\(\int \frac {8+e^{-3-60 x-2 x^2} x (x^2-60 x^3-4 x^4)}{x^3} \, dx\) [8784]

Optimal result

Integrand size = 34, antiderivative size = 19 \[ \int \frac {8+e^{-3-60 x-2 x^2} x \left (x^2-60 x^3-4 x^4\right )}{x^3} \, dx=1-\frac {4}{x^2}+e^{-3-2 x (30+x)} x \]

[Out]

exp(ln(x)-3-2*(30+x)*x)+1-4/x^2

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.63, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {14, 2326} \[ \int \frac {8+e^{-3-60 x-2 x^2} x \left (x^2-60 x^3-4 x^4\right )}{x^3} \, dx=\frac {e^{-2 x^2-60 x-3} \left (x^2+15 x\right )}{x+15}-\frac {4}{x^2} \]

[In]

Int[(8 + E^(-3 - 60*x - 2*x^2)*x*(x^2 - 60*x^3 - 4*x^4))/x^3,x]

[Out]

-4/x^2 + (E^(-3 - 60*x - 2*x^2)*(15*x + x^2))/(15 + 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}& = \int \left (\frac {8}{x^3}-e^{-3-60 x-2 x^2} \left (-1+60 x+4 x^2\right )\right ) \, dx \\ & = -\frac {4}{x^2}-\int e^{-3-60 x-2 x^2} \left (-1+60 x+4 x^2\right ) \, dx \\ & = -\frac {4}{x^2}+\frac {e^{-3-60 x-2 x^2} \left (15 x+x^2\right )}{15+x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {8+e^{-3-60 x-2 x^2} x \left (x^2-60 x^3-4 x^4\right )}{x^3} \, dx=-\frac {4}{x^2}+e^{-3-60 x-2 x^2} x \]

[In]

Integrate[(8 + E^(-3 - 60*x - 2*x^2)*x*(x^2 - 60*x^3 - 4*x^4))/x^3,x]

[Out]

-4/x^2 + E^(-3 - 60*x - 2*x^2)*x

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05

method	result	size
default	\(-\frac {4}{x^{2}}+{\mathrm e}^{\ln \left (x \right )-2 x^{2}-60 x -3}\)	\(20\)
risch	\(-\frac {4}{x^{2}}+x \,{\mathrm e}^{-2 x^{2}-60 x -3}\)	\(20\)
parts	\(-\frac {4}{x^{2}}+{\mathrm e}^{\ln \left (x \right )-2 x^{2}-60 x -3}\)	\(20\)
norman	\(\frac {-4+x^{2} {\mathrm e}^{\ln \left (x \right )-2 x^{2}-60 x -3}}{x^{2}}\)	\(24\)
parallelrisch	\(\frac {-4+x^{2} {\mathrm e}^{\ln \left (x \right )-2 x^{2}-60 x -3}}{x^{2}}\)	\(24\)

[In]

int(((-4*x^4-60*x^3+x^2)*exp(ln(x)-2*x^2-60*x-3)+8)/x^3,x,method=_RETURNVERBOSE)

[Out]

-4/x^2+exp(ln(x)-2*x^2-60*x-3)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {8+e^{-3-60 x-2 x^2} x \left (x^2-60 x^3-4 x^4\right )}{x^3} \, dx=\frac {x^{2} e^{\left (-2 \, x^{2} - 60 \, x + \log \left (x\right ) - 3\right )} - 4}{x^{2}} \]

[In]

integrate(((-4*x^4-60*x^3+x^2)*exp(log(x)-2*x^2-60*x-3)+8)/x^3,x, algorithm="fricas")

[Out]

(x^2*e^(-2*x^2 - 60*x + log(x) - 3) - 4)/x^2

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {8+e^{-3-60 x-2 x^2} x \left (x^2-60 x^3-4 x^4\right )}{x^3} \, dx=x e^{- 2 x^{2} - 60 x - 3} - \frac {4}{x^{2}} \]

[In]

integrate(((-4*x**4-60*x**3+x**2)*exp(ln(x)-2*x**2-60*x-3)+8)/x**3,x)

