\(\int \frac {e^{e^{10}} (e^{5+x} (-2+x)+4 e^5 x)}{5 x^3} \, dx\) [8911]

Optimal result

Integrand size = 28, antiderivative size = 28 \[ \int \frac {e^{e^{10}} \left (e^{5+x} (-2+x)+4 e^5 x\right )}{5 x^3} \, dx=\frac {e^{5+e^{10}} \left (-4+\frac {e^x}{x}\right )-x}{5 x} \]

[Out]

1/5*(exp(5)*(exp(x)/x-4)*exp(exp(10))-x)/x

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {12, 14, 2228} \[ \int \frac {e^{e^{10}} \left (e^{5+x} (-2+x)+4 e^5 x\right )}{5 x^3} \, dx=\frac {e^{x+e^{10}+5}}{5 x^2}-\frac {4 e^{5+e^{10}}}{5 x} \]

[In]

Int[(E^E^10*(E^(5 + x)*(-2 + x) + 4*E^5*x))/(5*x^3),x]

[Out]

E^(5 + E^10 + x)/(5*x^2) - (4*E^(5 + E^10))/(5*x)

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]]

Rule 2228

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[g*u^(m + 1)*(F^(c*v)/(b*c*
e*Log[F])), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} e^{e^{10}} \int \frac {e^{5+x} (-2+x)+4 e^5 x}{x^3} \, dx \\ & = \frac {1}{5} e^{e^{10}} \int \left (\frac {e^{5+x} (-2+x)}{x^3}+\frac {4 e^5}{x^2}\right ) \, dx \\ & = -\frac {4 e^{5+e^{10}}}{5 x}+\frac {1}{5} e^{e^{10}} \int \frac {e^{5+x} (-2+x)}{x^3} \, dx \\ & = \frac {e^{5+e^{10}+x}}{5 x^2}-\frac {4 e^{5+e^{10}}}{5 x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {e^{e^{10}} \left (e^{5+x} (-2+x)+4 e^5 x\right )}{5 x^3} \, dx=\frac {1}{5} e^{5+e^{10}} \left (\frac {e^x}{x^2}-\frac {4}{x}\right ) \]

[In]

Integrate[(E^E^10*(E^(5 + x)*(-2 + x) + 4*E^5*x))/(5*x^3),x]

[Out]

(E^(5 + E^10)*(E^x/x^2 - 4/x))/5

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75

method	result	size
parallelrisch	\(-\frac {{\mathrm e}^{{\mathrm e}^{10}} \left (4 x \,{\mathrm e}^{5}-{\mathrm e}^{5} {\mathrm e}^{x}\right )}{5 x^{2}}\)	\(21\)
norman	\(\frac {-\frac {4 \,{\mathrm e}^{5} {\mathrm e}^{{\mathrm e}^{10}} x}{5}+\frac {{\mathrm e}^{5} {\mathrm e}^{{\mathrm e}^{10}} {\mathrm e}^{x}}{5}}{x^{2}}\)	\(23\)
risch	\(-\frac {4 \,{\mathrm e}^{{\mathrm e}^{10}+5}}{5 x}+\frac {{\mathrm e}^{{\mathrm e}^{10}+5+x}}{5 x^{2}}\)	\(23\)
parts	\(-\frac {4 \,{\mathrm e}^{{\mathrm e}^{10}} {\mathrm e}^{5}}{5 x}+\frac {{\mathrm e}^{{\mathrm e}^{10}} {\mathrm e}^{5} {\mathrm e}^{x}}{5 x^{2}}\)	\(24\)
default	\(\frac {{\mathrm e}^{{\mathrm e}^{10}} \left ({\mathrm e}^{5} \left (-\frac {{\mathrm e}^{x}}{x}-\operatorname {Ei}_{1}\left (-x \right )\right )-\frac {4 \,{\mathrm e}^{5}}{x}-2 \,{\mathrm e}^{5} \left (-\frac {{\mathrm e}^{x}}{2 x^{2}}-\frac {{\mathrm e}^{x}}{2 x}-\frac {\operatorname {Ei}_{1}\left (-x \right )}{2}\right )\right )}{5}\)	\(58\)

[In]

int(1/5*((-2+x)*exp(5)*exp(x)+4*x*exp(5))*exp(exp(10))/x^3,x,method=_RETURNVERBOSE)

[Out]

-1/5*exp(exp(10))*(4*x*exp(5)-exp(5)*exp(x))/x^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {e^{e^{10}} \left (e^{5+x} (-2+x)+4 e^5 x\right )}{5 x^3} \, dx=-\frac {{\left (4 \, x e^{5} - e^{\left (x + 5\right )}\right )} e^{\left (e^{10}\right )}}{5 \, x^{2}} \]

[In]

integrate(1/5*((-2+x)*exp(5)*exp(x)+4*x*exp(5))*exp(exp(10))/x^3,x, algorithm="fricas")

[Out]

-1/5*(4*x*e^5 - e^(x + 5))*e^(e^10)/x^2

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {e^{e^{10}} \left (e^{5+x} (-2+x)+4 e^5 x\right )}{5 x^3} \, dx=- \frac {4 e^{5} e^{e^{10}}}{5 x} + \frac {e^{5} e^{x} e^{e^{10}}}{5 x^{2}} \]

[In]

integrate(1/5*((-2+x)*exp(5)*exp(x)+4*x*exp(5))*exp(exp(10))/x**3,x)

[Out]

-4*exp(5)*exp(exp(10))/(5*x) + exp(5)*exp(x)*exp(exp(10))/(5*x**2)

Maxima [C] (verification not implemented)

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

Time = 0.23 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {e^{e^{10}} \left (e^{5+x} (-2+x)+4 e^5 x\right )}{5 x^3} \, dx=\frac {1}{5} \, {\left (e^{5} \Gamma \left (-1, -x\right ) + 2 \, e^{5} \Gamma \left (-2, -x\right ) - \frac {4 \, e^{5}}{x}\right )} e^{\left (e^{10}\right )} \]

[In]

integrate(1/5*((-2+x)*exp(5)*exp(x)+4*x*exp(5))*exp(exp(10))/x^3,x, algorithm="maxima")

[Out]

1/5*(e^5*gamma(-1, -x) + 2*e^5*gamma(-2, -x) - 4*e^5/x)*e^(e^10)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {e^{e^{10}} \left (e^{5+x} (-2+x)+4 e^5 x\right )}{5 x^3} \, dx=-\frac {{\left (4 \, x e^{5} - e^{\left (x + 5\right )}\right )} e^{\left (e^{10}\right )}}{5 \, x^{2}} \]

[In]

integrate(1/5*((-2+x)*exp(5)*exp(x)+4*x*exp(5))*exp(exp(10))/x^3,x, algorithm="giac")

[Out]

-1/5*(4*x*e^5 - e^(x + 5))*e^(e^10)/x^2

Mupad [B] (verification not implemented)

Time = 13.67 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.64 \[ \int \frac {e^{e^{10}} \left (e^{5+x} (-2+x)+4 e^5 x\right )}{5 x^3} \, dx=-\frac {{\mathrm {e}}^5\,{\mathrm {e}}^{{\mathrm {e}}^{10}}\,\left (4\,x-{\mathrm {e}}^x\right )}{5\,x^2} \]

[In]

int((exp(exp(10))*(4*x*exp(5) + exp(5)*exp(x)*(x - 2)))/(5*x^3),x)

[Out]

-(exp(5)*exp(exp(10))*(4*x - exp(x)))/(5*x^2)