\(\int \frac {-100+200 e^3-100 e^6+e^{2 x} (-18 x^2+18 x^3)+e^x (90 x-30 x^2+e^3 (-90 x+30 x^2))}{225 x^5} \, dx\) [5808]

Optimal result

Integrand size = 62, antiderivative size = 28 \[ \int \frac {-100+200 e^3-100 e^6+e^{2 x} \left (-18 x^2+18 x^3\right )+e^x \left (90 x-30 x^2+e^3 \left (-90 x+30 x^2\right )\right )}{225 x^5} \, dx=5+\frac {\left (\frac {e^x}{5}+\frac {-1+e^3}{3 x}\right )^2}{x^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.64, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {12, 14, 2228} \[ \int \frac {-100+200 e^3-100 e^6+e^{2 x} \left (-18 x^2+18 x^3\right )+e^x \left (90 x-30 x^2+e^3 \left (-90 x+30 x^2\right )\right )}{225 x^5} \, dx=\frac {\left (1-e^3\right )^2}{9 x^4}-\frac {2 \left (1-e^3\right ) e^x}{15 x^3}+\frac {e^{2 x}}{25 x^2} \]

[In]

Int[(-100 + 200*E^3 - 100*E^6 + E^(2*x)*(-18*x^2 + 18*x^3) + E^x*(90*x - 30*x^2 + E^3*(-90*x + 30*x^2)))/(225*
x^5),x]

[Out]

(1 - E^3)^2/(9*x^4) - (2*E^x*(1 - E^3))/(15*x^3) + E^(2*x)/(25*x^2)

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}{225} \int \frac {-100+200 e^3-100 e^6+e^{2 x} \left (-18 x^2+18 x^3\right )+e^x \left (90 x-30 x^2+e^3 \left (-90 x+30 x^2\right )\right )}{x^5} \, dx \\ & = \frac {1}{225} \int \left (-\frac {100 \left (-1+e^3\right )^2}{x^5}+\frac {30 (-1+e) e^x \left (1+e+e^2\right ) (-3+x)}{x^4}+\frac {18 e^{2 x} (-1+x)}{x^3}\right ) \, dx \\ & = \frac {\left (1-e^3\right )^2}{9 x^4}+\frac {2}{25} \int \frac {e^{2 x} (-1+x)}{x^3} \, dx+\frac {1}{15} \left (2 (-1+e) \left (1+e+e^2\right )\right ) \int \frac {e^x (-3+x)}{x^4} \, dx \\ & = \frac {\left (1-e^3\right )^2}{9 x^4}-\frac {2 (1-e) e^x \left (1+e+e^2\right )}{15 x^3}+\frac {e^{2 x}}{25 x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.37 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {-100+200 e^3-100 e^6+e^{2 x} \left (-18 x^2+18 x^3\right )+e^x \left (90 x-30 x^2+e^3 \left (-90 x+30 x^2\right )\right )}{225 x^5} \, dx=\frac {\left (-5+5 e^3+3 e^x x\right )^2}{225 x^4} \]

[In]

Integrate[(-100 + 200*E^3 - 100*E^6 + E^(2*x)*(-18*x^2 + 18*x^3) + E^x*(90*x - 30*x^2 + E^3*(-90*x + 30*x^2)))
/(225*x^5),x]

[Out]

(-5 + 5*E^3 + 3*E^x*x)^2/(225*x^4)

