\(\int \frac {-4 x^3+e^5 (-6+18 x^2-18 x^4)}{x^2} \, dx\) [4919]

Optimal result

Integrand size = 26, antiderivative size = 33 \[ \int \frac {-4 x^3+e^5 \left (-6+18 x^2-18 x^4\right )}{x^2} \, dx=x^2+3 \left (-x^2+e^5 \left (\frac {2}{x}+x-x \left (-5+2 x^2\right )\right )\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {14} \[ \int \frac {-4 x^3+e^5 \left (-6+18 x^2-18 x^4\right )}{x^2} \, dx=-6 e^5 x^3-2 x^2+18 e^5 x+\frac {6 e^5}{x} \]

[In]

Int[(-4*x^3 + E^5*(-6 + 18*x^2 - 18*x^4))/x^2,x]

[Out]

(6*E^5)/x + 18*E^5*x - 2*x^2 - 6*E^5*x^3

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]]

Rubi steps \begin{align*} \text {integral}& = \int \left (18 e^5-\frac {6 e^5}{x^2}-4 x-18 e^5 x^2\right ) \, dx \\ & = \frac {6 e^5}{x}+18 e^5 x-2 x^2-6 e^5 x^3 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {-4 x^3+e^5 \left (-6+18 x^2-18 x^4\right )}{x^2} \, dx=\frac {6 e^5}{x}+18 e^5 x-2 x^2-6 e^5 x^3 \]

[In]

Integrate[(-4*x^3 + E^5*(-6 + 18*x^2 - 18*x^4))/x^2,x]

[Out]

(6*E^5)/x + 18*E^5*x - 2*x^2 - 6*E^5*x^3

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79

method	result	size
default	\(-6 x^{3} {\mathrm e}^{5}-2 x^{2}+18 x \,{\mathrm e}^{5}+\frac {6 \,{\mathrm e}^{5}}{x}\)	\(26\)
risch	\(-6 x^{3} {\mathrm e}^{5}-2 x^{2}+18 x \,{\mathrm e}^{5}+\frac {6 \,{\mathrm e}^{5}}{x}\)	\(26\)
gosper	\(-\frac {2 \left (3 x^{4} {\mathrm e}^{5}+x^{3}-9 x^{2} {\mathrm e}^{5}-3 \,{\mathrm e}^{5}\right )}{x}\)	\(28\)
norman	\(\frac {-2 x^{3}+18 x^{2} {\mathrm e}^{5}-6 x^{4} {\mathrm e}^{5}+6 \,{\mathrm e}^{5}}{x}\)	\(29\)
parallelrisch	\(-\frac {6 x^{4} {\mathrm e}^{5}-18 x^{2} {\mathrm e}^{5}+2 x^{3}-6 \,{\mathrm e}^{5}}{x}\)	\(30\)

[In]

int(((-18*x^4+18*x^2-6)*exp(5)-4*x^3)/x^2,x,method=_RETURNVERBOSE)

[Out]

-6*x^3*exp(5)-2*x^2+18*x*exp(5)+6*exp(5)/x

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.70 \[ \int \frac {-4 x^3+e^5 \left (-6+18 x^2-18 x^4\right )}{x^2} \, dx=-\frac {2 \, {\left (x^{3} + 3 \, {\left (x^{4} - 3 \, x^{2} - 1\right )} e^{5}\right )}}{x} \]

[In]

integrate(((-18*x^4+18*x^2-6)*exp(5)-4*x^3)/x^2,x, algorithm="fricas")

[Out]

-2*(x^3 + 3*(x^4 - 3*x^2 - 1)*e^5)/x

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int \frac {-4 x^3+e^5 \left (-6+18 x^2-18 x^4\right )}{x^2} \, dx=- 6 x^{3} e^{5} - 2 x^{2} + 18 x e^{5} + \frac {6 e^{5}}{x} \]

[In]

integrate(((-18*x**4+18*x**2-6)*exp(5)-4*x**3)/x**2,x)

[Out]

-6*x**3*exp(5) - 2*x**2 + 18*x*exp(5) + 6*exp(5)/x

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \[ \int \frac {-4 x^3+e^5 \left (-6+18 x^2-18 x^4\right )}{x^2} \, dx=-6 \, x^{3} e^{5} - 2 \, x^{2} + 18 \, x e^{5} + \frac {6 \, e^{5}}{x} \]

[In]

integrate(((-18*x^4+18*x^2-6)*exp(5)-4*x^3)/x^2,x, algorithm="maxima")

[Out]

-6*x^3*e^5 - 2*x^2 + 18*x*e^5 + 6*e^5/x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \[ \int \frac {-4 x^3+e^5 \left (-6+18 x^2-18 x^4\right )}{x^2} \, dx=-6 \, x^{3} e^{5} - 2 \, x^{2} + 18 \, x e^{5} + \frac {6 \, e^{5}}{x} \]

[In]

integrate(((-18*x^4+18*x^2-6)*exp(5)-4*x^3)/x^2,x, algorithm="giac")

[Out]

-6*x^3*e^5 - 2*x^2 + 18*x*e^5 + 6*e^5/x

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \[ \int \frac {-4 x^3+e^5 \left (-6+18 x^2-18 x^4\right )}{x^2} \, dx=18\,x\,{\mathrm {e}}^5+\frac {6\,{\mathrm {e}}^5}{x}-6\,x^3\,{\mathrm {e}}^5-2\,x^2 \]

[In]

int(-(exp(5)*(18*x^4 - 18*x^2 + 6) + 4*x^3)/x^2,x)

[Out]

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