}\, \operatorname {erf}\left (\frac {2 i \sqrt {5}}{3 x}\right )}{3}-{\mathrm e} \left (-{\mathrm e}^{\frac {20}{9 x^{2}}} x -\frac {2 i \sqrt {\pi }\, \sqrt {5}\, \operatorname {erf}\left (\frac {2 i \sqrt {5}}{3 x}\right )}{3}\right )\)	\(59\)
default	\(x^{2}-\frac {2 i {\mathrm e} \sqrt {\pi }\, \sqrt {5}\, \operatorname {erf}\left (\frac {2 i \sqrt {5}}{3 x}\right )}{3}-{\mathrm e} \left (-{\mathrm e}^{\frac {20}{9 x^{2}}} x -\frac {2 i \sqrt {\pi }\, \sqrt {5}\, \operatorname {erf}\left (\frac {2 i \sqrt {5}}{3 x}\right )}{3}\right )\)	\(59\)
parts	\(x^{2}-\frac {2 i {\mathrm e} \sqrt {\pi }\, \sqrt {5}\, \operatorname {erf}\left (\frac {2 i \sqrt {5}}{3 x}\right )}{3}-{\mathrm e} \left (-{\mathrm e}^{\frac {20}{9 x^{2}}} x -\frac {2 i \sqrt {\pi }\, \sqrt {5}\, \operatorname {erf}\left (\frac {2 i \sqrt {5}}{3 x}\right )}{3}\right )\)	\(59\)

[In]

int(1/9*((9*x^2-40)*exp(1/9*(9*x^2+20)/x^2)+18*x^3)/x^2,x,method=_RETURNVERBOSE)

[Out]

x^2+x*exp(1/9*(9*x^2+20)/x^2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.27 \[ \int \frac {18 x^3+e^{\frac {20+9 x^2}{9 x^2}} \left (-40+9 x^2\right )}{9 x^2} \, dx=x^{2} + x e^{\left (\frac {9 \, x^{2} + 20}{9 \, x^{2}}\right )} \]

[In]

integrate(1/9*((9*x^2-40)*exp(1/9*(9*x^2+20)/x^2)+18*x^3)/x^2,x, algorithm="fricas")

[Out]

x^2 + x*e^(1/9*(9*x^2 + 20)/x^2)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {18 x^3+e^{\frac {20+9 x^2}{9 x^2}} \left (-40+9 x^2\right )}{9 x^2} \, dx=x^{2} + x e^{\frac {x^{2} + \frac {20}{9}}{x^{2}}} \]

[In]

integrate(1/9*((9*x**2-40)*exp(1/9*(9*x**2+20)/x**2)+18*x**3)/x**2,x)

[Out]

x**2 + x*exp((x**2 + 20/9)/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 = 61, normalized size of antiderivative = 4.07 \[ \int \frac {18 x^3+e^{\frac {20+9 x^2}{9 x^2}} \left (-40+9 x^2\right )}{9 x^2} \, dx=\frac {1}{3} \, \sqrt {5} x \sqrt {-\frac {1}{x^{2}}} e \Gamma \left (-\frac {1}{2}, -\frac {20}{9 \, x^{2}}\right ) + x^{2} + \frac {2 \, \sqrt {5} \sqrt {\pi } {\left (\operatorname {erf}\left (\frac {2}{3} \, \sqrt {5} \sqrt {-\frac {1}{x^{2}}}\right ) - 1\right )} e}{3 \, x \sqrt {-\frac {1}{x^{2}}}} \]

[In]

integrate(1/9*((9*x^2-40)*exp(1/9*(9*x^2+20)/x^2)+18*x^3)/x^2,x, algorithm="maxima")

[Out]

1/3*sqrt(5)*x*sqrt(-1/x^2)*e*gamma(-1/2, -20/9/x^2) + x^2 + 2/3*sqrt(5)*sqrt(pi)*(erf(2/3*sqrt(5)*sqrt(-1/x^2)
) - 1)*e/(x*sqrt(-1/x^2))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.27 \[ \int \frac {18 x^3+e^{\frac {20+9 x^2}{9 x^2}} \left (-40+9 x^2\right )}{9 x^2} \, dx=x^{2} + x e^{\left (\frac {9 \, x^{2} + 20}{9 \, x^{2}}\right )} \]

[In]

integrate(1/9*((9*x^2-40)*exp(1/9*(9*x^2+20)/x^2)+18*x^3)/x^2,x, algorithm="giac")

[Out]

x^2 + x*e^(1/9*(9*x^2 + 20)/x^2)

Mupad [B] (verification not implemented)

Time = 15.31 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \frac {18 x^3+e^{\frac {20+9 x^2}{9 x^2}} \left (-40+9 x^2\right )}{9 x^2} \, dx=x\,\left (x+{\mathrm {e}}^{\frac {20}{9\,x^2}+1}\right ) \]

[In]

int(((exp((x^2 + 20/9)/x^2)*(9*x^2 - 40))/9 + 2*x^3)/x^2,x)

[Out]

x*(x + exp(20/(9*x^2) + 1))