\(\int \frac {5 x^2-3 e^3 \log (5)}{x^4} \, dx\) [4266]

Optimal result

Integrand size = 17, antiderivative size = 22 \[ \int \frac {5 x^2-3 e^3 \log (5)}{x^4} \, dx=\frac {5 x (-1+5 x)+\frac {e^3 \log (5)}{x}}{x^2} \]

[Out]

(5*x*(5*x-1)+ln(5)*exp(3)/x)/x^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {14} \[ \int \frac {5 x^2-3 e^3 \log (5)}{x^4} \, dx=\frac {e^3 \log (5)}{x^3}-\frac {5}{x} \]

[In]

Int[(5*x^2 - 3*E^3*Log[5])/x^4,x]

[Out]

-5/x + (E^3*Log[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 (\frac {5}{x^2}-\frac {3 e^3 \log (5)}{x^4}\right ) \, dx \\ & = -\frac {5}{x}+\frac {e^3 \log (5)}{x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {5 x^2-3 e^3 \log (5)}{x^4} \, dx=-\frac {5}{x}+\frac {e^3 \log (5)}{x^3} \]

[In]

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

[Out]

-5/x + (E^3*Log[5])/x^3

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68

method	result	size
default	\(-\frac {5}{x}+\frac {{\mathrm e}^{3} \ln \left (5\right )}{x^{3}}\)	\(15\)
gosper	\(\frac {{\mathrm e}^{3} \ln \left (5\right )-5 x^{2}}{x^{3}}\)	\(16\)
norman	\(\frac {{\mathrm e}^{3} \ln \left (5\right )-5 x^{2}}{x^{3}}\)	\(16\)
risch	\(\frac {{\mathrm e}^{3} \ln \left (5\right )-5 x^{2}}{x^{3}}\)	\(16\)
parallelrisch	\(\frac {{\mathrm e}^{3} \ln \left (5\right )-5 x^{2}}{x^{3}}\)	\(16\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {5 x^2-3 e^3 \log (5)}{x^4} \, dx=-\frac {5 \, x^{2} - e^{3} \log \left (5\right )}{x^{3}} \]

[In]

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

[Out]

-(5*x^2 - e^3*log(5))/x^3

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.64 \[ \int \frac {5 x^2-3 e^3 \log (5)}{x^4} \, dx=\frac {- 5 x^{2} + e^{3} \log {\left (5 \right )}}{x^{3}} \]

[In]

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

[Out]

(-5*x**2 + exp(3)*log(5))/x**3

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {5 x^2-3 e^3 \log (5)}{x^4} \, dx=-\frac {5 \, x^{2} - e^{3} \log \left (5\right )}{x^{3}} \]

[In]

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

[Out]

-(5*x^2 - e^3*log(5))/x^3

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {5 x^2-3 e^3 \log (5)}{x^4} \, dx=-\frac {5 \, x^{2} - e^{3} \log \left (5\right )}{x^{3}} \]

[In]

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

[Out]

-(5*x^2 - e^3*log(5))/x^3

Mupad [B] (verification not implemented)

Time = 11.95 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {5 x^2-3 e^3 \log (5)}{x^4} \, dx=\frac {{\mathrm {e}}^3\,\ln \left (5\right )-5\,x^2}{x^3} \]

[In]

int(-(3*exp(3)*log(5) - 5*x^2)/x^4,x)

[Out]

(exp(3)*log(5) - 5*x^2)/x^3