\(\int \frac {-5-10 x+8 e^{\frac {10+2 x}{1+x}} x-5 x^2}{x+2 x^2+x^3} \, dx\) [6646]

Optimal result

Integrand size = 39, antiderivative size = 20 \[ \int \frac {-5-10 x+8 e^{\frac {10+2 x}{1+x}} x-5 x^2}{x+2 x^2+x^3} \, dx=-e^{2+\frac {8}{1+x}}-\log \left (x^5\right ) \]

[Out]

-exp(2+8/(1+x))-ln(x^5)

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.128, Rules used = {1608, 27, 6874, 2262, 2240} \[ \int \frac {-5-10 x+8 e^{\frac {10+2 x}{1+x}} x-5 x^2}{x+2 x^2+x^3} \, dx=-e^{\frac {8}{x+1}+2}-5 \log (x) \]

[In]

Int[(-5 - 10*x + 8*E^((10 + 2*x)/(1 + x))*x - 5*x^2)/(x + 2*x^2 + x^3),x]

[Out]

-E^(2 + 8/(1 + x)) - 5*Log[x]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1608

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2240

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^n*(
F^(a + b*(c + d*x)^n)/(b*f*n*(c + d*x)^n*Log[F])), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2262

Int[(F_)^((e_.) + ((f_.)*((a_.) + (b_.)*(x_)))/((c_.) + (d_.)*(x_)))*((g_.) + (h_.)*(x_))^(m_.), x_Symbol] :>
Int[(g + h*x)^m*F^((d*e + b*f)/d - f*((b*c - a*d)/(d*(c + d*x)))), x] /; FreeQ[{F, a, b, c, d, e, f, g, h, m},
x] && NeQ[b*c - a*d, 0] && EqQ[d*g - c*h, 0]

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {-5-10 x+8 e^{\frac {10+2 x}{1+x}} x-5 x^2}{x \left (1+2 x+x^2\right )} \, dx \\ & = \int \frac {-5-10 x+8 e^{\frac {10+2 x}{1+x}} x-5 x^2}{x (1+x)^2} \, dx \\ & = \int \left (-\frac {5}{x}+\frac {8 e^{\frac {2 (5+x)}{1+x}}}{(1+x)^2}\right ) \, dx \\ & = -5 \log (x)+8 \int \frac {e^{\frac {2 (5+x)}{1+x}}}{(1+x)^2} \, dx \\ & = -5 \log (x)+8 \int \frac {e^{2+\frac {8}{1+x}}}{(1+x)^2} \, dx \\ & = -e^{2+\frac {8}{1+x}}-5 \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {-5-10 x+8 e^{\frac {10+2 x}{1+x}} x-5 x^2}{x+2 x^2+x^3} \, dx=-e^{2+\frac {8}{1+x}}-5 \log (x) \]

[In]

Integrate[(-5 - 10*x + 8*E^((10 + 2*x)/(1 + x))*x - 5*x^2)/(x + 2*x^2 + x^3),x]

[Out]

-E^(2 + 8/(1 + x)) - 5*Log[x]

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90

method	result	size
parts	\(-5 \ln \left (x \right )-{\mathrm e}^{2+\frac {8}{1+x}}\)	\(18\)
risch	\(-5 \ln \left (x \right )-{\mathrm e}^{\frac {2 x +10}{1+x}}\)	\(19\)
parallelrisch	\(-5 \ln \left (x \right )-{\mathrm e}^{\frac {2 x +10}{1+x}}\)	\(19\)
derivativedivides	\(5 \ln \left (\frac {8}{1+x}\right )-5 \ln \left (-8+\frac {8}{1+x}\right )-{\mathrm e}^{2+\frac {8}{1+x}}\)	\(36\)
default	\(5 \ln \left (\frac {8}{1+x}\right )-5 \ln \left (-8+\frac {8}{1+x}\right )-{\mathrm e}^{2+\frac {8}{1+x}}\)	\(36\)
norman	\(\frac {-x \,{\mathrm e}^{\frac {2 x +10}{1+x}}-{\mathrm e}^{\frac {2 x +10}{1+x}}}{1+x}-5 \ln \left (x \right )\)	\(42\)

[In]

int((8*x*exp((2*x+10)/(1+x))-5*x^2-10*x-5)/(x^3+2*x^2+x),x,method=_RETURNVERBOSE)

[Out]

-5*ln(x)-exp(2+8/(1+x))

Fricas [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {-5-10 x+8 e^{\frac {10+2 x}{1+x}} x-5 x^2}{x+2 x^2+x^3} \, dx=-e^{\left (\frac {2 \, {\left (x + 5\right )}}{x + 1}\right )} - 5 \, \log \left (x\right ) \]

[In]

integrate((8*x*exp((2*x+10)/(1+x))-5*x^2-10*x-5)/(x^3+2*x^2+x),x, algorithm="fricas")

[Out]

-e^(2*(x + 5)/(x + 1)) - 5*log(x)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {-5-10 x+8 e^{\frac {10+2 x}{1+x}} x-5 x^2}{x+2 x^2+x^3} \, dx=- e^{\frac {2 x + 10}{x + 1}} - 5 \log {\left (x \right )} \]

[In]

integrate((8*x*exp((2*x+10)/(1+x))-5*x**2-10*x-5)/(x**3+2*x**2+x),x)

[Out]

-exp((2*x + 10)/(x + 1)) - 5*log(x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {-5-10 x+8 e^{\frac {10+2 x}{1+x}} x-5 x^2}{x+2 x^2+x^3} \, dx=-e^{\left (\frac {8}{x + 1} + 2\right )} - 5 \, \log \left (x\right ) \]

[In]

integrate((8*x*exp((2*x+10)/(1+x))-5*x^2-10*x-5)/(x^3+2*x^2+x),x, algorithm="maxima")

[Out]

-e^(8/(x + 1) + 2) - 5*log(x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (19) = 38\).

Time = 0.34 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.20 \[ \int \frac {-5-10 x+8 e^{\frac {10+2 x}{1+x}} x-5 x^2}{x+2 x^2+x^3} \, dx=-e^{\left (\frac {2 \, {\left (x + 5\right )}}{x + 1}\right )} + 5 \, \log \left (\frac {2 \, {\left (x + 5\right )}}{x + 1} - 2\right ) - 5 \, \log \left (\frac {2 \, {\left (x + 5\right )}}{x + 1} - 10\right ) \]

[In]

integrate((8*x*exp((2*x+10)/(1+x))-5*x^2-10*x-5)/(x^3+2*x^2+x),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {-5-10 x+8 e^{\frac {10+2 x}{1+x}} x-5 x^2}{x+2 x^2+x^3} \, dx=-5\,\ln \left (x\right )-{\mathrm {e}}^{\frac {2\,x}{x+1}}\,{\mathrm {e}}^{\frac {10}{x+1}} \]

[In]

int(-(10*x - 8*x*exp((2*x + 10)/(x + 1)) + 5*x^2 + 5)/(x + 2*x^2 + x^3),x)

[Out]

- 5*log(x) - exp((2*x)/(x + 1))*exp(10/(x + 1))