3.55.65 \(\int \frac {-300 x^2+e^{\frac {2 (1+x)}{x}} (1-3 x-3 x^2)+e^{\frac {1+x}{x}} (-10+30 x+60 x^2)}{-25 x^2+225 x^3-675 x^4+675 x^5} \, dx\)

Optimal. Leaf size=28 \[ \frac {\left (-2+\frac {1}{5} e^{\frac {1+x}{x}}\right )^2}{2 (1-3 x)^2} \]

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Rubi [A]  time = 0.79, antiderivative size = 48, normalized size of antiderivative = 1.71, number of steps used = 6, number of rules used = 4, integrand size = 71, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {6688, 12, 6742, 2288} \begin {gather*} -\frac {2 e^{\frac {1}{x}+1}}{5 (1-3 x)^2}+\frac {e^{\frac {2}{x}+2}}{50 (1-3 x)^2}+\frac {2}{(1-3 x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-300*x^2 + E^((2*(1 + x))/x)*(1 - 3*x - 3*x^2) + E^((1 + x)/x)*(-10 + 30*x + 60*x^2))/(-25*x^2 + 225*x^3
- 675*x^4 + 675*x^5),x]

[Out]

2/(1 - 3*x)^2 - (2*E^(1 + x^(-1)))/(5*(1 - 3*x)^2) + E^(2 + 2/x)/(50*(1 - 3*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 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (10-e^{1+\frac {1}{x}}\right ) \left (30 x^2-e^{1+\frac {1}{x}} \left (-1+3 x+3 x^2\right )\right )}{25 (1-3 x)^3 x^2} \, dx\\ &=\frac {1}{25} \int \frac {\left (10-e^{1+\frac {1}{x}}\right ) \left (30 x^2-e^{1+\frac {1}{x}} \left (-1+3 x+3 x^2\right )\right )}{(1-3 x)^3 x^2} \, dx\\ &=\frac {1}{25} \int \left (-\frac {300}{(-1+3 x)^3}-\frac {e^{2+\frac {2}{x}} \left (-1+3 x+3 x^2\right )}{x^2 (-1+3 x)^3}+\frac {10 e^{1+\frac {1}{x}} \left (-1+3 x+6 x^2\right )}{x^2 (-1+3 x)^3}\right ) \, dx\\ &=\frac {2}{(1-3 x)^2}-\frac {1}{25} \int \frac {e^{2+\frac {2}{x}} \left (-1+3 x+3 x^2\right )}{x^2 (-1+3 x)^3} \, dx+\frac {2}{5} \int \frac {e^{1+\frac {1}{x}} \left (-1+3 x+6 x^2\right )}{x^2 (-1+3 x)^3} \, dx\\ &=\frac {2}{(1-3 x)^2}-\frac {2 e^{1+\frac {1}{x}}}{5 (1-3 x)^2}+\frac {e^{2+\frac {2}{x}}}{50 (1-3 x)^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 22, normalized size = 0.79 \begin {gather*} \frac {\left (-10+e^{1+\frac {1}{x}}\right )^2}{50 (1-3 x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-300*x^2 + E^((2*(1 + x))/x)*(1 - 3*x - 3*x^2) + E^((1 + x)/x)*(-10 + 30*x + 60*x^2))/(-25*x^2 + 22
5*x^3 - 675*x^4 + 675*x^5),x]

[Out]

(-10 + E^(1 + x^(-1)))^2/(50*(1 - 3*x)^2)

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fricas [A]  time = 1.87, size = 35, normalized size = 1.25 \begin {gather*} \frac {e^{\left (\frac {2 \, {\left (x + 1\right )}}{x}\right )} - 20 \, e^{\left (\frac {x + 1}{x}\right )} + 100}{50 \, {\left (9 \, x^{2} - 6 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^2-3*x+1)*exp((x+1)/x)^2+(60*x^2+30*x-10)*exp((x+1)/x)-300*x^2)/(675*x^5-675*x^4+225*x^3-25*x^
2),x, algorithm="fricas")

[Out]

1/50*(e^(2*(x + 1)/x) - 20*e^((x + 1)/x) + 100)/(9*x^2 - 6*x + 1)

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giac [B]  time = 0.19, size = 46, normalized size = 1.64 \begin {gather*} -\frac {\frac {600}{x} + \frac {e^{\left (\frac {2}{x} + 2\right )}}{x^{2}} - \frac {20 \, e^{\left (\frac {1}{x} + 1\right )}}{x^{2}} - 900}{50 \, {\left (\frac {6}{x} - \frac {1}{x^{2}} - 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^2-3*x+1)*exp((x+1)/x)^2+(60*x^2+30*x-10)*exp((x+1)/x)-300*x^2)/(675*x^5-675*x^4+225*x^3-25*x^
2),x, algorithm="giac")

[Out]

-1/50*(600/x + e^(2/x + 2)/x^2 - 20*e^(1/x + 1)/x^2 - 900)/(6/x - 1/x^2 - 9)