[Out]

x*exp(-2*x**2 - 60*x - 3) - 4/x**2

Maxima [C] (verification not implemented)

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

Time = 0.32 (sec) , antiderivative size = 151, normalized size of antiderivative = 7.95 \[ \int \frac {8+e^{-3-60 x-2 x^2} x \left (x^2-60 x^3-4 x^4\right )}{x^3} \, dx=\frac {1}{4} \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\sqrt {2} x + 15 \, \sqrt {2}\right ) e^{447} + \frac {1}{2} i \, \sqrt {2} {\left (-\frac {i \, {\left (x + 15\right )}^{3} \Gamma \left (\frac {3}{2}, 2 \, {\left (x + 15\right )}^{2}\right )}{{\left ({\left (x + 15\right )}^{2}\right )}^{\frac {3}{2}}} + \frac {450 i \, \sqrt {\pi } {\left (x + 15\right )} {\left (\operatorname {erf}\left (\sqrt {2} \sqrt {{\left (x + 15\right )}^{2}}\right ) - 1\right )}}{\sqrt {{\left (x + 15\right )}^{2}}} + 30 i \, \sqrt {2} e^{\left (-2 \, {\left (x + 15\right )}^{2}\right )}\right )} e^{447} + \frac {15}{2} i \, \sqrt {2} {\left (-\frac {30 i \, \sqrt {\pi } {\left (x + 15\right )} {\left (\operatorname {erf}\left (\sqrt {2} \sqrt {{\left (x + 15\right )}^{2}}\right ) - 1\right )}}{\sqrt {{\left (x + 15\right )}^{2}}} - i \, \sqrt {2} e^{\left (-2 \, {\left (x + 15\right )}^{2}\right )}\right )} e^{447} - \frac {4}{x^{2}} \]

[In]

integrate(((-4*x^4-60*x^3+x^2)*exp(log(x)-2*x^2-60*x-3)+8)/x^3,x, algorithm="maxima")

[Out]

1/4*sqrt(2)*sqrt(pi)*erf(sqrt(2)*x + 15*sqrt(2))*e^447 + 1/2*I*sqrt(2)*(-I*(x + 15)^3*gamma(3/2, 2*(x + 15)^2)
/((x + 15)^2)^(3/2) + 450*I*sqrt(pi)*(x + 15)*(erf(sqrt(2)*sqrt((x + 15)^2)) - 1)/sqrt((x + 15)^2) + 30*I*sqrt
(2)*e^(-2*(x + 15)^2))*e^447 + 15/2*I*sqrt(2)*(-30*I*sqrt(pi)*(x + 15)*(erf(sqrt(2)*sqrt((x + 15)^2)) - 1)/sqr
t((x + 15)^2) - I*sqrt(2)*e^(-2*(x + 15)^2))*e^447 - 4/x^2

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.32 \[ \int \frac {8+e^{-3-60 x-2 x^2} x \left (x^2-60 x^3-4 x^4\right )}{x^3} \, dx=\frac {{\left (x^{3} e^{\left (-2 \, x^{2} - 60 \, x\right )} - 4 \, e^{3}\right )} e^{\left (-3\right )}}{x^{2}} \]

[In]

integrate(((-4*x^4-60*x^3+x^2)*exp(log(x)-2*x^2-60*x-3)+8)/x^3,x, algorithm="giac")

[Out]

(x^3*e^(-2*x^2 - 60*x) - 4*e^3)*e^(-3)/x^2

Mupad [B] (verification not implemented)

Time = 15.29 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {8+e^{-3-60 x-2 x^2} x \left (x^2-60 x^3-4 x^4\right )}{x^3} \, dx=x\,{\mathrm {e}}^{-60\,x}\,{\mathrm {e}}^{-3}\,{\mathrm {e}}^{-2\,x^2}-\frac {4}{x^2} \]

[In]

int(-(exp(log(x) - 60*x - 2*x^2 - 3)*(60*x^3 - x^2 + 4*x^4) - 8)/x^3,x)

[Out]

x*exp(-60*x)*exp(-3)*exp(-2*x^2) - 4/x^2