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29

method	result	size
norman	\(\frac {\left (\frac {2 \,{\mathrm e}^{3}}{15}-\frac {2}{15}\right ) x \,{\mathrm e}^{x}+\frac {{\mathrm e}^{2 x} x^{2}}{25}+\frac {{\mathrm e}^{6}}{9}-\frac {2 \,{\mathrm e}^{3}}{9}+\frac {1}{9}}{x^{4}}\)	\(36\)
risch	\(\frac {30 \,{\mathrm e}^{3+x} x +9 \,{\mathrm e}^{2 x} x^{2}-30 \,{\mathrm e}^{x} x -50 \,{\mathrm e}^{3}+25 \,{\mathrm e}^{6}+25}{225 x^{4}}\)	\(37\)
parallelrisch	\(\frac {30 x \,{\mathrm e}^{3} {\mathrm e}^{x}+9 \,{\mathrm e}^{2 x} x^{2}-30 \,{\mathrm e}^{x} x -50 \,{\mathrm e}^{3}+25 \,{\mathrm e}^{6}+25}{225 x^{4}}\)	\(39\)
parts	\(-\frac {-\frac {4 \,{\mathrm e}^{6}}{9}+\frac {8 \,{\mathrm e}^{3}}{9}-\frac {4}{9}}{4 x^{4}}+\frac {2 \,{\mathrm e}^{3} \left (-\frac {{\mathrm e}^{x}}{2 x^{2}}-\frac {{\mathrm e}^{x}}{2 x}-\frac {\operatorname {Ei}_{1}\left (-x \right )}{2}\right )}{15}-\frac {2 \,{\mathrm e}^{x}}{15 x^{3}}-\frac {2 \,{\mathrm e}^{3} \left (-\frac {{\mathrm e}^{x}}{3 x^{3}}-\frac {{\mathrm e}^{x}}{6 x^{2}}-\frac {{\mathrm e}^{x}}{6 x}-\frac {\operatorname {Ei}_{1}\left (-x \right )}{6}\right )}{5}+\frac {{\mathrm e}^{2 x}}{25 x^{2}}\)	\(94\)
default	\(\frac {1}{9 x^{4}}-\frac {2 \,{\mathrm e}^{3}}{9 x^{4}}+\frac {{\mathrm e}^{6}}{9 x^{4}}+\frac {{\mathrm e}^{2 x}}{25 x^{2}}-\frac {2 \,{\mathrm e}^{x}}{15 x^{3}}-\frac {2 \,{\mathrm e}^{3} \left (-\frac {{\mathrm e}^{x}}{3 x^{3}}-\frac {{\mathrm e}^{x}}{6 x^{2}}-\frac {{\mathrm e}^{x}}{6 x}-\frac {\operatorname {Ei}_{1}\left (-x \right )}{6}\right )}{5}+\frac {2 \,{\mathrm e}^{3} \left (-\frac {{\mathrm e}^{x}}{2 x^{2}}-\frac {{\mathrm e}^{x}}{2 x}-\frac {\operatorname {Ei}_{1}\left (-x \right )}{2}\right )}{15}\)	\(98\)

[In]

int(1/225*((18*x^3-18*x^2)*exp(x)^2+((30*x^2-90*x)*exp(3)-30*x^2+90*x)*exp(x)-100*exp(3)^2+200*exp(3)-100)/x^5
,x,method=_RETURNVERBOSE)

[Out]

((2/15*exp(3)-2/15)*x*exp(x)+1/25*exp(x)^2*x^2+1/9*exp(3)^2-2/9*exp(3)+1/9)/x^4

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {-100+200 e^3-100 e^6+e^{2 x} \left (-18 x^2+18 x^3\right )+e^x \left (90 x-30 x^2+e^3 \left (-90 x+30 x^2\right )\right )}{225 x^5} \, dx=\frac {9 \, x^{2} e^{\left (2 \, x\right )} + 30 \, {\left (x e^{3} - x\right )} e^{x} + 25 \, e^{6} - 50 \, e^{3} + 25}{225 \, x^{4}} \]

[In]

integrate(1/225*((18*x^3-18*x^2)*exp(x)^2+((30*x^2-90*x)*exp(3)-30*x^2+90*x)*exp(x)-100*exp(3)^2+200*exp(3)-10
0)/x^5,x, algorithm="fricas")

[Out]

1/225*(9*x^2*e^(2*x) + 30*(x*e^3 - x)*e^x + 25*e^6 - 50*e^3 + 25)/x^4

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (20) = 40\).

