\(\int \frac {-15 x^3+e x (-20+3 x^2)}{e x^7} \, dx\) [495]

Optimal result

Integrand size = 23, antiderivative size = 16 \[ \int \frac {-15 x^3+e x \left (-20+3 x^2\right )}{e x^7} \, dx=\frac {-1+\frac {5}{e}+\frac {4}{x^2}}{x^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.12, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {12, 14} \[ \int \frac {-15 x^3+e x \left (-20+3 x^2\right )}{e x^7} \, dx=\frac {4}{x^5}+\frac {5-e}{e x^3} \]

[In]

Int[(-15*x^3 + E*x*(-20 + 3*x^2))/(E*x^7),x]

[Out]

4/x^5 + (5 - E)/(E*x^3)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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}& = \frac {\int \frac {-15 x^3+e x \left (-20+3 x^2\right )}{x^7} \, dx}{e} \\ & = \frac {\int \left (-\frac {20 e}{x^6}+\frac {3 (-5+e)}{x^4}\right ) \, dx}{e} \\ & = \frac {4}{x^5}+\frac {5-e}{e x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.12 \[ \int \frac {-15 x^3+e x \left (-20+3 x^2\right )}{e x^7} \, dx=\frac {4}{x^5}+\frac {5-e}{e x^3} \]

[In]

Integrate[(-15*x^3 + E*x*(-20 + 3*x^2))/(E*x^7),x]

[Out]

4/x^5 + (5 - E)/(E*x^3)

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.38

method	result	size
norman	\(\frac {4 x -\left ({\mathrm e}-5\right ) {\mathrm e}^{-1} x^{3}}{x^{6}}\)	\(22\)
risch	\(\frac {{\mathrm e}^{-1} \left (\left (5-{\mathrm e}\right ) x^{2}+4 \,{\mathrm e}\right )}{x^{5}}\)	\(22\)
default	\(\frac {4}{x^{5}}-\frac {1}{x^{3}}+\frac {5 \,{\mathrm e}^{-1}}{x^{3}}\)	\(24\)
parts	\(\frac {4}{x^{5}}-\frac {1}{x^{3}}+\frac {5 \,{\mathrm e}^{-1}}{x^{3}}\)	\(24\)
parallelrisch	\(\frac {\left (5 x^{5}-{\mathrm e}^{\ln \left (x \right )+1} x^{4}+4 x^{2} {\mathrm e}^{\ln \left (x \right )+1}\right ) {\mathrm e}^{-1}}{x^{8}}\)	\(38\)

[In]

int(((3*x^2-20)*exp(ln(x)+1)-15*x^3)/x^6/exp(ln(x)+1),x,method=_RETURNVERBOSE)

[Out]

(4*x-(exp(1)-5)/exp(1)*x^3)/x^6

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.31 \[ \int \frac {-15 x^3+e x \left (-20+3 x^2\right )}{e x^7} \, dx=\frac {{\left (5 \, x^{2} - {\left (x^{2} - 4\right )} e\right )} e^{\left (-1\right )}}{x^{5}} \]

[In]

integrate(((3*x^2-20)*exp(log(x)+1)-15*x^3)/x^6/exp(log(x)+1),x, algorithm="fricas")

[Out]

(5*x^2 - (x^2 - 4)*e)*e^(-1)/x^5

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.25 \[ \int \frac {-15 x^3+e x \left (-20+3 x^2\right )}{e x^7} \, dx=- \frac {x^{2} \left (-5 + e\right ) - 4 e}{e x^{5}} \]

[In]

integrate(((3*x**2-20)*exp(ln(x)+1)-15*x**3)/x**6/exp(ln(x)+1),x)

[Out]

-(x**2*(-5 + E) - 4*E)*exp(-1)/x**5

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.25 \[ \int \frac {-15 x^3+e x \left (-20+3 x^2\right )}{e x^7} \, dx=-\frac {{\left (x^{2} {\left (e - 5\right )} - 4 \, e\right )} e^{\left (-1\right )}}{x^{5}} \]

[In]

integrate(((3*x^2-20)*exp(log(x)+1)-15*x^3)/x^6/exp(log(x)+1),x, algorithm="maxima")

[Out]

-(x^2*(e - 5) - 4*e)*e^(-1)/x^5

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.44 \[ \int \frac {-15 x^3+e x \left (-20+3 x^2\right )}{e x^7} \, dx=-\frac {{\left (x^{2} e - 5 \, x^{2} - 4 \, e\right )} e^{\left (-1\right )}}{x^{5}} \]

[In]

integrate(((3*x^2-20)*exp(log(x)+1)-15*x^3)/x^6/exp(log(x)+1),x, algorithm="giac")

[Out]

-(x^2*e - 5*x^2 - 4*e)*e^(-1)/x^5

Mupad [B] (verification not implemented)

Time = 7.49 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {-15 x^3+e x \left (-20+3 x^2\right )}{e x^7} \, dx=\frac {\left (5\,{\mathrm {e}}^{-1}-1\right )\,x^2+4}{x^5} \]

[In]

int((exp(- log(x) - 1)*(exp(log(x) + 1)*(3*x^2 - 20) - 15*x^3))/x^6,x)

[Out]

(x^2*(5*exp(-1) - 1) + 4)/x^5