\(\int \frac {x+x^2+10 x^3+e^{2 e^5} (-5-x^2)}{x^2} \, dx\) [6430]

Optimal result

Integrand size = 29, antiderivative size = 27 \[ \int \frac {x+x^2+10 x^3+e^{2 e^5} \left (-5-x^2\right )}{x^2} \, dx=-5+e^{2 e^5} \left (\frac {5}{x}-x\right )+x+5 x^2+\log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {14} \[ \int \frac {x+x^2+10 x^3+e^{2 e^5} \left (-5-x^2\right )}{x^2} \, dx=5 x^2+\left (1-e^{2 e^5}\right ) x+\frac {5 e^{2 e^5}}{x}+\log (x) \]

[In]

Int[(x + x^2 + 10*x^3 + E^(2*E^5)*(-5 - x^2))/x^2,x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \int \left (1-e^{2 e^5}-\frac {5 e^{2 e^5}}{x^2}+\frac {1}{x}+10 x\right ) \, dx \\ & = \frac {5 e^{2 e^5}}{x}+\left (1-e^{2 e^5}\right ) x+5 x^2+\log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {x+x^2+10 x^3+e^{2 e^5} \left (-5-x^2\right )}{x^2} \, dx=e^{2 e^5} \left (\frac {5}{x}-x\right )+x (1+5 x)+\log (x) \]

[In]

Integrate[(x + x^2 + 10*x^3 + E^(2*E^5)*(-5 - x^2))/x^2,x]

[Out]

E^(2*E^5)*(5/x - x) + x*(1 + 5*x) + Log[x]

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04

method	result	size
default	\(5 x^{2}-{\mathrm e}^{2 \,{\mathrm e}^{5}} x +x +\frac {5 \,{\mathrm e}^{2 \,{\mathrm e}^{5}}}{x}+\ln \left (x \right )\)	\(28\)
risch	\(5 x^{2}-{\mathrm e}^{2 \,{\mathrm e}^{5}} x +x +\frac {5 \,{\mathrm e}^{2 \,{\mathrm e}^{5}}}{x}+\ln \left (x \right )\)	\(28\)
norman	\(\frac {\left (-{\mathrm e}^{2 \,{\mathrm e}^{5}}+1\right ) x^{2}+5 x^{3}+5 \,{\mathrm e}^{2 \,{\mathrm e}^{5}}}{x}+\ln \left (x \right )\)	\(34\)
parallelrisch	\(\frac {-{\mathrm e}^{2 \,{\mathrm e}^{5}} x^{2}+5 x^{3}+x \ln \left (x \right )+5 \,{\mathrm e}^{2 \,{\mathrm e}^{5}}+x^{2}}{x}\)	\(35\)

[In]

int(((-x^2-5)*exp(exp(5))^2+10*x^3+x^2+x)/x^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {x+x^2+10 x^3+e^{2 e^5} \left (-5-x^2\right )}{x^2} \, dx=\frac {5 \, x^{3} + x^{2} - {\left (x^{2} - 5\right )} e^{\left (2 \, e^{5}\right )} + x \log \left (x\right )}{x} \]

[In]

integrate(((-x^2-5)*exp(exp(5))^2+10*x^3+x^2+x)/x^2,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {x+x^2+10 x^3+e^{2 e^5} \left (-5-x^2\right )}{x^2} \, dx=5 x^{2} + x \left (1 - e^{2 e^{5}}\right ) + \log {\left (x \right )} + \frac {5 e^{2 e^{5}}}{x} \]

[In]

integrate(((-x**2-5)*exp(exp(5))**2+10*x**3+x**2+x)/x**2,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {x+x^2+10 x^3+e^{2 e^5} \left (-5-x^2\right )}{x^2} \, dx=5 \, x^{2} - x {\left (e^{\left (2 \, e^{5}\right )} - 1\right )} + \frac {5 \, e^{\left (2 \, e^{5}\right )}}{x} + \log \left (x\right ) \]

[In]

integrate(((-x^2-5)*exp(exp(5))^2+10*x^3+x^2+x)/x^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {x+x^2+10 x^3+e^{2 e^5} \left (-5-x^2\right )}{x^2} \, dx=5 \, x^{2} - x e^{\left (2 \, e^{5}\right )} + x + \frac {5 \, e^{\left (2 \, e^{5}\right )}}{x} + \log \left ({\left | x \right |}\right ) \]

[In]

integrate(((-x^2-5)*exp(exp(5))^2+10*x^3+x^2+x)/x^2,x, algorithm="giac")

[Out]

5*x^2 - x*e^(2*e^5) + x + 5*e^(2*e^5)/x + log(abs(x))

Mupad [B] (verification not implemented)

Time = 12.12 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {x+x^2+10 x^3+e^{2 e^5} \left (-5-x^2\right )}{x^2} \, dx=\ln \left (x\right )-x\,\left ({\mathrm {e}}^{2\,{\mathrm {e}}^5}-1\right )+\frac {5\,{\mathrm {e}}^{2\,{\mathrm {e}}^5}}{x}+5\,x^2 \]

[In]

int((x - exp(2*exp(5))*(x^2 + 5) + x^2 + 10*x^3)/x^2,x)

[Out]

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