Time = 0.10 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.89 \[ \int \frac {-100+200 e^3-100 e^6+e^{2 x} \left (-18 x^2+18 x^3\right )+e^x \left (90 x-30 x^2+e^3 \left (-90 x+30 x^2\right )\right )}{225 x^5} \, dx=- \frac {- \frac {4 e^{6}}{9} - \frac {4}{9} + \frac {8 e^{3}}{9}}{4 x^{4}} + \frac {15 x^{3} e^{2 x} + \left (- 50 x^{2} + 50 x^{2} e^{3}\right ) e^{x}}{375 x^{5}} \]

[In]

integrate(1/225*((18*x**3-18*x**2)*exp(x)**2+((30*x**2-90*x)*exp(3)-30*x**2+90*x)*exp(x)-100*exp(3)**2+200*exp
(3)-100)/x**5,x)

[Out]

-(-4*exp(6)/9 - 4/9 + 8*exp(3)/9)/(4*x**4) + (15*x**3*exp(2*x) + (-50*x**2 + 50*x**2*exp(3))*exp(x))/(375*x**5
)

Maxima [C] (verification not implemented)

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

Time = 0.23 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.36 \[ \int \frac {-100+200 e^3-100 e^6+e^{2 x} \left (-18 x^2+18 x^3\right )+e^x \left (90 x-30 x^2+e^3 \left (-90 x+30 x^2\right )\right )}{225 x^5} \, dx=-\frac {2}{15} \, e^{3} \Gamma \left (-2, -x\right ) - \frac {2}{5} \, e^{3} \Gamma \left (-3, -x\right ) + \frac {e^{6}}{9 \, x^{4}} - \frac {2 \, e^{3}}{9 \, x^{4}} + \frac {1}{9 \, x^{4}} + \frac {4}{25} \, \Gamma \left (-1, -2 \, x\right ) + \frac {2}{15} \, \Gamma \left (-2, -x\right ) + \frac {8}{25} \, \Gamma \left (-2, -2 \, x\right ) + \frac {2}{5} \, \Gamma \left (-3, -x\right ) \]

[In]

integrate(1/225*((18*x^3-18*x^2)*exp(x)^2+((30*x^2-90*x)*exp(3)-30*x^2+90*x)*exp(x)-100*exp(3)^2+200*exp(3)-10
0)/x^5,x, algorithm="maxima")

[Out]

-2/15*e^3*gamma(-2, -x) - 2/5*e^3*gamma(-3, -x) + 1/9*e^6/x^4 - 2/9*e^3/x^4 + 1/9/x^4 + 4/25*gamma(-1, -2*x) +
2/15*gamma(-2, -x) + 8/25*gamma(-2, -2*x) + 2/5*gamma(-3, -x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {-100+200 e^3-100 e^6+e^{2 x} \left (-18 x^2+18 x^3\right )+e^x \left (90 x-30 x^2+e^3 \left (-90 x+30 x^2\right )\right )}{225 x^5} \, dx=\frac {9 \, x^{2} e^{\left (2 \, x\right )} + 30 \, x e^{\left (x + 3\right )} - 30 \, x e^{x} + 25 \, e^{6} - 50 \, e^{3} + 25}{225 \, x^{4}} \]

[In]

integrate(1/225*((18*x^3-18*x^2)*exp(x)^2+((30*x^2-90*x)*exp(3)-30*x^2+90*x)*exp(x)-100*exp(3)^2+200*exp(3)-10
0)/x^5,x, algorithm="giac")

[Out]

1/225*(9*x^2*e^(2*x) + 30*x*e^(x + 3) - 30*x*e^x + 25*e^6 - 50*e^3 + 25)/x^4

Mupad [B] (verification not implemented)

Time = 12.77 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \[ \int \frac {-100+200 e^3-100 e^6+e^{2 x} \left (-18 x^2+18 x^3\right )+e^x \left (90 x-30 x^2+e^3 \left (-90 x+30 x^2\right )\right )}{225 x^5} \, dx=\frac {25\,{\left ({\mathrm {e}}^3-1\right )}^2+9\,x^2\,{\mathrm {e}}^{2\,x}+x\,{\mathrm {e}}^x\,\left (30\,{\mathrm {e}}^3-30\right )}{225\,x^4} \]

[In]

int(-((4*exp(6))/9 - (8*exp(3))/9 + (exp(2*x)*(18*x^2 - 18*x^3))/225 + (exp(x)*(exp(3)*(90*x - 30*x^2) - 90*x
+ 30*x^2))/225 + 4/9)/x^5,x)

[Out]

(25*(exp(3) - 1)^2 + 9*x^2*exp(2*x) + x*exp(x)*(30*exp(3) - 30))/(225*x^4)