\(\int (-\frac {320 e^{10}}{x^5}-\frac {320 e^5}{x^3}) \, dx\) [3435]

Optimal result

Integrand size = 17, antiderivative size = 13 \[ \int \left (-\frac {320 e^{10}}{x^5}-\frac {320 e^5}{x^3}\right ) \, dx=80 \left (1+\frac {e^5}{x^2}\right )^2 \]

[Out]

20*(exp(-3*ln(x)+5)*x+1)*(4*exp(-3*ln(x)+5)*x+4)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.31, number of steps used = 1, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \left (-\frac {320 e^{10}}{x^5}-\frac {320 e^5}{x^3}\right ) \, dx=\frac {80 e^{10}}{x^4}+\frac {160 e^5}{x^2} \]

[In]

Int[(-320*E^10)/x^5 - (320*E^5)/x^3,x]

[Out]

(80*E^10)/x^4 + (160*E^5)/x^2

Rubi steps \begin{align*} \text {integral}& = \frac {80 e^{10}}{x^4}+\frac {160 e^5}{x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.77 \[ \int \left (-\frac {320 e^{10}}{x^5}-\frac {320 e^5}{x^3}\right ) \, dx=-320 e^5 \left (-\frac {e^5}{4 x^4}-\frac {1}{2 x^2}\right ) \]

[In]

Integrate[(-320*E^10)/x^5 - (320*E^5)/x^3,x]

[Out]

-320*E^5*(-1/4*E^5/x^4 - 1/(2*x^2))

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.38

method	result	size
risch	\(\frac {80 \,{\mathrm e}^{10}}{x^{4}}+\frac {160 \,{\mathrm e}^{5}}{x^{2}}\)	\(18\)
norman	\(\frac {80 \,{\mathrm e}^{10}+160 x^{2} {\mathrm e}^{5}}{x^{4}}\)	\(19\)
default	\(\frac {80 \,{\mathrm e}^{10}}{x^{4}}+160 \,{\mathrm e}^{-3 \ln \left (x \right )+5} x\)	\(26\)
parallelrisch	\(\frac {80 \,{\mathrm e}^{10}}{x^{4}}+160 \,{\mathrm e}^{-3 \ln \left (x \right )+5} x\)	\(26\)
parts	\(\frac {80 \,{\mathrm e}^{10}}{x^{4}}+160 \,{\mathrm e}^{-3 \ln \left (x \right )+5} x\)	\(26\)

[In]

int(-320*x*exp(-3*ln(x)+5)^2-320*exp(-3*ln(x)+5),x,method=_RETURNVERBOSE)

[Out]

80*exp(5)^2/x^4+160*exp(5)/x^2

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \left (-\frac {320 e^{10}}{x^5}-\frac {320 e^5}{x^3}\right ) \, dx=\frac {80 \, {\left (2 \, x^{2} e^{5} + e^{10}\right )}}{x^{4}} \]

[In]

integrate(-320*x*exp(-3*log(x)+5)^2-320*exp(-3*log(x)+5),x, algorithm="fricas")

[Out]

80*(2*x^2*e^5 + e^10)/x^4

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.46 \[ \int \left (-\frac {320 e^{10}}{x^5}-\frac {320 e^5}{x^3}\right ) \, dx=- \frac {- 160 x^{2} e^{5} - 80 e^{10}}{x^{4}} \]

[In]

integrate(-320*x*exp(-3*ln(x)+5)**2-320*exp(-3*ln(x)+5),x)

[Out]

-(-160*x**2*exp(5) - 80*exp(10))/x**4

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \left (-\frac {320 e^{10}}{x^5}-\frac {320 e^5}{x^3}\right ) \, dx=\frac {160 \, e^{5}}{x^{2}} + \frac {80 \, e^{10}}{x^{4}} \]

[In]

integrate(-320*x*exp(-3*log(x)+5)^2-320*exp(-3*log(x)+5),x, algorithm="maxima")

[Out]

160*e^5/x^2 + 80*e^10/x^4

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \left (-\frac {320 e^{10}}{x^5}-\frac {320 e^5}{x^3}\right ) \, dx=\frac {160 \, e^{5}}{x^{2}} + \frac {80 \, e^{10}}{x^{4}} \]

[In]

integrate(-320*x*exp(-3*log(x)+5)^2-320*exp(-3*log(x)+5),x, algorithm="giac")

[Out]

160*e^5/x^2 + 80*e^10/x^4

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \left (-\frac {320 e^{10}}{x^5}-\frac {320 e^5}{x^3}\right ) \, dx=\frac {80\,{\mathrm {e}}^5\,\left (2\,x^2+{\mathrm {e}}^5\right )}{x^4} \]

[In]

int(- 320*exp(5 - 3*log(x)) - 320*x*exp(10 - 6*log(x)),x)

[Out]

(80*exp(5)*(exp(5) + 2*x^2))/x^4