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maple [A]  time = 0.18, size = 47, normalized size = 1.68

method	result	size
norman	\(\frac {-18 x^{3}+12 x^{2}-\frac {2 x \,{\mathrm e}^{\frac {x +1}{x}}}{5}+\frac {{\mathrm e}^{\frac {2 x +2}{x}} x}{50}}{x \left (3 x -1\right )^{2}}\)	\(47\)
risch	\(\frac {2}{9 \left (x^{2}-\frac {2}{3} x +\frac {1}{9}\right )}+\frac {{\mathrm e}^{\frac {2 x +2}{x}}}{50 \left (3 x -1\right )^{2}}-\frac {2 \,{\mathrm e}^{\frac {x +1}{x}}}{5 \left (3 x -1\right )^{2}}\)	\(49\)
derivativedivides	\(\frac {{\mathrm e}^{\frac {2}{x}+2}}{50}+\frac {3 \,{\mathrm e}^{\frac {2}{x}+2}}{25 \left (\frac {1}{x}-3\right )}+\frac {9 \,{\mathrm e}^{\frac {2}{x}+2}}{50 \left (\frac {1}{x}-3\right )^{2}}+\frac {12}{\frac {1}{x}-3}+\frac {18}{\left (\frac {1}{x}-3\right )^{2}}-\frac {12 \,{\mathrm e}^{\frac {1}{x}+1}}{5 \left (\frac {1}{x}-3\right )}-\frac {18 \,{\mathrm e}^{\frac {1}{x}+1}}{5 \left (\frac {1}{x}-3\right )^{2}}-\frac {2 \,{\mathrm e}^{\frac {1}{x}+1}}{5}\)	\(102\)
default	\(\frac {{\mathrm e}^{\frac {2}{x}+2}}{50}+\frac {3 \,{\mathrm e}^{\frac {2}{x}+2}}{25 \left (\frac {1}{x}-3\right )}+\frac {9 \,{\mathrm e}^{\frac {2}{x}+2}}{50 \left (\frac {1}{x}-3\right )^{2}}+\frac {12}{\frac {1}{x}-3}+\frac {18}{\left (\frac {1}{x}-3\right )^{2}}-\frac {12 \,{\mathrm e}^{\frac {1}{x}+1}}{5 \left (\frac {1}{x}-3\right )}-\frac {18 \,{\mathrm e}^{\frac {1}{x}+1}}{5 \left (\frac {1}{x}-3\right )^{2}}-\frac {2 \,{\mathrm e}^{\frac {1}{x}+1}}{5}\)	\(102\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-3*x^2-3*x+1)*exp((x+1)/x)^2+(60*x^2+30*x-10)*exp((x+1)/x)-300*x^2)/(675*x^5-675*x^4+225*x^3-25*x^2),x,m
ethod=_RETURNVERBOSE)

[Out]

(-18*x^3+12*x^2-2/5*x*exp((x+1)/x)+1/50*exp((x+1)/x)^2*x)/x/(3*x-1)^2

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maxima [B]  time = 0.41, size = 46, normalized size = 1.64 \begin {gather*} \frac {e^{\left (\frac {2}{x} + 2\right )} - 20 \, e^{\left (\frac {1}{x} + 1\right )}}{50 \, {\left (9 \, x^{2} - 6 \, x + 1\right )}} + \frac {2}{9 \, x^{2} - 6 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^2-3*x+1)*exp((x+1)/x)^2+(60*x^2+30*x-10)*exp((x+1)/x)-300*x^2)/(675*x^5-675*x^4+225*x^3-25*x^
2),x, algorithm="maxima")

[Out]

1/50*(e^(2/x + 2) - 20*e^(1/x + 1))/(9*x^2 - 6*x + 1) + 2/(9*x^2 - 6*x + 1)

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mupad [B]  time = 3.47, size = 28, normalized size = 1.00 \begin {gather*} \frac {\frac {{\mathrm {e}}^{\frac {2}{x}+2}}{50}-\frac {2\,{\mathrm {e}}^{\frac {1}{x}+1}}{5}+2}{{\left (3\,x-1\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((2*(x + 1))/x)*(3*x + 3*x^2 - 1) - exp((x + 1)/x)*(30*x + 60*x^2 - 10) + 300*x^2)/(25*x^2 - 225*x^3 +
675*x^4 - 675*x^5),x)

[Out]

(exp(2/x + 2)/50 - (2*exp(1/x + 1))/5 + 2)/(3*x - 1)^2

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sympy [B]  time = 0.19, size = 66, normalized size = 2.36 \begin {gather*} \frac {\left (- 900 x^{2} + 600 x - 100\right ) e^{\frac {x + 1}{x}} + \left (45 x^{2} - 30 x + 5\right ) e^{\frac {2 \left (x + 1\right )}{x}}}{20250 x^{4} - 27000 x^{3} + 13500 x^{2} - 3000 x + 250} + \frac {12}{54 x^{2} - 36 x + 6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x**2-3*x+1)*exp((x+1)/x)**2+(60*x**2+30*x-10)*exp((x+1)/x)-300*x**2)/(675*x**5-675*x**4+225*x**
3-25*x**2),x)

[Out]

((-900*x**2 + 600*x - 100)*exp((x + 1)/x) + (45*x**2 - 30*x + 5)*exp(2*(x + 1)/x))/(20250*x**4 - 27000*x**3 +
13500*x**2 - 3000*x + 250) + 12/(54*x**2 - 36*x + 6)

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