3.55.19 \(\int \frac {2+16 x+5 x^2+e^5 (10 x+5 x^2)}{2 x+x^2} \, dx\)

Optimal. Leaf size=22 \[ \log (x)+5 \left (9-e^5 (4-x)+x+\log (2+x)\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 17, normalized size of antiderivative = 0.77, number of steps used = 3, number of rules used = 2, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {1593, 1820} \begin {gather*} 5 \left (1+e^5\right ) x+\log (x)+5 \log (x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + 16*x + 5*x^2 + E^5*(10*x + 5*x^2))/(2*x + x^2),x]

[Out]

5*(1 + E^5)*x + Log[x] + 5*Log[2 + x]

Rule 1593

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

Rule 1820

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a +
b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2+16 x+5 x^2+e^5 \left (10 x+5 x^2\right )}{x (2+x)} \, dx\\ &=\int \left (5 \left (1+e^5\right )+\frac {1}{x}+\frac {5}{2+x}\right ) \, dx\\ &=5 \left (1+e^5\right ) x+\log (x)+5 \log (2+x)\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 17, normalized size = 0.77 \begin {gather*} 5 \left (1+e^5\right ) x+\log (x)+5 \log (2+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + 16*x + 5*x^2 + E^5*(10*x + 5*x^2))/(2*x + x^2),x]

[Out]

5*(1 + E^5)*x + Log[x] + 5*Log[2 + x]

________________________________________________________________________________________

fricas [A]  time = 0.56, size = 17, normalized size = 0.77 \begin {gather*} 5 \, x e^{5} + 5 \, x + 5 \, \log \left (x + 2\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^2+10*x)*exp(5)+5*x^2+16*x+2)/(x^2+2*x),x, algorithm="fricas")

[Out]

5*x*e^5 + 5*x + 5*log(x + 2) + log(x)

________________________________________________________________________________________

giac [A]  time = 0.13, size = 19, normalized size = 0.86 \begin {gather*} 5 \, x e^{5} + 5 \, x + 5 \, \log \left ({\left | x + 2 \right |}\right ) + \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^2+10*x)*exp(5)+5*x^2+16*x+2)/(x^2+2*x),x, algorithm="giac")

[Out]

5*x*e^5 + 5*x + 5*log(abs(x + 2)) + log(abs(x))

________________________________________________________________________________________

maple [A]  time = 0.22, size = 18, normalized size = 0.82

method	result	size
default	\(5 x \,{\mathrm e}^{5}+5 x +5 \ln \left (2+x \right )+\ln \relax (x )\)	\(18\)
norman	\(\left (5 \,{\mathrm e}^{5}+5\right ) x +5 \ln \left (2+x \right )+\ln \relax (x )\)	\(18\)
risch	\(5 x \,{\mathrm e}^{5}+5 x +5 \ln \left (2+x \right )+\ln \relax (x )\)	\(18\)
meijerg	\(-\ln \left (1+\frac {x}{2}\right )+\ln \relax (x )-\ln \relax (2)+2 \left (5 \,{\mathrm e}^{5}+5\right ) \left (\frac {x}{2}-\ln \left (1+\frac {x}{2}\right )\right )+2 \left (5 \,{\mathrm e}^{5}+8\right ) \ln \left (1+\frac {x}{2}\right )\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*x^2+10*x)*exp(5)+5*x^2+16*x+2)/(x^2+2*x),x,method=_RETURNVERBOSE)

[Out]

5*x*exp(5)+5*x+5*ln(2+x)+ln(x)

________________________________________________________________________________________

maxima [A]  time = 0.36, size = 16, normalized size = 0.73 \begin {gather*} 5 \, x {\left (e^{5} + 1\right )} + 5 \, \log \left (x + 2\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^2+10*x)*exp(5)+5*x^2+16*x+2)/(x^2+2*x),x, algorithm="maxima")

[Out]

5*x*(e^5 + 1) + 5*log(x + 2) + log(x)

________________________________________________________________________________________

mupad [B]  time = 3.37, size = 17, normalized size = 0.77 \begin {gather*} 5\,\ln \left (x+2\right )+\ln \relax (x)+x\,\left (5\,{\mathrm {e}}^5+5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*x + exp(5)*(10*x + 5*x^2) + 5*x^2 + 2)/(2*x + x^2),x)

[Out]

5*log(x + 2) + log(x) + x*(5*exp(5) + 5)

________________________________________________________________________________________

sympy [A]  time = 0.12, size = 17, normalized size = 0.77 \begin {gather*} x \left (5 + 5 e^{5}\right ) + \log {\relax (x )} + 5 \log {\left (x + 2 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x**2+10*x)*exp(5)+5*x**2+16*x+2)/(x**2+2*x),x)

[Out]

x*(5 + 5*exp(5)) + log(x) + 5*log(x + 2)

________________________________________________________________________________________