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

Optimal. Leaf size=20 \[ -e^{2+\frac {8}{1+x}}-\log \left (x^5\right ) \]

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Rubi [A]  time = 0.47, antiderivative size = 18, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 5, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.128, Rules used = {1594, 27, 6742, 2230, 2209} \begin {gather*} -e^{\frac {8}{x+1}+2}-5 \log (x) \end {gather*}

Antiderivative was successfully verified.

[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 1594

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 2209

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 2230

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 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 {-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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 18, normalized size = 0.90 \begin {gather*} -e^{2+\frac {8}{1+x}}-5 \log (x) \end {gather*}

Antiderivative was successfully verified.

[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]

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fricas [A]  time = 0.60, size = 18, normalized size = 0.90 \begin {gather*} -e^{\left (\frac {2 \, {\left (x + 5\right )}}{x + 1}\right )} - 5 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [B]  time = 0.23, size = 44, normalized size = 2.20 \begin {gather*} -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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [A]  time = 0.09, size = 19, normalized size = 0.95

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.38, size = 17, normalized size = 0.85 \begin {gather*} -e^{\left (\frac {8}{x + 1} + 2\right )} - 5 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-e^(8/(x + 1) + 2) - 5*log(x)

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mupad [B]  time = 0.10, size = 24, normalized size = 1.20 \begin {gather*} -5\,\ln \relax (x)-{\mathrm {e}}^{\frac {2\,x}{x+1}}\,{\mathrm {e}}^{\frac {10}{x+1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.15, size = 15, normalized size = 0.75 \begin {gather*} - e^{\frac {2 x + 10}{x + 1}} - 5 \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-exp((2*x + 10)/(x + 1)) - 5*log(x)

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